Fundamentals of Real Analysis [1 ed.] 9634635385 [PDF]

Zitiervorschau

Sterling K. Berberian

FUNDAMENTAL OF REAL ANALYSIS D

Jf I

y'

==

== f a.e., (£P)' == £q

jd(M x v) ==

II jdMdv

f(x, y), ~r(f * g) == (Tf)(x)

==

(! f)(~r g)

J K(x,y)f(y)dy

Springer

Universitext Editorial Board (North America):

S. Axler

F. W. Gehring K.A. Ribet

Springer New York Berlin Heidelberg Barcelona Hong Kong London Milan Paris Singapore Tokyo

Universitext Editors (North America): S. AxIer, F.W. Gehring, and K.A Ribet Aksoy/Khamli: Nonstandard Methods in Fixed Point Theory Andersson: Topics in Complex Analysis Aupetit: A Primer on Spectral Theory Berberian: Fundamentals of Real Analysis Booss/Bleecker: Topology and Analysis Borkar: Probability Theory: An Advanced Course CarIesooIGamelin: Complex Dynamics Cecil: Lie Sphere GeomeUy: Wilh Applications to Submanifolds Chao: lebesgue Inlegration (2nd ed.) Charlap: Bieberbach Groups and Flat Manifolds G1em: Complex Manifolds Withoul Potential Theory Cohn: AClassieallnviration 10 Algebraic Numbers and Class Fields Curtis: Abstract Linear Algebra Curtis: Matrix Groups DiBenedetto: Degenerate Parabolic Equations DUnca: Singularities and Topology of Hypersurfaces Edwards: A FonnaI Background to Mathematics I alb Edwards: A Formal Background to Mathematics 0 alb Foulds: Graph Theory Applications Friedman: Algebraic Surfaces and Holomorphic Vector Bundles Fuhnnann: A Polynomial Approach 10 Linear Algebra Gardiner: AFirs! Course in Group Theory GAnling/Tambour: Algebra for Compuler Science Goldblatt: Orthogonality and Spacetime GeomeUy GustafsonlRao: Numerical Range: The Foeid of Values of Linear OperalOrS and Matrices Hahn: Quadratic Algebras, Clifford Algebras. and Arithmetic Win Groups Holmgren: A FIrst Course in Discrete Dynamical SYSlems HoweITan: Non-Abelian Harmonic Analysis: Applications of SL(2, R) Howes: Modem Analysis and Topology HwnVMiller: second Course in Ordinary Differential Equations HurwitzlKritikos: Lectures on Number Theory Jennings: Modem GeomeUy with Applications JonesIMorrislPea.....n: Abstracl Algebra and Famous Impossibilities Kannan/Krueger: Advanced Analysis KellylMatthews: The Non-Euclidean Hyperbolic Plane Kostrikin: Introduction to Algebra LueckinglRubel: Complex Analysi X, idx(x) = x x E X). More generally, if A is a nonempty subset of X, the mapping A -+ X defined by x ..... x (x E A) is called the insertion mapping of A into X, denoted i A : A -+ X. In particular, ix = idx. The function CPA: X -+ {O,l} defined by

rv

CPA () x

lforxEA for x E CA

={ o

is called the characteristic function of A; CPA is often regarded as having values in the field IR of real numbers (cf. Chapter 4), but there is no law against letting 0, 1 be elements of any field, for example the field of two elements. 1.3.5. Definition. Given a pair of functions I: X -+ Y and g: Y -+ Z (g picks up where I leaves off), the function X -> Z defined by x ...... g(f(x)) is called the composite of 9 and I and is denoted go I (see Figure 4),

I

X --'---

II X; = (/) . iEI

What if all of the Xi are nonempty? The answer requires a full-fledged axiom of set theory: 1.4.5. Axiom of choice (AC). If (Xi)iEI is a family of nonempty sets then Xi -I(/), that is, there exists a family (Xi)iEI with Xi E Xi for all i; in other words, there exists a function f : I - t U Xi such that f(i) E Xi for all i E I.

n

14

1. Foundations

1.4.6. The Axiom of Choice is often presented in the following equivalent form: (AC') If X is a nonempty set, then it is possible to simultaneously select an element from each nonempty subset of X. Stated more formally, there exists a function c: P(X) - {(ll) --+ X such that c(A) E A for every nonempty subset A of X. Such a function c is called a choice function for X. {(AC') ~ (AC): With the notations of 1.4.4, let X = U Xi and let c be a choice function for X; then Xi = c(X i ) defines an element (Xi) of the product set IT Xi . (AC) ~ (AC'): If A = P(X) - {(ll), I = A and A has the identity indexing A ...... A (1.3.10), then any element c = (CA) of the product set

ITA

AEA

will serve as a choice function for X.}

1.4.7. Definition. Let (Xi)iEI be a family of nonempty sets, X = IT Xi the product set (nonempty, by the Axiom of Choice). Fix an index j E I. The j'th coordinate projection mapping of X is the mapping prj : X --+ X; that assigns to each element X = (Xi) of X its j'th coordinate, that is, pr;(x) = x; . 1.4.8. Theorem. With notations as in the preceding definition, (i) for every j E I, the projection mapping prj: IT Xi -+ X; is surjective,. (ii) for a pair of mappings f, g : Y -+ IT Xi, f

=g

¢}

pri

0

f

= pri 0 g

f01· all i E I.

Proof. (i) Let j E I. a E X;; the problem is to exhibit an element X = (Xi) of X = ITXi such that pr;(x) = a. Write A; = {a} and Ai = Xi for all i i' j; every element of the set IT Ai (nonempty, by the Axiom of Choice) meets the requirements for x. (ii) Here priof denotes the composite mapping y ...... pri(t(y)) (cf. Figure 6).

f y----=----. X

!

pri

Xi Figure 6

§1.5. Inverse Functions

15

For every x EX, x = (pr.(x»; thus the condition on the right side of the equivalence of (ii) means that pr.(J(y»)

= pr.(g(y»

for all y E Y and i E I,

that is, (pr.(J(y») tEl = (pr.(g(y»)) tEl

for all y E Y,

in other words f(y) = g(y) for all y E Y; that's also what the left side means. K. Godel showed (1939) that the Axiom of Choice is consistent with (Le., not disprovable from) the 'usual' axioms of set theory, and P. Cohen showed (1963) that it is independent of (Le., not provable from) the usual axioms. Thus, after a long career as a possible true theorem whose proof had not yet been found-or a possible false theorem for which a counterexample had not yet been constructed-the Axiom of Choice turned out, in fact, to be none of the above; now in semi-retirement as an axiom (lower case) in good standing, its fate reposes on the consistency (freedom from contradiction) of the rest of mathematics. 1 The reader will lind an excellent informal essay on these matters in Chapter 3 of 1. Kaplansky's Set theory and metric spaces [2nd edition, Chelsea, New York, 1977). In the present text the Axiom of Choice is used, and its use signaled explicitly, whenever (as in 1.4.8) it seems to be the appropriate tool for the discussion at hand.

1.5. Inverse Functions Recall that if f : X -+ Y is any function and B is a subset of Y, 1 (B} = {x EX: f(x} E B} (1.3.11); in this context, f- 1 is then a function that acts on subsets of Y to produce subsets of X, that is, f- I : 1'(Y} -+ 1'(X} . The kind of 'inverse' referred to in the section heading acts instead on points of Y to produce points of X. Whereas the set function f- I : 1'(Y) -+ 1'(X} exists for every function f: X -+ Y, the point function f- I : Y -+ X is defined only when f is bijective (1.5.3 below).

r

1.5.1. Lemma. For every function f : X 9 : Y -+ X such that fog 0 f = f·

-+

Y, there exists a function

Proof For each point y of the range f(X} of f, let Ay = {x EX: 1 ({y}}; since y E f(X} , A f(x} = y} = y is nonempty. We thus have

r

1 My own view (naive, as cautioned in the chapter introduction) is that the results that have been obtained using the Axiom of Choice are too beautiful to be thrown away without good cause. Innocent until proven guiltYi next case ....

1. Foundations

16

a family (AY)YEj(X) of nonempty subsets of X, indexed by the set f(X); by the Axiom of choice, there exists a function go: f(X) -> X such that go(Y) E A y for all y E f(X). Choose a point c E X and let g: Y -> X be the function defined by the formulas g(y)

= {go(y)

if y E f(X) if y ¢ f(X).

c

More precisely, as a set of ordered pairs, 9 = {(y,go(y)): Y E f(X)} U {(y,c): y

Let x EX; writing y therefore f(g(J(x)))

=

¢

f(X)}.

f(x) E f(X) , we have g(y)

,

0

1.5.2. Theorem. Let f: X -> Y. (i) f is surjective ¢} there exists a jv.nction 9 : Y fog = idy. (ii) f is injective ¢} there exists a function h : Y h 0 f = idx. (iii) If f is bijective, necessarily 9 = h. ~:

go(Y) E A y

= f(g(y)) = f(go(y)) = y = f(x) ,

thus 9 meets the requirements of the lemma.

Proof. (i),

=

->

X

->

X such that

such that

Assuming 9 has the indicated properties,

('If y E Y) y

= idy(y) = (J

0

g)(y)

= f(g(y))

E f(X) ,

whence f(X) = Y. (i), =>: By the lemma, there exists a function g: Y -> X such that fog 0 f = f , that is, (J 0 g) 0 f = idy 0 f; since f is surjective, it follows that fog = idy (§1.3, Exer. 2). (ii), ~: Suppose that h has the indicated properties. If f(x) = f(x') then h(J(x)) = h(f(x')); since h 0 f = idx, it follows that x = x'. (ii), =>: Let h: Y -> X be a function such that f 0 h 0 f = f, that is, f 0 (h 0 J) = f 0 id x ; since f is injective, it follows that h 0 f = id x . (iii) With 9 and h as in (i) and (ii), 9 = id x 0 9 = (h 0 f) 0 9 = h 0 (J 0 g) = h 0 idy = h. 0 1.5.3. Definition. With notations as in 1.5.2, if f is bijective then every 9 coincides with every h; the unique function g: Y -> X such that 9 0 f = id x and fog = idy is called the inverse of f and is denoted f-I. Since f-l(J(X)) = x ('If x E X) and f(J-I(y)) = Y ('If y E V), f- I undoes everything that f does, and vice versa. Some other properties of the inverse are given in the exercises.

1.5.4. Corollary. For sets X and Y, 3 injection X

->

Y

¢}

3 surjection Y

->

X.

§1.6. Equivalence Relations

17

Proof. ~: If / : X -+ Y is injective, then the mapping h: Y provided by (ii) of 1.5.2 is surjective (1.3.7). X/- defined by q(x) [x] is called the quotient mapping for the relation.

=

1.6.9. Examples. For the examples of 1.6.2 and 1.6.4, (1) 7l/R = 716 (the set of 'integers modulo 6'), (2) the elements of X/R are the sets of constancy of f, (3) the quotient mapping X -> X/R is the bijection x ...... {x} , (4) X/R consists of a single element (and the quotient mapping is constant), and (5) an element [I] of the quotient set can be identified with a (two-way) 'direction'. 1.6.10. Theorem. If X is a nonempty set and - is an equivalence relation in X, then the quotient mapping q: X -> X/- is surjective and - coincides with the equivalence relation deduced from q (1.6.2).

Proof. Straightforward. The essential message of the foregoing theorems is that equivalence relations, partitions, and sets of constancy of a function are three presentations of the same basic concept. 1.6.11. Let f: X -> Y be any function, - the equivalence relation deduced from f (1.6.2). Write :i; for the equivalence class of x EX, X = X/- for the quotient set, q: X -> X for the quotient mapping, and i : f(X) -> Y for the insertion mapping i = i fIX) (1.3.4); in particular, q is surjective and i is injective (cf. Figure 7; the lower arrow anticipates the next theorem).

X

f

Ii

ql X

Y

i Figure 7

f(X)

20

1. Foundations

1.6.12. Theorem. With the preceding notations, there exists a unique bijection j: X -+ f(X) such that f = i 0 j 0 q .

Proof. Uniqueness: If g, h : X -+ f(X) are functions such that i 0 9 0 q = i 0 h 0 q then, citing §1.3, Exer. 2, 9 0 q = h 0 q (because i is injective) and 9 = h (because q is surjective). Existence: Let u EX; the problem is to define j(u). If u = q(x) = q(y) then x ~ y, that is, f(x) = f(y) , therefore the formula j(u) = f(x) defines a function j: X -+ f(X) unambiguously. It is clear from the definitions t~at i(i(q(x))) = i(f(x)) = f(x) , and easy to check that j is bijective. { f is the inverse ofthe mapping z ..... f-I( {z}) (z E f(X)).} ¢ The formula f = i 0 j 0 q is called the canonical factorization of f (as the composite of a surjection, followed by a bijection, followed by an injection).

Exercise 1. Let X and Y be sets, R a relation in X and 8 a relation in Y. A mapping f: X -+ Y is said to be compatible with Rand 8 in case (for x, x' in X) xRx' f(x)8f(x'). If R and 8 are equivalence relations (1.6.1) in X and Y and if f : X -+ Y is compatible with Rand 8, then there exists a unique mapping g: X/R -+ Y 18 such that

'*

goQx

= Qy of,

where Qx: X -+ X/R and Qy: Y -+ Y 18 are the quotient mappings (1.6.8). The relation (*) is expressed by saying that the following diagram is commutative (the two ways of getting from X to Y 18 are equal):

X

X/R

f

----+1

9

Y

Y/8

1.7. Order Relations

1.7.1. Definition. A relation :5 in a set X is said to be a partial ordering of X (or to be an order relation in X) if it is reflexive, transitive and 'antisymmetric': (i) x :5 x for all x E X (reflexivity), (ii) x :5 y & y :5 z x:5 z (transitivity),

'*

§1. 7. Order Relations

21

(iii) x < y & y :5 x => x = Y (antisymmetry). A relation satisfying only (i) and (ii) is called a pre-ordering of X. Other notations in use for such relations are ~, c, :::>, -< , >-, etc. A simple ordering (or total ordering, linear ordering) is a partial ordering :5 for which any two elements are comparable, that is, (iv) ('v' x,y) x:5 y or y < x. A pre-ordered set is a pair (X,:5), where :5 is a pre-ordering of X; the terms partially ordered set and simply ordered set are defined similarly.

1.7.2. Examples. (1) In the set of all positive integers, the relation xly (x divides y) is a partial ordering (but not a simple ordering). (2) The relation A c B is a partial ordering of the set P(X) of all subsets of a set X (not a simple ordering when X has more than one element). (3) The usual relation :5 in the set of all positive integers is a simple ordering. (4) For every set X, the relation in X defined by X x X is a preordering of X (x :5 x' for all x,:If EX), called the trivial pre-ordering of X. When X has more than one element, the trivial pre-ordering is not a partial ordering. (5) Every equivalence relation (1.6.1) in a set X is a pre-ordering; the only eqUivalence relation in X that is a partial ordering is the identity relation (the relation :5 satisfying x :5 x for all x EX, and nothing else), defined by the diagonal b. of X x X. 1.7.3. Example. If (X,:5) is a pre-ordered set and A is a subset of X, the relation on A induced by :5 (cf. 1.2.5) is a pre-ordering of A. If the relation on X is a partial ordering (simple ordering), then so is the relation induced on A. A simply ordered subset of X is called a chain in X. For example, if X is partially ordered, A = {Xl> .. ' ,xn } C X and Xl :5 X2 :5 ... :5 X n ,then A is a chain. 1.7.4. Example. In a pre-ordered set (X,:5) , write x :5' y in case y :5 x (:5' is the 'reverse' of :5 in the sense of 1.2.7); the relation :5' is also a pre-ordering (called the dual of :5). When :5 is a partial order (simple order), then so is :5'. 1.7.5. Notations. In a pre-ordered set (X,:5), y ~ x is an alternative notation for x < y. If x:5 y and x t- y, we write x < y (or y > x). {CAUTION: If X is a set equipped with the trivial pre-ordering (1.7.2), then a < b and b < a for every pair a, b of distinct elements of X.} When < is a simple ordering, exactly one of the statements xy is true for any given pair of elements x, y (Law of trichotomy).

22

1. Foundations

1.7.6. Example. If (X,:5) is a partially ordered set and (Y,:5) is any pre-ordered set, then the relation (x, y) :5 (x', y') in X x Y defined by the condition «either x

< x', or x = x' & y :5 y'»

is a pre-ordering of X x Y, called the lexicographic pre-ordering (as in a dictionary of two-letter words). If both X and Y are partially ordered (simply ordered) then so is X x Y for the lexicographic pre-ordering. Another pre-ordering of X x Y is defined by the condition «x:5 x' & y:5 y'»; it is called the product of the given pre-orderings. If X and Y are partially ordered then so is X x Y , but the analogous statement for simple ordering is false. 1. 7.7. The study of pre-orderings is effectively reduced to the study of partial orderings by 'passing to quotients' in an appropriate way. Suppose (X,:5) is a pre-ordered set. For x, y EX, write x - y in case both x:5 y and y:5 x. It is easy to see that - is an equivalence relation on X. If u, v are elements of the quotient set X/-, say u = [xl and v = Iy] , we propose to define a relation u:5 v in case x:5 y; if also u = [x'] and v = [y'] , then x:5 y ¢} x' :5 y' , so the definition is unambiguous. The relation so defined in X/- is easily seen to be a partial ordering (called the quotient order relation); for example, if u:5 v and v:5 u then, with the preceding notations, x:5 y and y:5 x, therefore x - y and so u = v .

The next concepts pertain to relations between elements and subsets of an ordered set. 1.7.8. Definition. Let (X,:5) be a pre-ordered set, A C X and c EX. (i) If x:5 c ('If x E A), we say that c is an upper bound for A (or that c majorizes A, or that c is a majorant of A). Dually, (ii) If c:5 x ('If x E A), we say that c is a lower bound for A (or that c minorizes A, or that c is a minorant of A). If A has a majorant (minorant) in X, it is said to be bounded above (bounded below); A is said to be bounded if it is both bounded above and bounded below. 1.7.9. Remark. Suppose (X,:5) is a partially ordered set and a E A C X. If a majorizes (minorizes) A then it is uniquely determined by this property; it is called the largest (smallest) element of A. The following concept is more subtle: 1.7.10. Definition. Let (X,:5) be a pre-ordered set, A eX. An element a E A is said to be a maximal element of A if (x E A & x

~

a)

~

x

=a

§1.7. Order Relations

23

(in other words, A contains no element > a); dually, if a E A satisfies the condition

(x

E A

& x :=; a)

"*

x

=a

then a is said to be a minimal element of A.

1.7.11. Remark. In a partially ordered set (X,:=;), if a E A is the largest element of A then a is maximal in A. If (X,:=;) is simply ordered and if a E A is maximal in A, then a is the largest element of A. Similarly for "smallest" and "minimal". Thus, when X is simply ordered, the concepts of maximal element and largest element coincide (as do the concepts of minimal element and smallest element). 1.7.12. Examples. (1) In the field Q ofrational numbers, with the usual ordering, the set A = {r E Q: 0 < r < I} is bounded, but has neither a largest nor a smallest element. (2) For the usual ordering of 11", every nonempty subset has a smallest element (Principle 01 mathematical induction). (3) Let S be the set of all nonempty subsets A of II" such that A has at most 5 elements. Order S by the inclusion relation c. Every 5-element subset of II" is a maximal element of S, and every singleton in II" is a minimal element of S. . The rest of the section prepares the way for the discussion of well-ordered sets in §1.14; it can be deferred until then.

1.7.13. Definition. Let (X,:=;) and (Y,:=;) be pre-ordered sets, I : X -+ Y a function. We say that I is (1) an order morphism if x:=; x' I(x):=; I(x') , (2) an order isomorphism if I is bijective and both I and 1-1 are order morphisms, and (3) an order monomorphism if I is injective and x:=; x' # I(x) :=; I(x /). {For a possible definition of 'order epimorphism', see (iii) of Exercise 6.}

'*

Condition (1) says that I is compatible with the order relations in the sense of §1.6, Exercise 1. Condition (2) says that I is bijective and x 0 in Q, there exists an index N such that

Irm-r"l:5 t.

m,n ~ N ~ This is also expressed by saying that

(if t E Q, t > 0)

Irm

-

r,,1 :5 t

ultimately.

We write C for the set of all Cauchy sequences in Q. 1.8.8. Lemma. C is a subring of B containing the constant sequences.

Proof. The crux of the matter is to show that every Cauchy sequence (r,,) is bounded; this follows from the fact that Irm - r,,1 :5 1 from some index onward-say for m,n ~ N-and the inequality Ir,,1 :5lr" - rNI + IrNI. If (r n ) and (s,,) are both Cauchy, then the identity rmS m - r"s"

= rm(sm -

s,,) + (rm

-

r")s,,

shows that their product (r"s,,) is also Cauchy. The closure of C under addition follows at once from the triangle inequality (1.8.2). 1.8.9. Definition. A sequence (z,,) in Q is said to be null if

(V' t E Q, t > 0)

1z,,1:5 t

ultimately;

that is, for every rational t > 0, there exists an index N such that n ~ N ~ Iz"l:5 t. We write N for the set of all null sequences in Q. is null. {Proof: If t = MIN t (that is, N:5 Mn) holds, for

1.8.10. Example. The sequence (lin)

(M, N E IP) then the relation lin < example, for all n ~ N .}

1.8.11. Lemma. N is an ideal of C (and of B). Proof. It is obvious from the triangle inequality that N is an additive subgroup of C. If (b n ) is bounded and (z,,) is null, it follows from Ibnz,,1 = Ib"llz,,1 that (bnz,,) is null.

§1.8. Real Numbers

29

1.8.12. Definition. We write JR = C/N for the quotient ring of C modulo N, and (rn )' = (r n) + N for the coset of (r n ) E C; thus (rn) 1-+ (rn)' is the quotient mapping (a homomorphism of C onto JR). Our task is to show that JR is a complete ordered field. At any rate. it is elementary that JR is a commutative ring with unity element (1) + N . If r E Q and (r) is the constant sequence with all terms equal to r. we abbreviate (r)' = (r) +N to r'; thus. r 1-+ r' is a mapping Q -+ JR. 1.8.13. Lemma. The mapping r

1-+

r' (r

E

Q) is a ring monomorphism

Q-+JR. Proof The mapping r 1-+ r' is the composite of the homomorphism r 1-+ (r) of Q into C with the quotient homomorphism C -+ JR. If r' = 0' then (r) is null, therefore r = 0 (1.8.3). whence injectivity.

1.8.14. Remark. If (rn ) is a Cauchy sequence in Q. N is a positive integer, and (sn) is a sequence in Q such that Sn = r n for all n > N • then (sn) is also Cauchy and (r n )' = (sn)'; in other words, one can modify rk for k=I, ...• N without changing the element (rn )' of JR. 1.8.15. Lemma. JR is a field. Proof Let x E JR, x # 0; we seek an element y E JR such that xy = 1 (more precisely, 1'). Say x = (rn )'. Since x # O. (rn ) is not a null sequence, thus there exists a rational t > 0 such that Irnl fails to be ultimately < t. This means that Irnl > t 'frequently', that is. Irn.I;O: t for a sequence of indices nl < n2 < n3 < .... We assert that Irnl;O: t/2 ultimately. For. since (rn ) is Cauchy, there exists an index N such that Irm - rnl :::; t/2 for all m, n ;0: N; if k is an index such that nk;O: N then, for all n;O: N •

t :::; Irn• I < Irn• - rnl

+ Irnl :::; t/2 + Irnl,

whence Irnl > t/2. We are ready to define the required element y. Let (sn) be the sequence in Q defined by

Sn

=

0 forn O. Irn -rml :::; (t 2 /4)r (and therefore ISm -snl :::; r) for m and n sufficiently large. Let y = (sn)' . Since rns n = 1 for all n ;0: N , it follows that xy = 1 .

1. Foundations

30

So far, everything is relatively straightforward; the order relation in IR is more delicate. 1.8.16. Definition. For x E IR, we write x ~ 0 in case x = (Tn)' with Tn ~ 0 for all n. {It is the same to require that x = (Tn)' with Tn ~ 0 ultimately. } 1.8.17. RemaTks. (1) If x,yEIR and x>O, y~O,thenalso x+y~O and xy~O. (2) If x ~ 0 and -x ~ 0 then x = O. {Proof: If x = (Tn)' and -x = (sn)' with Tn ~ 0 and Sn ~ 0 for all n, then (Tn + sn) is a null sequence (because x + (-x) = 0); but 0 $ Tn $ Tn + sn, SO (Tn) is also a null sequence, whence x = 0 .} (3) If x E IR then either x ~ 0 or -x ~ O. {Proof: Suppose x = (Tn)' . If Tn ~ 0 ultimately then x ~ 0, and if Tn $ 0 ultimately then -x ~ O. It remains to consider the case that Tn < 0 frequently (say for the indices nl < n2 < n3 < ... ) and Tn > 0 frequently (say for the indices ml < m2 < m3 < ... ); we shall show that this implies x = 0, that is, (Tn) is null. Given any rational t > 0, choose an index N such that m,n ~ N =} ITm -Tnl $ t. Choose k so that mk ~ Nand nk ~ N; then ITm • - Tn. I $ t, thus

and it follows that

n

~ N

=}

ITnl $

IT n -

Tm.l + ITm.1

$ 2t

whence the nullity of (Tn) .} 1.8.18. Definition. For x, y E IR, write x $ y in case y - x ~ 0 in the sense of 1.8.16.

1.8.19. Lemma. FOT the Telation $ just defined, IR is an ordeTed field (in particulaT, $ is a simple oTdering of IR). Proof Immediate from 1.8.17. 6

0

**

1.8.20. RemaTk. For T E Q, T ~ 0 in Q T' ~ 0 in IR. {Proof: The implication =} is trivial. Assuming r' ~ 0 let us show that T ~ O. By assumption, T' = (Tn)' with Tn ~ 0 for all n. Suppose to the contrary that T < O. Since (T - Tn) is null, Tn - T < -T ultimately, whence (Tn)' $ 0, that is, T' $ 0; already T' ~ 0, so T' = 0, a contradiction.}

1.8.21. Lemma. If x E IR, x such that O 0 for all n, and (Tn) not a null sequence. Suppose to the contrary that no such T exists. Let T be

6

First course, Chapter I, §2.

§1.8. Real Numbers

31

any rational number> O. By supposition, r' ~ x, so r' - x = (sn)' with Sn ~ 0 for all n. Then (rn + Sn - r) is null, therefore r n + Sn - r ::; r ultimately, whence 0 ::; r n ::; r n + Sn ::; 2r ultimately. To summarize, (Vr E Q, r > 0) 0 O

{ -x if x < O.

{In view of 1.8.20, this definition is consistent with the earlier definition of absolute value for rationals, and Ir'l = Irl' for all r E Q.} The proof of 1.8.2 applies equally well for JR.

1.8.23. Definition. A sequence (Xk) in JR is said to converge to the limit x E JR if

(V

f

E JR,

f

> 0) IXk - xl ::;

f

ultimately.

{In view of 1.8.21, it suffices to consider f = r' with r > 0 rational.} Such limits x are unique by the usual elementary argument, and one writes Xk --+ x (as k --+ 00 ). 1.8.24. Lemma. If Xk

--+

x in JR and Xk

~

0 for all k, then x

~

O.

Proof. From the inequality Ilxkl-lxll ::; IXk -xl we see that IXkl --+ Ixl; but IXkl = Xk --+ x, therefore x = Ixl ~ 0 by the uniqueness of limits. 1.8.25. Lemma. If x = (r n )' E JR then rk

--+

x as k

--+ 00 .

Proof. For every rational t > 0, there is an index N such that Irm - rnl ::; t for all m, n ~ N. For each k ~ N, r k - x = (rk - rn) and Irk - r n I ::; t for all n ~ N, therefore Irk - xl ::; t' . Since f'S of the form t' suffice in the criterion for convergence (1.8.23), this shows that rk--+ x . 1.8.26. Theorem. (Cauchy's criterion) Every Cauchy sequence in JR converges to an element of JR.

Proof. Let (Xk) be a Cauchy sequence in JR (for the definition of 'Cauchy sequence', paraphrase 1.8.7 with Q replaced by JR ). For each index k, choose rk E Q with Irk - xkl ::; l/k' (1.8.25). Let t E Q, t > O. Since (Xk) is Cauchy, there is an index N such that

(V i,k

~ N)

Ix; -xkl::; t';

we can suppose further that 1/N ::; t (1.8.10 and 1.8.20). Then, for all i, k > N.

Irj - rkl ::; Irj - x;1

+ Ix; - xkl + IXk - rkl < Iii' + t' + l/k'

::; 3t' ,

1. Foundations

32

and, since Irj - rkl' = Irj - rkl, it follows from 1.8.20 that Irj - rk! ::; 3t. This shows that (rn ) is Cauchy in 1Qi. Let x = (rn )' = (r n ) + /If; since Ix - xkl ::; Ix - rkl and rk

--+

+ Irk -

xkl ::; Ix - rkl

x (1.8.25), it follows that Xk

--+

+ 11k'

x.

There is no further need to distinguish between a rational number rand its image r' in JR; henceforth we identify r with r', regard IQi as a subfield of JR, and call the elements of JR real numbers. If a, b E JR, a ::; b, we write [a, b] = {x E JR: a::; x ::; b} (called the closed interval with endpoints a and b). An easy consequence of Cauchy's criterion is the 'theorem on nested intervals': 1.8.27. Theorem. (Theorem on nested intervals) Suppose lan, bnJ is a sequence of closed intervals in JR such that

and such that bn - an --+ O. Then (i) n[an,bn ] = {c} for some c e JR, and (ii) if Xn e [an, bnJ for all n, then Xn --+ c. Proof. Suppose Xn e [an, bnJ for all n.1f m, n ~ N then Xm , Xn both belong to [aN,b N ], therefore IX m - xnl ::; bN - aN; since bn - an --+ 0, it follows that (x n ) is Cauchy, hence convergent (1.8.26), say X n --+ x. If also Yn e [an, bnJ for all nand Yn --+ Y ,then IXn - Yn I ::; bn - an shows that Xn - Yn --+ 0 , whence x = y. This proves both (i) and (ii).

1.8.28. Definition. A Dedekind cut of JR is a pair (A, B) of nonempty subsets of JR, with AuB = IR, such that a < b for all a e A and b e B.

It follows from the theorem on nested intervals that every Dedekind cut of IR is effected by an element of JR: 1.8.29. Theorem. If (A, B) is a Dedekind cut of IR, then there exists an element c E IR such that either A={XEJR: x::;c}, B={XEJR: x>c} or

A={xeJR: xc}. Proof. The crux of the matter is to show that either A has a largest element or B has a smallest element. Choose a I E A, bl E B; let us construct a closed subinterval la2,~) c [al,bll such that a2 E A, b2 E B and ~ - a2 = !(b l - all . Let CI = !(al + btl. If CI E A, let a2 = CI and ~ = bl ; the only alternative is that CI E B, in which case we take a2 = al and b2 = CI . Repeating the process, we get a sequence

§1.8. Real Numbers

33

of closed intervals

such that an E A, bn E B and bn - an ...... O. Let c be the unique real number such that n[an,bnl = {c} (1.8.27). In particular, an ...... c and bn -+ c. case 1: c E A If a E A then a < bn for all n, therefore a ::; c (1.8.24); thus A c {x E IR: x::; c} . On the other hand, if b E B then c < b (because c E A), thus B c {x E IR: x > c}. It then follows from the law of trichotomy for IR (cf. 1.7.5) that both inclusions are equalities. case 2: c E B One argues similarly that B = {x E IR: x;:: c} and A = {x E IR :

x b. On the other hand, B 'f IR; indeed, if xES then x-I ¢ B. Let A = IR - B (nonempty, by the preceding remark). If a E A and bE B then a < b (the alternative, a ;:: b E B, would place a in B). Thus (A, B) is a Dedekind cut of IR. Let c be the unique element of IR provided by 1.8.29; it will suffice to show that c E B. Assume to the contrary that c E A, that is, c does not majorize S; then S contains an element x with x > c. The element y = ~(x + c) belongs to B (because y > c) hence majorizes S; in particular, y;:: x , which contradicts the definition of y. 0 1.8.31. Terminology. A function whose range consists of real numbers is said to be real-valued; a function whose domain consists of real numbers is said to be a function of a real variable. For example, the function f : (0,11 ...... IR defined by f(x) = l/x is a real-valued function of a real variable.

Exercises 1. Show that the ordering of IR is Archimedean: if x, y E IR and x

> 0,

then there exists a positive integer n such that nx > y . {Hint: 1.8.21.} 2. Let x, Y E IR and let n be a positive integer. If x;:: 0 and y;:: 0, then

34

1. Foundations

(i) X < y ¢} x" < y" , (ii) X = Y ¢} x" = y" , (iii) X > y ¢} x" > y" . If n is odd, then the restriction that x and yare nonnegative can be dropped. 3. If a, b, c are nonnegative real numbers such that a $ b + c, then

abc

- F defined by

J'(x) = {f(X) x

for xEE for x E F - E.

is injective but not surjective. (1') This is (1) in contrapositive form. (2) Suppose E is infinite and f: E -> F is bijective. Let g: E -> E be an injection that is not surjective. The mapping fog 0 f- 1 : F -> F is injective (it is a composite of injectives) but not surjective (if it were, then f- 1 0 (f 0 9 0 f-l) 0 f = 9 would be surjective). (2') If f: E -> F is bijective, then so is f-l: F -> E, thus (2') is the contrapositive form of (2). (3) If E is infinite and f : E -> F is injective, then f defines a bijection E -> f(E) ,therefore f(E) is infinite by (2); then F:> f(E) is infinite by (1). (3') The contrapositive form of (3). (4), (4') Suppose f: E -> F is surjective. By 1.5.4 (the axiom of choice plays a role here), there exists an injection g: F -> E, thus E finite by (3'), and F infinite =>

=> F finite

E infinite by (3).

1.9.4. Theorem. A set E is infinite if and only if there exists an injection IP' -> E. Proof. "If': If there exists an injection f: IP' -> E then, since IP' is infinite (1.9.2), so is E by (3) of the preceding theorem. "Only if": Assuming E is infinite, let f: E -> E be an injection that is not surjective and choose a point z E E - f(E). Define 9 : IP' -> E by the formula g(n) = r(z) , where the 'composition powers' rare defined recursively by fl = f and fn+t = fn 0 f. We assert that 9 is injective. Assuming m, n E IP', m < n , we must show that g( m) oft g( n) . Let p = n - m; then

whereas g(m) = fm(z); assuming to the contrary that g(m) = g(n) , that is, fm(z) = fm(jI'(z» , it follows from the injectivity of f that z = fP(z) E f(E) , contrary to the choice of z. For each positive integer n, we write IP'n = {I, . .. ,n} (the set of all integers from 1 to n). The next target (1.9.13): A nonempty set E is finite if and only if it is bijective with some IP'n .

1. Foundations

36

1.9.5. Lemma. If S is a nonempty subset of J!> with no largest element, then S is infinite. Proof. For each a E S write S(a) = {k E S: k > a}; by assumption, S(a) '10. Define f: S -+ S as follows: for each a E S ,let f(a) be the smallest element of S(a). In particular, f(a) > a for all a E S. We show that S is infinite by verifying that f is injective but not surjective. f is injective: For, if a, b E S and a < b, then b E S( a), therefore f(a) :5 b < f(b) . f is not surjective: For, if z is the smallest element of S, then z :5 a < f(a) for all a E S, therefore z ¢ f(S).

1.9.6. Lemma. If A is a finite subset of J!>, then CA largest element.

= II" -

A has no

Proof. Write B = CA. Arguing contrapositively, let us show that if B has a largest element m, then CB is infinite. By assumption, B C {I, ... , m} = J!>m, therefore CB:::> CJ!>m = {k E J!>: k > m}; since ClI"m is infinite (consider the map k....., k + 1 ), so is its superset CB (1.9.3).

1.9.7. Lemma. If A is a finite subset of J!>, then CA is infinite. Proof. By the preceding lemma, CA has no largest element, therefore is infinite (1.9.5).

1.9.8. Lemma. J!> is not the union of two finite sets. Proof. Assuming J!> = AU B with A finite, we must show that B is infinite. We have B:::> J!> - A and J!> - A is infinite (1.9.7), therefore so is its superset B.

1.9.9. Lemma. If f : E f(A) is finite.

-+

F and A is a finite subset of E, then

Proof. The restriction of f to A defines a surjection A therefore f(A) is finite by (4) of 1.9.3.

-+

f(A) ,

1.9.10. Lemma. If A and B are finite sets, then Au B is finite. Proof. Let E = AuB (cf. 1.4.1) and assume to the contrary that E is infinite. By 1.9.4, there exists an injection f: J!> -+ E, therefore a surjection g: E -+ II" (1.5.4); then J!> = g(E) = g(A) u g(B) is the union of two finite sets (1.9.9), contrary to 1.9.8.

1.9.11. Lemma. For every positive integer n, II"n is finite. Proof. (by induction on n). 11"1 = {I} and II"n+l 1.9.2, (3) and the preceding lemma.

= II"n U {n + I}; cite

1.9.12. Lemma. Let m, nEil". If there exists a bijection II"n then m=n.

-+

Il'm,

§1.9. Finite and Infinite Sets

37

Proof. Let I: II"n -+ II"m be a bijection. We can suppose that n ~ m (if m ~ n, consider instead the inverse bijection I-I); then II"n:::> II"m. Consider the mappings

I

i

where i is the insertion mapping. The composite function i 0 I : II"n --> II"n is injective (because I and i are), therefore bijective (1.9.11); then (i 0 f) 0 /-1 = i is also bijective, therefore II"m = II"nand m = n (k is the largest element of II"k).

1.9.13. Theorem. A nonempty set is finite if and only if it is bijective with some II"n . Proof. "If': Suppose /: E --> II"n is bijective; since II"n is finite (1.9.11), so is E (1.9.3). "Only if": Arguing contrapositively, if E is a nonempty set that is not bijective with any II"n, we must show that E is infinite. By 1.9.4, it suffices to find an injection II" --> E, in other words, a sequence (x n ) in E such that n>--+ X n is injective (that is, the X n are pairwise distinct). An informal, 'recursive' argument for producing such a sequence is as follows. Choose XI E E and let EI = {xI}. Let n ~ 1 and assume already chosen distinct points Xl, ... , X n of E. Let En = {Xl>" . ,xn ) . Then i >--+ Xi is a bijection II"n --> En , so by hypothesis En # E; choose Xn+l E E - En. "And so on .... "3

1.9.14. Definition. If a set E is bijective with some II"n then n is unique (1.9.12); we then say that E has cardinality n and we write card E = n. Convention: card 0 = O. In particular, card II"n = n for all nEil". The use of the symbol card E in connection with infinite sets E is discussed in §1.12. In the next section, we study the infinite set II" in greater detail.

Exercises

1. For a nonempty set E, the following conditions are equivalent: (a) E is finite; (b) every surjection E --> E is injective. 2. Let E I , ... , E r be nonempty sets and let E that E is finite if and only if every Ek is finite.

= EI X

...

x

Er.

Show

3 For a formal (Le., honest) argument, see Hewitt and Stromberg (op. cit.). p. 22, (4.15). or P. R. Haimes. Naive set theory (Springer.Verlag, New York, 1974). pp. 60-61.

1. Foundations

38

3. (i) Every nonempty finite partially ordered set has at least one minimal element (1.7.10). (ii) A finite pre-ordered set need not have a minimal element. 4. As an alternative to Definition 1.9.1, call a set E infinite if there exists an injection II" --+ E, and finite if either E = (/) or there exists a surjection {I, ... , n} --+ E for some nEil". Work out a proof that a set is infinite if and only if it is not finite.

1.10. Countable and Uncountable Sets 1.10.1. Definition. A set E is said to be countable if either E = (/) or there exists a surjection II" --+ E. {In other words, either E = (/) or there exists a sequence (x n ) in E such that every point of E is equal to at least one term of the sequence.} If E is not countable it is said to be uncountable. (We shall see in 1.10.11 that the field JR of real numbers is uncountable.) 1.10.2. Remarks. 1. If E is countable (uncountable) and if f: E --+ F is bijective, then F is also countable (uncountable). 2. Every subset of a countable set is countable (hence every superset of an uncountable set is uncountable). {Proof: Suppose f: II" --+ E is surjective and A is a nonempty subset of E; we have to find a surjection II" --+ A. Choose a point a E A and let 9 : II" --+ A be the function such that g(n) = f(n) if f(n) E A, and g(n) = a if f(n) ¢ A. For every x E A, there exists an n with f(n) = x, therefore x = g(n); thus 9 is surjective.} 3. Every finite set is countable. {A nonempty finite set is bijective with some II"n (1.9.13), hence is countable by Remarks 1 and 2.}

On the way to characterizing the countable sets: 1.10.3. Lemma. Every infinite subset of II" is bijective with 11". Proof. Assuming A C II" infinite, we must construct a bijection f : II" --+ A . Define f(l) to be the smallest element of A and, recursively, f(n + 1) to be the smallest element of A - (f(I), ... ,f(n)} . It is obvious that f is injective. Better yet, it is strictly increasing. {Proof: f(2) > f(I) because f(I) is the smallest element of A and f(2) f. f(I); f(3) > f(2) because there are no elements of A between f(I) and f(2) , and none less than f(I). In general, there are no elements of A between f(i) and f(i+ 1) for any i, whence f(n+ 1) > f(n).} An easy induction then shows that f(n) > n for every n. To complete the proof, we need only show that f is surjective. Assume to the contrary that A - f(lI") contains some element k. In particular,

§l.IO. Countable and Uncountable Sets

39

k E A - {I(I), ... ,f(k)} , so k";? f(k + I) by the minimality of f(k + 1), whence the absurdity k + 1 $ f(k + 1) $ k. 0 1.10.4. Theorem. A set E is countable if and only if either (1) E is bijective with IP, or (2) E is bijective with IPn = {1, ... , n} for some positive integer n, or (3) E = 0 .

Proof "If": That a set satisfying (1) or (3) is countable is obvious from the definition, and a set satisfying (2) is countable by Remarks 1 and 2 of 1.10.2. "Only if": Suppose E f. 0 and f: IP -+ E is surjective; we have to show that E satisfies either (1) or (2). If E is finite, we are done (1.9.13). Suppose E is infinite. For every x E E let Ax = f- 1 ( {x}) (a nonempty subset of IP, since f is surjective) and let g(x) be the smallest element of Ax ; this defines a function g: E -+ IP', injective since Ax n A y = 0 when x f. y. Since 9 is injective, E is bijective with g(E); since E is infinite, so is g(E); thus g(E) is bijective with IP by the lemma, therefore soisE.O 1.10.5. Definition. In view of 1.10.4, a set that is bijective with IP' is said to be countab1y infinite (or denumerab1y infinite, or, simply, denumerable) . The rest of the section is devoted to examples of countable and uncountable sets. The following result is a ready source of uncountable sets: 1.10.6. Theorem. If E is a set and P(E) is its power set (1.4.1), then there does not exist a surjective mapping E -+ P(E) .

Proof {For finite sets, this is not news: If E has n elements then P(E) has 2n elements, and n < 2n . If E = 0 then P(E) = {0} is nonempty, so even if one admitted the empty mapping 0 -+ 1'(0), it would not be surjective.} Assume to the contrary that there exists a set E that admits a surjective mapping f: E -+ PiE) . Let A = {x E E: x

f/. f(x)}.

Since f is surjective, A = f(a) for some a E E. Either (i) a E A or (ii) a f/. A, but both alternatives lead to a contradiction: (i) if a E A = f(a) then a f/. A by the definition of A; (ii) if a f/. A = f(a) , then a E A by the definition of A. 0 1.10.7. Corollary. P(IP') is uncountable.

Proof By the theorem, there exists no surjection IP' -+ P(IP'). 0 1.10.8. Theorem. IP x IP is bijective with IP'.

1. Foundations

40

Proof. The mapping f: J!> x J!> -> J!> defined by f(m, n) = 2m 3n is injective, so J!> x J!> is bijective with its range A = f (J!> x J!» ; but A is infinite (the mapping m ...... 2m is an injection Jl' -> A), therefore A is bijective with J!> (1.10.3). 0

1.10.9. Corollary. If E and F are countable sets then so is Ex F. Proof. If E = 0 or F = 0 then Ex F = 0. Assuming E and nonempty, by hypothesis there exist surjections g: Jl' -> E and h: J!> F; the mapping Jl' x Jl' -> E x F defined by (m, n) ...... (g(m), h(n» surjective, and by the theorem there exists a surjection Jl' ..... Jl' x Jl', composition yields a surjection Jl' -> E x F. 0

F ..... is so

By an obvious induction, every finite product E I x ... x En of countable sets is countable. 1.10.10. Corollary. The field Q of rational numbers is denumerable. Proof. If S = {min: m, nEil'} (the set of all positive rational numbers), then Jl' C S and the mapping (m, n) ...... min is a surjection J!> x Jl' -> S, thus S is infinite and countable (1.10.8), therefore denumerable (1.10.3). Let f : Jl' -> S be a bijection and write r n = f(n) (n E Jl'). Then the sequence

exhausts Q and defines a bijection J!> ..... Q. 0 1.10.11. Corollary. The field JR of real numbers is uncountable. Proof. It suffices to show that (0, IJ is uncountable (1.10.2). For this, we need only show that every mapping f: J!> -> [0, 1J fails to be surjective. For any closed interval [a, bJ, let c and d be its points of trisection (a < c < d < b) and call la, c), [c, dJ, [d, bJ the closed thirds (left, middle and right, respectively) of [a, bJ. Define recursively a decreasing sequence II ::> 12 ::> h ::> ... of nondegenerate closed subintervals of [0, I} as follows: II is a closed third of (0, 1) such that f(l) 1- II (to be definite, we could choose II to be the left-most third with this property), and, recursively, In+! is a closed third of In such that f (n + 1) i In+! . By the theorem on nested intervals (1.8.27), the intersection In contains a point x (in fact, only one). For every n, we have x E In, therefore f(n) f x, thus x fails to belong to the range of f. 0

n

Exercises 1. A finite product of denumerable sets is denumerable.

2. For every infinite set E, P(E) is uncountable. {Hint: 1.9.4 and 1.10.7.} 3. The union of a sequence of countable sets is countable.

§1.11. Zorn's Lemma, Well-Ordering

41

1.11. Zorn's Lemma, the Well-Ordering Theorem

The Axiom of Choice (AC), introduced in the context of product sets (§1.4), already figures in our discussion of the algebra of functions (1.5.1, 1.5.2) and the mapping properties of finiteness and infiniteness (see (4) and (4') of 1.9.3). In this section we record several other statements, equivalent to the Axiom of Choice, that have many applications in algebra, topology and analysis. Ideally, such matters are taken up in a leisurely preliminary course l ; realistically, the reader who has had such a course is probably not reading this chapter. A detour into the foundations of mathematics (the homeland of the axioms of set theory) would run the risk of turning into a derailment; the present section is a skeletal compromise: the equivalent forms of the Axiom of Choice are stated for reference, omitting the proofs that they are equivalent but including some sample applications. First, a little motivation. One equivalent axiom (AC') already mentioned asserts the existence of 'choice functions' (1.4.6); here is a situation where no axiom is needed because a theorem is available: 1.11.1. Theorem. Every nonempty countable set admits a choice function. Proof. Assuming f : IP ---> E surjective (1.10.1) we seek a function c : P(E) - {12>} ---> E such that c(A) E A for every nonempty subset A of E. For each nonempty AcE, f-I(A) is a nonempty subset of IP; by the principle of mathematical induction (1.7.12), f-l(A) has a smallest element minf-l(A), and the formula c(A) = f(minr1(A)) defines a choice function for E. {The structure of the proof is that IP has a choice function (namely S ...... min S), therefore so does every functional image of IP.} 0 The property of the ordered set IP that makes the foregoing argument work is distilled in the following definition:

1.11.2. Definition. A partially ordered set (X, $) is said to be wellordered if every nonempty subset of X has a smallest (or "first") element; that is, if 12> i' SeX then there exists an element a E S such that a::; x for all xES. Such a relation ::; is called a well-ordering of X. 1.11.3. If E is a nonempty set that can be indexed (1.3.8) by a wellordered set, then E admits a choice function (cf. the proof of 1.11.1). The following property of well-ordered sets is the basis for inductive arguments in such sets: I Based, for example, on J. Kaplansky's Set theory and metric spaces [Chelsea, New York, 1977) or the first two chapters of E. Hewitt and K. Stromberg's Real and abstmct analysis [Springer-Verlag, New York. 196~1·

42

1. Foundations

1.11.4. Theorem. (Principle of transfinite induction) Let E be a wellordered set. If S is a subset of E satisfying the condition

(y < x

~

~

yES)

xES

then S=E. Proof. The meaning of (*) in words: if x is an element of E such that S contains every element of E that is smaller than x, then S must also contain x. {There is no harm if S = 0; this simply forces E = 0 .} Arguing contrapositively, it is the same to show that a proper subset of E cannot have the property (*). Let S be a proper subset of E and let x be the smallest element of E - S. If y < x then y If. E - S by the minimality of x, thus yES; S contains every element smaller than x but it does not contain x, SO S does not have the property (*).

The most accessible applications of transfinite induction are to the theory of well-ordered sets itself; numerous examples are given in §1.13. Georg Cantor (1845-1918), the founder of the theory of sets, conjectured (in 1883) that every set can be equipped with a well-ordering. This conjecture,

(WO) Every set can be well ordered is known as the well-ordering theorem. In 1904, Ernst Zermelo (18711953) proved that if the Axiom of Choice is admitted then Cantor's conjecture is true. Zermelo's theorem is the implication (AC) ~ (WO). The reverse implication is trivial (1.11.3), thus 1.11.5. (Zermelo's) Theorem. (AC)

¢}

(WO).

An equivalent axiom, frequently easier to apply than the Axiom of Choice or well-ordering, was proposed in 1935 by Max Zorn (1906-1993); this axiom, now known as Zorn's Lemma, is oonveniently stated in terms of the following concept:

1.11.6. Definition. A partially ordered set (X,:5) is said to be inductive if every simply ordered subset of X has an upper bound in X. Zorn's lemma: (ZL) A nonempty, inductive partially ordered set has at least one maximal element. To summarize, (AC)

¢}

(ZL)

¢}

(WO).

Two other eqUivalents of the Axiom of Choice are (H) Hausdorff's maximality principle, and (T) Thkey's lemma. Hausdorff's principle is the following proposition: (H) Let (X,:5) be any partially ordered set; let X be the set of all simply ordered subsets of X, and order X by inclusion. Then X has a

§l.ll. Zorn's Lemma, Well-Ordering

43

maximal element. (Briefly, every partially ordered set contains a maximal chain.) 'lUkey's lemma involves the following concept: 1.11.7. Definition. A set F of sets is said to be of finite character if, for a set A, AEF

$}

every finite subset of A belongs to F.

'lUkey's lemma: (T) Let F be a nonempty set of sets, order F by inclusion, and suppose that F is of finite character. Then F has a maximal element. Packaging it all into one statement, 1.11.8. Theorem. (AC)

$}

(ZL)

$}

(T)

~

(H)

~

(WO).

Variously called "axiom", "lemma", "principle", "theorem", each of these statements is in fact an axiom, consistent with and independent of the most widely accepted system of axioms for mathematics2 . The reader will find elegant and efficient proofs of these equivalences in the books of Kaplansky and Hewitt-Stromberg cited earlier. (The details are "elementary"-it is easy to follow the proofs step by step-but fiendishly ingenious.) The best way to get a feeling for the axioms is to work through some applications; we conclude the section with two such applications, the first to vector spaces (every vector space has a basis), the second for use later On in the theory of cardinality (given any two sets, one of them contains a copy of the other). 1.11.9. Theorem. Every vector space has a basis. Proof. # 1 (assuming the well-ordering theorem). Let V be a vector space. A subset A of V is (linearly) independent if nO element of A is a linear combination of the remaining elements of A, generating if every vector in V is a linear combination of elements of A, and a basis of V if it is both independent and generating. If V contains only the zero vector 8 then the empty set serves as basis. Assuming V '" {8} , well-order the nonempty set V - {8} and let

B

= {x E V -

{8}: x is not a linear combination of vectors < x };

we will show that (i) B is independent, and (ii) B is generating. (i) Assuming to the contrary that B is not independent, there exists a linear relation CI Xl + ... + c"xn = e with Xl, •• ·, x n distinct elements of Band Cl>"" c" nonzero scalars. If Xj is the largest of these vectors, then Xj is a linear combination of vectors < Xj ,therefore Xj ¢ B (by the definition of B), a contradiction.

21. Kaplan.ky, op. cit, p. 59.

44

1. Foundations

(ii) Assume to the contrary that some vector in V fails to be a linear combination of elements of B (hence is nonzero) and let x be the smallest such vector. In particular, x ¢ B, therefore x is a linear combination x = C1Xl + ... + CnXn with Xi < x for all i. By the minimality of x, every Xi is a linear combination of elements of B; but then so is x, a contradiction.

Proof. # 11 (assuming Thkey's lemma). As in the preceding proof, we can suppose that V is a vector space i {II} . Let S be the set of all independent subsets of V (for example, {x} E S for every nonzero vector x). It is clear from the definition of independence that S is of finite character, that is, a set A C V is independent if and only if every finite subset of A is independent. By Thkey's lemma, S has an element B that is maximal (for the inclusion relation). In particular, B is independent, so we need only show that it is generating. Indeed, if some vector x E V fails to be a linear combination of elements of B (so that, in particular, x ¢ B), then B U {x} is an independent set containing B properly, contrary to the maximality of B. 1.11.10. Before giving an application of Zorn's lemma, we will exhibit the

inductive sets that enter into the proof. First, some notations are in order. Given a pair of sets E and F, we consider the set F of all functions u : A -> F defined on subsets of A of E. A nonempty subset G of E x F is the graph of such a function u: A -> F if and only if it has the property

(x,y), (x,y') E G ~ y

= y',

in which case A=pr.G={x: (x,y)EG for some YEF} and, for x E A, u(x) is the unique element of F such that (x, u(x)) E G. We write Gu for the graph of a function u E F, and (; for the set of all graphs obtained in this way:

the mapping u...... G u is a bijection F -> (;. The set (; is partially ordered by inclusion (because P(E x F) is); explicitly, the meaning of G u C G v is that v is an extension of u, that is, the domain of u is contained in the domain of v (because prl G u C prl G v ) and, for every x in the domain of u, u(x) = v(x). We also consider the set F;nj

and write

(;;nj

= {u E F: u is injective}

for the corresponding set of graphs.

§I.ll. Zorn's Lemma, Well-Ordering

45

I.ll.ll. Lemma. With the preceding notations, the partially ordered sets Q and Qinj are inductive. Proof. We first prove that Q is inductive. Assuming C is a chain in Q, we must show that C has an upper bound in Q. Let H = UC be the union of all the sets G E C . Since H:> G for all G E C , it will suffice to show that H is the graph of a function. Suppose (x, y), (x, y') E H, say

(x, y) E G E C and (x, y') E G' E C.

Since C is a chain, either G c G' or G' c G; if, for example, G c G' , then (x, y) and (x, y') both belong to G', therefore y = y' because G' is the graph of a function. Finally, assuming in addition that C C Qinj , we must show that H is the graph of an injective function. Suppose (x,y), (x',y) E H; we are to show that x = x' . Say (x,y) E G E C and (x',y) E G' E C; if, for example, G C G', then x injective function.

= x'

because G' is the graph of an

I.ll.12. Theorem. If E and Fare nonempty sets, then either there exists an injection E ..... F or there exists an injection F ..... E. Proof (using Zorn's lemma). With notations as in the lemma, we know

that Qinj is inductive, so by Zorn's lemma it has a maximal element G. Say G is the graph of u: A ..... F. If A = E we are done: u is an injection E F. Similarly, if u(A) = F then u(a) ...... a (a E A) is an injection F E. Thus, we need only show that either A = E or u(A) = F. Assume to the contrary that A oF E and u(A) oF F. Choose elements x E E - A and y E F - u(A); then G U {(x,y)} is the graph of an injection Au {x} ..... F, which contradicts the maximality of G.

Exercises 1. Every finite simply ordered set is well-ordered.

2. Let (Xi)iEI be a family of partially ordered sets (1.7.1) and suppose that the index set I is well-ordered. Let X = Xi be the product set. For points x = (Xi), Y = (Yi) of X, define x:5 y to mean that either (1) x = y, or (2) x oF y and Xj < Yj for the smallest index j such that Xj oF Yj (cf. §I.7, Exercise 4). (i) The relation :5 in X is a partial ordering (called the lexicographic ordering). (ii) If the Xi are all simply ordered, then so is X.

n

46

1. Foundations

3. If X and Yare well-ordered sets, then X x Y is well-ordered for the lexicographic ordering. {Hint: If A is a nonempty subset of X x Y, then the set prl A of first coordinates of the elements of A is nonempty. Let Xl be the first element of prl A , let B be the set of all elements of A whose first coordinate is Xl, and let Yl be the first element of pr2 B. Argue that (Xi> ytl is the first element of A.}

4. If Xi> ... , Xn are well-ordered sets, then Xl x .. . xX n is well-ordered for the lexicographic order. {Hint: Exercise 3 and induction (cf. §1.7, Exercise 4).} 5. In Exercise 2, consider I = IJ> and Xn = {1,2} (n E IJ», all with the usual order, and let X = Xn be the product set, equipped with the lexicographic order. If en E X is the element with 2 in the n'th coordinate and 1's elsewhere, then el > e2 > e3 > ... ,thus X is not well-ordered.

n

1.12. Cardinality The concept of cardinality has been broached in connection with finite sets (1.9.14); our aim in this section is to reformulate the discussion in terms applicable to arbitrary sets. The discussion of cardinality for finite sets is expedited by the fact that we have at hand a full list of 'models' (up to a bijection) of such sets: 0, {I}, {1,2}, {l,2,3}, .... Thus, when we observe that a finite set E is bijective with 11"4 = {I, 2, 3, 4}, we already have in hand a symbol 4 that we can declare to be 'the number of elements of E' (or the 'cardinal number of E '). For infinite sets, we will have to make up such symbols as we go along. In the absence of symbols, the place to begin is with bijections.

1.12.1. Definition. For sets E and F, we write E ~ F if there exists a bijection E -+ F. {Thus E ~ F can be verbalized as 'E is bijective with F '; the term' E and F are equipollent' is also used.} Convention: P(E) defined by the formula

I{)(8) = Cg(Cf(8» for all 8 C E. With P(E) ordered by inclusion, I{) is order-preserving:

8

cT

=}

I{)(8)

c

I{)(T) .

{Reason: Mappings such as 8 ...... f(8) preserve inclusion, whereas complementation reverses inclusion.} Let

S

= {8 E P(E):

I{)(8)::> 8 }

(for example, (/) E S). Note that S is invariant under I{): if I{)(8) ::> 8 then I{)(I{)(8»::> I{)(8) , thus 8 E S I{)(8) E S. Let

'*

A=US=U{8: 8ES}; we will show that I{)(A) = A. For all 8 E S we have A::> 8, therefore lp(A) ::> I{)(8) ::> 8; thus lp(A) ::> US = A. This shows that A E S, therefore also I{)(A) E S, whence I{)(A) C US = A. 0 1.12.7. Corollary. If S is any set of sets, E ~ F is the equivalence relation in S defined by equipollence, and IE] denotes the equivalence class of a set E E S (so that [El = IF] ~ E ~ F), then the relation [E] :5 IF] defined in the quotient set E/~ by E ~ F is a partial ordering of E/~.

Proof. Immediate from 1.7.7, 1.12.5 and the theorem. 0 8uppose E ~ F. By 1.9.3, E is finite ~ F is finite. More precisely, if E ~ n>,. for some n then F ~ n>,. and cardE = n = cardF (1.9.14);

§1.l2. Cardinality

49

and if E = 0 then F = 0 and card E language to arbitrary sets:

= 0 = card F.

Let's extend this

1.12.8. Definition. If E - F we say that E and F have the same cardinality and we write card E = card F . What has been defined? A term ('same cardinality') and a notation ( card E = card F ). Implicit in this notation is that we are associating with each set E a symbol card E, and agreeing that two such symbols can be separated by the symbol "=" if and only if the sets to which the symbols correspond may be separated by the symbol "_" for equipollence. 3 In 1.9.14, cardIPn is in effect abbreviated to n.

1.12.9. Definition. Symbols card E (and abbreviations for them) are called cardinal numbers. If u is a cardinal number (that is, u abbreviates card F for some set F) and E is a set with card E = u (in other words, E - F), we say that E has cardinality u (or that u is the cardinal number of E). Thus, a set of cardinal numbers is the set of symbols card E associated with some set S of sets E (however, if E E S and F ~ E, S is not obligated to contain F). 1.12.10. Theorem. If u, v, w are cardinal numbers, then (i) u = U; (ii) u = v ~ v = U; (iii) u = v & v = w ~ u = w.

Proof. This is a rerun of 1.12.2.

Thus, in any set S of cardinal numbers, the relation u = v introduced in 1.12.8 and 1.12.9 is an equivalence relation. Here are two especially useful notations (abbreviations, from our perspective):

1.12.11. Definition. We write No for cardIP, and c for cardJR (called the cardinal of the continuum). The symbol No is read "aleph nought" (aleph is the first letter of the Hebrew alphabet). The reason for the subscript zero is that No is, in an appropriate sense, the first infinite cardinal number (more about this in the present section-see (iv) of 1.12.21-and in the next two sections). The only thing new in the following theorem is the language of cardinality: 1.12.12. Theorem. A set E is denumerable if and only if its cardinality is No. There is no proof for the preceding "theorem"; this is just Definition 1.10.5 in a new suit of clothes. So far we have just wrapped the language of 3This is called the l'axiom for cardinal numbers" in the book of P. Suppes [Axiomatic set theory, Van Nostrand, Princeton, 1960j reprinted Dover, New York, 1972].

50

1. Foundations

cardinality around the relation of equipollence for the relation of domination (~):

1.12.13. Definition. If cardE ).

E~F

(~);

now let's do the same

we write cardE:5 cardF (or cardF 2:

Note that u 2: 0 for every cardinal number u; for, if u = card E then by convention. In any set of cardinal numbers, the relation < is a partial ordering:

I/) ~ E

1.12.14. Theorem. If u, v, ware cardinal numbers, then (i) u = v ~ u:5 v; in particular, u:5 u; (ii) u :5 v & v :5 w ~ u:5 w; (iii) u :5 v & v :5 u ~ u = v.

Proof. (i), (ii) A rerun of 1.12.5. (iii) This is the Schroder-Bernstein theorem (1.12.6). 1.12.15. Definition. For cardinal numbers u and v we write u < v (as in 1.7.5) if u:5 v and u'l v (that is, u:5 v is true and u = v is false). The relation u < v is also written v> u. From this definition and (i) of 1.12.14, it is immediate that

u:5v

¢}

u No . Conversely, if card E > No then II" ;:$ E but II" 1- E, that is, E is infinite but not denumerable; if E were countable then (being not denumerable) it would be finite by 1.10.4, a contradiction. 0 The proof of the preceding theorem could be shortened slightly, at the cost of using the Schroder-Bernstein theorem and the Axiom of Choice. For

52

1. Foundations

example, the equivalence (iv) is the contrapositive form of the equivalence (iii); for, the left sides are each other's negations by definition (1.9.1), whereas the right sides are also each other's negations: the condition card E > No means that either card E > No or card E = No (as remarked in 1.12.15), so its negation is card E < No by trichotomy (1.12.20). Similarly, (v) is the contrapositive form of (ii). The proof of (ii) itself can be shortened: for E '10, the left side of (ii) says that there exists a surjection Il' -+ E, and the right side says that there exists an injection E -+ Il'; the two sides are therefore equivalent by 1.5.4 (whose proof requires the Axiom of Choice). The theorem suggests the following terminology: 1.12.22. Definition. A cardinal number u is said to be finite if u < No, infinite if u ~ No , countable if u $ No , and uncountable if u > No . The set N = {O, 1,2, 3, ...} of nonnegative integers is a set of abbreviations for the finite cardinals card 0 and card Il'n (n = 1,2,3, ... ); we may simply say that N is the set of finite cardinals.

Exercises 1. For nonempty sets E and F, E t F if and only if there exists a surjection E -+ F . 2. For a pair of nonempty sets E and F, there exists a bijection E -+ F if and only if there exist an injection E --+ F and a surjection E -+ F .

1.13. Cardinal Arithmetic, the Continuum Hypothesis If u and v are cardinal numbers, we propose to define uv, u + v and v" (in that order). Say u = card E, v = card F. If u and v are finite, say u = m = card Il'm' v = n = card Il'n' we expect uv = mn = card(ll'm x Il'n). In general, if E ~ E ' and F ~ F ' ,then E x F ~ E' X F ' , so the following definition is unambiguous (i.e., single-valued):

1.13.1. Definition. If u = card E and v = card F , we define uv

= card(E x

F) .

1.13.2. Remarks. (i) uv = 0 ¢> u = 0 or v = O. {Reason: Ex F = 0 ¢> E = 0 or F = 0.} (ii) uv = vu. {If E and Fare nonempty then E x F ~ F x E via the mapping (x, y) >-+ (y, x).} (iii) (uv)w = u(vw). {For nonempty sets E, F, G, (E x F) x G ~ Ex (F x G) via ((x, y), z) >-+ (x, (y, z)).} (iv) If u < v and u' $ Vi ,then uu' $ w' .

§1.l3. Cardinal Arithmetic

53

(v) If v> 0 then uv;::: u. {If bE F then E::::: Ex F via the mapping x ..... (x,b).} (vi) No No = No. { IP' x IP' ~ IP' by l.10.8.} If E and F are finite and E n F = {O,l} and A = {x E E: I(x) = 1}, then I is the characteristic function CPA of A; thus {O, l}E is the set of all characteristic functions of subsets of E, and P(E) ~ {O,l}E via the correspondence A ...... CPA. For this reason, the power set P(E) is sometimes written 2E . (iii) For nonempty sets, if E ~ E' and F ~ F' ,then FE ~ F'E' . This justifies the following definition: 1.13.12. Definition. Let u and v be nonzero cardinal numbers, say u = card E and v = card F; we define V U = card(FE) . Thus,

card(F E) = (card F)card E . Conventions: 0° = VO = 1 and OU = 0 (cf. 1.5.5): The only relation (/J C (/J xF but the only relation (/J C Ex (/J from E the number of qualifying 'functions' is 1

(recall that uf.O). {Rationale from (/J to F has domain (/J, to (/J does not have domain E; and 0, respectively.}

1.13.13. Examples. (i) If u = card E, then card P(E) = 2U (even if u = 0). {See 1.13.11, (ii).} (ii) (Cantor's theorem) u < 2U for every cardinal u (1.12.17). (iii) v u + w = VUV W • {For nonempty sets E, F, G, with EnG = 1/), every function I: E u G -> F corresponds to a pair (g, h) of functions g:E--+F, h:G->F via the formulas g=/IE, h=/IG.} (iv) (VU)W = V UW • {For nonempty sets E, F, G, every function I : G -> F(E, F) corresponds to a function 9 : E x G -> F via the formula g(x, z) = If(z)](x) (the value of the function I(z): E -> F at the point x E E).}

§1.13. Cardinal Arithmetic

57

1.13.14. Theorem. For nonempty sets E and F, FE is the product of card E copies of F, in the sense that there is a natural bijection

Proof. On the right side, we are considering the family of sets (F%)%EE with F % = F for all x e E. An element f: E - t F of the left side corresponds to the point (J(X»%EE ofthe product set on the right. When F = {O, I} , this yields the formula

1.13.15. Theorem. 2No = c.

Proof. By the preceding formula (with E = IP), 2No is the cardinal number of the set of all sequences (Xl> X2, X3,' ..) with X n = or 1 for all nelP. We show first that [0, 1] ~ lR, that is, the cardinality of the closed interval [0,1) is c. Since (0,1) ~ (-1,1) (via the mapping x>-+ 2x -1) and (-1, 1) ~ lR (via the mapping x>-+ x(I-X2)-1), we have (0, 1) ~ IR and need only show that [0, 1) ~ (0,1); the latter follows from

°

[0,11 ~ [!,~) C (0,1) C [O,lJ and the Schroder-Bernstein theorem. We are now reduced to proving that cp: {O,I}P

21 0) if E '" b E F, and we write 0 ~ {3 (or (3;:O: 0) if either 0 < Convention: 0 < 0 for every nonzero ordinal number 0 E i' etJ ); then 0 ~ 0 for every ordinal number o.

say 0 = ordE F(b) for some {3 or 0 = {3. (that is, when

If also 0 = ord E' and {3 = ord F' , so that E '" E' and F '" F' , it is clear that E is similar to an initial segment of F if and only if E' is similar to an initial segment of F', thus the relations 0 < {3 and 0 ~ {3 are well-defined. Also, the relations 0 < {3 and 0 = (3 are mutually exclusive, since 0 < 0 is ruled out by 1.14.10. We remark that if ord E < ord F then E '" F( b) for a unique b E F (1.14.12), and there is only one order isomorphism implementing the similarity E", F(b) (I.l4.3). 1.14.16. Lemma. The relation 0 ~ {3 of 1.14.15 is a partial ordering of every set of ordinal numbers; that is, for ordinal numbers 0, {3, 'Y, (i) 0 ~ 0, (ii) 0 ~ {3 & {3 ~ 'Y ~ 0 ~ 'Y, (iii) 0 < {3 & {3 ~ 0 ~ 0 = {3.

Proof. (i) In fact

= o. = {3 or

or 0 {3 = 'Y, the implication is obvious. Suppose o < 0, 0 < {3 and {3 < 'Y; say 0 = ord E, (3 = ord F, 'Y = Old G. By assumption, E '" F(b) and F '" G(c) for suitable elements band c; if d is the image of b under the order isomorphism that implements F '" G(c) , then d < c and F(b) '" (G(c»)(d) = G(d) , whence E '" G(d) , so that 0 < 'Y . (iii) Let 0 ~ {3 and {3 ~ 0 and assume to the contrary that 0 i' {3. Then 0 < {3 and {3 < 0, and the proof of (ii) yields the contradiction (ii) If

0

=0

0

0 ordEs = a. 1.14.33. Definition. With notations as in the preceding theorem, we call a the initial ordinal associated with the cardinal number u and we write a = inord u . For example, inord No = w by 1.14.30, (v); by the same token, w + 1 is not of the form inord u for any cardinal u (the only candidate for u is No, and it doesn't work).

1.14.34. Remark. Combining the preceding two definitions, we see that card(inord u) = u for every cardinal number u, whereas inord(card a) :'S a for every ordinal number a (with equality only when a is an initial ordinal). No is the first infinite cardinal and w = inord No (which might well be denoted wo) is the first infinite ordinal. Let us take the next step: 1.14.35. Theorem. There exists a smallest uncountable cardinal number (it is denoted I'll).

Proof. Consider c = 2No

= card P(J\l)

S = {card A: A

C

(see 1.13.15), let

P(J\l), A uncountable},

70

1. Foundations

(for example, c E S) and let N1 be the smallest element of S (1.14.25); say N1 = ord B ,where B c P(IIi) is uncountable. It will suffice to show that if u is any uncountable cardinal, then NI $ u. Assume to the contrary that u < N1 (1.12.20). Say u = card E. We know from u < NI that E ~ A for some A c B . Since E is uncountable, so is A, therefore cardA E Sand N1 $ cardA = cardE = u, contrary to u < N1 . Obviously N1 $ 2No ; the Continuum Hypothesis is the proposition that N 1 = 2No . In any case, inord N1 is clearly the smallest uncountable ordinal; we have arrived at "big omega" (also denoted WI): 1.14.36. Definition.

n=

inord N1 •

This train goes on forever, but this is where we get off.

Exercises E an order morphism. If there exists at least one point x E E such that f(x) $ x, then f has a fixed point (that is, f(z) = z for some z E E). {Hint: Let T = {x E E: f(x) $ x}, note that f(T) C T and consider the first element of T.} 1. Let E be a well-ordered set, f: E

-+

2. If E is a well-ordered set, then every nonempty subset of E that is

bounded above has a least upper bound. 3.

(0
IR. Explicitly, write a~ = -an and let b~, b' , c'n' c' be the quantities computed from (a~) analogous to the quantities bn , b, Cn, C computed from (an). Since {a~: k~n}={-ak: k~n}

we have b'n = -Cn, whence b' = -c; this is the first of the desired formulas, and the second follows from it by the substitution an ...... -an.

A useful characterization of the limit superior: 1.16.4. Theorem. Let (an) be a sequence in JR and let b = lim sup an .

Then (1) r (2) r

> b => an < r ultimately; < b => an > r frequently.

Moreover, these two properties determine b uniquely. Proof. "Ultimately" means "from some index onward"; "frequently"

means "for infinitely many indices".

§1.l6. Convergence in IR

81

Proof of (1): Let r be an extended real number with r > b (if b = +00 then no such r exists and the implication is vacuously true); we seek an index m such that ak < r for all k ~ m. Since r > b and b is the largest minorant of the set B = {bn : n E JP>}, r is not a minorant of B; therefore r > bm for some m. Since bm = sup{ ak: k ~ m} , we have k ~m ak :;:; bm < r, whence the assertion (1). Proof of (2): The negation of «an > r frequently" is «an :;:; r ultimately"; arguing contrapositively, let us show that

'*

an :;:; r ultimately Assuming ak :;:; r

for all k

k~m} r frequently by (2'); these two statements are contradictory. Arguing dually (or combining the preceding theorem with 1.16.3) we have 1.16.5. Theorem. Let (an) be a sequence in 1R, and let c = lim inf an' Then: (3) r < c an > r ultimately; (4) r > c an < r frequently. These two properties determine c uniquely.

'* '*

Always lim inf an < lim sup an ; when do we have equality? Theorems 1.16.4 and 1.16.5 yield the following criterion: 1.16.6. Theorem. For a sequence (an) in 1R, the following conditions are equivalent: (a) lim inf an = lim supan ; (b) there exists an extended real number a such that (i) r > a an < r ultimately, and (ii) r < a an > r ultimately. When the conditions hold,

'* '*

a

= lim inf an = lim sup an

(in particular, a is unique).

'*

(b): We are assuming that c = b; write a = c = b. Proof. (a) Condition (i) is (1) of 1.16.4, and condition (ii) is (3) of 1.16.5. (b) (a): Assuming a E IR satisfies (i) and (ii), we must show that a = b and a = c. To prove, for example, that a = b, it suffices to verify that a satisfies (1) and (2) of 1.16.4, and these are immediate from (i) and (ii).

'*

1. Foundations

82

1.16.7. Corollary. If (an) is a bounded sequence in JR, then liminfa n

= lim sup an

**

(an) is convergent in JR.

When this is the case, lim an = lim inf an = lim sup an .

n_oo

Proof. Say lanl ~ M for all n, where 0 ~ M < +00; then bn , en, b and c all belong to [-M, MI. Since (bn ) and (en) are bounded monotone sequences, we have bn --+ b and en -+ c (convergence in JR). =>: By assumption, c = b; write a = C = b. For every positive real number e, a - e < a < a + e; by condition (b) of the theorem, ultimately an - e < a < an + e , that is, Ian - al < e. This shows that an -+ a in JR. 0 and choose an index N such that, for every n > N, Ian - al ~ e, that is, a - e ~ an ~ a + e. It follows that n~N

=>

a-e~en~bn~a+e;

letting n -+ 00, we have a - e ~ c ~ b ~ a + e, and since this is true for every e > 0, we conclude that a = C = b. 0 The corollary points the way to a definition of convergence in JR: 1.16.8. Definition. A sequence (an) in JR is said to be convergent if

lim inf an = lim sup an ; the common value of the lim inf and lim sup is called the limit of the sequence (an), written lim

n_oo

an

(or briefly lim an ). If a = lim an ,the sequence (an) is said to converge to a, and one writes an-+a as n--+oo

(or simply an

-+

a).

1.16.9. Remarks. It is instructive to analyze the criteria (i), (ii) of 1.16.6 for convergence an -+ a, according to the value of a. case 1: a = +00. (i) is vacuous. (ii) says that if r < +00 then an > r ultimately; equivalently,

K E JR, K

> 0 => an > K ultimately

(and one need only consider positive integral values of K). case 2: a = -00. (ii) is vacuous. (i) says that if r > -00 then an < r ultimately; equivalently, K E JR, K > 0 => an

< -K ultimately.

§1.16. Convergence in lR

83

case 3: a E lR . Then a-I < a < a+ I, so a-I < an < a+ 1 ultimately. Thus, apart from a possible finite number of terms equal to ±oo, the sequence (an) is bounded. Dropping finitely many terms changes at most finitely many bn and en, and b, c not at all; this case is essentially "business as usual" for a convergent sequence in JR. 1.16.10. Example. Define a function f: [-I,ll f(-l) = -00, f(l) = +00 and

x

f(x) = 1 _ x 2 Since

l-l,lJ

f

-+

lR by the formulas

for Ixl < 1.

is bijective and order-preserving, for every sequence (x n ) in

we have

f(lim inf x n ) = lim inf f(x n ) , f(limsupx n ) = limsupf(x n ); it then follows from the definition of convergence in lR that f(x n ) -+ f(x) .

X n -+

x

**

Convergence in lR was defined by means of liminf and limsup; in turn, liminf and limsup can be analyzed in terms of convergence: 1.16.11. Theorem. Let (an) be a sequence in JR, c lim sup an, and let A

= {x E JR:

an.

-+

= liminfa n ,

b=

x for a subsequence (an.) of (an)}.

Then

{c,b} cAe [c,bl, thus b is the largest element of A , and c is the smallest. In particular, every sequence in lR has at least one convergent subsequence. Proof For the first inclusion let us show, for example, that b EA. case 1: b = +00. Then b > 1, so an > 1 frequently by (2) of 1.16.4. Choose nl with an. > 1 . But also b > 2 ,so an > 2 frequently; choose n2 > nl so that an, > 2. Continuing, we obtain a subsequence (an.) of (an) such that an. > k for all k, whence an. -+ +00; in particular, b = +00 EA. case 2: b = -00. Then -00 $ c $ b = -00, so c = b = -00 and an -+ -00; in particular, b = -00 EA. case 3: b E JR. For every e> 0, it is clear from 1.16.4 that b-e < an < b+e frequently; that is, for infinitely many n, an is finite and Ian - bl < e. Let f = 1 and choose nl so that lan, - bl < 1 . Let e = 1/2; since Ian - bl < 1/2

84

1. Foundations

frequently, we can choose nz > nl so that Ia". - bl < 1/2. Continuing, we obtain a subsequence (a".) of (a,,) such that Ia". - bl < l/k for all k, whence a". -+ b and b EA. Similarly, c E A (alternatively, use 1.16.3). Thus {c,b} C A. Finally, given x E A, we must show that c < x $ b. Let us show, for example, that x $ b. Assume to the contrary that x > b, and choose r so that b < r < x. Since bn 1 band b < r, there exists an index m such that bm < r . Then n~m

'*

a,,$bm 0

3. Given a power series (*) E:::Oakxk with real (or complex) coefficients ak, let L = IimsuPk>l!akl l / k . Prove: (i) L = 0 ~ lakl 1 / k -+ O~ (ii) L = +00 ~ lakl 1/ k is unbounded. (iii) If x # 0 and 1/lxl > L then H is absolutely convergent. (iv) If x # 0 and 1/lxl < L then (*) is divergent. (v) L = 0 ~ (*) is absolutely convergent for every x (in IR or in C). (vi) L = +00 ~ (*) converges only for x = O. Let R = 1/L, with the convention that 1/ + 00 = 0 and 1/0 = +00.

Then: (vii) L = I/R.

1

First course, p. 53, 3.7.7.

§1.16. Convergence in lR

85

(viii) Ixl < R => (*) is absolutely convergent. (ix) Ixl > R => (*) is divergent. (x) R = 0 # (*) converges only for x = o. (xi) R = +00 # (*) is absolutely convergent for every x (in IR or in iC). {Hints: (iii) If L < r < l/lxl then lakl 1/ k < r ultimately; compare (*) with the geometric series :Lr..o(rx)n. (iv) If l/lxl < L then lakl 1 / k > l/lxl frequently; infer that akxk f> 0 .}

CHAPTER 2

Lebesgue Measure

§2.1. §2.2. §2.3. §2.4. §2.5. §2.6.

Lebesgue outer measure on lR Measurable sets Cantor set: an uncountable set of measure zero Borel sets. regularity A nonmeasurable set Abstract measure spaces

One of the aims of the Lebesgue theory is to assign to each subset A of lR an element of [0. +001. to be thought of as the 'size' of A. in such a way that the size of a bounded interval is its length. and the function A ...... size of A

is well-behaved for as many sets A as possible. The facts are roughly as follows: (1) it is possible to assign a size to every subset of lR, but the function (*) is not well-behaved; (2) on a large class of subsets of lR (including the intervals) the function (*) is well-behaved. Thus, there is a trade-off between the extent of the domain of a size function and the goodness of its behavior. More precisely, to each subset A of lR there is assigned a nonnegative extended real number >'·(A). called the outer measure of A; certain subsets of lR are singled out and called measurable; the restriction of >'. to the class M of measurable sets is well-behaved and is called Lebesgue measure.

2.1. Lebesgue Outer Measure on lR

The point of departure is the concept of interval length: 2.1.1. Definition. If I is a bounded interval in lR, with endpoints a

and b (a $ b). the length of I is the nonnegative real number >'(1) b-a.

=

Built into the definition is an indifference as to whether or not the endpoints a.b belong to 1; for example, >'([a,bJ) = >.«a.b». In effect, the i!6

§2.1. Outer Measure on IR

87

finite set {a,b} (the set-theoretic gap between the two intervals) has been declared to be 'negligible'; the edifice of the Lebesgue theory rests on a precise definition of this word:

2.1.2. Definition. A set A c IR is said to be negligible if it can be covered by a sequence of open intervals whose total length is arbitrarily small. More precisely, for every < > 0 there exists a sequence (In) of open intervals such that 00

AC UIn n=l

00

and

E>'(In )

< •.

n=l

2.1.3. Remarks. (i) Since 11; = (a,a), coverings by finitely many intervals are permitted. (ii) The kind of interval (open, closed, semiclosed) is immaterial: in a covering by open intervals, including the endpoints does not change the lengths; in a covering by closed intervals I n = lan, bn ] of total length < • , the intervals can be enlarged to open intervals In = (an -6n , bn +6n ) with total length < 2. , for example by choosing 6n = ./2n+l . (iii) Every countable set in IR is negligible. For, if A = {an: nEil"} then the (degenerate) intervals I n = lan, an) form a covering of A by a sequence of closed intervals of total length 0, therefore A is negligible by the preceding remark. (iv) An example of an uncountable negligible set is given in §2.3. (v) A striking application of the concept of negligibilty is Lebesgue's criterion for Riemann-integrability: A bounded function f: [a, b] -+ IR is Riemann-integrable if and only if its set of discontinuities is negligible. This is proved in §5.13. A set A c IR is negligible if and only if the set of all sums L: >'(In ) , where (In) is a sequence of intervals with A C U In, has infimum O. This suggests a way of defining the size of an arbitrary subset of IR:

2.1.4. Definition. For every subset A of IR, the (Lebesgue) outer measure of A, denoted >'·(A) , is defined by the formula

where (In) varies over all possible sequences of open intervals of IR whose union contains A.

2.1.5. Remarks. (i) >.·(A) is the infimum of a nonempty subset of [0, +00) ; for example, this subset contains +00 since A C IR = U( -n, n). (ii) 0::; >.·(A) < +00 for every A C IR, thus >'.: P(IR) -+ 10, +ooJ. (iii) A c IR is negligible $} >.·(A) = O.

88

2. Lebesgue Measure

The basic formal properties of Lebesgue outer measure are gathered in the following theorem: 2.1.6. Theorem. If >" is Lebesgue outer measure (2.1.4) then (1) 0 ~ >"(A) ~ +00 for all A c Ill; (2) >"(0) =0; (3) >" is a monotone increasing set function, m the sense that A C B ~ >"(A) ~ >"(B); (4) >" is countably subadditive, in the sense that

for every sequence (An) of subsets of Ill; (5) >"(1) = >'(1) for every bounded interval I of Ill.

Proof (1) Already noted in 2.1.5. (2) 0 c (1,1) shows that 0 ~ >"(0) ~ 1 - 1. (3) Suppose A C B c Ill. If (In) is a sequence of open invertals with B C UI n , then also A C UIn , therefore >"(A) ~ E>'(I n ); varying the covering (In) of B, >"(A) ~ >'·(B). (4) If the sum on the right is +00, the inequality is trivial. Suppose E >"(An ) < +00; then >"(An ) < +00 for all n. Let £ > o. For each n, choose a sequence of open intervals Ink (k = 1,2, 3, ... ) such that 00

AnCUlnk

and

k=l

L >'(Ink) < >"(An ) + £/2" k=l

(possible because >"(A n ) is defined as an infimum and is finite). Then 00

U An C Ulnko n=1

n,k

therefore (see 1.15.8 for the maneuvers with infinite sums)

00

< L[>"(An )

00

+ £/2nJ = L >"(An ) + £,

n=1

n=1

and (4) results on varying £. (5) Let a and b be the endpoints of I, so that (a,b) C 1 C [a,b]

by the properties of >"

= (a, b) U {a,b};

already noted,

>"((a,b)) ~ >"(1) ~ >,'([a,bJ) ~ >" ((a, b) +>"({a,b})

= >"((a,b)) ,

§2.1. Outer Measure on IR

89

therefore )"(1)

(i)

= ).·(a,b» = ).·([a,b]}.

From (a,b) C (a, b) we see that ).·(a,b» ~ ).(a,b»

=b-

a, thus

)"(1) ~ b - a = ),(1);

(ii)

we need only show the reverse inequality. In view of (i), we can suppose that I = [a, b] ; assuming 00

C U (an, bn),

[a, bJ

n=l

we need only show that 00

b - a < ~)bn - an) n=l

(the infimum of such sums being equal to )"(1». By the Heine-Borel theorem, it follows from (*) that n

[a, bJ

U(ak' bk)

C

k=l

for some n; it will suffice to infer that n

b-a < ~)bk -ak).

(**)

k=l

We prove (**) by induction on n. For n = 1 it is obvious. Let n ~ 2 and assume that all's well with n - 1. Reordering the (ak' bk ) if necessary, we can suppose that a E (an, bn ), thus an < a < bn · case 1: bn > b. Then an < a ~ b < bn and (**) is obvious. case 2: bn ~ b. Then an < a < bn :5 b, so Ibm b] is disjoint from (an, bn ); but n

Ibn, b] C [a, bJ C

U(ak' bk), k=l

therefore

n-l

Ibn, b) C

U (ak' bk).

k=l

By the induction hypothesis, n-l

b - bn < ~)bk - ak), k=l

which. added to bn

-

a < bn

-

an , yields (**).

90

2. Lebesgue Measure

2.1.7. Corollary. Every subset of a negligible set is negligible, and a countable union of negligible sets is negligible. Proof Immediate from (3) and (4) of the theorem. 2.1.8. Corollary. If I is an unbounded interval, then A' (I) = +00 .

Proof If, for example, I = (-00, cJ then, for every positive integer n, I:> [c - n,cl, therefore .11'(1) ~ n by (3) and (5) of the theorem. The definition of outer measure specified coverings by open intervals, but the type of interval is not critical: 2.1.9. Corollary. For every subset A of IR,

A'(A) = inf {:~:>(Jn): A c

UJn } ,

where (In) runs over all countable coverings of A by bounded intervals that are closed (or left-closed and right-open, or right-closed and left-open, or a mixture of the four types). Proof Let A c IR and write a for the infimum on the right side. If A c U I n is a covering of the indicated type, then (by the theorem)

varying the covering, A'(A) ~ a. Suppose (I,,) is a sequence of open intervals with A C U In. Let I n be the interval of the contemplated type (closed, etc.) having the same endpoints as In. Then In C J" ,therefore A C U I n and

by the definition of a (and of A); varying the covering (I,,), we have a ~ A·(A). A key property of Lebesgue outer measure is its invariance under translation: 2.1.10. Theorem. A'(A + c) = A'(A) for all A C IR and c E IR. Proof The set A + c = {x + c: X E A} is the image of A under the order isomorphism x ...... x + c of IR onto IR. For an open interval I=(a,b), I+c=(a+c,b+c) and A(I+c) =.11(1). If (In) isasequence of open intervals with A C UIn , then A + c C U(In + c) and

varying the covering (In), A'(A + c) ~ A·(A). The inequality proves its own reverse: A'(A) = A' «(A+ c) + (-c» ~ A'(A + c), whence equality. For multiplication by c, there is a factor of proportionality:

§2.1. Outer Measure on lR

91

2.1.11. Theorem. ,x'(eA)

= lei' ,x'(A)

for all A

c

lR and e E lR.

Proof Here eA = {ex: X E A}. If A = 0 then cA = 0 and both sides of the asserted equality are O. Assume A f 0 . case 1: e = O. Then eA = {O}, ,x·(cA) = 0 and lei' ,x'(A) = 0 (even if ,x'(A) =

+00). case 2: e> O.

The argument is analogous to 2.1.10. Again x t-+ ex is an order isomorphism lR -+ lR; if I = (a, b) then d = (ca, cb) and ,x(d) = cA(I). If (In) is a sequence of open intervals with A C U In ,then cA C U cln and

(the last equality is immediate from the definition of such sums as suprema of finite subsums; cf. 1.15.5 or §1.15, Exercise 4); then (l/e),x·(cA) ::; ,x(In ) (even if-especially if-the sum on the right side is +00), and varying the covering yields (l/e),x·(cA)::; ,x·(A). Thus ,x'(eA)::; c,x'(A) and the inequality proves its own reverse: ,x' (A) = ,x'(I/e)(eA» ::; (l/e),x'(eA) . case 3: e=-l. Then eA = -A = {-x: x E A} and x t-+ -x is an order-reversing bijection lR -+ lR such that ,x(- I) = ,xCI) for all open intervals I = (a, b) ; it follows easily that ,x'(-A) = ,x·(A). case 4: e 0; by cases 3 and 2, ,x '(cA) = ,x.( -e)A) = (-e),x'(A) = lei' ,x' (A) . 0

r:

The pattern of the foregoing proofs: a property of ,x' for arbitrary sets A c lR is inferred from the analogous property of ,x for open intervals I. Here is a formal property of ,x that does not carry over to ,x': if I and J are bounded intervals such that I n J = 0 and I U J is an interval, then ,x(1 U J)

= ,xCI) + ,x(J) ;

for, if I UJ has endpoints a::; b, then I and J are obtained by splitting I U J at one of its points e, and (supposing I to be to the left of J) the asserted equality reduces to b - a = (c - a) + (b - e). However, the implication (0)

AnB=0

"*

>.O(AUB) = >.0 (A)

+ >.0 (B)

is in general false; a counterexample is given in §2.5, Exercise 2 (the Axiom of Choice is required!). This is expressed by saying that ,x' is not in general 'additive'. The remedy is to restrict ,x' to a smaller class of sets on which ,x' is well-behaved; this is the subject of the next section.

92

2. Lebesgue Measure Exercises 1. The union of a family of negligible sets need not be negligible; thus

countability is essential in 2.1.7. 2. If N c IR is negligible, then )"(A U N) = ).'(A) for every A C 1R. 3. If A and B are subsets of IR such that x < y for all x E A and y E B, then ).'(A U B) = ).'(A) + ).'(B).

2.2. Measurable Sets If a set A C IR is expressed as a union A = B U C of disjoint sets Band C, then the equation ).'(A) = ).'(B) + ).'(C) may fail (§2.5). However, if B and C are obtained by splitting A at some point e E IR , then all is well:

2.2.1. Theorem. If e E IR then e splits every subset A of IR additively, in the following sense: if B=An(-oo,e) and C=An[e,+oo) then ). '(A) = )"(B)

+ )"(C).

Proof. Suppose first that A is an open interval, say A = (a, b). If a < e < b then B = (a, c), C = [e, b) and the asserted equality reduces to b - a = (e - a) + (b - c); if e::S a then B = , C = (a, b) and the equality is trivial, and similarly if e;:: b. Now suppose A C IR is arbitrary. Write E = (-00, c), E' = [e, +00), so that

B=AnE, C=AnE'. If (In) is a sequence of open intervals with A C U In, then A nEe U(InnE) and the InnE are bounded intervals (possibly empty), therefore

).'(A n E)

::s L:>-(In n E);

).'(A n E')

::s L

similarly

)'(In n E') ,

and addition of these inequalities yields ).'(AnE) + ).'(AnE')::S L).(In ) by the first paragraph of the proof. Varying the covering (In), ).'(A n E) + ).'(A n E') < ).'(A); the reverse inequality follows from the subadditivity of ).' (2.1.6). 0

§2.2. Measurable Sets

93

This prompts a definition: 2.2.2. Definition. (C. Caratheodory). A set E e IR is said to be Lebesgue-measurable (briefly, measurable) if E splits every subset of IR additively, in the sense that "·(A) = "·(AnE)+"·(AnE') forall AelR, where E' = IR - E is the complement of E in IR. 2.2.3. Remarks. (i) Since A = (AnE)u(AnE/), the inequality "·(A) :s ". (A n E)+ ". (A n E/ ) always holds by the subadditivity of ".; to show that E is measurable, it suffices to verify the reverse inequality for all A e IR. (ii) An example of a nonmeasurable set is given in §2.5. (iii) For every real number c, the interval E = (-00, c) is measurable (2.2.1). (iv) E measurable => E' measurable. (v) Every negligible set is measurable. For, if "·(E) = 0 then, for every A e IR, "·(A n E) + "·(A n E/ ) = 0 + "·(An E') :S "·(A). (vi) If E is measurable, then so is E + c for every c E IR. {Proof: The function f: IR -+ IR defined by f(x) = x +c is a bijection that preserves outer measure (2.1.10); thus, for every A e IR, "·(A) = "·(r1(A») = "·(r1(A) n E) + "·(r1(A) n E') = ". [f(r1(A) n E)] + ". [J(r1(A) n E/)]

=

"·(A n f(E») + "·(A n f(E)'} ,

therefore f(E) is measurable.} (Vii) If E is measurable, then so is cE for every c E IR. {Proof: If c = 0 then cE = {O} or 0, so cE is measurable by (v). Assuming c'l 0, let f: lR -+ lR be the bijection f(x) = ex. The measurabiIityof f(E) follows from 2.1.11 by an argument similar to the one for (vi).} Summarizing, the class of measurable sets E e IR includes the intervals (-00, c) and the negligible sets, and is closed under complementation, translation and scalar multiplication. In the same vein: 2.2.4. Lemma. If E and F are measurable sets, then so are E U F, EnF and E-F; if, moreover, EnF=0, then

"·(An (Eu F») = "·(An E) + "·(AnF)

for all A e IR.

Proof It suffices to deal with E U F; the formulas E n F = (E' U F')' and E - F = E n F ' then finish the job. Let A e IR. We are to show that "·(A) = "·(An(EUF») + "·(An(EUF)'} .

94

2. Lebesgue Measure

Since E splits An (E U F) additively and An (EUF) nE = A nE, An (EU F) nE' = AnFn E', we have >"(A n (EU F)) = >"(A n E) + >.·(A n Fn E').

(*)

If E and F are disjoint (that is, FeE'), then (*) yields the equation in the statement of the theorem. In general,

>"(A) = >.·(A n E) + >"(A n E') = >"(A n E) + >,'(An E' n F) + >"(A n E' n F')

= >.' (A n (E U F)) + >.' (A n (E U F)') (the first equality because E splits A additively, the second because F splits An E' additively, the third by (*». 2.2.5. Theorem. If (En) is a sequence of measurable sets, then the sets 00

E=UEn

00

,

F=nEn

n=1

n=l

are also measurable. If, moreover, the En are pairwise disjoint, then 00

>"(A n E) =

L >"(A n En) n=1

for all A

c

IR .

Proof. Since F =

(U E~)' , we need only consider

E. The sets

E l , E l U E:!, E l U E:! U E3, ... are measurable by the lemma, with union E; thus, in proving E measurable, we can suppose that E l C E 2 C ~ C .... Then the sets E l , E 2 -EI> Ea-E:!, ... are measurable, with union E; changing notations again, we can suppose that the En are pairwise disjoint. Let A c IR. For each n, (i)

>.' (An

UEk) = t>"(AnEk)

k=l

k_l

(by induction on the lemma). Also, U~=l E k therefore (ii)

c

E,

50

(U~=l Ek )' :> E' ,

§2.2. Measurable Sets

95

adding (i), (ii) and citing the measurability of U~=l E k , we have n

>"(A) 2: L>,·(AnEk)+>,·(AnE'). k=l Since n is arbitrary, 00

>"(A) > L>"(AnEk) +>"(AnE') k=l 2: >"(AnE)+>"(AnE') 2:>"(A) (the second and third inequalities by the countable subadditivity of >"), whence equality throughout: 00

>"(A) = >,'(A n E) + >"(A n E') =

L >"(A n E

k)

+ >,'(An E').

k=l The first equality shows that E is measurable; replacing A by A n E , the second inequality yields 00

>.·(AnE)+O= L>'·(AnEk)+O. k=l 2.2.6. Remark. For future use, we note that if X is any set, the theorem and its lemma are valid with >" replaced by any set function p on P(X) satisfying the conditions (1)-(4) of 2.1.6, that is, (1) 0 '") for the set of all Lebesguemeasurable subsets of 1R. Since >'"(1) = >'(1) for all bounded intervals I (2.1.6), we may consistently define a function >. : M

-t

[0, +00]

by the formula >'(E) -

A"(E) for all E EM, that is, A = A"IM (the restriction of >'" to M); this set function >. is called Lebesgue measure on 1R.

The remaining corollaries depend only on Theorem 2.2.5 and its lemma, so they are valid for an outer measure p and its restriction to the class of p-measurable sets (2.2.6, 2.2.7). Lebesgue measure is 'countably additive' in the following sense: 2.2.10. Corollary. If (E,,) is a sequence of painuise disjoint measurable

sets, then A

(Q

E,,) =

~ >.(E,,) .

Proof In 2.2.5, put A = E (or A = 1R).

2.2.11. Corollary. If (Fn) is an increasing sequence of measurable sets with union F (briefly F n T F), then >'"(A n F n ) T A"(A n F) for all A c 1R; in particular, A(Fn) TA(F) . Proof Let E 1 = F 1 and En = F n - Fn-l for n

> 1. The En are

pairwise disjoint measurable sets with union F, therefore (2.2.5) >'"(AnF) =

~>'"(AnE,,) =s~p (~>'"(AnEk})

= sup>'" (A n n

U

Ek) = sup>'"(A n F n )

k=l

n

for all A cR. 2.2.12. Corollary. If (G n ) is a decreasing sequence of measurable sets

withintersectionG (brieflyG n ! G) andif>.(G t ) < +00, then>'"(AnG n )! >'"(A n G) for all A C 1R; in particular, >'(G n ) ! A(G). Proof Writing F n = Gl - Gn and F = G t - G, we have F n T F. Let A c 1R. By the preceding corollary, A"(A n F n ) T A"(A n F), where

A"(A n F) $ A"(G t )

< +00. For all n,

Gt=FnUGn=FUG

and F n nG n =FnG=0,

§2.2. Measurable Sets

97

therefore (2.2.4) 'x*(AnG,) 'x*(A n G,)

= 'x*(AnFn ) + 'x*(AnG n), = 'x*(A n F) + 'x*(A n G).

'x*(AnG n ) 'x*(A n G)

= 'x*(AnGt}- A*(AnFn), = 'x*(An Gt}- A*(An F),

By finiteness,

whence A*(A n G n ) ! 'x*(A n G).

0

Exercises 1. Give an example of an outer measure p on a set X, for which (/)

and X are the only p-measurable sets. {Hint: Consider a two-point set X = {a,b}.}

2. Every open set and every closed set in IR is (Lebesgue) measurable. {Hint: 2.2.8 and 2.2.5.} 3. Let (En) be a sequence of measurable sets in IR. (i) The sets

F E

= lim inf En = {x: = lim sup En = {x:

X

E En ultimately}

X E

En frequently}

are also measurable. {Hint: For the terminology, see 1.16.4. If F n = nk>n Ek, then F n TF.} (ii) 'x(lim infEn ) < lim inf 'x(En) . (iii) If A(U En) < +00, then A(lim sup En) ~ lim sup 'x(En ) . (iv) The inequality A(limsupEn) ~ limsupA(En) is in general false. {Hint: Try En = [n, n + I).} 4. Let (An) be any sequence of subsets of IR. (i) If An T A then A*(An) T A*(A) . (ii) A*(liminf An) ~ liminD*(An ). {Hint: (i) For every subset S of IR, there exists a measurable set E such that SeE and every measurable subset of E - S is negligible.'} 5. For A C [a,b] C IR, 'x*(A) is also called the exterior measure of

A in [a, b) and is denoted Ae(A); the interior measure of A in denoted 'xi (A) , is defined by the formula

la, b) ,

'xi (A) = (b - a) - Ae([a, b)- A) .

I

cr. the author. Mrosure and integration [Chelsea. New York. '9701. §8. Theorem

1.

2. Lebesgue Measure

98

In general, A;(A) $ A.(A); the set A is measurable if and only if A;(A) = A.(A) . 6. For a sequence of functions In : X -+ JR the functions lim sup In and lim inf In are defined by the formulas (limsup/n)(x) = limsup/n(x), (liminf In)(x) = liminf In(x).

If (En) is a sequence of subsets of a set X and if teristic function, then

(i)

lim sup 'PE..

= 'I'1im sup E.. ,

(ii)

lim inf 'PE..

= 'I'1im inl E...

'P

denotes charac-

(For the notations on the right side, see Exercise 3.)

2.3. Cantor Set: An Uncountable Set of Measure Zero

The Cantor set r is a negligible, closed subset of [0,1] with cardinality c (= card JR ); it is constructed by deleting the open middle third (!' ~) of [0,1), then deleting the open middle thirds of the two closed intervals that remain, 'and so on' (a set-theoretic passage to the limit). Establishing the right notation is 99% of the battle. If A = [a, bl is a nondegenerate closed interval, we write r(A) ('the rest of A ') for what is left of A after deleting the open middle third: r(A)

= (a,a+ W-a)] U (b - W -a),b),

a set whose Lebesgue measure is ~ A(A) . More generally, if A = Al U ... U A k is a finite union of pairwise disjoint, nondegenerate closed intervals Ai , we define r(A)

= r(A I ) U ... U r(Ak) ,

where the sets r(A;) have the meaning defined above; the right hand side is a set of the same sort as A, so the operation r on such sets A can be iterated, rn(A) (n = 1,2,3 ...) being defined recursively by the formulas rl(A) = r(A) , rn+l(A) = r(rn(A)) . The follOWing properties of this operation are easily checked: (1) A J r(A) J r 2 (A) J .... (2) A(rn(A)) = (~r A(A) , where A is Lebesgue measure. (3) rn(A) is a closed set l (it is the union of finitely many closed intervals).

1 First course, §4.2.

§2.3. Cantor Set

99

We write rOO (A) = n::"=1 r"(A) for the intersection of the decreasing sequence (r"(A»). (4) rOO(A) is a closed set. and >.(rOO(A)) $ >.(r"(A») shows that its Lebesgue measure is 0. The Cantor set is the result of applying this machinery to the closed unit interval: 2.3.1. Definition. The set r = r OO ([O.I)) is called the Cantor set; it is a closed set in lR with >'(r) = 0 . Let I

= [0, 1) . It is useful to refine the foregoing notations. We have r(I)

= [0, ~l u [~.11 = 10 UII,

where 10 is the 'left third' of I, and II is the 'right third' of I. In turn. 2 r (I)

= 100 U 101 U 110 U III •

where, for example. ho is the left third of II' For every n'ple Q (Q), ...• Q,,) with QI, ... , Q" E {O, I} , we recursively define

to be the left third of 1"'1"" ...0._1 if Q" = 0. and the right third if Q" = 1. Let us call Q = (QI, ...• Q,,) an index ofronk n and write IQI = n. For indices Q,13 let us write Q $ 13 in case IQI $ 1131 and Q; = 13; for i = 1,2, .... IQI (so to speak, Q is the 'initial IQI-ple' of 13). The following properties are easily verified: (5) If IQI = n then 10 is one of the 2" closed intervals that make up r"(I) ,and >'(1 0 ) = (6) If Q $ 13 then I",::> III . (7) If QI, Q2, Q3•... is a sequence of indices with QI < Q2 $ Q3 $ ... and if IQ" I = n for all n. then

W" .

n1 00

0 ,

= {y}

n=l

for some y E r. {Sketch of proof: The intersection is a singleton {y} by the theorem on nested intervals (1.8.27) and I",. C r"(I) for all n, therefore y E roo (I) = r .} (8) If IQI = 1131 and Q f. 13 then 10 n III = 0. {Consid~r the first coordinate in which Q and 13 differ; clearly 10 and III are contained in disjoint 'thirds' of some interval.} It remains to show that r has cardinality c = 21'(V):

A c V, V open} ,

is called outer regularity; so to speak, every set A c IR can be 'approximated from the outside' (more appropriately, 'from above') by open sets. The following property of Lebesgue measure, called inner regularity, says that every measurable set can be 'approximated from the inside' by compact sets5 : 2.4.18. Theorem. If E is a Lebesgue-measurable set, then >'(E)

= sup{>'(K):

K c E, K compact}.

Proof. Let c< be the supremum on the right side. The inequality >'(E) ~ c< is immediate from the monotonicity of >.. Suppose first that E is bounded, say E C C = la, b]. Let £ > O. Since >'(C - E) ::; >'(C) < +00, we know from outer regularity that there exists an open set V such that C - E c V and >'(V) ~ >'(C - E)

+£ =

>'(C) - >'(E) + £.

The set K = C - V = C n V' is closed and bounded, that is, compact, and from C - E c V we see that K = C - VeE, therefore >.(K) ::; c< by the definition of c'(C) ::; >'(V) + >.(K) < [>'(C) - >'(E) + £] + >.(K) , thus >'(E) < >'(K) + £ < c< + £; since £ is arbitrary, >'(E)::; c'(E) = c< in case E is bounded. Now let E be an arbitrary measurable set. If c< = +00 then >'(E) = c< is forced by the inequality >'(E) > c sup{ A(K): K compact} is discussed in §2.5, Exercise 1.

C

A, K

Exercises 1. If £ is the set of all closed (left-closed and right-open, left-open and right-closed) intervals in IR, then Sec) = B.

2. There are c Borel sets in IR, and 2< Lebesgue-measurable sets. (Sketch: Since M C P(IR) and card P(IR) = 2< (1.13.13), we have card M :$ 2< ; on the other hand, the Cantor set has cardinality c (2.3.2) and its subsets are all measurable (2.2.3, (v», consequently card M ~ 2'(G - F) = O. {Hint: Express E as the union of a sequence (En) of measurable sets of finite measure and apply Theorems 2.4.14 and 2.4.18 to each En.} (vii) Iteration of the 0 and u 'operations' can be pursued transfinitely to classify all Borel sets. 8 5. (H. Steinhaus9 ) If A is a Lebesgue-measurable subset of IR such that >'(A) > 0, then the difference set D = {a - b: a, b e A} is a neighborhood of 0, that is, (-0,6) c D for some 6 > O. {Sketch: By 2.4.18, one can suppose that A is compact. For every positive integer n, let Un={xelR: Ix-al'(A) = inf >'(U n ). Choose an index m such that >'(Um) < ~>'(A) and let 6=I/m.1f Izl'(Um - A n B) :::; >'(Um - A) + >'(Um - B) = 2>'(Um) - 2>.(A) < 3>'(A) - 2>'(A) = >'(A) < ~>'(Um), therefore An B

# 0 and a' = a + z for suitable a',a eA.}

6. (i) If X is any set, then the power set l'(X) is a commutative ring with unity element X for the operations A + B = (A - B) U (B - A) (called the symmetric difference of A and B, also denoted At:. B) and AB = AnB. Moreover, AA = A for all A e l'(X) (such rings are called

Boolean). (ii) A set S C l'(X) is an algebra of subsets of X if and only if XeS and S is a subring of the ring l'(X) of part (i). • K. Kuratowski, Topolagie. 1 [Monogralie Matematyczne, 2nd. edn., Warsaw, 1948), §26, p. 251, II: F. Hausdorff, Set theory (3rd. edn., Chelsea, New York, 1957), §IS. 9 Cf. E. Asplund and L. Bungart, A first course in integration [Holt, Rinehart and Winston, New York, 1966], p. 125, Theorem 3.3.9.

2. Lebesgue Measure

108

7. If £ is a set of subsets of a set X, the O'-algebra S(£) generated by £ may be obtained by a transfinite procedure. lo Let 0 be the first uncountable ordinal (1.14.36) and, for each ordinal a < 0, define £0 recursively as follows: E:o = £ U {0, X} and, for a > 0, £0

= (Up" (A 2 ) = >" (A3 ) = .... It is then easy to see that some An must be nonmeasurable; for, if every An were measurable then B would be measurable and

>'(B)

= >.

CQ

An)

= ~ >'(An) = ~ >'(A I )

(a constant term infinite series); since >'(B) > 0, necessarily >'(A I ) > 0, whence >'(B) = E::"=I >'(Ad = +00, a contradiction. Suppose 0 < >" (B) < +00 and (An) is a sequence of pairwise disjoint nonempty sets such that B = U::"= I An . A way to assure that the An all have the same outer measure is to arrange (if possible) that they are all translates of the same set A, say An = X n + A for all n (cf. 2.1.10). The terms of the sequence (x n ) will then be distinct (because the An are pairwise disjoint) and, writing D = {x n : n E J!>}, we have B = U::"=I (x n + A) = D + A. Indeed, every b E B has a unique representation b = x + a with xED and a EA. {If b = X n + a = X m + a' with a, a' E A, then b E X n + A = An and b E X m + A = Am, bEAn n Am , whence n = m, X n = X m and a = a'.} Since >"(B) > 0, B is uncountable (2.1.3, (iii»; but D is countable, so A must be uncountable. The 'direct sum' B = D + A = A + D effects a partition of B into countably many uncountable sets, B = U(x+A), xED

as well as a partition of B into uncountably many countable sets, B= U(a+D). aeA

110

2. Lebesgue Measure

Translates a + D remind us of cosets, and Q is a countable subgroup of the additive group IR. We are thus led to consider the quotient (additive) group IR/Q and the partition of IR into cosets x + Q. Since Q is countable and IR is not (1.10.10, 1.10.11), there are uncountably many cosets. From each coset u choose one representative au and let A = {au: U E IR/Q} be the resulting set of representatives. {For example, let -y: P(IR) - {} -. IR be a choice function for IR (1.4.6) and let au = -y(u) for all u E IR/Q.} Thus, A is a subset of IR such that, for every coset x + Q, A n (x + Q) is a singleton. Then

(*)

IR=A+Q.

For, every real number b belongs to some coset u (namely, u = b + Q), whence b-au E Q, say b-au = r E Q; then b = au +r E A+Q. Moreover, representations in (*) are unique: if a+r = a' +r' with a,a' E A and r, r' E Q, then a - a' = r' - r E Q, so a and a' belong to the same coset, whence a = a' and r = r'. Let Q = {r n : nEil"} be a faithful indexing of Q by II" (1.10.10). We have IR = Q + A =

U(rn + A) ; nEP

writing An = r n + A, we have a denumerable partition IR = U::'=l An with A" (An) constant. The only catch is that A(IR) = +00, so IR cannot play the role of B in the earlier discussion; nevertheless, it can be shown that A is nonmeasurable (Exercise 4), but it is simpler to playa 'bounded variation' on the preceding construction: 2.5.1. Theorem (G. Vitali1 ). There exists a subset of IR that is not Lebesgue-measurable.

prooF For x, y E (0,1) ,write x ~ y if x - y is rational; this defines an equivalence relation in (0,1). Note that {x-y: x,yE (0,1)}=(-1,1), thus x~y ~ x-YEQn(-I,I).Let n ...... r n be a faithful indexing of the denumerable set D = Q n (-1,1) . By the Axiom of Choice, there exists a subset A of (0, 1) such that A intersects each equivalence class in a singleton; let B = D + A = {rn

+ a: nEil", a E A}.

Writing An = r n + A, we have a denumerable partition B = with A"(A n ) = A"(A) for all n (2.1.10).

U::'=l An

1 Giuseppe Vitali (1875-1932) after the exposition of H. Kestelman, Modern theories oj integrotion rord Univ. Press, 1937; 2nd. revised edn., Dover, New York, 1960). pp. 90-91. 2 Patterned

[Ox-

§2.S. A Nonmeasurable Set

III

Note that (0,1) e B; for, if x E (0,1) and a is the element of A that belongs to the equivalence class of x, then x '" a, thus x - a E D and x E D+ A = B. On the other hand, since A e (0,1) and De (-1,1), we have B = A + D e (-1,2). Thus (0,1) e Be (-1,2),

:s

whence 1:S >"(B) 3. Finally, we assert that A is not Lebesgue-measurable. If, On the cOntrary, A were measurable, then the sets An = Tn + A and B = U::'=l An would be measurable, and, by the countable additivity of >., 00

>'(B)

=L

00

>.(An )

=L

n=l

8ince >'(B) > 0, necessarily >'(A) +00, a contradiction.

>'(A) .

n=l

> 0; but then >'(B) = L::'=l >'(A) =

Exercises 1. (i) If 8 is a subset of IR such that, for every A e 8,

>"(A)

= sup{>'(K):

K

e

A, K compact},

then 8 is Lebesgue-measurable. (ii) If S is a nonmeasurable set (2.5.1), then (*) must fail for some subset A of 8. {Hint: (i) We can suppose without loss of generality that >"(8) < +00. Show that there exists a Borel subset E of 8 (in fact, the UniOn of a sequence of compact subsets) having the same outer measure as 8, then repeat the argument to obtain a Borel subset F of 8 - E with the same outer measure as 8-E.Apply >.' to the inclusion 8:>EuF toconclude that 8 - E has outer measure o.} 2. Let B = U::'=l An as in the proof of Theorem 2.5.1. (i) There exists a positive integer n such that >"(U~l Ai) < L~l >'·(Ai). (ii) If m is the smallest positive integer such that >.• (U: 1 Ai) < L:l>"(Ai) and if 8 = U:~l~, T = Am, then 8 n T = (/) and >"(8 U T) < >"(8) + >'·(T). (iii) The set 8 of part (ii) is nonmeasurable. {Hints: (i) Consider the alternative. (ii) Argue that >"(U:~l Ai) = L:~l >'·(Ai). (iii) Use (ii) to show that 8 does not split 8 UT additively.} 3. There are 2< nonmeasurable subsets of IR.

112

2. Lebesgue Measure

4. The set A with IR = A+Q described in the remarks preceding 2.5.1 is nonmeasurable. {Hint: Since IR = UrEQ(r+A) and A'(r+A) = A'(A) for all r E Q, necessarily A'(A) > O. Let D = {a - b: a, b E A}; since D n Q = {O} by the choice of A, D can have no interior points, therefore A is nonmeasurable (§2.4, Exercise 5).}

2.6. Abstract Measure Spaces We note in this section a number of useful properties of Lebesgue measure that carry over to the general measure spaces defined in 2.4.12. Throughout the section, (X, S, Jl) is a fixed measure space. 2.6.1. Theorem. (1) Jl is finitely additive on S, that is, if E I , ... , E r are pairwise disjoint sets in S, then Jl(U~=1 E k ) = E~=l Jl(E k ) • (2) Jl is monotone, that is, if E, F are sets in S with E c F, then Jl(E) < Jl(F) . If, moreover, Jl(F) < +00, then Jl(F - E) = Jl(F) - Jl(E) .

Proof. (1) Write E I U ... UEr

= E I U ... UEr U(/)U(/)U ...

and cite the

countable additivity of Jl and the property Jl«(/) = O. (2) If E c F then F = E U (F - E) is a disjoint union; by finite additivity, Jl(F) = Jl(E) + Jl(F - E) ~ Jl(E) ; when Jl(F) is finite, the term Jl(E) can be transposed in the equality. 0 2.6.2. Corollary. If F and F n (n that F n TF, then Jl(Fn) TJl(F) .

= 1,2,3, ... )

are sets in S such

Proof. A simplification of the proof of 2.2.11, with A replaced by X and A' by Jl. 0

2.6.3. Corollary. If G and G n (n = 1,2,3, ... ) are sets in S that Gn ! G and if Jl(G.) < +00, then Jl(G n ) ! Jl(G) . Proof. A simplification of the proof of 2.2.12.

2.6.4. Corollary. For every sequence E:::I Jl(E k ). Proof. Writing F n Jl(F) by 2.6.2. Since

F2

= U~=l Ek

and F

(En)

such

0 in S,

= U:::I Ek ,

Jl(U;;"=1 E k )


(Ek) $ ~:>(Ek) k=l k=l

Exercises 1. For a sequence (En) of subsets of X, one defines

lim sup En liminfEn

= {x EX: = {x EX:

x E En frequently}, x E En ultimately}.

(i) Show that

In particular, if the En are in S then so are lim sup En and lim inf En . (ii) If the En belong to S then tt(Iim inf En) $ lim inf tt(En) (for the meaning of the right side, see §1.16). (iii) Suppose the En belong to Sand tt (~l Ek) < +00. Then tt(limsupE n ) in particular, if tt(En ) Young theorem).

~

~

limsuptt(En );

r > 0 for all n then tt(IimsupEn )

2. If E C F and tt(E) is finite, then tt(F - E)

= tt(F) -

~

r (Arzela-

tt(E) .

3. In the definition of a measure (2.4.12), the condition (2) that tt is countably additive may be replaced by the following pair of conditions: (2a) tt is finitely additive, and (2b) if (Fn) is an increasing sequence of sets in S with union F, then tt(F n) T tt(F) in IR, that is, (tt(Fn» is an increasing sequence with supremum tt(F). {Hint: Assuming (2a) and (2b), and given a sequence of pairwise disjoint sets En in S, contemplate the sequence F n = U~=l Ek with union F = U:=lEn .}

114

2. Lebesgue Measure

4. If S is a u-aIgebra of subsets of a set X, a set function p.: S -+ [0, +00] is a measure if and only if (i) p.( 0 there exists a fJ > 0 such that

*

*

Ix -

al

< fJ * I/(x) - l(a)1 < e.

If, for each e > 0, we define fJ. to be the supremum of all fJ > 0 for which the implication (*) is valid (conceivably fJ. = +00 -consider, for example, a constant function), then fJ. is the largest fJ E (0, +001 satisfying (*). The function e ...... fJ. conveys numerical information that is more pertinent to the rate 01 change of 1 than to its continuity. Finally, when we contemplate functions 1 that are continuous at every point a of JR, we are dealing with a function (a, e) ...... fJa •• (a E JR, e > 0) of two variables; clearly there is a lot of numerical baggage here that goes beyond the intuitive concept of continuity.

3. Topology

116

Here is a way of filtering out some of this numerical 'static'. To say that an -+ a means that for every open interval I containing a, an E I ultimately. 'Ib say that f is continuous at a means that if J is any open interval containing f (a) , then there exists an open interval I containing a such that f(I) c J; or, relaxing the notations a bit, if W is any set that contains an open interval containing f(a) , then f-l(W) contains an open interval containing a. In discussing functions f that are continuous at every point of 1R, one is naturally led to consider subsets U of R such that, for every x E U, U contains an open interval containing x; such subsets of IR are called open sets and open sets are the basis of the concept of topological space. General topological spaces are taken up in §3.3; in §§3.1, 3.2, we prepare the ground for the concept by looking at the precursors of the topological ideas in the context of metric spaces (a special kind of topological space in which there is assigned a numerical measure of 'distance' between any two points of the space).

3.1. Metric Spaces: Examples Metric spaces provide a unified setting for discussing a multitude of sequential convergence matters. Our discussion of examples of metric spaces rests on several inequalities, the first mainly of technical interest, the rest eminently practical. 3.1.1. Proposition. If a, b, c are real numbers ;::0: 0 such that a < b + c, then

abc

--0,

whence the inequality (1); the identity (R + 8)2 - [(a + c)2

+ (b + d)2) = 2(RS -

(ac + bd»)

shows that the inequalities of (1) and (2) are equivalent. 0

§3.1. Metric Spaces: Examples 3.1.3. Proposition. If

p

117

q

and

1

are positive real numbers such that 1

-+-=1 P q (in particular, p

>1

and q

> 1 ),

then

bq

aP

ab o. Consider the function

(0, +00) defined by t- q

tP

0);

is differentiable (therefore continuous), with IxI" E 1~(T), where Ixl"(t) = Ix(t)l" for all t E T, such a function x is said to be p-th power summable; it is zero for all but countably many points of T (1.15.9). The set of all x E 1l;(T) that are real-valued is denoted 1'(T). When T = {I, 2, ... ,n} , the metric

on C n is called the Minkowski p-metric, and the metric space (Cn , d,,) is called an n-dimensional complex Minkowski space. Similarly, (JRn , d,,) is called an n-dimension real Minkowski space. The symbol (I") is traditionally reserved for 1~(I!') (or 1l;(I!'», the space of p-th power summable sequences of real (or complex) numbers.

3.1.15. Example. When p = 2 in the preceding example, (JR n , d2) is called the n-dimensional Euclidean space and d 2 the Euclidean metric on JR n , whereas (Cn , d2 ) is called the n-dimensional unitary space. For p = 2 there are proofs of the triangle inequality simpler than the proof via Holder's inequality (consult any book on linear algebra).

Exercises

> x for all x E JR. (ii) If 'P is the function of 3.1.3, then 'P(t) > 1 for all t > 0 outside the closed interval [1/2,2]. {Hint: If t > 2 then t" > 2" > p.} 1. (i) The equation 2Z = x has no solution, therefore 2Z

2. If p and q are integers such that ~

+ ~ = 1 ,then

p

= q = 2.

§3.2. Convergence in Metric Spaces

3. With 3.1.11,

IIxli oo

and

IIxlip

123

(p ~ 1) defined for x E en as in 3.1.10 and

IIxII = p-oo lim IIxlip. 00

{Hint: One can suppose IIxli oo = 1. Let m be the number of coordinates Xk of x such that IXkl = 1. Then (lIxll p )P = m + t p , where t p -+ 0 as p -+ oo.}

4. With notations as in 3.1.14, if 1 ::; P < T then It (T) C IC(T) .

Ix(t)1 < 1

{Hint: If x E It(T) then

for all but finitely many t.}

a = (al>' .. , an), b = (bl> . .. , bn ), C= (Cl>' .. , c,,) are elements en such that lakl::; Ibkl + ICkl for all k, then lIallp::; IIbllp + IIcllp

5. If

of for all P

~ l.

6. Let (Xk, dk) (k = 1,2, ... , n) be pseudometric spaces and let X = Xl X, .. x Xn be the product set. For every pair of points x = (Xl> ... ,xn ) , Y = (Yl,· .. , Yn) of X, and for every real number p > 1, each of the formulas

d(x, y)

=

(~[dk(Xk' Yk)]P)

lip

d(x, y) = max dk(Xk, Yk) l:S;kSn

defines a pseudometric on X. In order that these be metrics on X, it is necessary and sufficient that the pseudometrics dl> ... , dn be sepamting in the following sense: if X,Y E X and xi- Y, then dk(Xk,Yk) > 0 for some k. 7. If (X k , d k ) (k = 1,2,3, ...) is a sequence of pseudometric spaces and X = n~l Xk is the product set, then the formula

d(x y) ,

=

f= ..!.- .

k=l

dk(Xk, Yk) 2k 1 + dk(Xk, Yk) ,

where x = (Xl>X2,X3,"') ' Y = (Yl>Y2,Y3,"') are points of X, defines a pseudometric on X. For d to be a metric, it is necessary and sufficient that the sequence dl> d2, d3, . .. be 'separating' (in the sense suggested by Exercise 6).

3.2. Convergence, Closed Sets and Open Sets in Metric Spaces The actual values of a metric can be important in geometrical situa.tions, but what often counts most in analysis is the notion of limit derived

3. Topology

124

from the metric. In a metric space, the core concept of limit is sequential convergence: 3.2.1. Definition. Let (X,d) be a metric space (3.1.4). A sequence (x n ) in X is said to be convergent (for d) if there exists a point x E X such that d(xn,x) -+ 0 as n -+ 00 (that is, for every l > 0, d(xn,x) < l ultimately). Such a point x is then unique; for, if also d(x n , Y) -+ 0 then

o :s d(x, y) :s d(x, x n ) + d(x n , y) -+ 0 shows that x = y. One says that and one writes

Xn

converges to the limit x (in X)

x = lim x n n-oo

,

X n -+ x as n -+ 00, or simply X n -+ x . Note that if (x n ) converges to x, then every subsequence (x n .) also converges to x.

or

3.2.2. Examples. (i) If X = IR and d is the usual metric on IR (3.1.6), then the concept defined in 3.2.1 is the usual concept of convergence in IR. (ii) In a discrete metric space (3.1.7),

xn

-+ X

¢}

Xn

= x ultimately;

:s

1 then 0 is the only available distance < l . Thus the only for, if 0 < l convergent sequences are the sequences that are ultimately constant. (iii) If d is a pseudometric on X and if D = d/(1 + d} (cf. 3.1.8), then

In particular, when d is a metric, X n -+

x for d

¢}

X n -+

x for D.

(iv) In C' with the sup-metric doo (3.1.10), a sequence is convergent if and only if , for each k = 1, ... ,r, the sequence of k'th coordinates is convergent in C. The notation gets a little messy: if X n = (anI> lln2, ... ,an.) and x = (al> ... , a.), so that doo(xn,x) = max Illnk - akl, l~k~.

then X n -+ x for doo if and only if, for each k = 1, ... , r, ank -+ ak in C. (v) If p;::: 1, d" is the Minkowski p-metric on C' (3.1.14) and doo is the sup-metric of the preceding example, then doo(x, y)

for all x, y in C

:s dp(x, y) :s r1/Pdoo(x, y)

(briefly, doo:S d p < r1/Pdoo ).

125

§3.2. Convergence in Metric Spaces {Proof: If x = (al. ...• a,.). y = (bl •...• br ) and j that laj - bj I = IIx - Ylloo • then

(lIx - Ylloo)p

= laj -

bjlP $

t;

lak - hiP $ r

is an index such

Cr;;~r lak -

bkl )

I' •

therefore IIx - Ylloo $ IIx - Yllp $ r1/pllx - Ylloo .} It follows that

xn

-+

x for doo

$}

X n -+

x for dp;

in view of Example (iv). convergence for dp means coordinatewise con-

vergence. These assertions hold a fortiori with C r replaced by JRr; in particular. in C = JR2 • a sequence is convergent for dp (or doo) if and only if its real and imaginary parts are convergent in JR. (vi) Let T be a nonempty set. doo the sup-metric on 8 = 8c(T). If xn.x E 8 then

xn $}

-+

x for doo

('v't

('v't

$}

> 0) 3N :;) n

one then says that

X n -+

~

> 0)

sup Ixn(t) - x(t)j $ t ultimately teT

~

N

Ixn(t) - x(t)1 $ t for all t E T;

x unifonnly on T. Since

Ixn(t) - x(t)1 $ doo(xn.x) for each t E T • it follows that

xn-+x uniformly

~

('v'tET) xn(t)-+x(t) in C;

so to speak. uniform convergence implies 'pointwise convergence'. {The converse is false (Exercise I).}} 3.2.3. Proposition. If d is a pseudometric on a set X. then

Id(x. y) - d(x'. y')1 $ d(x. x') + d(y. y') ' IaI x.Y.x.y "'X Jar m .

Proof. By the triangle inequality. d(x,y) $ d(x.x') +d(x'.y') +d(y'.y) • thus d(x. y) - d(x', y')

< d(x, x') + d(y. y');

interchanging x ..... x' and y ..... y' yields

d(x'. y') - d(x, y) $ d(x. x') whence the asserted inequality.

+ d(y, y') •

3.2.4. Corollary. In a metric space (X, d), if X n in X, then d(x n • Yn) -+ d(x. y) in JR.

-+

x and Yn

-+

Y

3. Topology

126

Proof. Id(xn,Yn) - d(x,Y)I:::; d(xn ,x) +d(Yn,Y)' 0 3.2.5. Definition. Let (X, d) be a metric space. A subset A of X is said to be closed in X (or to be a closed set) if, whenever the terms of a convergent sequence are in A, then the limit must also be in A; in symbols, o",EA} xEX ",,*xEA. an --+ X {Mnemonic: "You can't fight your way out of a closed set with a convergent sequence. n}

3.2.6. Example. (i) In ]R with the usual metric, every closed interval (a, bl is a closed set. {If a :::; X n :::; b and X n -+ x then a :::; x :::; b (1.8.24).} (ii) In a discrete metric space, every subset is a closed set (cf. 3.2.2, (ii». (iii) If (X, d) is a metric space and D = d/(1 + d) , then a subset A of X is closed for d if and only if it is closed for D (cf. 3.2.2, (iii». (iv) In any metric space, singletons A = {a} are closed sets. (v) If AI> ... , Ar are closed subsets of C (for the usual metric) then the product set Al x ... x Ar is a closed subset of C r for the metrics doo and d p (cf. 3.2.2, (iv), (v)). (vi) Let (X, d) be a metric space. For every c E X and every real number r > 0, the set Br(c) = {x EX: d(x,c):::; r} is a closed set. For, if X n -+ x and d(x n, c) :::; r for all n, then d(x, c) :::; r by 3.2.4 and part (i). For example, if B = Bc(T) as in 3.2.2, (vi), then, for every r > 0, the set

{x E B: Ix(t)l:::; r for all t E T} is closed for the sup-metric doo .

3.2.7. Definition. With notations as in (vi) above, the set Br(c) is called the closed ball in X with center c and radius r. As noted in 3.2.6, (vi), every closed ball in a metric space is a closed set. In ]R2 with the Euclidean metric (3.1.15), 'closed balls' in the sense of 3.2.7 are discs; in Euclidean 3-space, this use of the word 'ball' accords with its meaning in everyday language. The key formal properties of closed sets are as follows: 3.2.8. Theorem. In a metric space (X, d) , (I) (/) and X are closed sets; (2) the intersection of any family of closed sets is a closed set; (3) the union of any two closed sets is a closed set.

§3.2. Convergence in Metric Spaces

127

Proof. (I) The empty set is closed 'by default' (no sequence exists that contradicts the assertion that O. The

d(x,c) < r}

is called the open ball in X with center C and radius r. Note that CE Ur(c) C Br(c). 3.2.11. Examples. (i) In a discrete metric space (X, d), if 0 < r :5 I then Ur(C) = {c} forall CEX;if r>1 then Ur(c)=X for every c. (ii) In any metric space (X, d) , every closed ball is the intersection of a sequence of open balls:

n Ur+l/n(C) , 00

Br(c) =

n=l

because d(x,c):5 r # d(x,c) < r + lin for all n. {Exercise: Interpret this formula for the case that d is the discrete metric on X and, in turn, O 0 such that Ur(c) cA. (So to speak, not only is C in A, but a 'buffer zone' around c is contained in A; but this is just talk, and in a discrete metric space it is nonsense.) If c is interior to A, we also say that A is a neighborhood of c. We say that A is an open set in X if every point of A is an interior point:

('if c E A) 3 r > 0 3 Ur(c) C A (note that the size of r may vary with c); thus, A is open if and only if it is a neighborhood of each of its points.

3.2.13. Remark. With the preceding notations, (i) (/) and X are open sets (by default and trivially, respectively); (ii) if VI and Vz are neighborhoods of c, then so is VI n Vz (think min{ rl, rz} ); (iii) if V is a neighborhood of c and if W :::> V, then W is also a neighborhood of c; (iv) for every r> 0, Ur(c) is a neighborhood of c. Better yet: 3.2.14. Proposition. In a metric space (X, d) , every open ball Ur(c) is an open set.

Proof. Let x E Ur(c) ; we are to show that x is interior to Ur(c) , thus we seek an s > 0 such that U.(x) C Ur(c). Let s = r - d(x, c) , which IS > 0 because x E Ur(c). If y E U.(x) then d(y, c) :s; d(y, x)

+ d(x, c) < s + d(x, c) = r,

therefore y E Ur(c). (Draw a picture!) Here are the key properties of open sets: 3.2.15. Theorem. In a metric space (X, d) , (1) (/) and X are open sets; (2) the union of any family of open sets is open; (3) the intersection of any two open sets is open.

Proof. (1) Noted in 3.2.13. (2) Let (Ui)iEI be a family of open sets and let U = UiEI Ui . If x E U then x E Ui for some i, therefore Ui is a neighborhood of x, hence so is its superset U; thus U is a neighborhood of each of its points. (3) Let U and V be open sets. If x E UnV then x E U and x E V, so U and V are neighborhoods of x, hence so is U n v. The preceding theorem (properties of open sets) is hauntingly similar to Theorem 3.2.8 (properties of closed sets): to get the open set theorem from

§3.2. Convergence in Metric Spaces

129

the closed set theorem. we need only make the following substitutions: intersection - union union - intersection

-X X- closed - open. The first two of these transfomations are accomplished by complementation (De Morgan's formulas), as are the next two; so is the last: 3.2.16. Theorem. For a subset A of a metric space,

A is closed Od' and Od':::> 0d.

3.3.7. Corollary. (i) For any metric space (X, d), the metrics d and D = dl(l + d) are equivalent; thus every metric is equivalent to a metric with values in [0,1). (ii) On er, the metrics d.oo and dp (p ~ 1) of 3.1.10, (ii) and 3.1.14 are all equivalent. Proof Immediate from 3.2.2, (iii), (v) and the preceding corollary.

3.3.8. Definition. Let (X, 0) be a topological space, c E A eX. Guided by the case of metric spaces (3.2.12) we say that x is interior to A (or that A is a neighborhood of x) if there exists an open set U such that c E U c A (so to speak, an open set can be interpolated between c and A).

§3.3. Topological Spaces

133

3.3.9. Remarks. In a topological space (X, 0) : (i) A set U is open (that is, U EO) if and only if U is a neighborhood of each of its points. {"If": For each x E U choose an open set U z such that x E U z C U; then U = UzEu Uz is the union of a family of open sets. } (ii)If VI and V2 areneighborhoodsof x,thensois V l nV2 . {Reason: The intersection of two open sets containing x is an open set containing x.} (iii) If V ewe X and V is a neighborhood of x, then so is W. 3.3.10. Example. Let (X,d) be a metric space and let xEX.Forevery positive integer n, let V n = UI/n(x). Then the Vn are neighborhoods of x, and every neighborhood V of x contains Vn for some n. {For, if Ur(x) C V and lin < r then Vn C Ur(x) C V.} 3.3.11. Definition. In a topological space X, a fundamental sequence of neighborhoods of a point x is a sequence (Vn) of neighborhoods of x such that every neighborhood of x contains (at least) one of the Vn . If every point of X has a fundamental sequence of neighborhoods, then X is said to be first countable. Every metric space is first countable (3.3.10). There exist topological spaces in which no point has a fundamental sequence of neighborhoods (Exercise 3). 3.3.12. Definition. Let A be a subset of a topological space X. A point x of X is said to be adherent to A (or to be an 'adherent point' of A) if every neighborhood of x intersects A, that is, if V a neighborhood of x ~ Vn A # (/) . (It is the same to say that every open set containing x intersects A.) 3.3.13. Examples. (i) In JR with the usual (metric) topology, if I is an interval with endpoints a:$ b, the points adherent to I are precisely the points of the closed interval la, bl. {Proof: If, for example, x > b, and if r = x - b, then (x - r,x + r) is a neighborhood of x disjoint from I, therefore x is not adherent to I.} Every x E JR is adherent to the set Q of rational numbers (because every nondegenerate interval contains a rational number). (ii) In a metric space (X, d), x is adherent to A if and only if there exists a sequence (an) in A such that an -- x. {"Only if": Assuming x adherent to A, for every positive integer n the neighborhood UI/n(x) of x contains a point an of A.} (iii) In a discrete topological space (3.3.2, (iii», x is adherent to A if and only if x EA. {For, if x ¢. A then {x} is a neighborhood of x disjoint from A.} (iv) In a trivial topological space (3.3.2, (ii», every x E X is adherent to every nonempty subset A of x. {Reason: X is the only available neighborhood of x.}

3. Topology

134

3.3.14. Definition. If A is a subset of a topological space X, the set of all points x of X that are adherent to A is called the closure (or adherence) of A and is denoted A:

A = {x EX: x is adherent to A}. (The reason for the term 'closure' is clear from (5) of 3.3.16 below.)

3.3.15. Examples. (i) In IR with the usual topology, IQi = 1R, and if I is an interval with endpoints a:5 b, then i' = [a, b] (ef. 3.3.13, (i». (ii) For a subset A of a metric space (X, d) ,

A

= {x EX:

an

--+

x for some sequence (an) in A}

(cf. 3.3.13, (ii». (iii) In a discrete topological space, A

=A

for every subset A (3.3.13,

(iii». (iv) In a trivial topological space, A = X for every nonempty subset A (3.3.13, (iv». In a topological space X, A are its key properties:

A is a mapping 'P(X)

3.3.16. Theorem. Let X be a topological space, of X. Then:

--+

'P(X); here

A and B subsets

(1) Ac A; (2) A c B => A (3) A = A;

(4) A = A

~

c

B;

A is a closed set;

(5) A is the smallest closed set containing A;

(6) Au B - Au B . Proof. (1) If a E A and V is a neighborhood of a, then V n A # (/) (because a E V). (2) If every neighborhood of x intersects A then it intersects B. (3) By A w!.mean the closure of A, that is, A = (A)- . From A C A we have A C A by (2). To prove the reverse inclusion, assuming x E A and V a neighborhood of x, we must show that V n A # (/). We can suppose that V is open (3.3.8). Since x is adherent to A, V n A # (/); say y E VnA. Then y E A and V is also a neighborhood of y, therefore VnA#(/). (4) In any case A C A, so the problem is to show that

A

J

A

~

A is closed,

in other words,

CA c CA ~ CA is open.

§3.3. Topological Spaces

135

To say that CA e CA means that no point of CA is adherent to A; by the definition of adherent point, this means that

x E CA ~ some neighborhood of x is disjoint from A , in other words,

x E CA ~ some neighborhood of x is contained in CA, and it is the same to say that

x E CA ~ CA is a neighborhood of x. The latter implication says that CA is an open set (3.3.9, (i». (5) Since A = A :) A, A is a closed set containing A. On the other hand, if B is a closed set containing A then A e B = B, whence the minimality of A. (6) Since A e A and B e B we have Au B e Au B, and since Au B is closed (it is the union of two closed sets) it follows from (5) that AU B e AU B. On the other hand, A e AU B and B c Au B imply AeAuB and BeAUB,therefore AuBcAuB. 0

3.3.17. Example. (The metric topology 0/ IR) In §1.15 we introduced the notation JR = JR U {-00, +00} ; we are going to show that there is a metric topology on JR for which, among other things, JR is the closure of its subset JR (thus justifying the notation of §1.15). The first step is to note that JR is order-isomorphic to a closed interval. As in 1.16.10, define a function /: (-1,1)- JR by the formulas for x

-00

x

lex) =

= -1

for -1 0) 3 0 3 d(x, c) < c we need only show that X n E f-l(W) ultimately (3.2.19). Indeed, f(x n ) -> f(c) E W by (a), therefore f(x n ) E W ultimately (3.2.19).

§3.4. Continuity

139

(b) => (b'): If W is a neighborhood of f(c) , then the set V = f-1(W) is a neighborhood of c by (b), and f(V) = f(J-1(W») c W. (b')=>(c):1f 0 then the open ball W=U,(J(c») isaneighborhood of f(c) , so by (b') there exists a neighborhood V of c such that f(V) C W; any Ii > 0 such that U6(C) C V meets the requirements of (c). (c) => (a): Assuming X n --+ c we must show that f(x n ) --+ f(c) . Given any < > 0, choose {j > 0 as in (c); ultimately d(xn,c) < {j and therefore P(J(xn),f(c») < e. 0 Condition (b) suggests the generalization to topological spaces:

3.4.3. Definition. Let X and Y be topological spaces, f: X --+ Y, c EX. We say that f is continuous at c if it satisfies condition (b) of 3.4.2. {It is equivalent to say that f satisfies condition (b').} If f is continuous at every point of X, we say simply that f is continuous on X (or is a continuous mapping of X into V).

3.4.4. Theorem. Let X, Y, Z be topological spaces, let f: X --+ Y and 9 : Y --+ Z be mappings, and let x EX. If f is continuous at x, and 9 is continuous at f(x) , then the composite function go f : X --+ Z is continuous at x. Proof. Write y = f(x) and z = g(y) = g(J(x)) = (g 0 J)(x). If W is a neighborhood of z = g(y) in Z, then g-l(W) is a neighborhood of y in Y, therefore (goJ)-l(W) = f-1(g-1(W») is a neighborhood of x in X. 0

3.4.5. Theorem. Let X and Y be topological spaces, f: X -+ Y. The following conditions are equivalent: (a) f is continuous; (b) V open in Y => f-1(V) is open in X; (c) B closed in Y => f-1(B) is closed in X. Proof. The chain of implications (a) => (b) => (c) => (b) => (a) is not the shortest possible, but each implication has a brief conceptual proof. (a) => (b): Let V be an open set in Y. If x E r 1 (V) then f(x) E V; since V is a neighborhood of f(x) and f is continuous at x, f-1(V) is a neighborhood of x. Thus f-1(V) is a neighborhood of each of its points. (b) => (c): If B is a closed set in Y, then CB is an open set, therefore f- 1(CB) is open by (b); thus Cf- 1(B) = f-1(CB) is open, therefore r 1 (B) is closed. (c) => (b): The argument is similar to the preceding. (b) => (a): If x E X and W is a neighborhood of f(x) , choose an open set V with f(x) EVe W (3.3.8); then r 1 (V) is open by (b) 1 1 (V) C r1(W), therefore and x E (W) is a neighborhood of x. This shows that f is continuous at every x EX. 0

r

r

140

3. Topology

3.4.6. Corollary. II I : X --+ Y and g: Y --+ Z are continuous functions, then so is the composite function go I : X --+ Z. Froof. If W is an open set in Z, then 9 -I (W) is open in Y, therefore (g 0 f)-I(W) = I-I (g-I(W) is open in X. 0 3.4.7. Definition. Thpological spaces X and Y are said to be homeomorphic if there exists a bijection I: X --+ Y such that both I and I-I

are continuous; such a mapping is said to be a homeomorphism of X onto Y. 3.4.8. Example. The bijection I: [-1, 1) --+ IR of Example 3.3.17 is a homeomorphism for the indicated metric topologies.

Exercises 1. For a function I: X --+ Y between topological spaces, the following conditions are equivalent: (a) I is continuous; (d) I( A) C I(A) for every subset A of X. {Hint: Use criterion (c) of 3.4.5.}

2. If X is a trivial topological space and Y is discrete (3.3.2), then every continuous function I: X --+ Y is a constant function. {Hint: Each point of X has only one neighborhood.}

3. If Y is a discrete space then every continuous fu nction is constant. {Hint: rl({f(O)}) is both closed and open.}

I:

IR

--+

Y

4. Let (X, d) be a metric space, A a nonempty subset of X. For each x EX, the nonnegative real number

inf{d(x, a): a E A} is called the distance from x to A and is denoted d(x, A) . The formula I(x) = d(x, A) defines a function I: X --+ 1R. (i) I-I({O}) = A. (ii) I is continuous; indeed, I/(x) - I(y)l :5 d(x, y) for all x, y EX. {Hint: (ii) For all a E A, I(x) = d(x,A):5 d(x,a):5 d(x,y)+d(y,a).} 5. If A is a nonempty closed subset of a metric space (X, d) and if y is a point of X with y ¢ A, then there exists a continuous function 9 : X --+ [0, 1) such that 9 is zero at precisely the points of A, and g(y) = l. {Hint: With I as in Exercise 4, define g(x) = min{j(y)-I l(x),I}.} 6. Let (X, d) be a metric space and equip the product set X x X with the metric

D((x, y), (x', y'» = d(x, x') + dey, y') (cf. §3.1, Exercise 6, with n = 2 and p = 1).

§3.5. Limit of a Function

141

(i) (x n , Yn) ---+ (x, y) for D ¢> X n ---+ x and Yn (ii) d is a continuous function X x X ---+ JR. {Hint: (ii) Cf. 3.2.4.}

---+

Y for d.

7. Let A be a subset of a topological space X and let 'P A : X --+ JR be its characteristic function (1.3.4). Prove: (i) 'PA is continuous at a point a E A if and only if a is interior to A. (ii) The set of discontinuities of 'PA is the boundary 8A of A (cr. §3.3, Exercise 6).

3.5. Limit of a Function

Intuitively, continuity of a function f at a point c means that if x approaches c, then f(x) approaches f(c); the concept of limit merely asks that if x approaches c, then f(x) approaches something. A definition broad enough to encompass the one-sided limits and derivatives of functions of a real variable is so complex notationally as to require some preparation; the best place to start is to revisit the motivating example.· Consider a function 9 : [a, b) ---+ JR and a point c, a < c < b. The function 9 is said to be differentiable at c if the difference-quotient g(x) - g(c) x-c

has a limit L in JR as x approaches c, in the sense that for sequences (x n ) in la, b] , X n -+ C. In

:F c

=>

g(xn ) Xn

g(c) -c -

-+

L.

(Such an L is unique and is called the derivative of 9 at c.) Writing f(x)

= g(x) -

g(c) , x-c

we have a function f: B ---+ JR defined on the subset B [a, b], and the condition for differentiability reads X n --+ C, X n E

B

~

f(x n )

--+

= la, b) -

{c} of

L.

For right-differentiability, we require the existence of an L E JR such that X n ---+ C, X n

>C

~

f(x n )

---+

L;

here, the approximating sequences are confined to the subset A = (c, b] of B, and the condition now reads X n ---+ C, X n E

1

First course, Chapter 8, §l.

A

~

f(x n )

---+

L.

142

3. Topology

For left-differentiability, where X n < c is required, we consider instead A = [a, c). Thus, initially we have a metric space X = [a. b] , but the function whose limit is under consideration is defined only on a subset B of X. The point c of X at which the limit is contemplated happens not to belong to the domain B of I; it is, however, adherent to B (3.3.12) and to each of the subsets A considered above. The following diagram summarizes matters:

X U

I

B

IR

U

c

E

A

:>

A

For right-differentiability we take A = (c, b]. for left-differentiability A = [a, c) , and for differentiability A = B . With this example in mind, the following definition becomes digestible:

3.5.1. Definition. Let X and Y be metric spaces, B a subset of X, -+ Y a function defined on B and taking values in Y. Let A be a subset of B and let c E A (the closure of A in X). Schematically:

I :B

X

u

I

B

Y

u c

E

A

:>

A

If there exists a point y E Y such that X n -+ C, X n

we say that written

I

E A => I(x n )

-+

y,

has limit y as x tends to c through values in A, lim

%-c.%EA

I(x)

= y.

If A = B we write simply

lim I(x)

%_0

= y.

3.5.2. Remarks. (i) The existence or nonexistence of y is entirely determined by the restriction IIA of I to A. (ii) If such a y exists, then it is unique (3.2.1). (iii) If c E B one can drop down to B and dispense with X. In this case I(c) is defined. but if c ¢ A then I(c) is irrelevant to the existence of y. (iv) If c E A then I(c) is defined and the notion of limit brings nothing new to the table; it is just continuity of IIA at c:

§3.5. Limit of a Function

143

3.5.3. Theorem. With notations as in 3.5.1, suppose also that c EA. Then:

3

lim

x-c.xEA

~

f(x)

flA is continuous at c,

and in this case lim

x_c,xEA

f(x) = f(c).

Proof· =>: Let y E Y be the limit whose existence is assumed. Since c E A we are free to take the constant sequence X n = c in 3.5.1, therefore f(c) = y. Then f(x n ) -+ y = f(c) for every sequence X n E A converging to c, in other words f I A is continuous at c (where A is regarded as a metric space with the metric dlA x A it inherits from X). ~: By assumption, f(x n ) -+ f(c) for every sequence X n E A with X n -+ C, so f(c) meets the requirements for y in 3.5.1. There are also e,8 and neighborhood formulations of limit: 3.5.4. Theorem. With notations as in 3.5.1, let d be the metric on X, and p the metric on Y. Let y E Y. The following conditions are equivalent: (a) limz_c,zEA f(x) exists and is equal to y; (b) for every e > 0 there exists a 8 > 0 such that

< 8 => p(J(x),y) < e;

x E A, d(x,c)

(c) for every neighborhood W of y in Y, there exists a neighborhood V of c in X such that f(VnA) C W. Proof. (a) => (b): If, on the contrary, there is an e > 0 for which no suitable 8 exists, then for each positive integer n there exists a point X n E A such that d(xn,c) < lin but P(J(xn),y) 2': l . Then X n -+ C but f(x n ) f> y, contrary to (a). (b) => (c): Let W be a neighborhood of y and choose e > 0 so that W contains the open ball U,(y) (3.2.12). For this e, choose 8 as in (b); then the open ball U6(C) meets the requirements for V. (c)=>(a):Let XnEA, Xn-+c. Givenanye >O,let W=U,(y) and choose V as in (c); ultimately X n E V, therefore f(x n ) E W, that is, p(J(x n ), y) < l . This shows that f(x n ) -+ y. For functions of a real variable, the approach to c can be from either one or both sides. In the following definition, we restrict attention to the case that c is approachable through an adjacent interval from one or both sides; this covers the case of derivatives and is adequate for all of our applications:

3.5.5. Definition. Suppose X 3

= JR

lim

x-c,xEA

in 3.5.1, and suppose

f(x)

= y.

3. Topology

144

(i) If B is a 'deleted neighborhood' of c, that is, B:::> (c-r,c)U(c,c+r)

for some r > 0,

and if A=Bn(IR-{c})={xEB: xfc}, then we write instead lim

x-c, Z:FC

f(x) = y.

The possibility that c E B (so that B is a neighborhood of c) is not ruled out; it just does not figure in the definition either here, or in the following definitions (ii) and (iii). (ii) If B is a 'deleted right neighborhood' of c, that is, B:::> (c,c + r)

for some r > 0,

and if A=Bn(c,+OO)={xEB: x>c}, we write instead lim

x_c, x>c

f(x) = y

or

lim f(x) = y,

x-c+

and we say that f has right limit y at c, expressed concisely as f(c+) = y. (iii) If B is a 'deleted left neighborhood' of c, that is, B:::> (c-r,c)

for some r > 0,

and if A = Bn(-oo,c) = {x E B: x < c}, we write instead lim

x_c.x O. In view of 4.4.10, there exists a sequence (In) of ISF's such that 0 $ fn T f and I(Jn) Tf fdJ.L. Given anye > 0, let En

= {x:

fn(x) > e},

E

= {x:

f(x) > e};

the sets En and E are measurable by 4.1.8. Since fn T f pointwise, the sequence (En) is increasing with union E, briefly En TE, therefore J.L(En ) T J.L(E) . Clearly En C N(Jn) , so the sets En have finite measure. Moreover,

thus e 'PE., is an ISF and e J.L(En) $ I(Jn) $ f fdJ.L; passage to the limit in the inequality J.L(En ) $ (l/e) f fdJ.L yields J.L(E) $ (l/e) f fdJ.L < +00. The first assertion of the corollary now follows from the inclusion

{x: If(x)1

~

e}

C

{x: If(x)1 > e/2}.

The final assertion of the corollary is verified by the sequence F n = {x: l/n $ If(x)1 < n}. The next corollary dispels any ambiguity about the phrase 'integrable simple function': 4.4.15. Corollary. If f is a simple function, then the following conditions are equivalent: (a) f is an ISF in the sense of 4.3.3; (b) f is integrable in the sense of 4.4.7. For such a function f. f f dJ.L = I (J) .

170

4. Lebesgue Integral

r

Proof. (a) => (b): Writing I = 1+ (c!. 4.3.4), it follows from Remark 4.4.4 that I is integrable in the sense of 4.4.7 and that

J

I+dp. -

J

rdp. = I(j+) - I(f-) ,

in other words f I dp. = I (f) . (b) => (a): The problem is to show suming N(f) nonempty, that is, I the smallest nonzero value of I/(x)l; therefore N (f) has finite measure by

that N(f) has finite measure. Asnot identically zero, let c > 0 be then N(f) C {x: I/(x)1 > c/2}, the preceding corollary. 0

4.4.16. Corollary. II I is integrable and 9 is a measurable function such that 9 = I a.e., then 9 is integrable and f gdp. = f I dp. . Proof. The function h = Ig- II is measurable, h 2: 0 (everywhere) and h = 0 a.e.; let us show that h is integrable with integral O. Let (h n ) be a sequence of simple functions such that O:S h n T h; then h n = 0 a.e., therefore hn is an ISF with I(hn) = 0 (4.3.13). Thus h is integrable and

Since Ig - II = h is integrable, it follows from 4.4.12 that 9 integrable, therefore so is 9 = (g - I) + I; moreover,

J

(g - f)dp.
O} then it follows from the lemma that p.(E) = 0, that is, h $ 0 a.e., in other words I $ 9 a.e. By symmetry, 9 < I a.e. as well, therefore I = 9 a.e.

§4.5. Monotone Convergence Theorem

173

Exercises 1. If

f

~

0 is measurable and 9 > 0 is integrable, then f n 9 is

integrable. 2. If f and 9 are integrable functions such that IE fdJ.L all measurable sets E of finite measure, then f = 9 a.e.

= IE gdJ.L

for

3. The relation f = 9 a.e. is an equivalence relation in £1 = £MJ.L}. The set LI = Ll(J.L} of equivalence classes [f] is a vector space in a natural way, and [f] ..... J fdJ.L defines a linear form on L1 (positive in the appropriate sense, where [fl ~ 0 signifies that f ~ 0 a.e.).

4.5. Monotone Convergence Theorem, Fatou's Lemma

In the preceding section, the extension of the integral from the vector space £ of ISF's to the vector space £1 of integrable functions was acoomplished via approximation by monotone sequences of ISF's (Definition 4.4.1). The core result of the present section is that nothing new is obtained by repeating the process starting from £ 1 ; so to speak, the passage from C to £1 is a 'completion' process, analogous to the passage from the field Q of rational numbers to the field lR ofreal numbers (d. Theorem 1.8.26). Three highly useful corollaries (the monotone convergence theorem, dominated convergence theorem and Fatou's lemma) are readily derived from the core result, which is as follows: 4.5.1. Lemma. Let f : X ..... lR and suppose there exists a sequence (In) of integmble functions such that fn 1 f (pointwise on X) and such that the sequence of integmls I fndJ.L is bounded. Then f is integmble and

I fndJ.L 1 J fdJ.L.

Proof. At any rate, f is measurable (4.1.20). Replacing fn by fn - It for all n (and f by f - It), we can suppose that 0 ~ fn 1 f. For each n let (gnj) be a sequence of ISF's such that

o (b): For every n, h $ fn $ f a.e., therefore (4.4.19)

J

hdlJ. $

J

fndlJ. $

J

fdlJ. < +00.

4.5.4. Corollary. (Lebesgue's dominated convergence theorem) Let 9 be an integrable function and let Un) be a sequence of integrable functions such that IInl $ 9 a.e. for all n and such that the sequence (In(X)) is convergent for almost every x. Then, there exists an integrable lunction f such that f n -> I a.e. Necessarily J lin - IldlJ. -> 0, in particular J IndlJ. -> J fdlJ.· Proof Multiplying 9 and the In by 'PCE for a suitable null set E, we can suppose without loss of generality that Ifni $ 9 (everywhere on X) and that (Jn (x» is convergent for every x EX. Define f(x)

= lim In(x) n

for all x EX.

Since f is measurable (4.1.20) and If I $ g, itfollows that f is integrable (4.4.20). Then In - f is integrable, I/n- 11 $l/nl+11I $2g and Ifn - II -> o pointwise on X. The functions h n = lIn - 11 and h = 2g are integrable, with 0 $ h n $ hand h n -> 0 pointwise on X; our problem is to show that J hndlJ. -> O. For each n, define Hn

= suph;. i2::n

Since 0 $ H n $ hand H n is measurable (4.1.19), it follows that H n is integrable; moreover, for every x EX, Hn(x) ! lim sup h;(x) ;

.

= lim h;(x) =

0

(cf. 1.16.7), therefore J HndlJ. ! 0 by the monotone convergence theorem. Since 0 $ hn < H n , it follows that J hndlJ. -> o.

§4.5. Monotone Convergence Theorem

177

The next corollary serves as a lemma to the I Riesz-Fischer theorem' to be proved later on (Chapter 6, §7). 4.5.5. Corollary. (Fatou's lemma) 1/ (In) is a sequence functions such that fn ~ 0 a.e. for all n, and i/ limninf

0/ inte9rable

j In d/L < +00 ,

then there exists an inte9rable function f Moreover,

such that f

= lim inf /n

a.e.

j Id/L ~ limninf j Ind/L. Proof· Note that the values of the function lim inf / n may be infinite (4.1.16). The hypothesis on the sequence of integrals is that f fnd/L does not converge to +00 in the sense of 1.16.8. We can suppose without loss of generality that In ~ 0 (everywhere on X). Define 9n

= inf /; l.~n

(n

= 1,2,3, ...),

9

= sUP9n = lim.inf /; n 1.

(9 may have infinite values); the measurable functions 9n are integrable (because 0 ~ 9n ~ fn ) and 9n T9 pointwise. For each n,

i~n ~ 9n~/; ~ j9nd/L~j/;d/L' therefore

H

j9nd/L

~ l.2:n inf j/;d/L ~ lim.infj/;d/L < +00, 1.

thus the (increasing) sequence of integrals f 9nd/L is bounded; by the monotone convergence theorem, there exists an integrable function f such that 9n T f a.e. Already 9n T 9 a.e., so / = 9 a.e., that is, f = liminfn /n a.e. Moreover,

j fd/L = s~p j

9nd/L

~ limninf j fnd/L

by the monotone convergence theorem and the inequalities (*). The following convention is analogous to the notation for a divergent positive-term series:

4.5.6. Convention. If / is measurable, grable, we write

j fd/L=

L:::"=l an = +00

I > 0 a.e. and / is not inte-

+00;

178

4. Lebesgue Integral

this means that if (fn) is a sequence of simple functions such that 0:::; In T I a.e., then either some (hence every succeeding) In fails to be an ISF, or else the In are all ISF's but the sequence f Indp. is unbounded.

Exercise

1. The notational convention in 4.5.6 has the following consequences: (i) If I and 9 are measurable functions such that 0 < I :::; 9 a.e., then J I dp. :::; J gdp.. (ii) If I is measurable, then I E £1 AnBEC; (2) A E C => A' = X - A is the union of a finite number of pairwise disjoint sets in C. Then the algebra A(C) of subsets of X generated by C coincides with the set A of all finite disjoint unions of sets in C .

Proof. It is obvious that C cAe A(C) ; to prove that A(C) c A it will suffice to show that A is an algebra of sets. According to Definition 2.4.1, we must show that tl> E A and that A is closed under complementation and finite unions. Suppose E, F E A, say m

E=

UA

n

i ,

i=1

where the Ai (the B j

(a)

)

F=

UB

j ,

j=1

are pairwise disjoint sets in C. Then

En F

= U Ai n B j i.;

EA

§4.6. Monotone Classes

179

because the mn sets A;nB j belong to C by (1) and are pairwise disjoint; thus A is closed under finite intersections. Since the A; belong to A by (2), it then follows that m

(b)

E'=nA;EA. i=l

From (a) and (b) we conclude that 0 = EnE' E A, and EuF =(E'nF')' E

A.O 4.6.2. Example. Let la, b) be a fixed closed interval of IR and let C be the set of all subintervals (not necessarily closed) of [a, b]. If I, J E C then In J E C and la, b)- I is either a subinterval of [a, b] or the union of two disjoint subintervals. Thus C satisfies the conditions (1), (2) of the preceding proposition. It follows that the set of all finite disjoint unions of subintervals of la, b] is the algebra A(C) of subsets of [a, bJ generated by C. 4.6.3. Example. Let C and V be sets of subsets of X and Y, respectively, each satisfying the conditions (1), (2) of 4.6.1, and let P = {C x D: C E C, DE V}.

The formulas

(C 1 x Dd n (C2 x D2) (X x Y) - (C x D)

= (C 1 nC2) x (D 1 n D2) = [(X - C) x Y] u [C x (Y -

D)J

show that the set P of subsets of X x Y satisfies the conditions (I), (2) of 4.6.1. In particular, if C and V are algebras of subsets of X x Y, then the set of all finite disjoint unions of 'rectangles' C x D with 'sides' C E C, D E V is an algebra of subsets of X x Y (namely, the algebra generated by the rectangles C x D). The 'monotonicity' referred to in the section heading is a key concept in passing from algebras of sets to u-a1gebras; the definition is as follows:

4.6.4. Definition. Let X be a set, T C P(X) a set of subsets of X. We say that T is monotone (or that T is a 'monotone class') if it is closed under monotone sequential unions and intersections: (E,,) a sequence in E" TE or E,,!E

T}

=>

E E

T.

4.6.5. Example. If J.L and v are two finite measures defined on a u-algebra S, then the set {E E S: J.L(E) = v(E) } on which J.L and v agree is a monotone class: it is closed under increasing sequential unions by a general property of measures (2.6.2) and under decreasing sequential intersections by a property of finite measures (2.6.3).

4. Lebesgue Integral

180

It is obvious that every ,,-algebra is monotone. Conversely, every monotone algebra is a ,,-algebra (because every countable union is the increasing union of its sequence of finite 'partial unions'). The following result is more subtle: 4.6.6. Theorem. (Lemma on monotone classes)1

If

A eTc P(X) ,

where A is an algebra of subsets of X and T is a monotone class, then S(A) c T. Proof In words, every monotone class that contains an algebra A of sets also contains the ,,-algebra generated by A; this means that S(A) is the smallest monotone class containing the algebra A. There exist monotone classes containing A (for example, S(A) or P(X)); let M be the intersection of them all. Clearly M is itself a monotone class containing A, and is the smallest such class. In particular,

A

c

MeT and A

c

M

c

S(A) .

To prove that S(A) C T, it will suffice to show that S(A) = M; already M c S(A) , so we need only show that S(A) eM. For this, it will suffice to show that M is an algebra; being monotone, it will then be a ,,-algebra containing A and therefore containing S(A). We are reduced to proving that M is closed under finite unions and complementation. To this end, we consider pairs of subsets A, B of X such that

A-B, B-A and AuB belong to M. For example, the condition (*) is satisfied if A E A and B EA. For any set A C X we write

K:(A)

= {B eX:

A and B satisfy (*)}.

Since the condition (*) is symmetric in A and B, we have (i)

B E K:(A)

¢}

A E K(B) ,

and, by the remark following (*),

(ii)

A E K:(B) for all A, B EA.

The key to the rest of the proof is the observation that

(iii)

for every subset B of X, K(B) is a monotone class.

1 P.R. Halmos. Measure theory [Van Nostrand. New York, 1950j reprinted. SpringerVerlag, New York, 1974j, §§4, 5.

§4.6. Monotone Classes

181

For, if (An) is a monotone (increasing or decreasing) sequence of sets in K:(B) , SO that in particular An-B, B-An, AnUB EM for all n,

then 'passage to the limit' implies that A and B satisfy (*); for example, if An TA then An-BTA-B, B-AnlB-A, AnUB TAuB,

that A - B, B - A , AU B belong to M (because M is a monotone class). If B E A then A C K(B) by (ii); since K(B) is monotone, it follows that M C K:(B) . Thus,

50

B E A => M

(iv)

c

K(B) .

If A E M and B E A, then A E K:(B) by (iv), therefore BE K:(A) by (i); thus A EM=> A

c

K(A) ,

and, since K(A) is monotone, we conclude that A EM=> Me K(A).

In other words, A, B EM=> A - B, B - A, Au B EM.

In particular, M is closed under finite unions and (because X E A eM) complementation. 4.6.7. Corollary. Let C be a nonempty set of subsets of a set X, satisfying the conditions (I), (2) of 4.6.1, and let S = S(C) be the a-algebra generated by C. If J.l qnd II are finite measures on S such that

J.l(A)

= II(A)

for all A E C,

then J.l = II on S. Proof. Let T = {E E S: J.l(E) = v(E)}; our problem is to show that SeT. Let A be the algebra generated by C. Obviously S(A) = S(C) = S, and it is clear from 4.6.1 that ACT; since T is a monotone class (4.6.5), we conclude from 4.6.6 that S(A) cT. The corollary can be extended to certain not necessarily finite (but not too far from finite!) measures: 4.6.8. Corollary. Let X be a set and let C be a set of subsets of X

with the following properties: (1) A, BEC => AnBEC; (2') A, B E C => A - B is the union of a finite number of pairwise disjoint sets in C;

182

4. Lebesgue Integral

(3) X is the union of a sequence of sets in C. Let S = S(C) be the CT-algebra generated by C and suppose that J.L and v are measures on S such that

J.L(A) = v(A) Then J.L

=v

< +00 for all

A E C.

on S.

Proof. Note that the pair of conditions (1), (2') is weaker than (i.e., implied by) the pair (1), (2) of 4.6.1 (because A - B = An B'). Note also

that if A, B E C then it follows from (2') that the set Au B = (A - B) U B is a finite disjoint union of sets in C. More generally, if AI, ... , An is a finite list of sets in C, then AI U... U An is a finite disjoint union of sets in C. The proof is by induction on n: assuming inductively that Au ... U An-I = CI U ... U C r with the C; pairwise disjoint sets in C, the set Al u ... uAn = (C I u = [(C I U

=

UCr)UAn uCr ) - AnI UA n

[U(C; -An)]

UAn

1=1

is a pairwise disjoint union of sets in C by the condition (2'). By the condition (3), there exists a sequence (An) of sets in C such that X = U::"=I An. The idea of the rest of the proof is to reduce to the case of finite measures by applying 4.6.7 in each of the sets An. Writing Bn = U~=I A k , we have Bn nET E for every E E S, so it will suffice to show that J.L(B n n E) = v(B n n E) for all n and for all E E S. Since Bn is known to be a finite disjoint union of sets in C, we need only show that (*)

J.L(AnE) = v(AnE) for all A E C and E E S.

Fix a set A E C . The desired relation (*) calls attention to the class of sets AnS={AnE: EES}, which is easily seen to be a CT-algebra of subsets of A; moreover, the restrictions of J.L and v to A n S are finite measures. On the other hand, the class AnC={AnB: BEC} clearly satisfies the conditions (I), (2) of 4.6.1 relative to the set A, and we know that J.L = v on A n C; in view of 4.6.7, it follows that J.L = v on the CT-algebra SA = SA(AnC) of subsets of A generated by the class

§4.6. Monotone Classes

183

An C . To verify the condition (*), which says that J.L need only show that An S = SA , that is,

= 1/

on An S , we

(**)

The argument for this is valid for every every set C of subsets of X and for every subset A eX. For, An S(C) is a q-a1gebra on A containing A n C , therefore An S(C) ::> SA(A n C);

on the other hand, the set {E eX: An E E SA(A n C)} is obviously a q-algebra on X containing C, hence also containing S(C) , therefore AnS(C)CSA(AnC). 4.6.9. Example. The class C of all bounded intervals (open, closed, semiclosed) in IR clearly satisfies the conditions (1) and (2') of the preceding corollary, and the q-a1gebra S(C) generated by C is the class B = B(IR) of Borel sets in IR (2.4.5). The restriction J.L = AlB of Lebesgue measure A to the q-a1gebra B is a measure that assigns to each bounded interval its length; by the preceding corollary, J.L is the only such measure on B. Indeed, any two measures on B that are equal and finite on C must be identical:

4.6.10. Corollary. If B = B(IR) is the class of Borel sets in IR, I is the set of all open intenJais in IR, and J.L, 1/ are measures on B such that J.L(I) then J.L

= 1/

= 1/(1) < +00

for all I E I,

on B.

Proof If J is any closed (or semi-closed) interval in IR, then there exists a sequence (In) of open intervals with In 1 J, therefore J.L(J) = I/(J) by 2.6.3. Thus J.L = 1/ on the set C of all bounded intervals, therefore J.L = 1/ on B by Corollary 4.6.8.

Exercises 1. Corollary 4.6.10 remains true with "open" replaced by "closed" or by "semiclosed" .

2. Let S be a q-a1gebra of subsets of IR containing the class B of Borel sets, and let J.L be a measure on S assigning to each open interval its length. Then J.L = A on S n M ,where A is Lebesgue measure and M is the class of all Lebesgue-measurable sets. In particular, A is the only measure on M that assigns to each open interval its length. {Hint: 4.6.10 and 2.4.15.}

184

4. Lebesgue Integral

4.7. Indefinite Integrals

The 'indefinite integrals' discussed in this section are a generalization to abstract measure spaces of the classical concept (see the remark following 4.4.22). The context is a fixed measure space (X,S,Jl) and, as in 4.4.22, we write

h =J IdJl

'PE/dJl

for every integrable function I E £ 1(Jl) and every measurable set E E S . The term 'a.e.' is understood to be relative to this measure space (§4.2). 4.7.1. Definition. For each

IE £1(Jl)

we define a real-valued function

I·Jl:S ..... 'R

on the u-algebra S of measurable sets by the formula (J . Jl)(E)

=

h

IdJl.

The 'set function' I· Jl is called the indefinite integral associated with the Jl-integrable function I. Our goal in this section is to prove some general properties of indefinite integrals and to apply them to the classical special case of Lebesgue measure. In the next section we characterize the set functions I· Jl associated with a finite measure Jl (the core case of the Radon-Nikodym theorem, to be proved in greater generality in Chapter 9). We begin with the simplest formal properties of indefinite integrals: 4.7.2. Theorem. Let I,g E £1(Jl) and let c E 'R. (i) (J + g) . Jl = I· Jl + g. Jl and (cf)· Jl = c(J· Jl) ; thus, the correspondence I - I· Jl is a linear mapping £1(Jl) ..... .1'(S,IR). (ii) I . Jl = 9 . Jl ~ I = 9 a.e. (iii) I . Jl ? 0 ~ I? 0 a.e. (iv) I· w:5. g. Jl ~ I 75, 9 a.e. (v) I . Jl = 1+ . Jl - 1- . Jl, where the set functions 1+ . Jl, 1- . Jl are positive. (vi) II E and F are measumble sets with E n F = 1]), then (J . Jl)(E U F) = (J . Jl)(E) + (J . Jl)(F) . (vii) II E E Sand Jl(E) = 0, then (J. Jl)(E) = O. Proof (i) Indefinite integrals I· Jl belong to the vector space .1'(S, 'R) of real-valued functions S ..... 'R. whence the possibility of forming the indicated sums and scalar multiples of them. The formulas in question are immediate consequences of the identities

'PE(J + g)

= 'PEl + 'PEg,

and the linearity of integration.

'PE(c/)

= C('PEf)

§4.7. Indefinite Integrals

185

(ii) The equivalence is immediate from 4.4.16 and 4.4.24. (iii), ~: The implication is a consequence of 4.4.19. =>: Let E = {x: I(x) < O}. By assumption, IE IdlJ. = (f ·IJ.)(E) ~ O. However, since 'PEl < 0 by the definition of E, we have also IE IdlJ. $ 0, therefore IE fdlJ. = O. Then IE( - f)dlJ. = 0 and -I > 0 on E, therefore IJ.(E) = 0 by 4.4.23, in other words, I ~ 0 a.e. (iv) The equivalence is clear from (i) and (iii). (v) The equality follows from applying (i) to the decomposition I = r - r (4.4.12), and the functions r·IJ., I-·IJ. are positive by (iii). (vi) Since E and F are disjoint, 'PEuF = 'PE + 'PF ; the asserted equality then follows from the linearity of integration. (vii) This follows from the fact that 'PEl = 0 a.e. (relative to the mea,. sure IJ. ). It follows form (vi) that the set function I· IJ. is 'finitely additive' in the sense of 2.6.1, (1). Better yet:

the indefinite integral I· IJ. is countably additive in the sense 01 2.4.12: il (En) is a sequence 01 pairwise disjoint measurable sets, then 4.7.3. Theorem. For every

I

E £I(IJ.) ,

and the infinite series on the right converges absolutely. In particular, il I ~ 0 a.e., then I· IJ. is a (finite) measure on S. Proof Let F n = U~=I Ek, F = U~I Ek, SO that F n 1 F. Suppose first that I ~ 0 a.e., so that I·IJ. ~ 0 by (iii) of the preceding theorem. Then 'PF. I 1 'PF I a.e., so by the monotone convergence theorem (4.5.3),

J

'PF•.fdlJ. 1

J

'PF IdlJ..

that is, (f ·1J.)(Fn ) 1 (f ·IJ.)(F). By (vi) of the preceding theorem, we have n

(f ·1J.)(Fn ) = L(f ·1J.)(Ek) , k=1

thus n

(f. IJ.)(F)

= sup(f ·1J.)(Fn ) = n

lim L(f ·1J.)(Ek); n-oo k=l

this proves the desired formula and shows that I· IJ. is a finite measure (2.4.12). In the general case, write I = g - h with 9 and h integrable functions ~ 0 (for example, g = 1+ , h = 1- ). Citing the preceding case, we

4. Lebesgue Integral

186 have

U . IL}(F) = (g. IL)(F) -

(h· IL}(F)

00

00

= ~)g .1L)(En) - ~)h. IL}(En) n=l

n=l

00

=

00

L[(g 'IL}(En) -

(h '1L)(En)]

= LU ·IL}(En).

n=l

n=l

Moreover, the inequality IU· IL)(En)1 :5 (g. 1L)(En) + (h . 1L)(En) shows that the convergence is absolute. (Alternatively, the series converges for every permutation of its terms-Le., for every permutation of the En a property that characterizes absolute convergence.) 0 4.7.4. Corollary. Let A be an algebra 01 sets such that S is the (j -algebra generated by A. II f and 9 are integrable functions such that

L =i fdlL

gdlL

lor all A E A,

then f = 9 a.e. Proof. Let h = I - g. We know that fA hdlL = 0 for all A E A, and the problem is to show that h = 0 a.e. Write h = u - v with u, v integrable and ;::: O. Then

L =i udlL

vdlL for all A E A,

and we are to show that u = v a.e. By the preceding theorem, u· IL and v . IL are finite measures on S, and we know from (*) that u· IL = v . IL on A, therefore U'IL = V'IL on S(A) = S by 4.6.7, whence u = v a.e. by (ii) of Theorem 4.7.2. 0

In the next corollary, we specialize to the case that (X, S,IL) is the Lebesgue measure space (JR, M, >') of 2.4.13: 4.7.5. Corollary. If >. is Lebesgue measure on IR and il I and 9 are Lebesgue-integrable functions such that

tId>' then I

=9

=t

gd>' lor all a,b E IR

(a:5 b),

a.e.

Proof. As in the preceding corollary, write I - 9 = u - v, where u, v are ;::: 0 and integrable. By hypothesis, the finite measures u· >. and v· >. are equal on the set of all closed intervals in JR, and our problem is to show that u = v a.e.

§4.7. Indefinite Integrals

187

Since singletons are negligible for >., we see that u· >. and v· >. are also equal on the set I of all open intervals; it follows that u· >. = v . >. on the u-algebra 8 = S(I) of Borel sets of IR (4.6.10). If E is any Lebesgue-measurable set, we may write E = BUN, where B is a Borel set, N is negligible and B n N = 0 (2.4.17); we know that (u· >')(B) = (v· >')(B) and, by (vii) of 4.7.2, (u· >')(N) = (v· >.)(N) = 0, therefore (u· >')(E) = (v· >')(E) by the additivity of measures. Then u = v a.e. by (ii) of 4.7.2. 0 The following example will be the focus of the next chapter:

Example. Let (IR,M, >') be the Lebesgue measure space (2.4.13) and fix a closed interval la, b] in IR (a < b). The intersections of the sets of M with [a, b] form au-algebra M o of subsets of [a, b] , 4.7.6.

Mo = M n

la, b] = {E n [a, b]: E EM} .

Moreover, Mo is the class of all sets in M that are contained in la, b],

Mo = {E EM: E C [a, bj} ; this follows from the fact that la, b] E M and M is closed under finite intersections. Similarly, if 8 is the u-algebra of Borel sets in IR, then the class 80

= 8 n [a, b] = {B n la, b]:

B E 8}

is a u-algebra of subsets of [a, b], and

Bo = {B E 8: Bela, bj}. If C is the class of all bounded intervals in IR, then 8 is the u-algebra generated by C, 8 = S(C) (cr. 4.6.9). Clearly the class Co = C n [a, b] is the class of all subintervals of [a, b] and, by an argument in the proof of 4.6.9, the u-a1gebra of subsets of la, b] generated by Co is Bo:

Sla.b)(C n la, b))

= S(C) n [a, b] = 8 n [a, b] = 80.

The restriction of Lebesgue measure to Mo is a finite measure; let us for a moment denote it by >.0: >.0 = >'IMo. The finite measure space (la, b], Mo, >'0) is called Lebesgue measure on [a, b] , and a function f: [a, b] -+ IR is said to be Lebesgue-integrable if f E .c1(>.0). To simplify the notation, we shall drop the subscript 0; thus, if f is a Lebesgueintegrable function on [a, b], we shall write simply

J

fd>', or

for the integral

f fd>.o , and if

b

r fd>',

fa

or

1.

[a.b)

fd>'

[e, d] is a closed subinterval of [a. b) then

4. Lebesgue Integral

188

the notation

replaces the ponderous

J

'P[c,d!fd>.o,

where 'Plc.d1 : [a, b] of la, b) .

-+

lR is the characteristic function of

Ie, dJ

as a subset

4.7.7. Corollary. If f: [a,b]-+ lR is Lebesgue-integrable and if

1% fdA = 0 then f

=0

for all x E [a, b) ,

a.e.

Proof Write f = u - v with u, v Lebesgue-integrable and > 0 on la, b]. Our assumption is that

1% ud)' = 1% vdA

for all x E

la, b]

and the problem is to show that u = v a.e. With notations as in the preceding example, consider the measures u· A and v· A on the u-algebra Mo of Lebesgue-measurable subsets of la, b) . By hypothesis, u·). = v· A on the set of intervals [a, xl, a:5 x :5 b. Both measures obviously vanish on singletons, so from the formula

(c, dJ

= [a, dl- [a, c]

we see that u·). = v . A on the set Co of all subintervals of [a, b] , and it follows that u· A = v . A on the u-algebra Bo generated by Co (4.6.2, 4.6.7). Finally, u· A = v .). on M o by the argument in Corollary 4.7.5, and u = v a.e. by 4.7.2, (ii). 0

Exercises 1. A function f:

la, b]-+ lR

is Lebesgue-integrable in the sense of 4.7.6 if and only if the function g: lR -+ lR defined by

g(x) = {f(X) for x E [a, b] o for x E lR - [a, b] is integrable with respect to the Lebesgue measure A on lR. That is, in the notations of 4.7.6, and, in this case,

f E .c 1 (Ao) ¢} f fd>.o = f gdA .

9E

.c 1 (A);

§4.8. Finite Signed Measures

189

2. Let f: [a, b] --+ JR be Lebesgue-integrable. (i) Show that the function x ...... fdA is continuous on [a, b] . {Hint: It suffices to consider X n Tx and x n ! x .} (ii) If IdA = 0 for all x in a dense subset of [a, b] ,then f = 0 a.e. (iii) If fdA = 0 a.e. then f = 0 a.e. and fdA = 0 for all x. {Hint: A negligible set has empty interior; cf. §3.3, Exercise 5.}

f:

f: f:

f:

3. (i) As in Example 4.7.6, let Bo be the set of all Borel subsets of the closed interval [a, b] , and let J.l be any measure on Bo . If f is a J.l-integrable function such that

1 x

fdJ.l

=0

for all x E [a, b] ,

then f = 0 a.e. (relative to J.l). {Hint: Write I = u - V with u, V J.l-integrable and > 0, and let T = {B E 130: (u· J.l)(B) = (v· J.l)(B) } . If a ~ c ~ d < b, then the formula (c, d] = [a, d] - la, c] shows that (c, d] E T. Infer that T contains every singleton {d} (d E [a, bJ) , hence it contains the set Co of all subintervals of la, b] ; conclude that if A is the algebra of subsets of [a, b) generated by Co, then AcT. Then cite 4.7.4.} (ii) It suffices to assume in (i) that fdJ.l = 0 for X = b and for X in a dense subset of (a, b).

f:

f is a Lebesgue-integrable function on JR, c is a fixed point of JR, x and fe fdA = 0 for all X (or for all X in a dense subset of JR), then f = 0 a.e. 4. If

{Hint: The convention is that when that if a ~ b then

t

fdA

={

X

< c,

fdA - [

f; fdA = - f: fdA. Argue

fdA,

by considering separately the three cases c < a c.}

~

b, a

~

c < b, a

~

b
'

for all

X E

-+

IR such that

[a, b].

The function f is determined essentially uniquely by F, in the following sense: a Lebesgue-integrable function 9 on [a, bJ represents F as in (b) if and only if 9 = f a.e. In particular, f can be taken to be a Borel function. Proof. (b)

'*

'* (a): This is immediate from 5.1.13.

(a) (b): By the Jordan decomposition, it suffices to consider the case that F is increasing (5.1.14). Consider first the case that F is strictly increasing. Writing I = la, b] and J = F(I) = [F(a), F(b)] , we have a homeomorphism F: I -+ J. It follows that the correspondence E ..... F(E) is a bijection between the Borel sets of I and those of J that preserves the set-theoretic operations. Let S be the a-algebra of Borel sets of I and define v: S -+ [0, +ooJ by the formula forall EES; clearly v is a finite measure on S, and v« >. by 5.1.16. By the RadonNikodym theorem (4.8.11) there exists a function f: [a, bJ -+ IR, measurable with respect to S (hence a Borel function) and integrable with respect to >., such that v(E)=>'(F(E»

v(E)

=

L

f d>'

for all E E S.

In particular for E = [a,x], a $ x $ b, we have F(E) = [F(a),F(x)J and v(E) = >.(F(E» = F(x) - F(a) , thus F(x) - F(a) =

f

J,c,%1

f d>'

for all x E [a, bl .

Suppose now that F: la, bl -+ IR is any increasing AC function. Applying the preceding case to the strictly increasing AC function F I (x) = F(x) + x, there exists a >.-integrable Borel function II: [a, bl -+ IR such that FI(x) - Ft(a)

= f

J(c,%)

II d>'

for all x E [a, bJ,

that is, F(x) - F(a)

= -(x -

a) +

1% II d>'

for all x E [a, b] , thus the function f = II - 1 meets the requirements of the theorem. Finally, if f and 9 are Lebesgue-integrable functions on [a, b] , each of which represents F as in (b), then

f

J1C,%]

f d>'

= F(x) -

F(a)

=f

J1C,%] for all x E [a,b] , therefore f = 9 >.-a.e. (4.7.7). 0

gd>'

§5.3. limsup, liminf of FUnctions; Dini Derivates

215

Exercise 1. If A is Lebesgue measure on

la, bJ and

v is a finite measure defined on the 80rel sets of [a, bl , such that A(8) = 0 ~ v(8) = 0, then, by the Radon-Nikodym theorem (4.8.11) there exists a Lebesgue-integrable Borel function f: [a, bl -+ IR such that v(8) = fB fdA for all 80rel sets 8 C [a, bJ. Deduce this (circularly, of course!) from 5.2.l. (Hint: Define a function F : [a,b] -+ IR by the formula F(x) = v([a, x]) and argue that F is absolutely continuous.!}

5.3. limsup, Iiminf of Functions; Dini Derivates The main mission of this section is to define the 'Dini derivates' of a function 9 : la, bl -+ IR; these will play a key role in the proof that a function of bounded variation is differentiable almost everwhere. As usual, it is worthwhile to see things in a larger perspective, so the definition of derivate is preceded by a general discussion of the underlying techniques (useful in other situations as well). In §l.16, the concept of limit of a sequence of extended real numbers was created by fusing two more general concepts, limit superior and limit inferior (applicable to all sequences, not just those that have a limit). In an analogous way, we are going to dissect the concept of limit of a function, defined on a subset of a metric space and taking values in the extended reals IR, into limit superior and limit inferior. The general setup is the same as that for limits of functions (3.5.1). We have a metric space (X, d) and a function f defined on a subset 8 of X. In §3.5, f was allowed to take values in any metric space; here, we require f to have values in IR, thus f: 8 -+ IR. We are interested in the behavior of f(x) as x approaches a point c EX, so c will at least have to be approximable by points of 8, that is, c E 8. We also want the option of restricting the way in which x approaches c (for example, if X = IR we might want to require x to approach c from the left or from the right); in other words, we may want to specify a subset A of 8 and require x to approach c while remaining in A, so we will want c to be adherent to A. Thus, the framework for the discussion is as follows: (X, d) is a metric space, AC8cX,

f:8-+1R,

cEA.

1 cr. the author, Measure and integration [Chelsea, New York, 19701, p. 149, §43, Theorem 1, or P.R. Halmos, .'.·!ee$vr~ Ih!'o'r'fj fSp,:,ir'lt;er. 1974]. p. 181 , Theorem 43.D.

5. Differentiation

216 Schematically,

x U

B

f

U

ceA

~

A

5.3.1. Definition. With the preceding notations, for each neighborhood V of c in X,wehave VnA#0;wewrite

f3v = sup{J(x): x e VnA} and define f3 to be the infimum of the f3v as V varies over the set of all neighborhoods of c:

f3 = inf {f3v: V a neighborhood of c}. The extended real number f3 is called the limit superior of f(x) as x approaches c through values in A, written

f3 = limsup f(x)

= inf ( sup f(x») ,

z_c,zEA

V

zEvnA

where V runs over the set of all neighborhoods of c in X. Similarly, writing

= inf{J(x): x e "'I = sup{'YV : V a

'YV

VnA}, neighborhood of c},

we call "'I the limit inferior of f(x) as x approaches c through values in A, written "'I

= z-c.zEA lim inf f(x) = sup ( inf f(x»). V zEvnA

If V and W are neighborhoods of c such that V c W ,then f3w and 'YV;:: "yw • Adapting the notations of §1.16, we write

f3v ! f3 and A special case: If A = B

'YV

f3v:5

1"'1 as V! c.

= X , we write simply f3 = limsupf(x), z-e "'I = lim inf f(x). z-e

That's a lot of machinery, but what is going on is very simple: as we shall see in the next theorem, f3 is the largest element of JR that is the limit of f(x n ) for some sequence (x n ) in A with X n --+ c (and "'I is the smallest).

§5.3. limsup, Iiminf of Functions; Dini Derivates

217

5.3.2. Lemma. If (Vn) is a sequence of neighborhoods of c such that diam Vn --> 0 (cf. 3.2.20), then f3v.. --> 13 and "tV.. -+ 'Y in 1R.

Proof. If 13 = +00 then f3v = +00 for all V, so the assertion about the 13's is trivial. 8uppose 13 < +00. Assuming r > 13, we have to show that 13:$ 13v.. < r ultimately. Choose a neighborhood V of c such that 13 < 13v < r (possible because 13 is defined as a greatest lower bound). 8ince c is interior to V and diam Vn --> 0, Vn C V ultimately, say for n;:: N; then 13:$ f3v.. :$ 13v < r for all n;:: N. This shows that f3v.. -+ 13; the proof that "tV.. -+ 'Y is similar. 5.3.3. Theorem. With notations as above, let 8 be the set of all Q E IR such that f (x n ) -+ Q for some sequence X n E A with X n -+ c. Then

13 and 'Y belong to 8, 13 is the largest element of 8, and 'Y is the smallest; thus {oy,13} C 8 C where ['Y, 131

= {Q

h,13I,

E 1R: 'Y:$ Q :$ 13}. In particular, 'Y:$ 13, that is, liminf f(x):$ limsup f(x). x-c. :tEA

x-c, xEA

Proof. 80 to speak, the assertion is that 13 is the largest (extended real) number that f(x) can be made to approach as x approaches c through points of A, and 'Y is the smallest. Let V n = U1/n(c) = {x: d(x,c) < l/n}; by the lemma, f3v.. -+ 13 and "tV.. -+ 'Y. Let us show, for example, that 13 E 8 (the proof that 'Y E 8 is similar). case 1: 13 = +00. Then 13v = +00 for every neighborhood V of c. In particular, f3v.. = +00, so there is a point X n E Vn n A such that f(x n ) > n. Clearly X n -+ c and f(x n ) -+ +00 = 13, thus 13 = +00 E 8. case 2: 13 = -00. For each n, choose a point X n E Vn n A; then X n -+ c and f(x n ):$ f3v.. --> -00, whence it is clear that f(x n ) -+ -00, so that 13 = -00 E 8. case 3: 13 E IR . Then f3v.. E IR ultimately, say for n;:: N . For 1:$ n < N , choose any XnEVnnA;for n;::N choose xnEVnnA sothat 13vn

-

l/n < f(x n ) :$ f3v..

(possible because 13v.. is a least upper bound). Then X n -+ c and f(x n ) --> 13, thus 13 E 8. We now show that 13 is the largest element of 8 (the proof that 'Y is the smallest is similar). Assuming Q E 8, we have to show that Q:5 13· This is trivial if 13 = +00 or if Q = -00; thus we can suppose that

218

5. Differentiation

< a and


a for a sequence (x n ) in A such that X n -> c. Since a > -00, f(x n ) > -00 ultimately. Fix k;::: N . Since X n -> C we have X n E Vk ultimately, so -00

(3

+00.

< f(x n )

-00

::;

{3v.

< +00

for all sufficiently large n; it follows that a E IR and, passing to the limit as n -> 00, we have -00 < a ::; {3v•. Since k;::: N is arbitrary, a < (3 by the lemma. As noted in Example 3.3.17, IR can be equipped with a metric d' compatible with the concept of sequential convergence introduced in §1.16, that is, an -> a in the sense of 1.16.8 if and only if d'(a n , a) -> O. The particular metric d' need not be specified; what counts is the property just mentioned. Viewing IR as a metric space, the present section fits into the general framework of limits in §3.5; the order structure of IR yields the following criterion for f to have a limit in the sense of §3.5:

5.3.4. Theorem. With f: B equivalent: (a) '"( = (3, that is,

->

IR as above, the following conditions are

liminf f(x)

%-C, :tEA

=

limsup f(x);

x_c, xEA

(b) the limit

lim

x-c. :tEA

f(x)

exists (in the sense of 3.5.1). When this is the case, the three elements of IR in question are equal; expressed concisely, lim f

= lim inf f = lim sup f

at the point c. Proof. (b)

'* (a): Let a

= x-c,lim:tEA f(x);

this means (3.5.1) that if X n E A and X n -> c, then f(x n ) -> a, thus the set S of 5.3.3 is a singleton, namely S = {a}. By 5.3.3, b,{3} c {a}, so '"(={3=a. (a) (b): Write a for the common value of {3 and '"(. By 5.3.3, S = {a}. Assuming x n E A and X n -> c, we have to show that f(x n ) -> a. At any rate, the sequence (f(x n has a limit superior {3' and a limit inferior 'Y' in the sense of §1.16; in view of the definition of limit given there (1.16.8), the problem is to show that {3' = '"(' = a. By 1.16.11,

'*

»

§5.3. limsup, liminf of Functions; Dini Derivates

219

there exists a subsequence (x n .) of (x n ) such that f(x n .} --+ fJ'; since xn• --+ c, we have fJ' E S = {,,} ,thus fJ' = ". Similarly, '"I' = ".

5.3.5. Example. Let X = IR with the usual metric d(x, y} = and let f: B -+ IR ,where B c IR. Let c E IR and suppose that B :::> (c,c + r)

for some r

Ix - yl

> 0,

so that c is approachable from the right by points of B. Writing A=Bn(c,+oo), we have c E A (in the event that c E B, we have just masked it out), and

the foregoing machinery is applicable. In the present situation, the limits superior and inferior of 5.3.1 are denoted limsup f(x}

and

lim sup f(x}

and

x-c, x>c

liminf f(x} ,

x-c, x>c

or, more concisely, %_c+

lim inf f(x} . %_c+

According to 5.3.4, these two numbers are equal if and only if lim

x-c, xEA

f(x}

exists, that is (in the notations of 3.5.5), f(c+} exists, in which case

f(c+) = lim sup f(x} = lim inf f(x}. %_c+

x-c+

One might reasonably write

f(c+} = lim sup f(x} , %-c+

f(c+} = lim inf f(x} , %-c+

sO that f(c+) exists

**

f(c+) = f(c+) ,

in which case f(c+} = f(c+} = f(c+}; this would place too much strain on the notation and we shall not do so. (A minor variation on this idea is in common use for the 'derivates' to be defined shortly, but in that context the signs are positioned so that it is easier for the eye to sort them out.) Similarly, if B :::> (c - r,c) for some r > 0, the numbers lim sup f(x} %_c_

and

liminf f(x} %-c-

are defined in the expected way: we set A = B n (-00, c) and again apply the machinery of 5.3.1. Thus, these two numbers are equal if and only if

5. Differentiation

220

f (c-) exists in the sense of 3.5.5, in which case all three numbers are equal. Taking into account the criterion of 3.5.6 for the existence of a (twosided) limit, we have:

5.3.6. Theorem. Let B c JR, f: B --> R, c E JR, and suppose that B:J (c-r,c)U(c,c+r) for some r > O. In order that lim

s-c. zelc

f(x)

exist (in the sense of 3.5.5), it is necessary and sufficient that the four numbers

lim sup f(x) ,

lim inf f(x) ,

lim sup f(x) ,

liminf f(x) ,

x-c+ x-c-

%_c+

%-c-

be equal, in which case all five numbers are equal. B

5.3.7. Definition. Let g: [a, bJ = [a, bl - {c} and define f: B f(x)

--> -->

R, a < b, and let c E [a, bl. Write IR by the formula

= g(x) -

g(c) . x-c

Of course the values of f are in JR, but we are being consistent with the foregoing notations; some of the numbers we are about to associate with f may be infinite. If c E [a, b) then c is approachable from the right by x E B and we define (0+ g)(c)

= lim sup f(x) = lim sup g(x) %_c+

(o+g)(c)

%_c+

= lim inf f(x) = lim inf g(x) :1:_ O. The set of discontinuities of f in (a. b) is the union of the sets

Sn

= {x E (a. b) :

f(x+) - f(x-)

> l/n}

(n = 1.2.3, ... ). so it suffices to show that each Sn is finite. To simplify the notation, let r > 0 and let

S = {x E (a,b): f(x+) -f(x-) > r}; we need only show that S is finite. Suppose Xl •... , Xn are points of S with Xl < X2 < ... < x n . Choose points Cl ..... c",c,,+l in (a, b) such that Cl < Xl. x;-l < e; < X; for i = 2•... , n. and Xn < c,,+l' For i = 1•...• n we have e; < X; < e;+l , whence f(e;)

:s;

f(x;-)

:s;

:s;

f(e;+Il.

:s; f(e;+l)

- f(e;);

f(x;+)

so that f(x;+) - f(x;-)

summing over i and telescoping the sum on the right side. we have n

Llf(x;+) - f(x;-)]:s; f(c,,+l) - f(Cl):S; feb) - f(a). i=l

whence feb) - f(a) > nr. This shows that n is bounded, so S cannot be infinite. 0 The following theorem is somewhat more general than our immediate needs in this section (a proof of Theorems A and B of the introduction) but one of its corollaries (5.4.7) is needed for the proof of Theorem C (a.e. differentiability of monotone functions) in §5.12. 4 5.4.2. Theorem. Suppose f: la, bJ .... IR satisfies the conditions (i) f(x) < lim inft-x+ f(t) for all X E [a, b), (ii) limsup'_:r_ f(t) :s; f(x) for all X E (a.b]. and let N = {x E (a, b) : (D+ f)(x) < O}. If f (N) has empty interior, then f is increasing; if. in addition, N has empty interior, then f is strictly increasing.

4ft seems paradoxicaJ that a theorem whose conclusion is monotonicity should figure in the proof of a theorem about monotone functions, which is to say that McShane's proof of Theorem C is a very cunning proof.

5. Differentiation

224

Proof. {Note that (i) and (ii) are necessary conditions for f to be increasing. If f is continuous on (a, b) then (i) and (ii) hold trivially, with equality. To say that N has empty interior means that every point of N is a boundary point; equivalently, [a, bJ - N is dense in [a, b] , in other words, the set of all points x with (D+ f)(x) > 0 is dense in (a, b).} Assuming f satisfies (i)-(ii) and f(N) has empty interior, wemust show that f is increasing. Suppose to the contrary that there exists a pair of points c, d with a $ c < d $ b and f(c) > f(d). Since f(N) has empty interior, it cannot contain the interval (J(d), f(c)) ; choose a point k E (J(d) , f(c)) that does not belong to f(N) , that is,

f(d) < k < f(c)

and

(V x E N) f(x) f k.

Consider the set

S={xE[c,d): f(x»k}; for example, c E S. Defining 5 = sup S, we have c $ 5 $ d. We will show that each of the alternatives f(s) > k, f(s) < k, f(s) = k leads to a contradiction. (1) If f(s) > k then 5 f d (because fed) < k), therefore 5 < d. For every x E (5, d) we have

x> hence x

1- S, therefore

5

= supS,

f(x) < k by the definition of S. Thus sup f(x) $ k; %E(',d)

since (5, d) is a deleted right neighborhood of 5, it follows from the definition of limit superior (as an inf of sups) that sup f(x) $ k < f(s) ,

lim sup f(x) $ %_.5+

:tE(.9,d)

therefore also lim inf f(x) < f(s) , %-$+

which contradicts hypothesis (i). (2) If f(s) < k then 5 f c (because f(c) > k), therefore c < s. For each r such that c $ r < 5 = sup S, there exists a point t E S with r < t $ 5, and we have f(t) ~ k by the definition of S; necessarily t f 5 (because f(s) < k by supposition), thus r < t < 5, that is, t E (r, 5) . It follows that sup f(x) :r::E(r.s)

~

k;

§5.4. Criteria for Monotonicity

225

taking the infimum of the left sides over alI possible r, we have lim sup f(x) 2: k > f(s) ,

x-,-

which contradicts hypothesis (ii). (3) If f(s) = k then f(d) < f(s) < f(c); in particular, sf- d, sf- c, so c < S < d. For every X E (s,d) we have x E [c,d] but x ¢ S (because x> s = supS), therefore f(x) < k by the definition of S, that is, f(x) < f(s); thus f(x) - f(s) x-s

0 for all but countably many x in (a, b), then f strictly increasing on [a, b] .

is

Proof. By assumption, the set

N = {x

E

(a, b): (Df)(x) $ O}

is countable, therefore so is f(N) , thus both N and f(N) have empty interior. 5.4.7. Corollary. Suppose f: [a, bJ -> IR satisfies conditions (i) and (ii) of Theorem 5.4.2, and let D be one of D+, D+, D-, D_ . (1) If (Df)(x) ?: 0 for all but countably many x in (a, b) , then f is increasing. (2) If (DJ)(x) = 0 for all but countably many x in (a, b) , then f is constant. Proof. (1) As in the proof of 5.4.5, write fr(x) = f(x) + rx for r > O. It is clear that each fr satisfies the conditions (i) and (ii) (because rx is continuous; cf. 5.3.3). Moreover, (Dfr)(x)

= (DJ)(x) + r ?: 0 + r > 0

for all but countably many x, so every fr is (strictly) increasing by the preceding corollary, therefore f is increasing by the argument in 5.4.5. (2) In the proof of (2) of 5.4.5, replace "a.e." by "at all but countably many points". The next corollary includes Theorem B of the introduction: 5.4.8. Corollary. Suppose f : [a, bl -> IR is continuous and let D be one of D+,D+,D-,D_. (1) If D f ?: 0 at all but countably many points, then f is increasing. (2) If D f = 0 at all but countably many points, then f is constant. Proof. Conditions (i) and (ii) of Theorem 5.4.2 are trivially verified by f, so the present corollary is immediate from 5.4.7.

Corollary 5.4.7 also yields a criterion for a monotone function to be strictly monotone: 5.4.9. Corollary. Let f : [a,b] -> IR be increasing (so that all four derivates of f are ?: 0), let D be one of D+, D+, D-, D_ and let A

= {x E (a,b):

(DJ)(x) > O}.

Then f is strictly increasing on [a, b] if and only if A is dense in [a, b] . Proof· Arguing contrapositively, let us show that (assuming f increasing)

f not strictly increasing

**

A not dense in [a, b].

§5.5. Semicontinuity

229

'*: Assuming f increasing but not strictly, the argument at the end of the proof of Theorem 5.4.2 shows that the set N = (a, b)-A has nonempty interior, consequently A is not dense. r, the set {x EX: fix) > r} is a neighborhood of e. {Informally, every inequality fie) > r persists on a neighborhood of e.}

1

Metric spaces are sufficient for our applications.

5. Differentiation

230

5.5.2. Example. Let A eX. The characteristic function f = 'PA of A is l.s.c. at every point of X - A. If c e A, then f is l.s.c. at c ¢} c is interior to A. {Proof: If c e X - A then f(c) = 0; for every r < f(c) = 0, the set {x: f(x) > r) = X is a neighborhood of c. Suppose c e A, so that f(c) = 1; if r < f(c) = 1 then the set {x: f(x) > r} is either A or X according as r ~ 0 or r < 0, guaranteed to be a neighborhood of c if and only if A is a neighborhood of c.} 5.5.3. Remarks. In the notations of Definition 1, if f(c) = -00 then f is l.s.c. at c 'by default' ( f(c) > r is impossible). If f is l.s.c. at c, then so is af (a> 0).

5.5.4. Theorem. Let f; : X ..... JR (i e I) be a family of junctions, c eX, and f = sup f; the upper envelope of the family, that is, f(x) = sup{f.(x): i e I} for all x eX. If every f; is lower semicontinuous at c, then so is f· Proof. Assuming f(c) > r e JR, we have to show that {x: f(x) is a neighborhood of c. Since f(x)

r}

e I,

we have f(x)

>r

¢}

{x: f(x)

f;(x)

> r for some

i

e I,

> r} = U{x: f.(x) > r} . iEI

Since c belongs to the left side of (*), it belongs to some term on the right, say ce{x: !;(x) >r}. In particular, !;(c»r;since!; isl.s.c.at c, {x: !;(x) > r} is a neighborhood of c, therefore so is the left side of (*). 0 For finite families, the analogous result holds for lower envelopes: 5.5.5. Theorem. If each of the functions f; : X -+ JR (i = 1, ... , n) is lower semicontinuous at c eX, then so is their lower envelope f = inf(ft, ... , fn). Proof. Suppose f(c) > r e JR. For i = 1, ... , n, f;(c) ~ f(c) > r, therefore {x: f.(x) > r} is a neighborhood of c by the lower semicontinuity of f; at c; thus n

{x: f(x) > r}

= n{x:

f;(x) > r}

&=1

is the intersection of finitely many neighborhoods of c.

It is time to drop the other shoe:

0

§5.5. Semicontinuity

231

5.5.6. Definition. With notations as in Definition 5.5.1, f is said to be upper semicontinuous (u.s.c.) at c if, for every real number r such that f(c) < r, the set {x: f(x) < r} is a neighborhood of c. {So to speak, every inequality f(c) < r persists on a neighborhood of c.} 5.5.7. Remarks. (1) f is u.s.c. at c ¢* - f is I.s.c. at c. {Cf. 1.15.4, (i). } (2) If f(c) = +00 then f is u.s.c. at c. (3) Let A eX. The characteristic function f = 'PA is u.s.c. at every point of A. If c E X - A, then f is u.s.c. at c ¢* c is interior to X - A (that is, exterior to A). (4) If f is u.s.c. at c, then so is af (a> 0). In view ofthe above remark (1), Theorems 5.5.4 and 5.5.5 yield the 'dual' statements: 5.5.8. Theorem. Let fi: X -+ JR (i E I) be cE X. (i) If every Ii is upper semicontinuous at lope inf Ii of the family. (ii) If I = {I, ... ,n} and if every Ii is then so is the upper envelope sup(h, ... , fn)

a family of functions and let

c, then so is the lower enveupper semicontinuous at c, of the family.

Upper and lower semicontinuity are a 'dissection' of continuity in the following sense: 5.5.9. Theorem. For a function f : X -+ JR and a point c EX, the following conditions are equivalent: (a) f is continuous at c; (b) f is both lower and upper semicontinuous at c.

"*

Proof. (a) (b): We are assuming that f is continuous at c. Suppose first that f(c) E JR. If r < f(c) < s then the interval (r, s) is a neighborhood of f(c) , therefore the set {x: r < f(x) < s}

= r1(r,s))

is a neighborhood of c; since

(0)

{x: r r} is a neighborhood of c; thus

f

= r1(r, +00])

is also I.s.c. at c.

5. Differentiation

232

Finally, if f{c) = -00 then f is I.s.c. at c. For every r E JR, f{c) < r and [-00, r) is a neighborhood of -00 = f{c) , therefore

{x: f{x)

< r} = r l (1-00, r))

is a neighborhood of c; thus f is also u.s.c. at c. (b) ~ (a): We are assuming that f is both I.s.c. and u.s.c. at c. Suppose first that f{c) E 1R. If r < f{c) < 5 then (r,s) is a basic neighborhood of f{c) in 1R; from (*) we see that rl({r,s)) is the intersection of two neighborhoods of c, hence is a neighborhood of c. This shows that f is continuous at c. If f{c) = +00 then f{c) > r for every r E JR and the sets (r, +00] are basic neighborhoods of +00 = f{c); by assumption, the sets

rl({r,+oo])={x: f{x»r} are neighborhoods of c, whence the continuity of f at c. Similarly, if f{c) = -00 and r E JR then f-I(I-oo,r)) = {x: f{x) r} is a neighborhood of c, therefore f. is continuous at c.


JR is said to be lower semicontinuous (I.s.c.) on X if it is I.s.c. at every point of X; it is said to upper semicontinuous (u.s.c.) on X if it is u.s.c. at every point of X. 5.5.11. Example. For a subset A of X, CPA I.s.c. CPA

U.S.C.

~

A open,

~

A closed

(see the remarks following Definitions 5.5.1 and 5.5.6).

5.5.12. Theorem. The following conditions on a function f: X -> JR are equivalent: (a) f is lower semicontinuous on X; (b) for every real number r, {x: f{x) > r} is an open set in X.

Proof. (a) ~ (b): Let r E JR, A = {x: f{x) > r}. For every c E A, A is a neighborhood of c (Definition 5.5.1), thus every point of A is an interior point. (b) ~ (a): Let c EX. For every real number r < f{c), the set {x : f{x) > r} is open, hence is a neighborhood of c, thus f is I.s.c. at c. 5.5.13. Corollary. A junction f: X -> JR is upper semicontinuous on X if and only if, for every real number r, {x: f{x) < r} is an open set. 5.5.14. Theorem. Let /; : X -> JR (i E I) be a family of functions. (I) If every fi is I.s.c. on X, then so is sup /;; if, moreover, I finite, then inf fi is also I.s.c. on X.

loS

§5.5. Semicontinuity

233

(2) If every fi is u.s.c. on X, then so is inf Ii; if, moreover, I is finite, then sup Ii is also u.s.c. on X. Proof Immediate from Theorems 5.5.4, 5.5.5, 5.5.8 and the defini-

tions. In §5.3, liminf and lirnsup were defined in the context of a metric space; inspection of Definition 5.3.1 shows that the concepts carry over verbatim to functions defined on a topological space. (However, the proofs of the results in §5.3 were based on sequential convergence in metric spaces, so we must be careful not to cite these results without revisiting the proofs.) The form of the definitions we need here are as follows: 5.5.15. Definition. Let f: X

--+

JR and let c EX. For each neighbor-

hood V of c in X, we define 'YV = inf f(x) , ",ev

fJv

= sup f(x) . ",eV

Letting V vary over the set of all neighborhoods of c in X. we define "Y

= sup{ 'YV:

V a neighborhood of c}

and call it the limit inferior of lim inf f(x) %_c

f at

c. written

= "y = sup (inf f(X») V zeV

.

Similarly, we define /3 = inf{/3v: V a neighborhood of e} and call it the limit superior of fate, written lim sup f(x) %-c

= /3 = inf (sup f(X») V :cEV

As noted following 5.3.1, 'YV and fJv are monotone functions of V (increasing and decreasing, respectively), and we write 'YV T"Y and

fJv l/3 as V le.

In the next theorem, we shall see that upper and lower semicontinuity relate to limsup and liminf as continuity relates to limit. 5.5.16. Lemma. If f: X --+ JR and e EX. then liminff(x) %_c

< f(e)

$limsupf(x). %-c

Proof For any two neighborhoods V and W of e, we have

inf f(x) $ f(e) $ sup f(x). xEV

2:EW

5. Differentiation

234

In the notations of Definition 5.5.15, ftc) is an upper bound for the 'YV, therefore "Y ~ f(c); similarly ftc) is a lower bound for the (3w, so ftc) ~ (3. 0 5.5.17. Theorem. If f : X -> IR and c EX, the following conditions are equivalent: (a) f is lower semicontinuous at c; (b) lim inf",_c f(x) = f(c).

Proof. (a) => (b): Let m

= lim inf f(x); "'_c

in view of the lemma, we need only show that m;::: f(c). This is trivial if m = +00 or if ftc) = -00, so we can suppose that m < +00 and ftc) > -00. If ftc) > r E IR then, by (a), the set V = {x: f(x) > r} is a neighborhood of c, so that

r

~

inf f(x)

",EV

~

m

(by the definition of m as a sup of infs); thus r ~ m for every real number r < ftc) ,therefore ftc) ~ m. (b) => (a): If ftc) = -00 then (a) holds trivially. Suppose ftc) > -00. Assuming ftc) > r E IR, we have to show that the set W = {x: f(x) > r} is a neighborhood of c. Citing (b), we have sup (inf V

"'EV

f(X»)

> r,

where V runs over the set of all neighborhoods of c, thus there exists a neighborhood V such that inf f(x)

",EV

then V of c. 0

c

{x: f(x)

> r;

> r} = W, therefore W is also a neighborhood

Dually, 5.5.18. Corollary. Iff: X -> IR and c EX, the following conditions are equivalent: (a) f is upper semicontinuous at c; (b) limsup",_c f(x) = f(c). 5.5.19. Corollary. If f : X -> IR and c EX, the following conditions are equivalent: (a) f is continuous at c; (b) liminf",_cf(x) = limsup",_cf(x). When the conditions are verified, the number in (b) is equal to f(c).

§5.5. Semicontinuity

Prool· (a)

~

235

(b): Immediate from 5.5.9, 5.5.17 and the preceding corol-

lary. (b) ~ (a): In view of 5.5.16, it is immediate from (b) that

liminf/(x) = I(c) =limsup/(x); %-c

%_c

thus, I is both lower and upper semicontinuous at c (5.5.17 and 5.5.18), hence continuous at c (5.5.9). 0 The applications of the next theorem to derivates will play an important role in the proof that indefinite integrals are a.e. antiderivatives (§5.9). Recall that if I, 9 : X -+ lR then I + 9 is defined except at the points x where I(x) and g(x) are both infinite and of opposite signs (1.15.4, (iv) . 5.5.20. Theorem. Let I, 9 : X -+ lR be functions such that I everywhere defined on X and let c be any point 01 X. Then: (i) liminf",_c(f + g)(x) ;::: lim inf",_c I(x) + lim inf",_c g(x) , (ii) limsup",_c(f + g)(x) < limsup",_c I(x) + limsup",_c g(x) , provided that the right members are defined.

+9

is

Proof The stipulation at the end ofthe statement is that the (undefined) sums (+00) + (-00) and (-00) + (+00) do not occur on the right side. It will suffice to prove (i), for (ii) can then be deduced by applying (i) to - I and -g. Define Ct = lim inf I(x) , (J = lim inf g(x) , 'Y = lim inf(f + g)(x). %-c

%-c

%-c

By assumption Ct + /3 is defined; the problem is to show that 'Y ;:::

Ct + (J .

Let us first dispose of some special cases: the inequality (*) is trivial if 'Y = +00, or if one (or both) of Ct,(J is -00. Thus we can suppose that

Ct > -00, (J > -00, 'Y < +00. For every neighborhood V of c, write Ctv = inf I(x) , zEV

f3v = zEV inf g(x) , "IV = inf [/(x) + g(x)) xEV

(not to be confused with the notations in Definition 5.5.15); thus,

CtvTCt, /3vT/3 and "IVT'Y as Vic. Since Ct that Ctv and (1)

> -00 and (J > -00, there exist neighborhoods V of c such > -00 and f3v > -00; for such V, the sum Ctv + f3v is defined +00 > 'Y ;::: "IV ;::: Ctv + f3v > -00.

5. Differentiation

236

Let (Vn) be a sequence of neighborhoods of c such that Qv..

> -00 and sup QV.. = Q; n

replacing Vn by VI n ... n Vn , we can suppose that Vnl

and

QV.. TQ.

Similarly, there exists a sequence (Wn) of neighborhoods of c such that Wn

1

13w.. T{3.

and

Replacing both V n and Wn by V nnwn , we can suppose that QV n

TQ and f3v.. T{3.

By (1), we have (2)

+00

> 'Y ~ "tV..

~ QV.. +

f3v..

for all n. Since the right member of (2) is increasing and bounded above, it is clear that neither Q nor {3 can be +00, thus both are in JR and passage to the limit in (2) yields 'Y ~ Q + {3. 0 5.5.21. Corollary. With notations as in the theorem, suppose that f is everywhere defined on X. (1) If f and 9 are l.s.c. at c, then so is f + 9· (2) If f and 9 are u.s.c. at c, then so is f + 9 .

+9

Proof. (1) In particular, f + 9 is defined at c. In view of 5.5.17, this means that the right side of (i) in the preceding theorem is defined, and

lim inf(f + 9)(X) ~ f(c) x-c

the reverse inequality holds by 5.5.16, thus (2) Apply (1) to - f and -9· 0

+ 9(C);

f + 9 is l.s.c. at c by 5.5.17.

We now apply some of this machinery to difference-quotient functions: 5.5.22. Definition. Let f: [a, bl define

(D f)(x)

=

--+

JR, a < b. For every x

E [a,

bJ ,

lim inf f(t) - f(x) , t- x

t-x. t,ox

(Df)(x)

=

limsup f(t) - f(x) , t-%, t~% t - x

cal1ed, respectively, the lower derivate and the upper derivate of at x.

f

The liminf and limsup in the above definition are applied to a function of t defined on the subset la, bl - {x} ofthe metric space [a, b], and x is not an isolated point, that is, x is adherent to [a, b)- {x} ; we are thus in

§5.5. Semicontinuity

237

the general framework of 5.3.1. The results of §5.3 are therefore applicable here; in particular: 5.5.23. Theorem. With notations as in the preceding definition (5.5.22), f is differentiable at x e (a, b) if and only if (D f)(x)

= (Df)(x) e IR,

in which case its derivative f'(x) is the common value of the upper and lower derivates of f at x. Proof. This is immediate from 5.3.4 and the definitions. 5.5.24. Theorem. With notations as in 5.5.22 and 5.3.7, (i) (D f)(x) = min{(D_f)(x), (D+f)(x)} , (ii) (Df)(x) = max{(D- f)(x), (D+ f)(x)} , for every x e (a, b) . A t the endpoints a and b, (iii) (D f)(a) = (D+f)(a) , (D f)(b) = (D_f)(b), (iv) (Df)(a) = (D+ f)(a) , (Df)(b) = (D- f)(b) . Proof (i) Let x e (a, b) and write g(t)

= f(t) -

f(x) t-x

for all t

e

[a,b] - {x}.

By 5.3.3, (D f)(x)

= min{a e

IR: g(t n )

-+

(D_f)(x) = min{a e IR: g(t n )

-+

(D+f)(x)

= min{a e

IR: g(t n )

a with t n i- x, t n

a with t n < x, t n -+ a with t n > x, t n

-+

x},

-+

x},

-+

x};

the equality in (i) is immediate from the fact that if t n i- x for all n, then either t n < x frequently or t n > x frequently (or both). (ii) The proof is similar to (i). (iii), (iv) These equalities are immediate from the definitions, since approach is possible only from one side. 5.5.25. Theorem. If f, 9 : [a, b]

-+

IR, then

D(f - g) ;?: D f - Dg at every point of [a, b] for which the difference on the right side is defined. Proof. Write h = -g, so that f - 9 = f

+ h, and let

x

e

[a, b]. For all

t i- x, (f - g)(t) - (f - g)(x) = f(t) - f(x) t-x t-x

+ h(t) -

h(x) ; t-x

5. Differentiation

238

by the proof of (i) of 5.5.20 (with neighborhoods V replaced by deleted neighborhoods V - {x) , possible because x is not an isolated point), we have lim inf (f - g)(t) - (f - g)x) t_%. t#:x t- x

> lim inf - t_%. t,fx

f(t) - f(x) t- x

+

lim inf h(t) - h(x) t- x

t_x, t¢%

at every x for which the right side is defined, thus

D(f -g) at every such x.

~ Df +D(-g)

= Df -

Dg

5.5.26. Theorem. If f, 9 : [a, b}

->

IR then

D+(f-g)~D+f-D+g

at every point of [a, b) where D+ 9 is finite. The same inequality holds with D+ replaced by D- and [a, b) by (a, b] . Proof. Suppose (D+g)(x) E IR. Writing f

(D+ f)(x)

= lim sup f(t) t-x+

g)

+ g, we have

f(x)

t - x

. {(f = Itmsup '-z+

= (f -

g)(t) - (f - g)(x) t - x

+ g(t) -

g(X)}

t - x

< lim sup (f - g)(t) - (f - g)(x) + lim sup g(t) - g(x) -

t_x+

t- x

t-%+

t -

x

by the proof of (ii) of 5.5.20 (with neighborhoods replaced by deleted right neighborhoods), valid because the sum on the right side of the inequality is obviously defined. Thus

since the last term on the right side is finite, it can be transposed to yield the desired inequality. The second assertion of the theorem follows on replacing deleted right neighborhoods by deleted left neighborhoods in the foregoing argument.

Exercise 1. With notations as in Theorem 5.5.23,

f is right-differentiable at a

if and only if (D f)(a) , (Df)(a) are equal and finite, in which case f;(a) is the common value of the upper and lower derivates of f at a. Similarly for left-differentiablity at b.

§5.6. Semicontinuous Approximations

239

5.6. Semicontinuous Approximations of Integrable Functions Throughout this section, >. denotes Lebesgue measure either on IR or on the closed interval la,bl; £1 = £1([a,bJ,>.) is the class of Lebesgueintegrable functions I: la, bl -+ IR. If A c la, b) we write 'PA for the characteristic function of A, as a function on la, bl . The following approximation theorem is for application in §5.9 (in the proof that the indefinite integral of I E £1 has derivative I(x) almost everywhere): 5.6.1. Theorem. II I E £1 and f > 0, there exist Iunctions h E £1 and k: la, bJ -+ IR U {+oo} with the Iollowing properties: (i) I :5 h a.e., (ii)Jhd>':5 JId>'+f, (iii) k is lower semicontinuous and 1:5 k everywhere on la, bl , (iv) h = k a.e. Prool. Informally, every integrable function I admits a lower semicontinuous 'cover' k that is equal a.e. to an integrable function h whose

integral is as close as we like to that of I. The proof is by reduction to special cases. We consider, successively, (a) I = 'PE the characteristic function of a measurable set E CIa, bl ; (b) I simple and ~ 0; (c) I integrable and ~ 0; and (d) I integrable (the general case). (a) Suppose 1= 'PE, E a Lebesgue-measurable subset of [a, b). By the regularity of Lebesgue measure on IR (2.4.14, 2.4.18) and the finiteness of >'(E) , there exist a closed set K and an open set U in JR such that

KCE

cU

and

>'(U - K) < f

(of course U is not required to be a subset of la, bl). As noted in the preceding section, the characteristic function of U (as a function on JR) is l.s.c. on JR, therefore its restriction to la, b] is l.s.c. on the (metric) topological space la, bl ; we denote the restricted function by h, thus h = 'Punla.b) . Writing A = U n la, bl ' we have K C E c A , therefore 'PK

and

J

h d>'

:5 'PE

=

I :5 h =

'PA

= >.(A) < >'(U) :5 >.(K) + >.(U < >'(K) + f < >'(E) + f =

J

K)

I d>'

+ f;

setting k = h , the requirements of the theorem are fulfilled. Note also that 'PK is u.s.c., 'PK:5 I and

J

'PK d>' = >'(K)

~ >'(U)->'(U-K) > >,(U)-f ~

J

hd>'-f

~

J

I d>'-f,

240

5. Differentiation

thus

J'PKdA> JIdA-Eo (b) Suppose

I is simple and

~

O. Write

1= CI'PE, + ... + Cn'PE

n

,

where the E; are pairwise disjoint Lebesgue-measurable subsets of [a, b] and e; > 0 for all i. For each i, choose a closed set K; and an open set U; in lR such that K;

Write A;

c

E;

= U; n [a, bl 9 = CI'PK.

Then 0:;; 9 :;;

(h - 9) d>. =

U;

and

t

A(U; - K;)

< f/nc; .

h = CI'PA,

+ ... + Cn'PA

and let

+ ... + Cn'PK

I < h,

J

c

n

,

n

•

h is l.s.c. (5.5.21) and e;A(A; - K;) :;;

i=l

t

e;A(U; - K;)

i=l


J IdA-Eo (c) Suppose I ~ 0 (and IE .e l ). Choose a sequence (In) of (integrable) simple functions such that 0:;; In T I. By the preceding case, there exist simple functions 9n and hn such that

o:;; 9n :;; In

:;; h n ,

9n is u.s.c., hn is l.s.c. and

Define Gn

= SUp(9b··., 9n),

Hn = sup(h b ... ,hn );

G n is u.s.c. and H n is l.s.c. (5.5.8 and 5.5.4), both are simple functions, and 0:;; 9n :;; G n < sup(fI,· .. .In)

= In:;; h n :;; H n .

In particular, 0:;; In - G n :;; hn - 9n , therefore (1)

0:;; J In dA - J Gn dA:;; J (h n - 9n)dA

< f/2 n ;

§5.6. Semicontinuous Approximations

241

it follows that lim JGnd>- = n-oo lim Jln d>- = Jld>-.

n-oo

{The second limit exists by the monotone convergence theorem, so the first limit exists and is equal to it by (I).} It is elementary that n

o$

Hn - Gn $ ~)hi - gi) . i=l

{The crux of the matter is that if ai, {3i E JR and ai $ {3i (i then

= 1, ... ,n),

n

max {3i - max ai $ ~)(3i - ai) ; i=l

for, if max {3i

= {3j

and max ai

= ak ,then ak ~ aj

and

n

{3j -

ak $ (3j - aj $ L({3i - ai)'} i=l

Thus,

0$ J(Hn - Gn)d>-

$

t

J(hi - gi)d>- < tE/2 i < E,

i=l

i=l

therefore,

(2)

J Hnd>-< J Gnd>-+E$ J Ind>-+E$ J Id>-+E

for all n; since H nT, by the monotone convergence theorem there exists an hE £1 such that (3)

H n T h a.e.,

and since H n > 0 for all n, we can suppose (by modifying h on a negligible set, if necessary) that h ~ 0 everywhere on [a, bJ . From (3) we have

J Hnd>- T J hd>-; passing to the limit in (2), we have

Jhd>-$ Jld>-+E. Define k = sup H n ; then k: [a, b] k = h a.e. by (3). Moreover,

[0, +00], k is I.s.c. (5.5.4) and

1= sup In $ supHn = k,

5. Differentiation

242

thus f ::; k (everywhere on [a, b]). Since k = h a.e., it follows that f :5 h a.e. This completes the proof for the case that f ~ (d) Consider now the general case that f E £1. Write f = It - 12 with It, 12 integrable and ~ 0. Applying the preceding case (c) to It and £/2, there exist functions hi E £1, k l : la, b] --+ 10, +00], such that hi ~ 0, kl is l.s.c., 1t:5 k l everywhere on [a, b], hi = k l a.e. (hence It ':5

(4)

JIt

d>' + £/2.

Also, applying case (c) to 12 and £/2, the proof of (c) shows that there exists a simple function 92, with 0:5 92 < 12 and 92 u.s.c., such that

J(12 -

(5)

92)d>. < £/2.

Then -92 is l.s.c., hence so is k l - 92 (5.5.21), and

f

=

It - 12 < It -

92

= kl

= hi -

< kl

-

92 .

Define k

-

92, h

92 .

Then k: [a,b! --+ lRu {+oo} is l.s.c., f ::; k (everywhere), hE £1, k = h a.e. (because k l = hi a.e.), hence f::; h a.e.; moreover, h- f

= (hi -

therefore

J

(h - f)d>'

=

92) - (II - h)

J

(hi - ft)d>. +

= (hi -

J(12 -

It) + (12 - 92) ,

92)d>' < £/2 + £/2

by (4) and (5), thus h and k meet the requirements ofthe theorem. 5.6.2. Remark. For each positive integer n, let £ = l/n and choose functions hn, k n satisfying the conditions (i)-(iv) of the theorem. From (i) and (ii) we see that J I hn - f I d>' --+ 0. A neater way of packaging this result is as follows. Call a function k : la, b] --+ IR inte9mble if there exists a function h E £1 such that k = h a.e., and define the integml of k to be the integral of h. The theorem can then be stated succinctly as follows: Every f E £1 is the limit in mean of a sequence of lower semicontinuous integrable functions that are ~ f.

5.7. F. Riesz's "Rising Sun Lemma"

Riesz's lemma (which we shall use once and only once, in the next section) is part of the technical preparation for the proof that indefinite integrals

§5.7. Rising Sun Lemma

243

are a.e. primitives (§5.9). The following structure theorem for open sets in IR is needed before we can state Riesz's lemma: 5.7.1. Lemma. Every nonempty open set U in IR is the union U = UIn of a countable family of pairwise disjoint intervals that are open sets. Proof. {The intervals In are permitted to be unbounded (possible for at most two values of n); for example, if U = IR - {1,2} then U = (-00,1) U (1,2) U (2,+00) is the representation of U promised in the lemma. We reserve the term 'open interval' for intervals of type (a, b) with endpoints a, b E IR , whence the locution "intervals that are open sets". The intervals making up such a decomposition of U are unique (Exercise 1), but this fact is not needed in our application.} For x, y E U, write x '" y if the closed interval with endpoints x and y is contained in U (equivalently, there exists an interval I such that x, y E leU). The relation'" is an equivalence relation in U (for transitivity, note that the union of two intervals with a common point is an interval). Let IC be the set of all equivalence classes for "'. At any rate, the sets in IC are pairwise disjoint. claim 1: Every K E IC is an interval. Given x, y E K with x ~ y, it suffices to show that lx, yj C K.l Since x '" y, we know that lx, yJ cU. If z E lx, yj then lx, z) C [x, yj C U shows that z '" x E K, therefore z E K; thus lx, yj c K. claim 2: Every K E IC is an open set. Since K is an interval in IR, we need only show that it has no largest element and no smallest element. Assume to the contrary, for example, that K has a largest element b. Since b E U , there exists a 6 > 0 such that [b - 6, b + 6j c U; then b + 6 '" b E K ,therefore b + 6 E K , which contradicts the maximality of b.

For each K E IC choose a rational number rK E K. Since the sets in IC are pairwise disjoint, K >-> rK is an injective mapping IC - 0 IQ, whence the countability of IC. {A slightly more formal argument: For each K E IC, K n IQ '" C!J. Consider the family (K n IQ)KEK of nonempty subsets of IQ; by the Axiom of Choice, there exists a mapping f : IC - 0 IQ such that f (K) E K n IQ for all K E IC. Since the sets K n IQ are pairwise disjoint, f is injective, therefore card K ~ card IQ = No·) 0 In the context of a function 9 : la, b) - IR, let us say that a point x E (a, b) is a peak point ifthe restriction of 9 to lx, bj takes its maximum value at the left endpoint x, in other words, g(t) ~ g(x) for all t E (x, bj. If x E (a, b) is not a peak point, let us say that x is topped to the right; this means that there exists a point t E (x, bJ such that g(t) > g(x). 1 First

course. p. S9, Theorem 4.1.4.

244

5. Differentiation

5.7.2. Theorem. (F. Riesz's "Rising sUn lemma") Let g: la, b] --+ IR be a continuous function and let E be the set of all points in (a, b) that are "topped to the right",

E = {x

E

(a, b): g(t)

= {x

E

(a, b): g(t) - g(x) t-x

> g(x) for some t > x} > 0 for some t E (x, bl} .

Then: (i) E is an open set in 1R. (ii) E = (/) ~ 9 is a decreasing function. (iii) If E # (/) then, writing E = U(a", bn ) as in the lemma, where the (a", bn ) are pairwise disjoint, we have g(a,,):5 g(b n ) for all n.

Proof (The second formula for E indicates that it is a gauge of the 'slope' of g. In (iii), g(a,,) = g(bn ) for all except possibly one value of n (Exercise 2).} (i) Assuming c E E we have to show that c is interior to E. By assumption, there exists a point t E (c, bl such that g(t) > g(c). c-
0 such that a < c - < < c + < < t and g(x) < g(t) for all x E (c - x (that is, the point on the graph at x is the left end of some chord of slope > r ). Define g: [a, bl -+ IR by g(x)

= I(x) -

rx

(x E la, b)) ;

for t > x, we have

I(t~ =~(x) > r

$}

I(t) - I(x)

> ret -

x)

$}

get) > g(x) ,

therefore

G = {x E (a, b): get) > g(x) for some t E [a,b] with t > x}. We are now in the framework of the "Rising sun lemma" of the preceding section; in particular. G is an open set in IR.

§5.8. Continuous Increasing Functions

°

247

°

If G = then 9 is decreasing, therefore 0+ 9 :5 0 on [a, b); then 0+ f :5 T on [a, b), so E = and the assertion of the theorem is trivially verified. If G '1O, then

= U(an,bn )

G

for a countable (possibly finite) family of pairwise disjoint open intervals such that g(a n ) < g(bn) for all n, that is,

T(bn - an) :5 f(b n ) - f(an)

for all n.

It follows that for each n, n

(*)

n

T ~)bi - ail :5 L[f(b i ) - f(ai») < f(b) - f(a). i=l

i=l

{The second inequality is obtained by inserting into the sum non-negative terms corresponding to the gaps between the intervals (at. b1 ), ... , (an, bn ) , then telescoping the sum. For example, assuming (at. bl ), •.• , (an, bn ) to be arranged from left to right on the number line, f(a2) - f(b l ) 2: 0, f(a3) - f(~) 2: 0, etc., by the monotonicity of f.} Since (*) holds for every n, we have

T L(bi - ail :5 f(b) - f(a) , that is,

f(b) - f(a) 2: TA(G). To complete the proof, we need only show that G:> E . Let x E E. By the definition of E, lim sup f(t) - f(x)

> T.

t-x

t_x+

By Theorem 5.3.3, there exists a sequence (t n ) such that t n and

f(tn) - f(x) t n -x if n is any index for which

-+

(0+ f)(x)

> x, t n -+ x

> T;

",--,f(c..:;tn~) -:f,--,(X....!-) > T, t n -x

then g(t n )

> g(x) ,therefore

x E G.

5.8.2. RemaTk. A similar argument shows that if

F = {x

E

(a, b): (0- f)(x) > T}

then f(b) - f(a) 2: TAO (F) . {One could also deduce this by applying the theorem to the function f" of the proof of 5.4.3. }

5. Differentiation

248 5.8.3. Corollary. If f: [a, bl

--+

IR is continuous and increasing, r > 0

and

A={xE (a,b): (Df)(x) >r}, then f(b) - f(a) ~ !rA*(A). Proof {Again, A Can be shown to be a Borel set, but this is not needed for our application in the next section.} From 5.5.24, we know that for every x E (a,b), (DJ)(x)

= max{(D+ f)(x), (D- J)(x)} ,

thus

A

= {x E (a, b):

> r or (D- f)(x) > r} >r}U{x: (D-f)(x) > r};

(D+ f)(x)

= {x: (D+J)(x)

in the notations of the theorem and the remark following it, A therefore

A*(A) $ A*(E)

+ A*(F)

$ 2[f(b) - f(a)Jlr.

=EUF,

5.9. Indefinite Integrals are a.e. Primitives

A function F : [a, bl --+ IR is called an a.e. primitive of f : [a, bl --+ IR if, for almost every x E (a,b), F is differentiable at x and F'(x) = f(x). The purpose of this section is to prove that if f is Lebesgue-integrable then its indefinite integral F is an a.e. primitive of f. The context: A is Lebesgue measure on the closed interval [a, bl, a < b, and £.1 = £.l([a,b],A) is the class of all Lebesgue-integrable functions on

[a, bJ . 5.9.1. Lemma. Suppose hE £.1 and k: [a, b] --+ IR is a lower semicontinuous function such that k = h a.e. Let H be the indefinite integml of h,

H(x)

= 1% hdA

(a $ x $ b).

Then D H ~ k everywhere on [a, bl . Proof Recall (Definition 5.5.22) that, for all x E (a, b) , (QH)(x)

=

liminf H(t) - H(x) . t- x

t_%, t¢z

By 5.5.24,

(D H)(x)

= min{(D_H)(x) , (D+H) (x)}

§5.9. a.e. Primitives

249

for all x E (a, b), and (DH)(a) = (D+H)(a), (QH)(b) = (D_H)(b).

Given x E [a, b), we are to show that (D H)(x) 2: k(x). We can suppose that k(x) > -00. Suppose first that a ~ x < b. Let r be a real number such that k(x) > r. Since k is l.s.c. on [a, b], we know (Definition 5.5.1) that the set

v = {t E

[a, b]: k(t)

> r}

is a neighborhood of x in the (metric) topological space [a, b] . For almost all t E V n (x, +00) we have h(t) = k(t) > r, therefore

l'

hd>'

~ r(t -

x)

for every t E V n (x, +00), that is, H(t) - H(x) 2: r(t - x) for all t E V n (x, +00) . Thus H(t) - H(x) > r t-x -

for all t E V n (x, +00);

it follows from the definition of liminf (as a sup of infs) that r
r conveys no information about points other than x, but semicontinuity projects the inequality k(x) > r into an entire neighborhood of x, and carries h along with it almost everywhere.} 5.9.2. Remark. Note (for use in §5.12) that the 'dual' of the lemma is also true: If hEr}, k: [a, b] -+ IR is upper semicontinuous and k = h a.e., then the function H: [a, b) -+ IR defined by H(x)

=

l

z

hd>'

5. Differentiation

250

satisfies DH:s k everywhere on [a, bl. {It suffices to apply the lemma to the functions -h and -k (integrable and l.s.c., respectively), noting that - H is the indefinite integral of - h .} 5.9.3. Theorem.

II I: (a, b] -> lR is Lebesgue-integrable and F: (a, b) -> lR

is defined by F(x)

then F'

=I

= [Id>.

a.e.

Proof The claim is that, for almost every x E (a, b), F is differentiable at x and F'(x) = I(x). The first target is to prove that D F 2: I a.e. We have to show that the set

A = {x: rnF)(x) - I(x) < O} is negligible; given r E lR, r > 0, it will suffice to show that the set Ar = {x: (D F)(x) - I(x)

< -r}

is negligible (A is the union of the sequence of sets A lin for n = 1,2,3, ...). Given any f > 0, it is enough to show that >'*(Ar ) < f. With the notations f and r established above, by Theorem 5.6.1 there exist functions h E £} and k: [a, bl -> lR ,with k lower semicontinuous, such that h = k a.e., 1:S k everywhere on [a, bl, and

1 1 b

hd>'


I d>' + ~rf.

lR be the indefinite integral of h,

H(x) =

1% hd>'

(a:s x

:s b).

Since h - I 2: 0 a.e. and H - F is the indefinite integral of h follows that H - F is an increasing function. Let

Er

= {x E (a, b):

[D(H - F)](x)

I,

it

> r}

(note that E r is a function of H, therefore of f); since H - F is continuous and increasing,

(H - F)(b) - (H - F)(a) 2: ~r>'*(Er) by Corollary 5.8.3 of the preceding section, in other words

(2)

(H - F)(b) 2: ~r>'*(Er).

On the other hand, the inequality of (1) says that (H - F)(b) < ~re;

§5.9. a.e. Primitives

251

combined with (2), this shows that

).·(Er ) < f.

(3)

Now, Er is a function of f but A r is not. We are going to show that Ar c Er , whence ).·(Ar ) ~ ).·(Er ) < f, completing the proof that A is negligible. Arguing contrapositively, assuming x ¢ E r we will show that x ¢ Ar , that is,

illF)(x) - I(x)

~ -T.

Consider F = H - (H - F); by 5.5.25,

W

DF>DH-D(H-n

at every point of [a, bl where the difference on the right side is defined. Since H - F is increasing and x ¢ Er , we have (5)

o ~ (D(H -

F)J(x)

~

T,

thus the subtraction in (4) is permissible at x and it follows from (4) and (5) that (D F)(x) ~ (D H)(x) - [D(H - F»)(x) ~ (D H)(x) - T; but (D H)(x) > k(x) by the lemma, and k(x) ~ I(x) by the choice of k, thus (D F)(x)

~

I(x) -

T,

as we wished to show. We now know that A is negligible, in other words,

DF > 1 a.e.

(6)

Applying (6) to - f (whose indefinite integral is -F), we have

D(-F)

~

-I a.e.,

-(DF)

~

- f a.e.,

that is,

thus

DF < f a.e.

(7)

From (6), (7) and 5.3.3, we have

f

~

DF

< DF :5 f a.e.,

therefore

D F = DF = f a.e.; in view of 5.5.23, this means that for almost every x E (a, b), F is differentiable at x and F'(x) = f (x). 0

252

5. Differentiation

5.9.4. Corollary. If F: [a, b] -+ lR is absolutely continuous, then F is differentiable almost everywhere.

Proof. By the representation theorem 5.2.1, the function F - F(a)l is an indefinite integral. 0 Remarks. In many expositions, the almost everywhere differentiability of absolutely continuous functions precedes that of indefinite integrals, that is, the Corollary precedes (and is a lemma to) the Theorem. The strategy in such expositions is to prove the Corollary directly, without the aid of the machinery of integration theory. The direct proofs employ instead a recursive application of the 'Rising sun lemma' (to prove that every continuous function of bounded variation is differentiable a.e. l ) or Vitali's covering theorem (to prove that every function of bounded variation is differentiable a.e. 2 ); once one renounces the use of integration techniques, the full force of absolute continuity seems to provide no simplification. The key to the approach in this chapter is (a) an early proof that an AC function maps negligible sets to negligible sets (5.1.16), (b) use of the Radon-Nikodym theorem to represent AC functions via indefinite integrals (5.2.1), and (c) the use of semicontinuous approximations to simplify the proof of the a.e. differentiability of indefinite integrals3 .

5.10. Lebesgue's "Fundamental Theorem of Calculus"

This section is essentially a "Scholium", a gathering together of earlier results into a memorable form. As in the preceding section, the context is Lebesgue measure on an interval [a, bl. 5.10.1. Theorem. The following conditions on a function f: [a,b] --+ lR are equivalent: (a) f is Lebesgue-integrable; (b) there exists an absolutely continuous function F: [a, bl --+ lR such that F' = f a.e. When the conditions are satisfied, necessarily

F(x)

= F(a) +

l

z

f d>"

for all x E

la, b] ,

1 B. Sz.-Nagy, Introduction to reo.! functions and orthogonal ezpansions [Oxford, 1965J, pp. 107, 204. 2 H. L. Royden, Real analysis 13rd OOn., Macmillan, 1988], p. 106, proof of Lemma 9; also E. Hewitt and K. Stromberg, Real and abstract analysis [Springer, 1965], p. 264, 17.12 and p. 275, 18.3. Reversing the order of events in these expositions, a proof of the almost-everywhere differentiability of a function of bounded variation, based on Theorem 5.9.3, is given in §12 of the present chapter (Corollary 5.12.8 below). 3 Following E. J. McShane, Integration IPrinceton, 1944], p. 198, 33.3.

§5.11. Measurability of Derivates

253

F is uniquely determined by I up to an additive constant, and tId>' = F(b) - F(a).

(b): The indefinite integral F of I meets the requirements of (b) by 5.2.1 and the theorem of the preceding section; if also G is absolutely continuous and G' = I a.e. then (G - F)' = 0 a.e., therefore G - F is constant by 5.4.5 (one could also prove this using 5.2.1). Then, for all x E [a, bl , G(x) - F(x) = G(a) - F(a) = G(a) , thus

Proof. (a)

~

G(x)

= G(a) + F(x) = G(a) + [

I d>',

and in particular

G(b)

= G(a) + tId>..

(b) ~ (a): By 5.2.1, F - F(a)1 is the indefinite integral of a Lebesgueintegrable function g: la, bJ -+ 1R, and F' = g a.e. by the theorem of the preceding section; thus I = F' = g a.e., therefore I is also Lebesgueintegrable. 5.10.2. Theorem. The lollowing conditions on a function F: [a, bl -+ IR are equivalent: (a) F is absolutely continuous; (b) there exists a Lebesgue-integrable function I: [a, bl -+ IR such that

F(x)

= F(a) + 1~ I d>'

lor all x E la, bl·

When the conditions are satisfied, F' = I a.e. and I is a.e. unique (in the sense that any two such functions I are equal a.e.).

Proof. The equivalence of (a) and (b), and the essential uniqueness of I, are proved in 5.2.1. With notations as in (b), F' = I a.e. by the theorem of the preceding section.

5.U. Measurability of Derivates of a Monotone Function This section is technical preparation for the Lebesgue decomposition theorem of the next section. The 'official' definition of measurable function given in §4.1 requires the function to be real-valued; although the derivates of a monotone function may be infinite-valued, the following theorem 1 shows that they are 'measurable' in an appropriate sense: 1

E. J. McShane, Integration [Princeton. 1944]. p. 194, 32.1.

254

5. Differentiation

5.11.1. Theorem. Let f: all four derivates of f are k n D+ f defined by

(k n D+ f)(x)

la, bl ..... JR

be an increasing function (so that 0). For each real number k, the function

~

= min{k, (D+ f)(x)}

(a < x :5 b)

is a bounded Borel function (hence is Lebesgue-integrable). The same is true with D+ replaced by D+. D- or D_. Proof. The convention is that (0+ f)(b) = (O+f)(b) = 0 and (0- f)(a) = (D_f)(a) = O. To simplify the notations, we extend f to an increasing function on JR by defining f(x) = f(a) for x < a and f(x) = f(b) for x > b. (Note that the 'honest' derivates of the extended function at a and b agree with the preceding conventions.) It will suffice to prove the assertion of the theorem (except for integrability) for the extended function, then restrict to the closed interval la, bl. For each real number a > 0, define functions 0, x E IR and let 0 < $ < a. For each rational number r such that s < r < or, we have f(x + r) ~ f(x + $) , thus t/J",(x) ~ f(x + r) - f(x) ~ f(x + $) - f(x) ; r

r

letting r ..... $+ in the right-hand member, we have

t/J",(x) > f(x

+ s) -

f(x) ,

s and the validity of this inequality for all $ E (0, a) implies that t/J",(x) ~ cp",(x) . The reverse inequality was noted earlier. In the proofs of the following claims, it is sometimes convenient to use the original formula for CP",. sometimes that for t/J""

§5.1l. Measurability of Derivates

255

claim 2: 'Po -+ 0+ f pointwise as a and an -+ 0 as n -+ 00 , then 'Po.. (x)

-+

0+. More precisely, if an

>0

(0+ f)(x) in lR for every x E lR.

-+

Fix x E lR. Write Vn = (x - 0n,X + an) (n = 1,2,3, ...) and A (x, +00); since diam Vn = 20n -+ 0, Lemma 5.3.2 is applicable.

=

From claims 1 and 2, we see that t/lo

-+

0+ f pointwise as a

-+

0+

.

Moreover, t/lo ! 0+ f pointwise as a ! 0, in the following sense: if an > 0 , an ! 0 and x E lR, then the sequence t/lo.. (x) is decreasing, with infimum (0+ f)(x).

claim 3: If a > 0 and c E lR, then the set E

= {x:

t/lo(x):5 c}

is a Borel set. For every r E Q n (0, a), define gr : lR

gr(x)

-+

= f(x + r) -

lR by the formula

f(x) ;

r

note that gr is a linear combination of two increasing functions of x. It is easy to see that every increasing function on lR is Borel. {For example, every inverse image f-1((t,+00» (t E lR) is an interval (because f(x) > t ~ f(y) > t for all y 2: x), hence is a Borel set; thus f is Borel.} It follows that gr is a Borel function. By definition, t/lo is the upper envelope of the family (gr) , that is,

t/lo(x)

=

gr(x)

sup

for all x,

rEQn(O,o)

therefore

t/lo(x) :5 c

~

gr(x):5 c for all r E Q n (0, a) ;

=

n

in other words, E

{x: gr(X):5 c} ,

rEQn(O,o)

thus E is the intersection of a countable family of Borel sets, hence is a Borel set. It follows from claim 3 that the set 00

{x: t/lo(x) < c} =

U {x:

t/lo(x):5 c - lin}

n=1

is also a Borel set, therefore so is its complement

{x: t/lo(x) > c} .

5. Differentiation

256

claim 4: For every real number c, the set {x: (D+ f)(x) 2: c} is a Borel set.

For, by the remarks following claim 2, we have "'I/n ! D+ f pointwise as n - 00 , therefore (D+ f)(x) > c

¢}

"'1/n(X) 2: c for all n;

it follows that

n{x: 00

{x: (D+ f)(x) 2: c} =

"'1/n(X) 2: c}

n=l

is the intersection of a sequence of Borel sets. Now let k E IR. If k :; 0 then k n D+ f is the constant function k (because D+ f 2: 0), a bounded Borel function in good standing. If k > 0 then 0:; k n D+ f :; k ,so k n D+ f is certainly bounded; moreover, for every real number c, {x: (knD+f)(x) 2: c}

= {x:

k 2: c and (D+f)(x) 2: c};

this set is empty if k < c, and if k 2: c it is equal to {x: (D+ f)(x) 2: c}. thus, in view of claim 4, it is always a Borel set. This completes the proof that if f: la, b] - IR is an increasing function, then k n D+ f is a bounded Borel function for every k E IR. We can infer that k n D- f is Borel by using the '*-trick' of Corollary 5.4.3: writing x·=a+b-x and j*(x)=-f(x·),wehave (D- f)(x·)

= (D+ j*)(x)

for all x E

la, b] .

Since j* is increasing, k n D+ j* is a Borel function by what we have already proved; since x 1--+ x· is a homeomorphism of la, b] onto itself, and (k n D- f)(x)

= min{k, (D- f)(x)} = min{ k, (D+ j*)(x·)} = (k n D+ j*)(x·) ,

it is clear that k n D- f is also a Borel function. The analogous assertions for D+ and D_ are left as exercises (they are not needed for the application in the next section). 0 5.11.2. Remark. The conclusion of the theorem is also true for decreasing functions and for continuous (but not necessarily monotone) functions. 2

2 McShane, loco cit.

§5.12. Lebesgue Decomposition

257

Exercises 1. Every function f : la, bJ -+ R of bounded variation is a bounded Borel function. {Hint: Jordan decomposition.}

2. Complete the proof of 5.11.1 for D+ and D_.

5.12. The Lebesgue Decomposition of a Function of Bounded Variation

The theorem in question (proved in 5.12.9 below):

Every function F: la, bl -+ R of bounded variation can be written as a sum F = G + H with G absolutely continuous and HI = 0 a.e. (in particular, F is differentiable a.e.). Such a representation is essentially unique: all others are of the form F = (G + el) + (H - el), where el is a constant function. Included in this result is Theorem C of the remarks at the beginning of §5.4: Every increasing function F: [a, b] -+ R is differentiable a.e. We begin the proof with a general observation on limits. As in Definition 5.3.1, let (X,d) be a metric space, A c B C X, f: B -+ R, c E A, and let g: B -+ R be another function: X U

8

f,g I

R

U

c

A :J

E

A

(In §5.3, the functions were allowed to have infinite values; the motive for requiring finite values is to simplify the algebra.) 5.12.1. Lemma. With the preceding notations, suppose that 9 has a finite limit lim

%_c. :rEA

g(x)

= L E JR.

Then limsup [f(x) %_C,

:rEA

+ g(x)] = limsup f(x) + L %-C,

and similarly with limsup replaced by liminf.

::z:EA

5. Differentiation

258

Proof. The assumption L E R assures that the sum on the right side exists (and that L can be transposed freely). To simplify the notations, we write briefly limsup(/ + 9}

= limsupf + L

for the equation to be verified. Let S

= {s E IR:

f(x n }

->

s for some sequence

Xn

E A with

X n ->

c}.

lim sup f is the largest element of S. Similarly, lim sup (/ + 9) is the largest element ofthe set

As noted in 5.3.3,

T

= {t E IR:

(/ +9)(Xn }

->

t for some sequence Xn E A with Xn

If (x n ) is a sequence in A with Xn and

f(x n } + 9(xn } thus, t E T

->

->

c and if t E IR, then 9(X n }

t

c}.

->

L

t - L;

t - L E S, that is, T={s+L: SES},

and the first assertion of the lemma reduces to the observation that max T = max S + L. For the second assertion, replace lim sup by lim inf and max by min in the preceding argument. Here is an application of the preceding lemma to derivates: 5.12.2. Lemma. If f,9: tiable at x, then

la, b]-> IR,

x E la, b} and 9 is right-differen-

[D+(/ + 9}](x}

= (D+ f)(x) + 9~(X} , ID+(/ + 9}I(x} = (D+f)(x) + 9~(X} . The analogous relations hold for D- and D _ , assuming 9 left-differentiable at a point x E (a, bl . Proof. For t

#x,

=(f_+--,g:..:..)(.:...;t)c--_(;.:..f_+....:::9.:..,:)(-,-x} t-x

= f (t) - f (x) + g(t} t-x

g(x} . t-x'

by hypothesis, the second term on the right has a finite limit g~(x} as t -> x+ , thus the asserted formulas follow from the preceding lemma (with A = (x, bl). In what follows, >. denotes Lebesgue measure on a closed interval [a, bl , and £1 = £1 ([a, b], >.) is the class of Lebesgue-integrable functions on

la, bl·

§5.12. Lebesgue Decomposition

259

5.12.3. Lemma. If gE£,l is bounded, say Igi :5 M < +00, and if G is the indefinite integral of g, then IDGI :5 M for D = D+, D+,

D-, D_.

Proof If x, t E [a, b] and t IG(t) - G(x)1

=

> x, then

l'

gd)"

< 1'19Id).. < M(t - x),

thus G(t) - G(x)

-x

t

ail and \0 denotes characteristic function. Since mi > 0 (recall that I ~ 1), each term of g

SCI. N. Bourbaki, General topology. VoL 1 (Addison-Wesley, Reading, 1966J. Ch. IV. §6, no. 2. 6 H. L. Royden, Real analysis 13rd. . $ l f. d>' $ l r d>' $ l hd>', that is,

b b s(u) < l f. d>' $l r d>' $ 8(u) , where s('

b r d>'

-b f(t)dt.

is Riemann-integrable, so the extremeties of (*) are

t' = 0,

r -

r -

and from f. :::: 0 we infer further that f. = 0 a.e. (4.4.21, (3)). Thus f. = f = r a.e.; if e is any point where f.(e) = r(e) , then f is continuous at e (5.5.19). A corollary ofthe proof of "(a) 5.13.8. Corollary. If f :

la, bj

'* (b)":

-+

IR is Riemann-integrable, then f

also Lebesgue-integrable and

b l f(t)dt =

r

t

is

f d>..

Proof In the notations of (*) of the preceding proof, f = E .c 1 = .c 1 ([a, bj, >.) ,therefore f E.c 1 and

r

a.e. and

lbfd>.= l brd>'; by equality in (*),

t t r d>' =

f(t)dt.

A Riemann-integrable function is bounded and Lebesgue-integrable (5.13.8). In the Lebesgue theory, essentially bounded functions are on an

5. Differentiation

270

equal footing with bounded ones, and an essentially bounded measurable function on [a, b] is Lebesgue-integrable; for such functions, there is a wellrounded "Fundamental theorem of calculus": 5.13.9. Theorem. The following conditions on a junction f: [a, b] --+ IR are equivalent: (a) f is essentially bounded and Lebesgue-measurable (hence Lebesgueintegrable on [a, b] ); (b) there exists a junction F: [a, b] --+ IR satisfying a Lipschitz condition IF(x) - F(y)1 :::; Klx - YI, such that F I = f a.e. Any two such functions F differ by a constant, and F(x) in particular

= F(a) + 1% f

t

f d)"

d)"

for all x E [a, bJ ,

= F(b) -

F(a).

Proof (a) ~ (b): If Ifl :::; K a.e. and F is the indefinite integral of then F I = f a.e. (5.9.3) and, if a:::; x < y :::; b, then IF(y) - F(x)1

={

f d)" :::; {

f,

Ifld)" :::; Klx - yl·

(b) ~ (a): Suppose K is a real number> 0 such that IF(x) - F(y)1 :::; Klx - yl for all x, y in [a, b]. It follows that F is absolutely continuous (5.1.10, (vi)), so there exists a Lebesgue-integrable function 9: [a, bJ --+ IR such that F' = 9 a.e. (5.10.2); it is clear from the Lipschitz condition that IF'I:::; K, therefore 191:::; K a.e., thus 9 is essentially bounded. By assumption, F I = f a.e., therefore f = 9 a.e.; thus f is also Lebesguemeasurable and essentially bounded. The last assertions of the theorem are immediate from 5.10.1. Inspection of the above proof yields the following characterization of the 'primitives' that occur in the theorem: 5.13.10. Corollary. For a function F: [a, b] --+ IR , the following conditions are equivalent: (a) F is the indefinite integral of an essentially bounded, Lebesguemeasurable junction f: [a, b] --+ IR ; (b) F(a) = 0 and F satisfies a Lipschitz condition IF(x) - F(y)l :::; Klx-yl· Is there a "Fundamental theorem of calculus" for Riemann-integrable functions? The best we can squeeze out of the theorems of this section is the following: Let f: [a, b]

--+

IR be Riemann-integrable. Then

§5.13. Riemann Integral

271

(I) the indefinite integral F(x)

= 1% f(t)dt

(a =5 x =5 b)

of f is absolutely continuous (even Lipschitz) and satisfies F' = f a.e.; (2) if also 0: [a, b] --+ IR is absolutely continuous and 0' = f a.e., then 0 differs from F by a constant, therefore

= O(a) + 1% f(t)dt

O(x)

for all x E [a, b].

The only thing in view resembling an 'integral-free' characterization of 'primitives' for the Riemann theory is condition (b) of the following neartautology: For a function F: [a, b] --+ IR, the following conditions are equivalent: (a) F is the indefinite integml of a Riemann-integrable function on [a, b] ; (b) F(a) = 0, F is absolutely continuous, and there exists a bounded function f : [a, b] --+ IR, with negligible set of discontinuities, such that pI = f a.e. With notations as in (b), f is Riemann-integrable and F is its indefinite integml.

The "Fundamental theorem of calculus" that the Riemann integral would like to enjoy seems to be preempted by the essentially bounded measurable functions (5.13.9 and 5.13.10); what is missing in the Riemann case is a condition on F, stronger than Lipschitz, that does not give the show away like the above condition (b).

Exercises 1. (i) The function F: [O,IJ --+ IR defined by F(x) = (I - X2 )1/2 is continuous on [0, I] and differentiable on (0, I), but it is not the primitive of a continuous function f: [0, IJ --+ IR. {Hint: F' is unbounded.} (ii) Let g: [0, IJ --+ IR be the (Riemann-integrable) function defined by

sin(I/x) g(x) = {

and let F: [0, I)

--+

°

for x E (O,IJ for x

=

°

IR be the indefinite integral of g, F(x)

= 1% g(t)dt

(0 =5 x < I) .

272

5. Differentiation

Then F is continuous on [0, 1) ,differentiable on (0,1), and F I is bounded, but F is not a primitive of a continuous function I: [0,1] - JR. {Hint: g(O+) does not exist.} 2. Let

I: la, b] -

lR be Riemann-integrable, F its indefinite integral,

c E [a, b). (i) If 1 has a right limit L at c, then F is right-differentiable at C and F:(c)=L. {Hint: Redefining 1 at c, one can suppose that 1 is right-continuous at c.} (ii) The converse of (i) is false. For example, if 1 : [0,21 - lR is the characteristic function of the set A = {I -lin: nell'} U {I

+ lin:

nell'},

then 1 is Riemann-integrable, its indefinite integral F is identically zero, F'(l) = 1(1) but neither 1(1+) nor /(1-) exist. 3. If 1 : [a, bl - lR is any bounded function, then its set of discontinuities is the union of a sequence of closed sets (Le., is an Fer set in lR). {Hint: With notations as in Definition 5.13.5, argue that the set {x E [a,b]: j*(x) - I.(x) ~ lin} is a closed set (cf. 5.5.13).}

CHAPTER 6

Function Spaces

§6.1. §6.2. §6.3. §6.4. §6.5. §6.6. §6.7. §6.8. §6.9.

Compact metric spaces Uniform convergence, iterated limits theorem Complete metric spaces L'

Real and complex measures L"" L" (1 < p < +00) C(X)

Stone-Weierstrass approximation theorem

The emphasis of the present chapter is on metric spaces whose elements are functions (or equivalence classes of functions). The main examples arise in topological or measure-theoretic contexts; the first three sections prepare the way with the necessary topics in topology and metric spaces.

6.1. Compact Metric Spaces The concept of compactness to be defined shortly is motivated by the following property of a closed interval: 6.1.1. Theorem. (Heine-Borel)' Let [a, b] be a closed interval in IR and suppose C is a class of open sets in IR such that each point of [a, b] belongs to at least one of the sets of C, briefly la, b] cue. Then

[a,b]

c

UI U ... UU n

for a suitable finite list U I, ... ,Un of sets in C.

Proof. Let S be the set of all points x of [a, b] such that the closed interval [a, xl is contained in the union of a finite number of sets in C. Since [a,a] = {a} C U for some U e C, we have a e S; our problem is to show that b e S . 'Heinrich Eduard Heine (1821-1881), Emile Borel (1871-1956).

273

6. Function Spaces

274

The set S is nonempty and bounded; let M = sup S. Obviously a:5 M :5 b. We will show that b E S by proving that (1) M E S, and (2) M = b. (1) Since M E [a, bl, by assumption there exists a set VEe such that MEV; since V is open, we have [M - f, M + fl c V for a suitable f > O. We note for use in the proof of (2) that f can be taken to be as small as we like. Choose xES so that M - f < X b; for, the alternative M +f :5 b would imply (by the preceding inclusion) that M + f E S, contrary to the fact that every

since lx, M] C [M - f, M

element of S is:5 M . (2) The foregoing argument shows that b < M + f for arbitrarily small f ,therefore b:5 M ; already M:5 b, so b = M E S. For use in more general situations later on, we separate out several concepts involved in the Heine-Borel theorem in a form applicable in general topological spaces (not necessarily derived from a metric):

6.1.2. Definition. Let X be a topological space and let A be a subset of X. A set C of subsets of X is said to be a covering of A if A c UC, that is, if each point of A belongs to at least one of the sets in C; if 'D c C and 'D is also a covering of A, then 'D is said to be a subcovering of C. (The language is regrettably awkward; 'D is a subset of C but it is A that gets covered.) A covering C of A consisting of a finite number of sets is called a finite covering of A; a covering C of A whose elements are open sets in X is called an open covering of A. Expressed in the foregoing language, the Heine-Borel theorem asserts that (in the topological space IR) every open covering of a closed interval la, b] has a finite subcovering. This prompts the next definition:

6.1.3. Definition. A subset A of a topological space X is said to be quasicompact if every open covering of A has a finite subcovering; expressed in the notation of indexed families, this means that if (Ui)iEI is a family of open sets in X such that A C UiEI U i , then there exists a finite subset J of I such that A C UiEJ U i . If X is a quasicompact subset of itself thcn it is called a quasicompact space.

§6.1. Compact Metric Spaces

275

Reformulated in terms of closed sets, quasicompactness has the flavor of an 'induction principle' (from finite to infinite): 6.1.4. Theorem. The following conditions on a topological space X are equivalent: (a) X is quasicompact; (b) if (F')'EI is a family of closed sets in X such that n'EJ F, f 0 for every finite subset J of I, then n'EI F, f 0 . Proof Stated contrapositively, condition (b) says that, for a family (F')'EI of closed sets,

nF,=0

=>

iEI

nF, = 0

for some finite J

c I,

iEJ

in other words,

UCF, =X iEI

=>

UCF, = X

for some finite J

c

I.

iEJ

In view of the duality between closed sets and open sets (3.3.1), it is clear that (b) is equivalent to the assertion that every open covering of X has a finite subcovering, which is also the meaning of (a). The hypothesis in condition (b) can be expressed by saying that the family (F')'EI has the finite intersection property (every finite subfamily has nonempty intersection); condition (b) then says that every family of closed sets with the finite intersection property has nonempty intersection. 6.1.5. Corollary. If X is a quasicompact topological space and if (Fn) is a sequence of nonempty closed sets in X such that F 1 ::> F2 ::> F3 ::> ... , then n::"=I F n f 0. Proof It is obvious that the family (Fn)nEP has the finite intersection property. The definition of compactness requires quasicompactness and one extra condition: 6.1.6. Definition. A topological space is said to be separated (or to be a Hausdorff space 2 ) if, for every pair of distinct points x and y of the space, there exist open sets U and V such that x E U , Y E V and Un V = 0 (so to speak, distinct points can be separated by means of disjoint open sets-or, equivalently, by means of disjoint neighborhoods of the points). A topological space is said to be compact if it is both quasicompact and separated. 6.1.7. Remarks. (i) Every metric space (X,d) is separated for the topology Od derived from its metric. {Proof: If x f y and if r = ~d(x, y) , 2

After Felix Hausdorff (1868-1942).

276

6. FUnction Spaces

then the open balls U = Ur(x) and V = Ur(y) are disjoint neighborhoods of x and y respectively; for, the existence of a point z E U n V would imply that d(x,y) ~ d(x,z) + d(z,y) < r + r = d(x,y).} Thus, for a metric space, the concepts of compactness and quasicompactness coincide. (ii) A quasicompact space need not be compact (consider a two-point set equipped with the trivial topology (3.3.2». (iii) Let (X,O) be a topological space and let A be a subset of X. The class

o n A = {U n A:

U EO}

of subsets of A is easily seen to be a topology on A; it is called the relative topology on A induced by 0 (cf. §3.3, Exercise 7). One also writes 0 A = OnA, and (A,OA) is called a (topological) subspace of (X, 0).1f C is an open covering of A in the sense of 6.1.2, then en A = {U n A: U E C} is a class of open subsets of A (for the relative topology) whose union is A; it follows easily that A is a quasicompact subset of X (in the sense of 6.1.3) if and only if A is a quasicompact space for the relative topology. (iv) With notations as in (iii), if X is separated then A is separated for the relative topology (if U and V are disjoint, then so are UnA and V nA). 6.1.8. Definition. A subset A of a topological space (X,O) is said to be compact if, for the relative topology induced by 0, A is a compact topological space, that is, if (A, 0 A) is a compact space in the sense of Definition 6.1.6. 6.1.9. Examples. (1) Let X be a separated topological space (for example, a metric space) and let A be a subset of X. In view of (iii) and (iv) of 6.1.7, A is a compact subset of X if and only if it is a quasicompact subset of X. In particular, the Heine-Borel theorem asserts that every closed interval [a, bl is a compact subset of IR; in other words (cf. Exercise 2), la, bJ is a compact metric space for the usual metric (x, y) ...... Ix - yl. (2) If (x n ) is a convergent sequence in a metric space X, say x n ..... x, then the set

A

= {x}U{x n

:

n

= 1,2,3, ... }

is a compact subset of X. {Hint: An open set containing x contains all but finitely many of the X n (cf.3.2.19).} The Weierstrass-Bolzano theorem (cf. 1.16.11) states that every bounded sequence in IR has a convergent subsequence. In particular, every sequence in a closed interval la, b] of lR has a convergent subsequence, whose limit is in la, bl because la, b] is a closed subset of lR.

§6.1. Compact Metric Spaces

277

6.1.10. Definition. A metric space is said to have the WeierstrassBolzano property if every sequence in the space has a convergent subsequence.

The main goal of this section is to prove that a metric space is compact (for the topology derived from its metric) if and only if it has the Weierstrass-Bolzano property. Some of the most important metric space concepts figure in the proof (total boundedness, separability, completeness); the proof is organized in a series of lemmas, interspersed with the definitions of these concepts and some examples. Half of the equivalence is disposed of by the first lemma: 6.1.11. Lemma. If (X, d) is a compact metric space, then every sequence in X has a convergent subsequence. Proof. Let (x n ) be a sequence in X. For each index n, let

An

= {Xk: k

> n}.

The sets An are nonempty and AI:::> A2 :::> A3 :::> .... Since the closure operation preserves inclusion (3.3.16), we have

Al :::> A2 :::> A3 :::> ... ; by compactness, the intersection of the sets An is nonempty, say x E n::,,= I An· We will show that x is the limit of a suitable subsequence of (x n ). Since x is adherent to AI , there exists an index n I > 1 such that d(xn" x) < 1; then, since x is adherent to An., there exists an index n2 > nl such that d(xn.,x) < 1/2. Continuing recursively, we obtain a sequence of indices nl < n2 < n3 < ... such that d(x n., x) < 11k, thus (xn.) is a subsequence of (x n ) with d(xn.,x) -> 0 as k -> 00. This proves that every compact metric space has the Weierstrass-Bolzano property. Before proving the reverse implication, let us note a property of compactness that motivates the next definition: 6.1.12. Remark. If (X, d) is a compact metric space then, for every ( > 0, there exists a finite list of points YI •... , Yr in X such that each point of X is within ( of at least one of the Yi. that is, r

X

= U U,(Yi) i=1

(of course r. and the points YI, ... , Yr, will in general depend on (). {Proof: The open balls U,(y) , Y EX, constitute an open covering of X; pass to a finite subcovering.}

6. FUnction Spaces

278

6.1.13. Definition. Let (X, d) be a metric space and let ( > O. An (-net in X is a finite subset F of X such that X=

U U.(y). yEF

Thus, if F = {Yl,' .. ,Yr} , then every point of X is within ( of a least one of the points Yi. The metric space (X, d) is said to be totally bounded if it has an (-net for every (> 0 (it clearly suffices that there exist a ~-net for every positive integer n). For example, every compact metric space is totally bounded (6.1.12), but the converse is false (cf. Exercise 5). If x, Y E Ur(a) then d(x, y) ~ 2r by the triangle inequality. This prompts the next definition:

6.1.14. Definition. Let (X,d) be a metric space, A a nonempty subset of X. We say that A has finite diameter if there exists a real number K ;0: 0 such that d(x,y) ~ K for all X,Y E A;

more precisely, the diameter of such a set, denoted diam A, is defined to be the infimum of all such K, diamA

= inf{K:

d(x,y) ~ K

for all X,Y E A},

and it is clear from the definition of suprema that diam A = sup{ d(x, y) :

x, YEA}. 6.1.15. Examples. (i) In a metric space, every ball (open or closed) ohadius r has diameter :5 2r , and every subset of finite diameter is contained in some ball. (ii) In a discrete metric space (3.1.7) every open ball of radius r ~ 1 has diameter O. (iii) A metric space (X, d) is totally bounded if and only if, for every ( > 0, X is the union of finitely many sets of diameter < ( . The next definition is a generalization to metric spaces of a concept familiar from elementary analysis:

6.1.16. Definition. A sequence (x n ) in a metric space (X,d) is said to be a Cauchy sequence if d(x m , x n ) -+ 0 as m, n -+ 00, in the following sense: for every (> 0 there exists an index N such that d(x m , x n ) < f for all m, n ~ N . Every convergent sequence is Cauchy; for, if d(x n , x) -+ 0 then d(x m , x n ) ~ d(x m , x) + d(x, x n ) < ( provided that d(x m , x) < (/2 and d(x n , x) < £/2. The converse is false; for example, in the open interval X = (0, +00) equipped with the usual metric d(x, y) = Ix - YI, the

§6.1. Compact Metric Spaces sequence x n limit in X.

= 1In

279

is Cauchy (because it is convergent in IR) but has no

6.1.17. Lemma. If (X, d) is a metric space in which every sequence has a Cauchy subsequence, then the space is totally bounded.

Proof. (The converse is also true-see 6.1.24 below.) We argue contrapositively: assuming that X is not totally bounded, let us construct a sequence (x n ) in X that has no Cauchy subsequence. By assumption, there exists an E > 0 for which no E-net exists; that is. every finite subset of X fails to be an E-net. Thus, for every finite subset F of X. there exists a point x E X such that d(x, y) ~ E for all y E F. The construction of (x n ) proceeds as follows. Choose any point Xl in X. Since {Xl} is not an E-net. there exists a point X2 such that d(X2. Xl) ~ Eo Since {Xl. X2} is not an E-net. there exists a point X3 such that d(X3. xil ~ E and d(X3' X2) ~ Eo Continuing in the obvious recursive way. we obtain a sequence (x n ) such that d(x m • x n ) ~ E whenever m # n. a sequence that can have no Cauchy subsequence. 6.1.18. Definition. A metric space is said to be separable if it has a countable dense subset. that is. a countable subset A such that A = X . For example. the real number field IR equipped with the usual metric is separable. with the rational field IQi as a countable (1.10.10) dense subset (1.8.25). An uncountable discrete metric space is not separable. 6.1.19. Lemma. Every totally bounded metric space (X. d) is separable.

Proof. For each positive integer n. let Fn be a ~-net in X. The set A = U::"=l F n is countable; we will show that it is dense in X. It suffices to show that every open ball Ur(x) has nonempty intersection with A. Choose n so that ~ < r. Since Fn is a ~-net. there exists a point y E F n such that d(x. y) < lin; then y E Ur(x) (because ~ < r) and yEA (because FnCA),thus Ur(x)nA#. 6.1.20. Definition. Let X be a topological space and let B be a set of open sets in X; B is said to be a base for the topology of X (or for the open sets of X) if every open set is a union of sets in B; equivalently. U open,

X

EU

'*

3VEB 3

X

E V cU.

So to speak. the sets of B 'pry into every neighborhood': between any point X and any of its neighborhoods. one can interpolate one of the sets of the base B. 6.1.21. Lemma. Every separable metric space has a countable base for the topology derived from the metric.

6. FUnction Spaces

280

Proof. Let A = {ak: k = 1,2,3, ... } be a countable dense subset of the space (6.1.18) and let B = {Ul/n(ak): n,k e Il'}

be the set of all open balls, centered at the ak, with radii l/n (n e Il'). Clearly B is a countable set (cf. 1.10.8) of open sets; we will show that it is a base for the topology. Let U be an open set and let x e U; we are to interpolate a set of B between x and U. Choose r > 0 so that Ur(x) C U, let n be a positive integer such that ~ < ~, and let k be an index such that d(ak,x) < l/n (possible because A is dense). Then

x e U1/n(ak) C U; for, if ye U1/n(ak) then d(y,x) ~ d(y,ak)+ d(ak' x)

1

1

2

< -n + -n = -n < r,

thus y e Ur(x) cU. ¢ The property of having a countable base in fact characterizes the separable metric spaces (Exercise 7). Proving compactness entails finding finite subcoverings; finding a countable subcovering, which can be a valuable intermediate step, is available in every space with a countable base for open sets: 6.1.22. Lemma. (Lindelofs theorem) In a topological space with a countable base for the open sets, every open covering of the space has a countable subcovering. Proof Let B be a countable base for the open sets of the topological space X and let U be any open covering of X; we seek a countable

subcovering Uo of U. Let Bo={VeB: VcU for some UeU};

since Bo C B, Bo is countable (1.10.2), say Bo = {Vn : nell'}. For each positive integer n, choose a set Un e U with Vn C Un (possible by the definition of Bo) and let Uo = {Un: nell'}; Uo is a countable subset of U, and we need only show that it is a covering of X. Let x eX; we seek an index n such that x e Un. Choose U e U with x e U (U is a covering of X) and let Ve B with x eVe U (B is a base for the topology); then V e Bo by the definition of Bo, thus V = V n for some n, and finally x e V n C Un. ¢ We can now characterize compact metric spaces as the metric spaces having the Weierstrass-Bolzano property: 6.1.23. Theorem. The following conditions on a metric space (X, d) are equivalent:

§6.1. Compact Metric Spaces

281

(a) X is compact (jor the topology 0 d derived from the metric d ); (b) every sequence in X has a convergent subsequence. Proof. (a) => (b); This is Lemma 6.1.11. (b) => (a); Since convergent sequences are Cauchy, we know that every sequence in X has a Cauchy subsequence, therefore X is totally bounded (6.1.17), hence separable (6.1.19), hence there is a countable base for the open sets (6.1.21). Given any open covering U of X, we seek a finite subcovering. By Lindelof's theorem (6.1.22) we can suppose that U is countable, say U = {Un; n E !P'} . For every positive integer n, let Vn=UIU ... UUn ;

we know that V n TX and it will suffice to show that V n = X for some n. Assume to the contrary that no such n exists, that is, X - V n 'f (/) for aU n. For each n select a point X n E X - Vn . By hypothesis, the sequence (x n ) has a convergent subsequence, say x n• ..... x. By monotonicity, V T X, so x E V nj for some j; since Vn; is open, x n • E V nj ultimately. Choose any k such that k > j and x n • E V nj . Then

n.

contrary to

Xn•

E X - Vn.'

The totally bounded spaces are characterized by a 'Cauchy' variant of the Weierstrass-Bolzano property; 6.1.24. Theorem. The following conditions on a metric space (X, d) are equivalent; (a) X is totally bounded; (b) every sequence in X has a Cauchy subsequence. Proof (b) => (a); This is Lemma 6.1.17. (a) => (b); Let (x n ) be a sequence in X; assuming X is totally bounded, we seek a Cauchy subsequence (x n .). Given any £ > 0, X is expressible as a finite union of open balls of radius £/2, hence of diameter :$ £. It follows that every subset A of X is the union of finitely many sets of diameter :$ £; if, moreover, X n E A for infinitely many n, then one of the terms of such a union must contain X n for infinitely many n. Summarizing, if A is a subset of X containing X n for infinitely many n, then, given any £ > 0, A has a subset B of diameter :$ £ that contains X n for infinitely many n. We now construct a sequence (Ak) of subsets of X such that the k'th term x n • of the desired Cauchy subsequence will be drawn from A k • By the preceding paragraph (with A = X and £ = 1) there exists a subset Al of X such that diam Al :$ 1 and X n E Al frequently. Similarly (with A = Al and £ = 1/2) there exists a subset A2 of Al such that diam A2 < 1/2 and X n E A2 frequently. One continues recursively

282

6. Function Spaces

in the obvious way, obtaining a sequence Al ::> A2 ::> A3 ::> . .• such that diam A k ~ 1I k and such that, for each k, x... E Ak for infinitely many n. The desired subsequence (x.... ) of (x...) is now constructed as follows. Choose any index nl such that x.... E AI. Then choose any index n2 > nl such that x..., E A 2 . Recursively, choose nk > nk_1 such that x.... E Ak. The resulting subsequence (x.... ) is Cauchy. For, given any f > 0, there is an index k such that ;. < f; for every pair of indices i, j ?: k, we have

therefore d(x..." x nj ) ~ diam An. ~ link < f.

As remarked following 6.1.13, 'compact => totally bounded'. What can be added to total boundedness to convert the implication => into an equivalence

¢}

? 'Completeness' does the job:

6.1.25. Definition. A metric space is said to be complete if every Cauchy sequence in the space is convergent to a point in the space.

The classical example of a complete metric space: the real number field lR, equipped with the usual metric (x, y) ..... Ix - yl (cf. 1.8.26). Other examples are given in the exercises, and complete metric spaces are studied in greater depth in Section 3 of this chapter. 6.1.26. Theorem. The following conditions on a metric space (X,d) are equivalent: (a) X is compact Uor the topology 0d derived from the metric d); (b) (X, d) is complete and totally bounded. Proof. (a) => (b): By the remark following 6.1.13, X is totally bounded; we are to show that every Cauchy sequence (x n ) is convergent. By 6.1.11, (x n ) has a convergent subsequence, say x.... --> x, and it will suffice to show that X n --> x. Given any e > 0, choose an index N such that d(x n , x m ) < f/2 for all n, m ?: N, then choose an index k such that both d(x n ., x) < f/2 and nk?: N ; then

for all n?: N . (b) => (a): By Theorem 6.1.23, we need only show that every sequence (x n ) has a convergent subsequence. By total boundedness, (x n ) has a Cauchy subsequence (6.1.24) which, by completeness, is convergent. Compactness and completeness figure prominently in the rest of the book; we record here a theorem concerning each of these concepts, both for application in the next section. The first is a mapping property of quasicompactness:

§6.1. Compact Metric Spaces

283

6.1.27. Theorem. If f: X -> Y is a continuous mapping between topological spaces X and Y and if A is a quasicompact subset of X, then its image f(A) is a quasicompact subset of Y. Proof Assuming (Vi)iEI is a family of open sets in Y with f(A) c UiEI Vi , we seek a finite subset J of I such that f(A) C U jEJ Vj . We have

A c rl(UEI Vi)

= UiEI f-I(V i );

since A is quasicompact and the f-I(V i ) are open sets in X (3.4.5), there exists a finite subset J of I such that A C UjEJ f- I (Vj) , in other words f(A) C UjEJ Vj . 0 6.1.28. Remark. Every nonempty subset A of a metric space (X,d) can itself be regarded as a metric space simply by restricting the given distance function d to pairs of points of A (cf. Exercise 2). Equipped with the restricted metric d A = dlA x A, A is called a metric subspace of X. 6.1.29. Theorem. Let (X, d) be a metric space and let A be a nonempty subset of X, regarded as a metric subspace of X (6.1.28). (1) If A is a complete metric subspace of X, then A is a closed set in X. (2) If X is a complete metric space and A is a closed subset of X, then A is a complete metric subspace of X. Proof. (1) If an E A and dean, x) -> 0, then (an) is Cauchy in X, hence Cauchy in A,henceconvergenttosomepoint a of A, d(an,a) -> 0; then x = a E A by the uniqueness of limits (3.2.1), thus A is a closed set in X (3.2.5). (2) Assuming A is a closed set in X, suppose that (an) is a Cauchy sequence in A, that is, d(a m , an) -> 0 as m, n -> 00; then (an) is also Cauchy in X, so by hypothesis there exists a point x E X with dean, x) -> 0, and, since A is closed, necessarily x EA. Thus, every Cauchy sequence in A is convergent in A. 0

Exercises

1. Let X be an infinite set and declare a subset U of X to be open if either U = 0 or X - U is finite. This defines a topology on X for which X is (1) quasicompact, but (2) not separated. {Hint: (1) If (Ui)iEI is an open covering of X and if j is an index such that Uj l' 0, then X - Uj is covered by finitely many of the Ui . (2) C(U n V) = CU u CV ; infer that if U and V are nonempty open sets, then U n V is infinite.}

284

6. Function Spa0./3>0 such that

ad(x. y) ~ d'(x. y) ~ /3d(x, Y) for all x. Y E X (so that. in particular. d and d' are equivalent metrics by Corollary 3.3.6), then (X. d) is complete (totally bounded, compact) if and only if (X. d') is complete (totally bounded. compact). (iii) Same as Part (i). with d defined by ) lip

r

d(x. y)

= ( f;ldk(Xk. Yk)jP

•

where p > 1 is a constant. {Hint: Cf. Example 3.2.2. (v).} (iv) Let (Xk,dk)kEP be a sequence of (nonempty) metric spaces and. as in §3.1. Exercise 7. let d be the metric on the product set X = Xk defined by

n;:l

d(x y) •

=

f: .!...

k=l

dk(Xk. Yk) 2k 1 + dk(Xk, Yk)

for X = (Xk) , Y = (Yk) in X. Then (X.d) is complete (totally bounded. compact) if and only if every (X k•dk) is complete (totally bounded. compact). 11. Every closed subset A of a quasicompact space X is quasicompact. {Hint: {X - A} is an open covering of X - A }. See also §8.1, Exercise 1. f

12. A metric space (X. d) is totally bounded if and only if. for every > 0, X is the union of finitely many sets of diameter ~ f.

6.2. Uniform Convergence, Iterated Limits Theorem When two limiting operations are applied successively to a function (or to a family of functions), the outcome in general depends on the order in which they are applied. The 'iterated limits theorem' is a situation in which the

6. Function Spaces

286

order does not matter, provided that one of the limiting operations takes place 'uniformly' in the sense to be discussed in this section.

6.2.1. Definition. Let T be a nonempty set, (Y, p) a metric space, In: T --> Y (n = 1,2, 3, ...) a sequence of functions on T with values in Y, and let I: T --> Y be another such function. We say that: (i) In --> I pointwise on T (or that I is the pointwise limit of the sequence In) if, for each t E T, In(t) --> I(t) in the metric space Y. This means (cf. 3.2.1) that for each t E T and for every e > 0 there exists an index N = Nt,. (depending on t and e) such that p(Jn(t),/(t» :5 e for all n ~ N; formally, ('It E T) ('Ie> 0) 3 N = Nt,. ;) n ~ N

'*

p(Jn(t),f(t» :5 e.

(ii) In --> I uniformly on T (or that I is the uniform limit of the sequence In) if, for every e > 0, there exists an index N = N. (depending on e) such that

n ~N

'*

p(Jn(t),f(t»:5 e for every t E T;

formally, (Ve>0)3N=N.;) n~N

'*

p(Jn(t),f(t»:5e (VtET).

In the perspective of (i), for every e > 0 an index N can be found that 'works' at every point t E T.

6.2.2. Remark. If In --> I uniformly, it is clear that In the converse is false (Exercise 1).

-->

I pointwise;

6.2.3. Example. The Weierstrass polynomial approximation theorem affirms that if I: la, b) --> IR is any continuous real-valued function on a closed interval, then there exists a sequence of polynomial functions Pn : la, bj --> IR such that Pn --> I uniformly on la, b]; M.H. Stone's generalization of Weierstrass' theorem will be proved later in this chapter (§6.9). 6.2.4. Definition. With notations as in 6.2.1, the sequence of functions In : T --> Y is said to be pointwise Cauchy on T if, for each t E T, the sequence (Un(t» is Cauchy in (Y,p); formally, ('It E T) ('Ie> 0) 3 N = Nt,. ;) m, n ~ N

'*

p(Jm(t),fn(t» :5 e,

also expressed by saying that, for each t E T, p(Jm(t),fn(t» m,n

-->

0 as

-+ 00.

The sequence Un) is said to be uniformly Cauchy on T if, for every e > 0, there exists an index N such that for m, n ~ N, p(Jm(t),fn(t» :5 e for all t E T, equivalently, SUPtET p(Jm(t), In(t» :5 e whenever m, n ~ N; formally, ('Ie >0) 3N=N. ;) m,n~N

'*

p(Jm(t),fn(t»:5e (VtET),

§6.2. Uniform Convergence

287

also expressed by writing supp(Jm(t),fn(t»

->

0

as m,n

-> 00.

'ET

6.2.5. Remark. A pointwise convergent (uniformly convergent) sequence of functions is pointwise Cauchy (uniformly Cauchy). We thus have the diagram of implications uniformly convergent

,/

'\.

pointwise convergent

uniformly Cauchy

'\.

,/ pointwise Cauchy

The following lemma leads up to a 'Cauchy criterion' for uniform convergence. 6.2.6. Lemma. With notations as in 6.2.1, the following conditions are equivalent: (a) Un) is uniformly convergent; (b) Un) is uniformly Cauchy and pointwise convergent.

Proof (a) =} (b): Noted in 6.2.2 and 6.2.5. (b) =} (a): Suppose Un) is uniformly Cauchy and fn By the triangle inequality in Y,

for all indices m, n and for all t E T. Given any N such that (i)

m,n > N

=}

p(Jm(t),fn(t») $

f

f

->

f pointwise.

> 0, choose an index

for all t E Tj

it will suffice to show that (ii)

n ~N

=}

p(Jn(t),f(t») $

f

for all t E T.

Fix an index n ~ N and fix a point t E T. For all m from (.) and (i) that

p(Jn(t), f(t» $ passing to the limit as m (ii). 0

-> 00,

f

~

N , it follows

+ p(Jm(t), f(t»);

we have p(Jn(t), f(t» $

f

+

0, whence

6. Function Spaces

288

When the metric space (Y, p) of 6.2.1 is complete (6.1.25), every pointwise Cauchy sequence (In) is pointwise convergent to a function f, namely the function f: T -+ Y defined by

f(t)

= n-oo lim fn(t)

(t E T).

The same is true with "pointwise" replaced by "uniformly": 6.2.7. Proposition. If T is a nonempty set and (Y,p) is a complete metric space, then every uniformly Cauchy sequence fn : T -+ Y (n = 1,2, 3, ...) is uniformly convergent.

Proof This is immediate from the preceding remarks and Lemma 6.2.6.

0

z::..

6.2.8. Example. (Weierstrass M-test) Let Mk be a convergent series with terms Mk ~ 0 and let (Uk) be a sequence of real-valued (or complex-valued) functions defined on a set T such that, for each index k, IUk(t)1 :5 Mk for all t E T. Then the sequence n

Sn=LUk

(n=I,2,3, ...)

k=.

z::..

is uniformly convergent (and the series udt) is said to be uniformly convergent on T). For, if m < n then the inequality

ISm(t) - sn(t)1

=

n

n

L

Uk(t) :5 L

k=m+l

Mk,

k=m+l

valid for all t E T, shows that the sequence (sn) is uniformly Cauchy (hence uniformly convergent by the preceding corollary). It follows that the series IUk I is also uniformly convergent (and one says that the series is 'uniformly and absolutely convergent'). As an application, consider a power series Z:;;"=o aktk ,where an, a., a2, ... is a sequence oheal (or complex) numbers. If c is a real (or complex) number such that the series Z::'o akck converges absolutely, then the series Z::'o akt k is uniformly and absolutely convergent on the closed interval (closed disk) ItI :5 Icl. {Here T = {t: ItI < Icl} and Mk = lakckl·} .

z:

If, in the notations of Proposition 6.2.7, the ranges of the functions in question have finite diameter, then the uniform convergence can be expressed in terms of a metric. It is useful to have a short name for such functions:

6.2.9. Definition. A function f: T -+ Y with values in a metric space (Y, p) is said to be bounded if its range f(T) has finite diameter (6.1.14),

§6.2. Uniform Convergence

289

that is, if sup p(J(s), f(t))

< +00.

8,tET

We write B = B(T, Y) for the set of all such functions f (with the understanding that the concept of boundedness depends on the specific metric p on Y). In applications of this concept, it will be useful to know that the union of finitely many sets of finite diameter has finite diameter: 6.2.10. Lemma. If A and B are sets of finite diameter in the metric space (Y, p), then their union A U B also has finite diameter.

Proof Let M and N be real numbers ;::: 0 such that p(a, a') :::; M

and

p(b, b') < N

for all a, a' E A and all b, b' E B. Fix a pair of points a' E A, b' E B. Given any x, yEA U B , we assert that

p(x, y) :::; M

+ p(a' , b') + N;

this is obvious if x, yEA or if x, y E B, whereas if x E A and y E B then

p(x,y):::; p(x,a /) +p(a',b') +p(b',y):::; M +p(a',b/ ) +N. Here is an important situation in which boundedness comes free of charge: 6.2.11. Proposition. If T is a quasicompact topological space (6.1.3) and (Y, p) is a metric space, then every continuous function f: T - t Y is bounded.

Proof By Theorem 6.1.27, f(T) is a quasicompact subset of Y. If, for every y E Y .. UII is the open ball with radius 1, then (UII}YeY is an open covering of Y, hence of f(T) , therefore f(T) CUll' U ... U U II, for a suitable finite list of points Yl> .•. , Yr of Y; since each UII has finite diameter (:::; 2) it follows from the lemma that f (T) has finite diameter. The next lemmas will show that the uniform convergence of bounded functions can be expressed in terms of a suitable metric; this leads to the construction of many useful complete metric spaces of functions. Until further notice, we fix the notations of Definition 6.2.9: T is a nonempty set, (Y, p) is a metric space, and B = B(T, Y) is the set of all bounded functions f: T - t Y.

6. Function Spaces

290

6.2.12. Lemma.. If f,g E B then

sUPP(f(t),g(t)) < +00. tET

Proof. By assumption, the sets f (T) and g(T) have finite diameter, therefore so does f(T) U g(T) (6.2.10); it follows that sup p(f(s), g(t))

< +00,

8.teT

and the inequality of the lemma is obtained by specializing to s

= t.

¢

6.2.13. Definition. With notations as in the lemma, we define D(f,g)

= supp(f(t),g(t») tET

(sometimes also denoted Doo(f, g». In view of the next lemma, we may call D the sup-metric on the set B: 6.2.14. Lemma. With the foregoing notations, D is a metric on the set B of bounded functions.

Proof. At any rate, the values of D are nonnegative real numbers, and D(f, g) = D(g, f) by the symmetry of the metric p. Similarly, D(f, f) = 0 by the corresponding property of p. If f # 9 then f(t) # g(t) for at least one point t, whence D(f,g) > O. Finally, if f,g, hE B then p(f(t),h(t)) $ p(f(t),g(t)) + p(g(t),h(t)) $ D(f,g) + D(g,h) for all t E T, whence D(f,h) $ D(f,g) + D(g,h). Thus, D meets the requirements (i)-(iv) of Definition 3.1.4. ¢

6.2.15. Remark. If fn 6.2.4 and 6.2.13 that

E

B (n

= 1,2,3, ...) it is clear from

(fn) is uniformly Cauchy

0 as m, n

Definitions ->

00,

and if fEB then

fn

->

f uniformly on T

0 as n

->

00.

Thus, for sequences of bounded functions, 'uniformly Cauchy' and 'uniformly convergent' are nothing more than the concepts of 'Cauchy' and 'convergent', respectively, in the metric space (B, D). When the metric space (Y, p) is complete, so is the metric space (B, D): 6.2.16. Theorem. If T is a nonempty set, (Y, p) is a complete metric space, B = B(T, Y) is the set of all bounded functions on T with values in Y, and D is the sup-metric on B (6.2.14), then the metric space (B, D) is also complete.

Proof. Assuming (fn) is a sequence in B with D(fm, fn) -> 0 as m, n -> 00, we seek a function fEB such that D(fn, f) -> O. Since

§6.2. Uniform Convergence

291

(fn) is uniformly Cauchy (6.2.15), hence pointwise Cauchy, and since Y is complete, we can define a function f: T -+ Y by f(t) = lim fn(t) n-oo

(t E T) .

By definition, fn -+ f pointwise, therefore fn -+ f uniformly (6.2.6). To complete the proof, we need only show that f is bounded. Choose an index N such that p(JN(t),f(t» :5 1 for all t E T

(possible because fn -+ f uniformly). Since fN is bounded, there is a positive number K such that p(JN(t), fN(t'») :5 K

for all t, t' E T.

Then, for all t, t' E T, p(J(t),f(t'») :5 p(J(t),fN(t»

+ p(JN(t),fN(t'») + P(JN(t'), f(t'»

:51+K+l, thus f(T) has finite diameter.

6.2.17. Example. If Y = lR with the usual metric, then B(T, lR) is also denoted BIR (T) ,and D is the metric

D(f,g) = sup If(t) - g(t)J = lET

IIf - glloo

considered in Example 3.1.10. Since lR is complete by Cauchy's criterion (1.8.26), by the preceding theorem the set BIR(T) of all bounded realvalued functions defined on the set T is complete for the sup-metric. The same is true for the set Bc(T) = B(T, q of bounded complex-valued functions. When the set T is a topological space, we can consider functions f : T -+ Y that are continuous (at particular points, or on all of T). The following theorem shows that the property of continuity is preserved under uniform limits: 6.2.18. Theorem. Let T be a topological space, (Y, p) a metric space, f and fn (n = 1,2,3, ...) functions on T with values in Y, and let c be a point of T. If (1) each fn is continuous at c, and (2) fn -+ f uniformly on T, then f is also continuous at c. Proof The following argument recurs so frequently that it might be called the 'classical t/3 proof'. Given any t > 0, we seek a neighborhood

V of c such that p(J(t), f(c» < t

for all t E V

292

6. FUnction Spaces

(cf. 3.4.3). By the uniformity of the convergence, there exists an index N such that

P(JN(t), f(t)) $ E/3 for all t E T. It follows that for every t E T ,

p(J(t),f(c)) $ P(J(t),fN(t)) + p(JN(t),fN(C)) + P(JN(C),f(C)) $ E/3 + P(JN(t),fN(C))

+ E/3.

By the continuity of f N at c, there exists a neighborhood V of c such that, for every t E V, P(JN(t),fN(C)) $ E/3, whence, by substitution in the preceding inequality,

p(t(t), f(c)) $ E/3 + E/3 + E/3

for all t E V.

6.2.19. Corollary. Let T be a topological space, (Y,p) a metric space, T3 = T3(T, Y) the set of all bounded functions on T with values in Y, and let c be a point of T. Then, the set

{f E T3: f is continuous at c} is a closed subset of T3 for the sup-metric. Proof This is immediate from the preceding theorem and Remark 6.2.15. 6.2.20. Definition. If X and Y are topological spaces, we write C(X, Y) for the set of all continuous functions f: X -+ Y. When Y = JR or Y = iC (with the usual absolute-value metric) we also write CUl(X) = C(X, JR) and CdX) = C(X, iC) . 6.2.21. Corollary. Let X be a topological space, (Y, p) a metric space, T3 = T3(X, Y) the set of all bounded functions equipped with the sup-metric D (6.2.14), and C = C(X, Y) the set of all continuous functions (6.2.20). Then: (1) T3 n C is a closed subset of the metric space T3. (2) If (Y, p) is a complete metric space then the set T3 n C of bounded continuous functions, equipped with the sup-metric, is also a complete metric space.

Proof (1) The set

T3nc

=

n {f E T3: f

is continuous at x}

zEX

is, by the preceding corollary, the intersection of a family of closed sets in T3, hence is itself a closed set (3.2.8). (2) If, moreover, Y is complete, then so is T3 (6.2.16), hence so is its closed subset T3 n C (6.1.29).

§6.2. Uniform Convergence

293

6.2.22. Corollary. With notations as in the precedin9 corollary, assume in addition that X is quasicompact. Then: (1) C is a closed subset 01 8. (2) II (Y, p) is a complete metric space then the set C, equipped with the sup-metric, is also a complete metric space.

Prool. By the quasicompactness of X, C c 8 (6.2.11), so the present corollary simply restates the conclusions of the preceding one. 6.2.23. Examples. If X is quasicompact then the spaces CIl(X) and

Cc(X) are complete for the sup-metric III - 91100 . In particular, for every closed interval [a, b) the spaces Cil [a, b] and Cc[a, b] are complete for this metric. The next theorem is important for integration theory in topological spaces 1; it is a (rare!) situation in which pointwise convergence implies uniform convergence: 6.2.24. Theorem. (Dini's theorem) Let X be a quasicompaet space, I E C = CIl(X) and In E C a sequence such that In -+ I pointwise on X. If. moreover, 1I:5 h :5 fa :5 . .. , then In -+ I unilormly on X.

Proof. By the preceding corollary, C is complete for the sup-metric D(9, h) = 119 - hll oo . Let 9n = I - In ; we have 9n 10 pointwise, and the problem is to show that 9n -+ 0 uniformly, that is, 119nll00 -+ O. Given any £ > 0 , we seek an index N such that 119n 1100 :5 £ for all n ;::: N. For each x E X, 0:5 9n(x) < £ ultimately, that is, x E 9,:;-1(-£,£» ultimately. Each of the sets Un

= 9;1(_£,£»

(n

= 1,2,3, ...)

is open in X by the continuity of 9n, and U 1 C U 2 C U3 C ... because 91 ;::: 92 ;::: 93 > .... By the preceding remark, each x E X belongs to some (hence to all subsequent) Un, thus 00 X= Un.

U

n=l

Since X is quasicompact,

X = Un.

U ... U

Un.

for a suitable finite set of indices; writing N = max{ nlo' .• ,n,.} , we have X = UN (because the sequence Un is increasing) and X = Un for all n;::: N (for the same reason). Thus, if n;::: N then X = 9;1(-£,£», in other words 9n(X) < £ for all x E X, whence 119nll00 :5 £ (in fact, < £).

E. Hewitt and K. Stromberg, Real and abslracl analysis [Springer-Verlag, New York, 1965), p. 115, (9.6) and p. 205, (13.40). 1 Cf.

6. Function Spaces

294

The next application of uniformity gives conditions under which the order of two limiting operations can be reversed; the basic setup is the one already employed in the definition of the limit of a function (3.5.1): 6.2.25. Theorem. (Iterated limits theorem) Let (X, d) be a metric space, (Y, p) a ccmplete metric space. Suppose

ACBCX, cEA,

(i) and let

(ii)

f : B -+ Y

fn: B -+ Y (n

and

be functions such that fn (iii)

f uniformly on A, that is,

-+

fnlA

-+

flA

uniformly.

Finally, suppose each fn has a limit Yn values in A, that is (cf. 3.5.1),

(iv)

3

lim

x-c, xEA

= 1,2,3, ... )

fn(x)

= Yn

EY

as x approaches c through (n

= 1,2,3, ...).

Then: (1) The sequence (Yn) is convergent in Y, and (2) f has a limit equal to limn_co Yn as x approaches c through values in A, that is,

3

lim

x_c, xEA

f(x)

= n-oo lim Yn;

thus,

lim

x_c, xEA

(lim fn(x») n-oo

= n_oo lim ( x_c, limxEA fn(x»).

Proof. It is the latter formula that gives the theorem its name. It is helpful to have a picture of the setup underlying (iv):

X U

In

B

Y

3

Yn

U C

E

A

~

A

We consider two cases, according as

C

does or does not belong to A.

case 1: c EA. In this case, statement (iv) says that for each n, Yn = fn(c) and fnlA is continuous at c (for the restricted metric dlA x A; cf. 3.5.2). Then Yn = fn(c) -+ f(c} and, by Theorem 6.2.18, flA is continuous at c, thus

3

lim

x_c, xEA

f(x)

= f(c) = n-oo lim Yn'

§6.2. Uniform Convergence

295

case 2: c~ A. (I) Since Y is complete, to prove (I) it suffices to show that the sequence (Yn) is Cauchy. Given any e > 0, we seek an index N such that P(Ym, Yn) < e for all m, n ;::: N. By (iii) there exists an index N such that

'*

m. n ;::: N

P(Jm{x),/n{x))::; e for all x E A

(cf. 6.2.6). Fix a pair of indices m, n > N. For all x EA.

(*)

P(Ym,Yn) < P(Ym' 1m(x)) + p(Jm{x),fn{x)) + p(Jn{X),Yn)) ::; P(Ym./m{x)) + e + P(Jn{x),Yn)'

Since C E A we may choose a sequence (Xk) in By (iv), P(Ym./m{Xk)) - 0 and P(Yn,/n{Xk)) Replacing x by Xk in (*) and letting k - 00, 0+ e + O. as we wished to show. (2) Let Y = limn_co Yn and, for each n, define the formula {Yn In (x) = In (x)

A such that Xk - c. 0 as k - 00 (3.5.1). we have P{Ym, Yn) ::;

In: Au {c} - Y by

forx=c for x E A.

If XkEAU{C} and Xk-C then In{Xk)-Yn=/n{c) as k-oo. thus f n is continuous at c. Similarly. define f: A U {c} - Y by _ I{x)

= {Y f{x)

forx=c for x E A.

Since InJA - IIA uniformly and In{c) = Yn - Y= I{c) , it is clear that In - I uniformly on Au {c}; by case 1, 3

lim %_C,

and, since

xEAU{c}

I{x) = I{c) = Y.

I = I on A we see that 3

lim

x-c. :z:EA

f{x) = y.

6.2.26. Corollary. Let In : [a. b] - lR (n = 1,2,3, ... ) and I: [a, b] - lR be lunctions such that In - f uniformly on [a, b], and let a::; c < b. II every In has a right limit at c then so does I. and I (c+ ) = limn_co fn{c+) .

Proof Here X = B = [a,b] and A = (c,b]. We know that In uniformly on A and, for every n, 3

lim

x-c, xEA

In{x) = In{c+) E lR.

I

6. Function Spaces

296

so by the iterated limits theorem, the sequence (tn(c+» and

3

lim %-C,

zEA

=

f(x)

lim fn(c+).

is convergent

0

Tl-OO

6.2.27. Corollary. (Term-by-term-differentiation) Let Sn: [a, b]--> IR be a sequence of functions such that, for every n, Sn is differentiable on [a, bl (one-sided at the endpoints). Assume that there exists a function t : [a, b] -+ IR such that s:, --> t uniformly on [a, bl ; finally, assume that there exists a point c E [a,b] such that the sequence (Sn(c» is convergent. Then, the sequence (sn) is uniformly convergent on [a, b], the limit function S = lim Sn is differentiable on [a, b] (one-sided at the endpoints) and s' = t. Thus,

(lim sn)'

= Iims~.

Proof. Note that each Sn is continuous (even differentiable) on [a,b]. Convention: we are writing s:'(a) for the right-derivative (sn)~(a), and s~(b) for the left-derivative (sn);(b). Since the space C = Cllda,b] is complete for the sup-metric (6.2.23), to prove that (sn) is uniformly convergent we need only show that it is uniformly Cauchy. Let f > O. Since the sequence of derivatives (s:') is uniformly Cauchy (indeed, uniformly convergent to t) there is an index N such that

(i)

m,n

~

N

=*

Is:"(x) -

s~(x)1

:::; 2(b ~ a)

for all x E [a,b].

While we are at it, we can suppose that also m, n ~ N

(ii) Fix m, n

(.)

~

=*

ISm(c) - sn(c)1 :::;

l

"2 .

N; it will suffice to show that ISm (x) - sn(x)1 :::;

l

for all x E [a, b].

Fix x E [a, b] . If x = c the inequality of (.) holds by (ii). Suppose, for example, that x > c (if x < c the argument is similar). By the mean value theorem applied to the function Sm - Sn on the interval [c, x] , there exists a point {E (c, x) such that (Sm - sn)(x) - (sm - sn)(c) = (sm - sn)'({) . (x - c);

it then follows from (i) that

(iii)

I(Sm - sn)(x) - (sm - sn)(c)1 :::; 2(b ~ a) . (x - c) l

:::; 2(b _ a) . (b - a)

f

= "2 '

§6.2. Uniform Convergence

297

thus I(sm - sn)(x)1 :5 I(sm - Sn)('!=) - (sm - sn)(c)l l

+ I(sm -

sn)(c)1

l

p(J(x), f(x')) < f,

that is, any two points in X that are within 6 of each other have images in Y that are within f of each other. Formally,

(Vf > 0) 36> 0 3

x,x' E X, d(x, x') < 6 => p(J(x),f(x'))

0, we are to show that p(J(xn , f(x~» < £ ultimately. Choose {j > 0 as in 6.3.2, then choose an index N such that d(xn,x~) < {j for all n 2: N; by the choice of {j, p(J(xn ), f(x~» < £ for all n 2: N. (b) ~ (a): We argue contrapositively: assuming f is not uniformly continuous, let us construct a pair of sequences (X n ) , (x~) in X for which the implication in (b) fails. By assumption, there exists an £ > 0 such that, for every {j > 0, the implication of Definition 6.3.2 fails; in particular, for each positive integer n, the implication fails for {j = lin, thus there exists a pair of points X n , x~ in X such that d(xn,x~) < lin but p(J(x n ), f(x~» 2: £. Then d(x n , x~) ..... 0 but p(J(x n ), f(x~» f> o. 0 The function f: (O,I] ..... 1R defined by f(x) = sin(llx) is continuous on its domain, but the right limit f(O+) fails to exist (for example, there are sequences X n ..... 0 and Yn ..... 0 in (0,1] with f(x n ) = 0 and f(Yn) = 1 for all n); thus, no definition of f(O) can render f continuous on the closed interval [0, 11. It is easy to see directly that f fails to be uniformly continuous, but the following theorem-a capital application of uniform continuity-gives an interesting roundabout way of seeing it. 6.3.5. Theorem. If f : A ..... Y is a uniformly continuous function defined on a dense subset A of a metric space (X, d) and taking values in a complete metric space (Y, p), then there exists a unique continuous function 7 : X ..... Y such that flA = f; moreover, the function 7 is uniformly continuous.

Proof. Uniqueness. Assuming 9 and h are continuous functions from X into Y whose restrictions to A are equal to f (hence to each other), we are to show that 9 = h on X. Writing

= {x EX: g(x) = h(x) }, and that 9 = h on B; we are to show that B

we know that A c B The set B is closed in X; for, if continuity of 9 and h,

g(x)

Xn

E Band x n .....

B =X. x E X then, by the

= lim g(x n ) = lim h(x n ) = h(x) ,

so that x E B. It follows that X = A c B, whence 9 = h on X.

6. Function Spaces

302

Existence. The function f: X graph G (cf. 1.3.1). Let G

= ((x,y) E X x

--+

Y will be defined by specifying its

Y: 3 an E A with

an --+ x and f(a n ) --+ y};

to show that G is the graph of a function, we must show that for each x E X there exists one and only one y E Y such that (x, y) E G. Given any x EX, choose a sequence (an) in A with an --+ x (possible by the density of A); since (an) is Cauchy and f is uniformly continuous, the sequence (J(a n ) is also Cauchy (6.3.4) and therefore convergent in Y, thus (x, y) E G with y = lim f(a n ). Note that if (a~) is any other sequence in A with a~ --+ x, then necessarily f(a~) --+ y; for, d(a~,an)

therefore p{J(a~), f(an))

< d(a~,x) + d(x,an) ..... 0 + 0, --+

0 (6.3.4), thus

p{J(a~), y) $ p{J(a~).!(an))

+ p{J(an), y))

--+

0

+ O.

It follows that y is the only point of Y for which (x, y) E G. We may therefore define a function f: X ..... Y by

f(x)

= y,

where (x,y) E G,

and the above argument shows that if (an) is a sequence in A with E X then f(x) = lim f(a n ). In particular, if a E A and (an) is the constant sequence an = a, then

an ..... x

f(a)

= lim f(an) = f(a) ,

thus ] is an extension of f. To complete the proof, we need only show that f is uniformly continuous on X. Given any E > 0, choose 6 > 0 so that

'*

a,a' E A, d(a,a') < 6

p{J(a),J(a')) < Eo

Assuming x, x' E X with d(x, x') < 6, it will suffice to show that p(J(x),J(x') $ E. Let (an) and (a~) be sequences in A such that an ..... x and a~ -+ x'. From d(x,x') < 6 and the inequality d(an,a~) $ d(a,.,x)

+ d(x, x') +d(x',a~)

we see that d(an, a~) < 6 ultimately, therefore p{J(an).!(a~)) ultimately (by the choice of 6). Thus, for all sufficiently large n,

P(J(x),J(x') < p(J(x),f(an)


0 small enough so that the closed ball Br (x) with radius r and center x is contained in V, and such that r < 1/2; then the open ball Ui = Ur(x) has diameter :5 1 and U i C Br(x) C V.} Since Ai has empty interior, it cannot contain U i , thus the open set Ui n (X - Ai) is nonempty. Arguing as above, there exists an open set U2 such that 0i-U 2 cU2 cU i n(X-A i ) and

diamU 2 :51/2.

Again, A2 cannot contain U2, so that U2 n (X - A 2) is a nonempty open set; let U3 be an open set such that

o i- U3 C U3 C U2 n (X -

A2) and

diam U3 :5 1/3.

Continuing recursively, we construct a sequence of nonempty open sets Uh U2, U3,···, where Un+ i is chosen so that

o i- Un+l C Un+l C Un n (X -

An) and

diam Un+l :5 1/(n + 1).

In particular Un+i C Un C Un; thus, the sequence of closed sets F n = Un satisfies the conditions in (b) of Theorem 6.3.1, consequently

n Un = {x} 00

n=1

for a suitable point x. Then x E U1 C V and, for every positive integer n,

x E Un + i C Un

n(X -

An) C X -

An,

thus x E Vnn::"=l(X-An) = Vn(X-A) and the proofof(l) is complete. (2) Given a meager open set U in X, we are to show that U is empty. By (I), X - U = X; but X - U is closed, whence X - U = X, U = 0. This theorem has a corollary with many important applications in functional analysis: 6.3.10. Corollary. (Uniform boundedness principle) Let (X,d) be a complete metric space and let £ C C(X,IR), that is, £ is a set of continuous real-valued junctions defined on X. Assume that £ is pointwise bounded on X, in the sense that for each point x EX, the set of values £(x)

= {f(x): f

E

£}

is a bounded set of real numbers. Then £ is uniformly bounded on some nonempty open set, that is, there exists a nonempty open set U in X such that the set of restrictions

£IU

= {flU: f E£}

is uniformly bounded (i. e., bounded for the sup-norm).

6. Function Spaces

306

Proof. By assumption, for each point x E X there exists a real number M z > 0 such that

I/(x)1 :5 M z for all fEE:; we seek a nonempty open set U in X and a real number M that

If(x)1 :5 M

> 0 such

for all x E U and for all fEE:,

that is, in the notations of 3.1.10,

IIfIUII"" :5 M

for all fEE:.

For each positive integer n, let

An = {x EX: If(x)l:5 n for all fEE:}

=

n{x

EX: If(x)l:5 n}

fEE

=

n r1(I-n,n)); fEe

from the continuity of the functions fEE:, we see that An is the intersection of a family of closed sets (3.4.5) and is therefore closed. Moreover,

for, if x E X and m is a positive integer such that m

If(x)1 :5 M z :5 m

~

Mz , then

for all fEE:,

whence x E Am. Since X = U::'=l An is a Baire space (6.3.9), hence is nonmeager, there exists an index M such that AM is not rare; if U is a nonempty open set such that U C AM = AM (for example, U = int AM ), then

If(x)1 :5 M

for all x E U and fEE:,

thus the functions fEE: are uniformly bounded on U.

0

Completeness is clearly a useful property for a metric space to have; it is a consolation prize to incomplete spaces that every metric space can be regarded as a dense subspace of a complete metric space: 6.3.11. Theorem. (Completion) If (X,d) is any metric space, then: (Exist~nce) There exist a complete metric space (X, d) and a mapping f : X -+ X such that 1° f is distance-preserving: d(J(x), f(y)) = d(x, y) lor all x, y in X, and 2° f has dense range: I (X) = X.

§6.3. Complete Metric Spaces

307

(Uniqueness) If also (Z. p) is a complete metric space and g: X -+ Z is a distance-preserving mapping with dense mnge, then (Z. p) may be identified with (X. d), in the sense that there exists a distance.preserving bijection h : X -+ Z that carries f(X) into g(X). more precisely, h(J(x)) = g(x) for all x EX. In order that (X. d) be complete, it is necessary and sufficient that f be surjective. Proof A distance-preserving mapping is also said to be isometric (or to be an isometry). Existence. The const.ruction of X from X imitates Cantor's method for constructing the reals from the rationals (§1.8). Let C be the set of all Cauchy sequences s = (x n ) in X; for s = (x n ) and t = (Yn) in C, we write s = t in case x n = Yn for all n. An equivalence relation ~ in C is defined by writing 5 ~ t in case d(x n •Yn) -+ 0 (the transitivity of ~ follows from the triangle inequality for d). We write s = {r E C : r ~ s} for the equivalence class of s in C, and X for the set of all equivalence classes:

X=

C/~ = {s: 5 E C};

thus 5 ...... S is the quotient mapping C -+ X . Given u. v E X we are to define a distance d(u, v) . Say u = s. v = t. where s = (x n ) and t = (Yn) are Cauchy sequences in X. The sequence (d(x n •Yn)) is Cauchy (therefore convergent) in IR. as we see from the inequality (cf. 3.2.3) Id(x m •Ym) - d(x no Yn)1 < d(x m •Xn)

+ d(Ym. Yn) .

To justify defining d(u.v)

= limd(xn,Yn)

we must check that the limit is independent of the particular representatives 5, t of the equivalence classes u,v: if also 1L = s'. v = t' • where s' = (x~) and t' = (y~) • then the equality limd(xn.Yn)

= limd(x~.y~)

follows from the inequality Id(x n •Yn) - d(x~. y~)1

< d(x n •x~) + d(Yn. y~)

and the fact that d(xn,x~) -+ O. d(Yn. y~) -+ O. The function d: X x X -+ IR defined by the formula (*) is a metric on X. For. if 1L = (x n ) , V = (Yn). W = (zn) then. passing to the limit in the relations d(x n •Xn)

= O.

d(x n •Yn) ?: O. d(x n •Yn)

d(x n • Zn) :5 d(x n , Yn)

= d(Yn. Xn ).

+ d(Yn, Zn),

6. Function Spaces

308 we see that

d(u,u) =0, d(u,v)~O, d(u, v) = d(v,u) , d(u,w) $ d(u, v)

+ d(v,w) ,

and if d(u, v) = 0 then d(x n , Yn) -> 0, whence (x n ) ~ (Yn) , u = v. If x E X we write (x) for the constant sequence x, x, x, ... ; the mapping f: X -> X defined by

= (x)

f(x)

(x E X)

is isometric since, for x, Y EX, d(J(x),J(y» = limd(x, y) = d(x, y). In particular, f is injective; for, if f(x) = f(y) then d(x, y) = d(J(x), f(y» = 0, whence x = y. To prove that f(X) is dense in X, we need only observe that if u = (x n ) E X then f(x n ) -> U; indeed, if E > 0 and N is chosen so that d(x m , x n ) $ E for all m, n ~ N, then

m~ N

=}

d(J(x m ), u) = lim d(x m , Xk)
u. Consider first the special case that Un E f(X) for all n. Then Un = f(x n ) for a unique X n EX. Since f is isometric, the sequence (x n ) is also Cauchy; writing u = (x n ) , we have Un -> U by the argument in the preceding paragraph. In the ~eneral case, for each index n we may choose a point V n E f(X) with d(vn , un) < l/n. The sequence (vn ) is also Cauchy, since

+ d(Urn, un) + d(Un, Vn) < lim + d(Urn, Un) + lin -> 0 as m, n

d(vm , vn) $ d(vm , Urn)

By the special case considered first, (vn ) has a limit u in follows from the computation d(Un, u) $ d(Un, Vn)

-> 00.

X, and

Un -> u

+ d(vn , u) < lin + d(vn , u) -> o.

Uniqueness. Suppose (Z, p) and g: X -> Z have the properties indicated in the statement of the theorem; we then have a diagram id X X

h

Z

§6.3. Complete Metric Spaces

309

for which we seek an isometric bijection h: X -+ Z such that h(f(x» = g(x) for all x EX. {The completed diagram will then be 'commutative' in the sense that the two ways of getting from X to Z will coincide: h 0 f = 9 = go id.} We be/pn by defining a mapping he: f(X) -+ Z on the dense subset f(X) of X, by means of the formula he(J(x»)

= g(x)

(x E X);

since x is uniquely determined by f(x) , there is no ambiguity in the definition of he (one says that he is well-defined). The mapping he is isometric since, for all x, Y EX, p(he(f(x»),he(f(y»))

= p(g(x),g(y» = d(x,y) = d(f(x),f(y»).

In particular, he is uniformly continuous; since it takes values in a complete metric space, it is extendible to a uniformly continuous mapping h: X -+ Z (6.3.5). In fact, h is isometric; for, if u, v E X and (x n ), (Yn) are sequences in X such that f(x n ) -+ u and f(Yn) -+ v, then h(J(x n »)-+ h(u) and h(f(Yn» -+ h(v) by the continuity of h, thus d(u, v)

= lim d(f(x n ), f(Yn» = limd(xn,Yn) = lim p(g(xn ), g(Yn»

= lim p ( he(f(xn»,ho(f(Yn») = lim p ( h(f(xn»).h(f(Yn») = p(h(u), h(v»). By assumption, h(f(X» = he(f(X») = g(X) is a dense subset of Z, so all the more h(X) is dense in Z. However, X is complete and h is isometric, so h(X) is a complete metric subspace of Z and is therefore a closed set in Z (6.1.29); h(X) = h(X) = Z, so h is surjective. {Note, incidentally, that h is uniquely determined by the property h 0 f = g, since h is continuous and f(X) is dense in X.} Finally, if X is already complete then f(X) = X by the argument used in proving h surjective. Conversely, if f is surjective then it is an isometric bijection, so the completeness of X implies that of X. 6.3.12. Definition. With notations as in the preceding theorem, the complete metric space (X, d) is called the completion of the metric space (X, d), and the isometric mapping f: X --> X is called the embedding of X into X. 6.3.13. Remarks. Two metric spaces that are connected by an isometric bijection-such as (in the above notations) the set X equipped with the metric d and the set f(X) equipped with the restricted metric

310

6. Function Spaces

dlf{X) X f{X)-are essentially 'equal' as metric spaces. They are candidates for 'identification': t~row away (X, d) ,retain (J{X), dlf{X) x f{X)) , and rename f{X) and dlf{X) x f{X) to be X and d. The net effe~t !s that (X, d) can be regarded as a dense subset of a complete metric (X, d) such that the metric d is obtained by restricting d to pairs of points of X; after this identification, the embedding mapping of the preceding definition becomes the insertion mapping i: X ---+ X, i{x) = x (V' x E X) .

Here are a few applications of the concept of completion: 6.3.14. Proposition. A metric space is totally bounded if and only if its completion is compact. Proof. By the preceding remarks, we may regard the given metric space (X, d) as a dense subset of a complete metric space (Z, p), with d = piX x X. If Z is compact then it is totally bounded (6.1.12), therefore every subspace of Z-in particular X-is totally bounded. (Ce. 6.1.24 for a proof by overkill. Alternatively, given any e > 0, write Z = Al U ... U An with diam Ai :5 e/2 for all i, then consider the sets Ai () X .) Conversely, if X is totally bounded then its closure Z is also totally bounded: given any e > 0, choose an e/2-net F for X; each point of Z is within f/2 of some point of X, hence within f of some point of F. Thus Z is complete and totally bounded, therefore compact (6.1.26). 0

6.3.15. Proposition. Let (X,d) be a complete metric space and let A be a nonempty subset of X, regarded as a metric subspace of (X, d) (6.1.28). Then the completion of the metric space A may be identified with its closure in X (concisely, A = A ). Proof. Let us write d A and d X for the restrictions of d to A x A and A x A, respectively. The metric space (A, d X ) is complete (6.1.29); moreover, A is a dense subset of A, so by the "uniqueness" part of Theorem 6.3.11, the insertion mapping f: A ---+ A, f{a) = a (V' a E A) , is extendible to an isometric bijection A ---+ A. 0

6.3.16. Corollary. A subset A of a complete metric space X is a totally bounded metric subspace of X if and only if its closure A is compact. Proof. We have the equivalences

A totally bounded

$}

A

compact

$}

A compact

by Propositions 6.3.14 and 6.3.15, respectively. 0 The rest of the chapter is devoted to some important examples of complete metric spaces.

§6.4. L'

311

Exercises 1. Let X = (0,1] with the usual metric and let F n = (O,I/n) (n = 1,2,3, ... ). The F n form a decreasing sequence of nonempty closed sets in X with diam F n --+ 0, but F n = 0. There is no conflict with Theorem 6.3.1.

n

2. Condition (b) in the theorem on nested closed sets (6.3.1) requires that (i) the F n are closed sets, (ii) F n 1, and (iii) diamF n --+ O. If any of these three conditions is omitted, then the implication (a) ~ (b) fails. {Hint: In X = IR with the usual metric, contemplate the sequences (i) F n = (0, lIn), (ii) F n = [n, n + I/nl, (iii) F n = In, +00) .} 3. In a compact metric space (which is complete, by 6.1.26) there is a shorter proof of the 'nested closed sets property'. {Hint: 6.1.5.} 4. (i) The function f: IR --+ IR defined by f(x) = x 2 is continuous but not uniformly continuous (for the usual metric of IR). (ii) For a bounded example, contemplate the function f: (O,IJ --+ IR,

= sin(I/x). 5. If X = Y = IR with the usual metric and f: IR --+ IR is the function f(x) = x 2 , then f satisfies condition (c) of 6.3.4 (because IR is complete f(x)

and f is continuous) but f is not uniformly continuous.

6.4. L' Topology, measure theory and functional analysis grew up together in the first half of the 20th century, so it is not surprising that some of the most useful examples of metric spaces of analysis are based on integration over a measure space. This section is devoted to such an example, a complete metric space associated with the class of integrable functions relative to a measure space. For the rest of the section, (X,S,J.L) is a fixed measure space; thus, X is a set, S is a IT-algebra of subsets of X, and J.L is a measure defined on the IT-algebra S (2.4.12).

We begin by bringing complex-valued functions on board: 6.4.1. Definition. A function f : X --+ C is said to be measurable (with respect to the IT-algebra S) if its real and imaginary parts

1

Im f =2i(f-7) (regarded as functions X --+ IR) are measurable with respect to S in the sense of Definition 4.1.3, in other words, if f = g + ih with g, h: X --+ IR measurable in the sense of 4.1.3.

6. Function Spaces

312

It is clear from the definition that the correspondence 9 ..... 9 + Oi maps the set of all measurable functions 9 : X -> IR onto the set of all measurable functions f: X -> iC that are real-valued (that is, for which f(X) C IR) and that this correspondence preserves the algebraic operations (pointwise sums, products, and scalar multiples by real scalars). 6.4.2. Proposition. If f, 9: X -> iC are measurable, c E iC and then the functions f + 9, cf, f9 If I" are also measurable.

Q

>0

Proof. Write f = !I + ih and 9 = 91 + i92 with !I, 12, 91 , 92 real-valued, and suppose c = a + ib with a, b E IR. Then f + 9 = (!I + h) + i(91 + 92) , cf = (at. - bh) + i(ah + bh), f9 = (!l91 - 1292) + i(!l92 + h9d ,

IfI" = (1t.1 2 + 1121 2)"/2 ; the real and imaginary parts of the functions on the right side are measurable by Theorems 4.1.9 and 4.1.13. 0 In particular, the measurable complex-valued functions form a vector space over the field of complex numbers. As in the real-valued case, the pointwise limit of a sequence of measurable complex-valued functions is measurable: 6.4.3. Proposition. If (In) is a sequence of measurable complex functions, f: X ---+ iC and f n -> f pointwise, then the limit function f is

also measurable. Proof. Since Refn -> Ref and Imfn is immediate from Corollary 4.1.20. 0

->

Imf pointwise, the assertion

Recall that .c~ = .c~(X,S,tl) denotes the real vector space of realvalued functions f: X ---+ IR that are integrable with respect to tl (4.4.7).

6.4.4. Definition. A complex function f: X -> iC is said to be integrable with respect to the measure tl if its real and imaginary parts are integrable in the sense of 4.4.7, in other words, if Re f, 1m f E .c~. We write .c~

= L:t(X,S,tl)

for the set of all such functions f, and the integral (with respect to tl) of such a function, denoted J fdtl, is defined by the formula

J J f dtl =

(Re f)dtl

+i

J

(1m f)dtl .

If 9 E.c~ then the function f: X ---+ iC defined by f = 9 + Oi belongs to .c~, and the integral J fdtl just defined coincides with J9dtl as

§6.4. L1

313

defined in 4.4.9. Identifying 9 with 9 + Oi, we may regard ,c~ as the set of all f E,Cb with f(X) C JR.

,Cb

6.4.5. Proposition. linear form on it.

is a complex vector space and f

1-+

J fdJ1.

is a

Proof This follows easily from the case of real-valued functions (4.4.11) and the formulas in the proof of Proposition 6.4.2. 6.4.6. Proposition. The following conditions on a complex function f : X -+ iC are equivalent: (a) f E 'cb; (b) f is measurable and If I E ,C~ . Moreover, J fdJ1.1 ~ J IfldJ1. for every f E 'cb·

I

Proof (a) => (b): Assuming f E 4, write f = g+ih with g, h E'c~ (6.4.4). Then f is measurable (6.4.1), therefore so is If I (6.4.2), and

1/1 = (l + h 2 )1 / 2

:=:;

Igi + Ihl

(to check the inequality, square both sides); since (4.4.11), so is If I (4.4.5).

Igi + Ihl

is integrable

I

To verify the asserted inequality, write J fdJ1.1 = c J fdJ1. with lei = 1 (if J fdJ1. = 0 let c = 1; if 0 = J fdWF 0 let c = 101/0). Say c = a+ib (a, bE JR) . Then

!

fdJ1. = =

! !

cfdJ1. = ![(a g - bh) + i(ah + bg)]dJ1. (ag - bh)dJ1. + i

fdJ1.

= = :=:;

(ah

+ bg)dJ1.;

J(ah + bg )dJ1. = 0

since the left side is real, necessarily

!

!

! ! !

and

(ag - bh)dJ1. (ag - bh)dJ1. lag - bhldJ1.

(cf. 4.4.12). Then

lag - bhl :=:; Ja 2 + b2. J g2 + h 2 = Icl·lfl = If I by Cauchy's inequality in JR2 (or by 3.1.12 with p = 2), applied after evaluating the functions at each point of X; combining the preceding inequalities, we have

!

fdJ1. :=:;

!

lag - bhldJ1.:=:;

!

IfldJ1..

6. Function Spaces

314

(b) => (a): With / as in (b), write / = g+ih, where 9 = Re/ and h = 1m /. By hypothesis, 9 and h are measurable (6.4.1); moreover,

therefore 9 (6.4.4). 0

E £~

(4.4.5, 4.4.12) and similarly h

E £~,

In particular, £~ is closed under the operation f ......

thus /

E £~

1/1 .

6.4.7. Definition. For / E £~ we write II/III =

J

I/Id/lo,

called the £I-norm of /. 6.4.8. Proposition. If /, 9 E £~ and c E C, then (1) II/III ~ 0, (2) lie/lit = 1c11l/111 , (3) II/ + gill II/III + IIglll .

:s

Proof. Properties (1) and (2) are clear from the definition, and (3) follows

from integrating the inequality

If + gl 5 1/1 + Igl· 0

6.4.9. Definition. A real-valued function x ...... IIxll defined on a (real or complex) vector space V is said to be a seminorm if it satisfies the

conditions (1)-(3) of the above proposition (with e restricted to real scalars in the case of a real vector space). If, in addition, (4) IIxll > 0 for every nonzero vector x, then the function x...... IIx II is called a norm on V, IIxll is called the norm of the vector x, and the pair (V,III1)-V equipped with the norm function x ...... IIxll-is called a normed vector space (or, briefly, a normed space). 6.4.10. Remark. If x ...... IIxll is a seminorm on the vector space V, then

the formula d(x, y) = IIx - yll defines a pseudometrie on V (3.1.4); in particular, the triangle inequality for d follows on applying property (3) of the preceding definition to the sum x - y = (x - z) + (z - y). The correspondence / ...... II/III of 6.4.7 is a seminorm on £~ (6.4.8); it is not in general a norm, since II/III = 0 whenever / = 0 a.e. (4.4.21), which does not in general imply that / = O.

6.4.11. Definition. We write N = N(X,S,/lo) for the set of all measurable functions / : X --+ C such that / = 0 a.e. (equivalently, Re f = o a.e. and 1m / = 0 a.e.). Such functions are called null functions (rela,. tive to the given measure space); they are the integrable functions whose absolute value has integral 0:

§6.4. LI

315

6.4.12. Proposition. The set N

01

01 null lunctions is a linear subspace

.cf;, and

N

= {j E .cf;:

1I/11i

= O}.

Proof. If lEN then III is measurable (6.4.2) and Ifl

=

0 a.e.,

therefore III is integrable and J I/ld/-' = 0 (4.4.21). Thus I E Cl; (6.4.6) and 1I/11i = O. Conversely, if I E.cl: and J I/ld/-' = 0, then III = 0 a.e. (4.4.21), whence lEN. Thus the stated formula for N is verified. In particular, N contains the zero function. If 1,9 E N and c E iC then III + 911i :5 1I/11i + IIgll1 = 0 and IIc/lli = lei 1111" = 0, therefore I + 9, cl EN; thus N is a linear subspace of .cf;. In particular, for

I, 9 E.cf; ~

1=9 a.e.

we have

I-gEN

~ III-gill =0.

The vector space .cl:, equipped with the pseudometric (f, g) ...... III - 9111 derived from the seminorm I ...... 11/111 (d. 6.4.10), is 'complete' in the following sense:

6.4.13. Theorem.

II (fn) is a sequence in 111m -

In II I

---+

0

.cl: such that

as m,n

---+ 00

(such a sequence is said to be 'Cauchy in mean' or 'fundamental in mean '), then there exists a function I E.cf; such that II/n - llli

---+

0

as n

---+ 00

(and one says that In ---+ I 'in mean '). II also 9 E then 1=9 a.e. (so to speak, I is 'a.e. unique ').

cf; and

IIIn - 9111

---+

0,

Prool. Passing to a subsequence (d. the proof of 6.1.26), we can suppose that II/n+l Write

10 = 0

Inl" :5 r

n

(n

= 1,2,3, ...).

and, for every positive integer n, define

hn

n

n

k=1

k=2

= L Ilk - Ik-II = Ihl + L Ilk - Ik-II·

The functions hn are integrable (§4.4) and form an increasing sequence,

6. Function Spaces

316 h"

T;

moreover, the sequence of integrals

!

h"dJl

= IIMl +

t

f h"dJl

is bounded,

IIIk - Ik-llh

k=2

n-l = IIhlll

+

E IIlk+l -

/kill

k=l n-l

:5 IIMl +

E 2-

k

< IIhlh + 1 ,

k=l so by the monotone convergence theorem (4.5.3) there exists an integrable function h such that h" T h a.e. Let E be a null set on whose complement hn(x) Th(x); replacing In by 'PCEIn, we can suppose that hn(x) Th(x) for every point x of X. Note that n

IInl

=

n

E

EUk - Ik-l) :5 Ilk - Ik-ll k=l k=l

= h" :5 h

for all n; we will show that the sequence Un) is pointwise convergent. For every positive integer n, write Sn = In - In-l ; then, for every point

x E X, n

E ISk(x)1 = hn(x) :5 h(x) , k=l so the series E~l Sk(X) is convergent (absolutely). Deline a function I : X -+ iC by the formula

n

= n-oo lim E[!k(x) -

Ik-l(x))

k=l

thus In

-+

= n_oo lim fn(x),

I pointwise on X; in particular, I is measurable (6.4.3) and II(x)1 =

00

00

k=l

k=l

E Sk(X) :5 E ISk(x)1 :5 h(x)

for all x E X. It follows from III :5 hE.c1t that III E.clt (4.4.19), therefore IE.ct (6.4.6); and, for every positive integer n, lIn - II :5 IInl

+ III :5 IInl + h :5 2h E .cit;

since lIn - II -+ 0 pointwise on X, it follows from the dominated convergence theorem (4.5.4) that f lIn - IldJl -+ o. This proves the existence of a function I E.ct such that IIIn - 1111 -+ O.

§6.4. L'

317

If also 9 E.c1; and IIIn -

gil,

-+

0, then

III - gill $ III - Inllt therefore III - gill a.e.

= 0, whence I -

+ IIIn - gilt 9

=0

-+

0,

a.e. (6.4.12), that is,

I =9

The message of the preceding theorem is that the vector space 4, equipped with the pseudometric dU,g) = III - gilt, is 'complete' in the sense that if a sequence Un) in .c1; satisfies the Cauchy condition dUm, In) -+ 0, then dUn, J) -+ 0 for a suitable IE.c1;; the 'limit function' I is 'a.e. unique', in the sense that if also 9 E.c1; and dUn, g) -+ 0, then the difference function I - 9 belongs to the linear subspace N of null functions in .c1; (6.4.12). To pass from the pseudometric space (.c1;, d) to a complete metric space, we need only pass to the quotient vector space .cVN :

6.4.14. Definition. The quotient vector space

.cVN

is denoted

L1;(X, S, IL) , briefly L1;. If u E L1;, say u = I + N, we write also u = j; if also u = !I-that is, if I - 9 EN-then I = 9 a.e. (6.4.11), so we may define

which depends only on the coset u and not on the particular function that represents it. Thus, by definition, IIjlll = IIIIIt .

I

6.4.15. Proposition. The correspondence u ...... lIulh ol 6.4.14 defines a norm on L1;; that is, il u, v E Ll: and c E C, then (1) lIulll;::: 0, (2) II cull I = Icillulit , (3) lIu + vllt $ lIullt + IIvlll' and (4)lIulll>0 *> ui'O. Proof (1)-(3) are immediate from 6.4.8. (4) Say u = j; then lIulh = II III I ,so

lIullt thus lIulh

=0

=0

*>

*>

IIIIII

=0

*>

lEN (6.4.12)

*>

I = 0;

u = 0, whence (4).

In other words, (L1;, II II tl is a normed space (6.4.9). By the same arguments as in §3.1, the formula d( u, v) = lIu - vllt defines a metric on L1;; in fact, it is a complete metric:

6.4.16. Proposition. II (un) is a sequence in L1; with lIum - Unlll

-+

0 as m,n -+

then there exists u E L1; such that II Un -

U

III

-+

00,

0.

6. FUnction Spaces

318

Proof. Immediate from 6.4.13.

6.4.17. Definition. A (real or complex) normed space that is complete for the metric (x, y) .... IIx - yll deduced from the norm is called a Banach

space. Propositions 6.4.15 and 6.4.16 can be expressed as follows: 6.4.18. Theorem. For every measure space (X,S,JL) , the quotient vector space

Lf: = .cf:(x,S,JL)/N{X,S,JL) is a Banach space for the norm

u .... lIulll

of 6.4.14.

The superscript in the symbol Lf: alludes to the fact that it is the first power of III that is being integrated when one defines lIilll = IIflll = J IfldJL. In §6.7 below, consideration of IfI P , p > 1, leads to Banach spaces L~ analogous to (indeed, generalizing) the finite-dimensional "Minkowski spaces" of 3.1.11. We conclude this section with the complex analogue of a result in §4.7: 6.4.19. Proposition. Let A c S be an algebra of sets such that the 0' -algebra generated by A is S. If f, g E.cf: and if

i then f

=g

fdJL

=

i

gdJL

for all A E A,

a.e.

Proof. If E E S then 'PEl E.cf: follows from the case of real-valued functions (4.4.18) and Definition 6.4.4; following 4.4.22, one defines JEfdp. to be J 'PE f dp. . Assuming f, g E.cf: satisfy the condition of the hypothesis, let h = I - 9 ; our assumption is that

i

hdp.=O

for all AEA,

and we seek to show that h = 0 a.e. Writing h = u + iv with u, v E .c~ , it is clear from Definition 6.4.4 that it suffices to consider the case that h is real-valued; but then h = 0 a.e. by Corollary 4.7.4.

Exercises 1. If V is a vector space (real or complex) and if x .... IIxll is a seminorm on V (6.4.9), then the set N = {x E V: IIxll = O} is a linear subspace

§6.5. Real and Complex Measures of V and the formula vector space V IN .

IIx + Nil

319

=

IIxll

defines a norm on the quotient

2. If I and 9 are measurable complex-valued functions on a measurable space (X, S) , then the function h defined by I(x} h(x} =

{

g(~}

when g(x} f 0 when g(x}

=0

is also measurable. More generally, if r is any measurable complex-valued function on X, one can require that h(x} = r(x} whenever g(x} = o. . {Hint: § 4.1, Exercise 3} 3. A measurable complex-valued function I can be written as where u is measurable and lu(x}1 = 1 for all x. {Hint: Apply Exercise 2 with 9 = III and r = 1 .}

1= ulII,

4. The set BV[a, b] of all functions I: la, b] ---> lR of bounded variation, equipped with the pointwise linear operations, is a (real) Banach space for the norm 11111 = II(a}1 +V~I ,as well as for the norm lilli' = 1111100+ V~I (§5.1, Exercise 4).

6.5. Real and Complex Measures A measure is a function J.I.: S ---> [0, +ooJ , defined on a a-algebra S, that is countably additive and vanishes at the empty set (2.4.12). In this section we consider the analogues for set functions with values in IR or in iC; the slightly more delicate case of values in the extended reals IR is deferred until the final chapter, the delicacy being that the values +00 and -00 cannot both be taken on by a particular 'extended-real-valued measure' (§9.1).

6.5.1. Definition. Let S be a a-algebra of subsets ofaset X. A complex measure on S is a function v: S ---> iC that is countably additive in the sense that

whenever (En) is a sequence of pairwise disjoint sets in S. If, moreover, v is real-valued, it is called a real measure (or, as in §4.8, a finite signed measure) on S.

6.5.2. Remarks. In the following remarks, v, Il-, ..• are complex measures on a a-algebra S. 1. A complex measure on S is an element of the vector space .1'(S, iC) of complex-valued functions on S, whence the possibility of performing

6. Function Spaces

320

linear operations on complex measures. Since the sum and scalar multiples of complex measures are themselves complex measures (by the properties of term-by-term sums and scalar multiples of convergent series of complex numbers), the complex measures on S form a linear subspace of F(S, q; for example, if JL and v are complex measures on S, (E,,) is a sequence of pairwise disjoint sets in S, and E = U::'=l E" , then (JL

+ v)(E) = JL(E) + v(E) =

00

00

n=1

n=1

L JL(E,,) + L

v(E,,)

00

= L(JL(E,,) + v(En)J n=1 00

= L(JL + v)(E,,) n=1

(in particular, the last series is convergent 1 ), therefore JL + v is a complex measure. 2. If v is a complex measure on S, then so is the complex conjugate function v, defined by

v(E)

= v(E)

(E E S);

the countable additivity of v follows from that of v and from the continuity of complex conjugation in C. It follows that v is uniquely expressible as a linear combination v = p + iC1 of real measures p and C1, namely 1

p= 2(v+v)

and

1

C1

= 2i(v-v),

called the real and imaginary parts of v. 3. If JL and v are (positive) finite measures on S, then p = JL - v is a real measure on S. Conversely, every real measure is a difference of positive measures (4.8.8), but not necessarily uniquely since, for example,

JL -v = 2JL- (JL+v). 4. When v is a real measure, the convergent series on the right in Definition 6.5.1 is absolutely convergent (it is 'commutatively convergent', since U::'=l E" is invariant under every permutation of the indices2 ). In view of Remark 2, the same is true for every complex measure v. 5. v(0) = 0 (let En = 0 for all n). 6. v is finitely additive (by Remark 5 and countable additivity); it follows that v is subtractive, that is, v(F - E) = v(F) - v(E) when F ~ E, as one sees by applying v to the disjoint union (F - E) u E = F (cf. 2.6.1). 1 cr. FiTS! couTse, p. 183, Theorem 1O.2.l. 'cr. E. Landau, Differential and integral calculus

[Chelsea, New York, 1951), p. 158, Theorem 217; W. Rudin, Principles of mathematical analysis [3rd edn., McGraw-Hill, New York, 1976J, p. 76, Theorem 3.54.

§6.5. Real and Complex Measures

7. If En T E then v(En ) countable disjoint union E

321

-+

= E I U (E:l -

similarly, En! E implies v(En)

v(E) , as one sees by applying v to the Ed U (Ea - E 2) U ... -+

;

v(E) (cf. 2.6.2, 2.6.3).

Two complex measures on S that agree on a generating subalgebra are identical: 6.5.3. Theorem. Let A be an algebra of subsets of a set X (2.4.1), and let S be the C1.algebra generated by A (2.4.4). If VI and V2 are complex measures on S su.ch that

then VI

= V2

on S.

Proof. (Cf. 4.6.7.) Let T = {E E S: vI(E} = v2(E}}. By assumption AcT, and T is a monotone class by the preceding Remark 7, therefore SeT by the Lemma on monotone classes (4.6.6). 6.5.4. Corollary. Let [a, bJ be a closed interval in IR, a < b, and let 8 0 be the set of all Borel sets in IR that are contained in [a, b]. If VI and V2 are complex measures on 8 0 such that

VI ([a, x]) then VI

= V2

= v2([a, x])

for all x E [a, bl ,

on Bo.

Proof. Let Co be the set of all subintervals of [a, bl , and let Ao be the algebra of subsets of [a, b] generated by Co· We know that Co C Ao c 8 0 and that 80 is the C1·algebra of subsets of [a, bl generated by Co (noted in Example 4.7.5), therefore 8 0 is also the C1-algebra generated by Ao; in view of the preceding theorem, we need only show that VI = V2 on Ao, and, since Ao is the set of all finite disjoint unions of elements of Co (4.6.2), it will suffice (by the additivity of the Vi) to show that VI (I) = v2(1} for every subinterval 1 of [a, bJ . case 1: 1 = (c,d] , where a:5 c:5 d:5 b. Then 1 = [a, d)- [a, cl ,and VI (I) = v2(1) follows from the subtractivity of the Vi (Remark 6 of 6.5.2). case 2: 1 = [a,d} , where a < d:5 b. Let a < en < d with en T d. Then [a, en) T [a, d), and VI (I) = v2(1} follows from Remark 7 of 6.5.2. case 3: 1 a singleton. If 1 = {a} then 1 = [a, al and VI (I) = v2(1} by hypothesis, whereas if 1 = {d} with a < d :5 b then VI (I) = v2(1} follows from the formula

{d} and case 2.

= [a,d] -

[a,d)

6. Function Spaces

322 The remaining cases then follow from the formulas

(c,d) [c,d) [e, d]

= (c,d) - {d}, = {e} U (e,d) , = [e, d) U {d}.

6.5.5. Corollary. Let Bo be as in the preceding corollary, let p. be any measure on Bo , and suppose f E .ct(p.) is such that

1% fdp. = 0 where, by definition, to p.).

for all x E [a,b] ,

J: fdp. = J'P[a,%lfdp..

Then f

=0

a.e. (relative

Proof. Since f = g + ih with g, h E .c~{p.), we can clearly suppose that f E .c~{p.); writing v = f· P. (4.7.1), we know v is a real measure on S (4.7.3) and, by assumption, v{[a, xl) = 0 for all x E [a, bJ . By the preceding corollary, v = 0 on Bo , therefore f = 0 a.e. (4.7.4).

Exercises 1. (i) If p is a real measure on the u-a1gebra S, then there exist a finite measure p. on S and a function f E .cMp.) such that p = f·p. (d. 4.7.1). (ii) Extend (i) to complex measures. {Hint: (i) Write p as a difference p = P.l - P.2 of finite measures, let p. = P.I + P.2, and apply the Radon-Nikodym theorem (4.8.11) to P.I,P. and to P.2, p..}

2. Let (X, S, p.) be a measure space. A complex measure v on S is said to be absolutely continuous with respect to p., written v« p., in case v vanishes on the null sets for p., that is, E E S, p.{E) (cf. 4.8.6). (i) If v =

VI

+ iV2

with

VI

=0

and

V2

~

v(E)

=0

real measures on S, then

(ii) If f E .ct(p.) then the set function f· p. : S -+ iC defined by (f. p.)(E) = J'PEfdp. is a complex measure on S such that f· p. « p.. (iii) If the measure p. is finite and if v is a complex measure on S such that v« p., then there exists a function f E .c~ (p.) such that v = f . p. . {Hint: Cf. 4.8.12.}

§6.6. L""

323

6.6. 1'>0

Throughout this section, (X, S, fJ.) is a fixed measure space (2.4.12), eventually specialized to the example (la, bj, Mo, Ao) of Lebesgue measure on a closed interval (4.7.6).

6.6.1. Definition. A function I : X -+ C is said to be essentially bounded (with respect to fJ.) if there exists a real number M ~ 0 such that III < M a.e. (with respect to fJ.). Such a number M is called an essential bound for I (more aptly, for III). 6.6.2. Lemma. Every essentially bounded lunction has a smallest essential bound.

Proof. Suppose

I: X

-+

of all essential bounds for

C is essentially bounded and let S be the set

I,

= {M ~ 0: 1/1:5 M a.e.}; . Let M = inf S ; it will suffice to show that

S

by assumption, S f; 0 M ES. Choose a sequence M n E S with M n -+ M. For each index n, let En E S be a set of measure zero such that 1/1:5 M n on E~ = X - En . Then E = U::"=l En is a measurable set of measure zero, and 00 x E E' = E~ ;0} I/(x}l:5 M n for all n ;0} I/(x}l:5 M,

n

n=l

thus M is an essential bound for

I,

in other words, M E S.

6.6.3. Definition. If I : X -+ C is essentially bounded, the smallest essential bound for I (6.6.2) is denoted 11/1100 and is called the essential supremum of I (more aptly, of III). CAUTION: In another context, 11/1100 stands for the supremum of III (cf. 3.1.10). 6.6.4. Definition. The set of all functions I: X -+ C that are measurable (with respect to S) and essentially bounded (with respect to fJ.) is denoted L:C'(X,S,fJ.} ,

briefly L:C' or, when it is necessary to indicate the measure in question,

L:C'(fJ.} . As in 6.4.11, we write N = N(X,S,fJ.} for the set of all measurable functions I: X -+ C such that I = 0 a.e.

6.6.5. Proposition. L:C' is a subalgebra 01 the algebra F(X, C) 01 all For all I, g E complex-valued /unctions on X, and N is an ideal 01 L:'C and e E C, (I) 11/1100 ~ 0, (2) lIellioo = lei 11/1100 ,

Ce .

324

6. FUnction Spaces

(3) 111+ glloo ~ 11/1100 + IIglloo, (4) 11/1100 = 0 ¢} lEN, (5) II/glloo ~ 1l/11001lg1l00. In particular, the mapping I ...... 11/1100 is a seminorm on the complex vector space L.C' .

Proof The algebra operations in :F(X, iC) are the pointwise operations; for example, (fg)(x) = I(x)g(x) for all x EX. If I, 9 E L.C' and e E C, then the functions 1+ g, el and I 9 are measurable by Proposition 6.4.2. (1) Obvious from Definition 6.6.3. (2) If e = 0 then el = 0 E L.C' and the equality is obvious. Suppose efO.Then le/(x)1 = lell/(x)1 ~ lei 11/1100 a.e. (because 11/1100 is an essential bound for I), therefore el E lIe/li oo ~ lei 11/1100 . It follows that 11/1100 = lIe- 1(ef)lloo ~ le-11Ile/Il 00 ,

L.C'

and

therefore Iclll/iloo ~ lIe/li oo . (3) I/(x) + g(x)1 ~ I/(x) + Ig(x)1 ~ 11/1100 + IIglioo a.e., therefore 1+ 9 E L.C' and III + gil 00 ~ 11/1100 + II gil 00 . The message of (1)-(3) is that I ...... 11/1100 is a seminorm on L.C' (6.4.9). (4) If lEN then 1/1=0 a.e., therefore III ~ 0 a.e., whence 11/1100 ~ 0; in view of (1), 11/1100 = O. This shows that N c L.C' (as a linear subspace). Conversely, if I E L.C' and 1If1100 = 0 then III < 0 a.e. (6.6.3), therefore I = 0 a.e., thus lEN. (5) I/(x)g(x)1 = I/(x)llg(x)1 ~ 1l/1100llgll00 a.e., therefore Ig E L.C' and II/glloo ~ 11/1100 II gil 00 . From (4) and (5) we see that if IE L.C' and 9 EN then Ig E N, therefore N is an ideal of L.C'.

6.6.6. Definition. The quotient algebra

L.C'/ N

is denoted

LC'(X, S,!,) , briefly LC' or LC'(!') , the operations on cosets being given by the formulas

(f +N) + (g +N) = (f + g) +N e(f +N) = el +N (I +N)(g+N) = Ig+N for all I,g E L.C' and e E C. If IE

L.C'

we write

j=I+N for the image of I under the quotient homomorphism

L.C' -> LC .

§6.6. Leo

325

If u E LC' ,say u 0; since

11/1100

= i = g, then

= IIU -

g)

+ glloo

1- 9 E N and therefore 111- glloo = :5 III - glloo

+ IIglioo = IIglioo

and similarly IIglloo:5 11/1100, so that 11/1100 = II gil 00 , it follows that the definition lIuli oo = 11/1100 is independent of the particular representative I of the coset u. It then follows from Proposition 6.6.5 (by the same argument as in 6.4.15) that u ...... lIuli oo is a norm on the complex vector space LC'. In fact:

6.6.7. Theorem. LC'(JL) is a Banach space lor the norm u ...... lIuli oo .

Prool. The problem is to show that every Cauchy sequence in LC' is convergent (6.4.17). The crux of the matter is as follows: assuming Un) is a sequence of functions in £.C' such that 111m - In 1100 -+ 0 as m, n -+ 00, we seek a function IE £.C' such that II/n - 11100 -+ O. For each pair of indices m, n E IP let E mn E S be a set of measure zero such that 11m (x) - In(x)1 :5 111m - Inlloo

for all x E E:"n ;

then E = U:,n=l E mn has measure zero and, for each x E E', the inequalities 11m (x) - In(x)1 :5 111m - fnlloo show that

(fn(X)) is a Cauchy sequence of complex numbers. Define

I:X-+C by

I(x)

={

lim In(x)

for x E E'

o

forxEE.

Since 'PE'ln -+ I pointwise on X, the limit function (4.1.20). For all x E E' and for every pair of indices m, n ,

(*)

I is measurable

I/(x) - In(x)1 :5 I/(x) - Im(x)1 + I/m(x) - In(x)1 :5 I/(x) - Im(x)1 + 111m - In 1100 .

Given any e > 0, choose an index N such that m, n

>N

~

111m - Inlloo :5 e.

Fix a pair of indices m, n ;::: N . For each x E E' , it follows from (*) that

I/(x) - In(x)1 :5 I/(x) - Im(x)1 + e; keeping n fixed and letting m

-+ 00,

we have

I/(x) - In(x)1 :5 0 + e.

326

6. Function Spaces

This shows that I - In E C'C, hence also

1= (f - In)

+ In

E C'C ,

and that III - Inlloo :=:; f; since the inequality holds for an n;:: N, we have shown that 111- Inlloo -+ O. The next results explore the relation between the Banach spaces L'C(IL) and Lt(IL). 6.6.8. Lemma. II I is integrable and 9 is essentially bounded and measurable, then Ig is integrable and IIlgllt :=:; 11/11 1IIglioo . Proof We are assuming IE Ct and 9 E (6.4.2), III is integrable (6.4.6) and

Ce. Since

Ig is measurable

I/gl :=:; 1/IIIglioo a.e., it fonows that

IIgl

IIMI =

is integrable (4.4.20), therefore so is

J

I/gldIL :=:; IIglioo

I9

J

I/ldIL = IIgliooll/llt·

(6.4.6), and

6.6.9. Theorem. II 9 E C'C then the lormula

Ti =

J

(f

IgdIL

E

Ct)

defines a continuous linear lorm on the Banach space Lt. Proof It is clear from the lemma that T is a wen-defined linear form on Lt, where "wen-defined" means that for u = E Lt, Tu depends only on the equivalence class u and not on the particular function I representing u. Moreover, by Proposition 6.4.6 and the lemma,

i

ITil =

J

IgdIL :=:;

thus ITul:=:; lIullt 1191100 for an u E

J

1/91d IL:=:; 11/1111191100,

L1: . Thus, for an

u, v E Lf:,

lTv - Tul = JT(v - u)1 :=:; IIv - ulltll91100,

whence the continuity of T: IIUn-ulll-+O

~

ITun-Tul-+O.

Is the converse of the theorem true, that is, is every continuous linear form T on Lt given by the formula of the theorem for a suitable function 9 E C'C ? The answer depends on the measure space, and may be "no".1 For Lebesgue measure, the answer is "yes"; for simplicity, we limit the fonowing

1 ce. E. Hewitt and K. Stromberg, Real and abstract analysis (Springer. New York, 1965), p.349, (20.17).

§6.6. Loo

327

discussion to a closed interval la, b], a LMa, b], with notations as in 4.7.6:

< b, and to the real Banach space

>. = Lebesgue measure on IR M = {E c 1R: E Lebesgue-measurable}

= {E EM: >'0 = >.IMo.

Mo

E cia, b]

Form the real Banach space U = L~(Ia, b], Mo, >'0) (cf. 6.4.18). Bending the notation, we sometimes abbreviate >'0 to >.. 6.6.10. Theorem. (Riesz representation theorem2 ) With the preceding notations, if T : LI -> IR is a continuous linear form on U, then there exists an essentially bounded measurable function g : [a, b] -> IR such that

Tj

=

f

fgd>.

for all f E t:}.

Proof. We show first that there exists a real number M (1)

ITul

:s Mllulll

for all

U E

LI

~

0 such that

.

Assume to the contrary that no such M exists. Then each positive integer n fails to have the property required of M, so that there exists a Un E U such that ITUnI > nllunill. In particular, TUn f. 0, therefore Un f. O. Writing

we have

IIvn lh = lIn ITvnl =

and

~llunlllllTUnI > ~lIunllll. nllunlll = n n

1;

thus Vn -> 0 but TVn f> 0 = TO, contrary to the continuity of T at 0 ELI. For the rest of the proof, fix a number M ~ 0 satisfying (1). {Incidentally, there is a smallest such M, easily seen to be equal to sup ITul ,where u varies over all elements of U such that lIulll 1; this supremum is called the norm of T and is denoted IITII.} For every x E la, b] , let

:s

'P(a,zl :

[a, b]

->

IR

be the characteristic function of the subinterval (a, xl of la, b) . Then IR by the formula F(x)

, cr.

= T'

= M>.(x,yj) = M(y - x) = M[y - xl. whence the Lipschitz condition (2). Since F is Lipschitz, it is absolutely continuous (5.1.10, (vi)). By Lebesgue's Fundamental theorem of calculus (5.10.2), there exists a function 9 E £ 1 such that F' = 9 a.e. and F(x)

= 1% gd>' + F(a)

for all x E [a,b].

It follows from the Lipschitz condition (2) that

F(y) - F(x) < M y-x -

for all x, y in [a, bJ with x # y, consequently Ig[ :=:; M a.e., thus 9 E £00 . Moreover, if a:=:; x :=:; y :=:; b then TtP(%,yl

= F(y) =

F(x)

=1

J

Y

gd>' -1% gd>'

['P(a,y) - 'Pla,%!l 9 d>'

=

J

'P(%,y)g d>.;

if I is any of the four possible intervals with endpoints x and y, then 'P(%,y) = 'PI a.e., and we have shown that (3)

TtPi

=

J

'PIg d>'

for every subinterval I of [a, b].

For every Lebesgue-measurable set E /.II(E) =

J

'PEgd>',

c [a, b] , define /.I2(E) = TtPE.

Since /.11 = 9 . >'0 (cf. 4.7.1), we know that /.11 is a real measure on Mo (cf.4.7.3).

§6.6. L''''

329

claim: 112 is also a real measure on Mo. Assuming E = U::"= 1 E,. is the union of a sequence of pairwise disjoint sets in M o , we are to show that 00

112(E)

=L

112(E,.) .

n=1

Since

2:::"=1 >'(En )

= >'(E) ~ b - a n

L >.(E

>'(E) -

< +00, we see that

00

k)

L

=

k=1

>'(E k )

-+

as n

0

-+

00,

k=n+l

that is. J['PE - 2:~=1 'PE.ld>'

-+

0; thus n

II'PE -

L 'PE.lh

-+

0,

k=1

whence 2:~=1 '(B - E) = 0 (2.4.15); replacing B by B n la, bl , we can suppose that B E 80. Then 'PE = 'PB a.e. (with respect to >'0), thus .

for every simple function I: [a, bl -+ lR (relative to Mo), such functions being automatically integrable with respect to the finite measure >'0. Our problem is to show that the equation (6) holds for every IE L:M>.o); in view of 4.4.7, we can suppose that I > O. If (In) is a sequence of simple functions (relative to M o ) such that 0 :S In T I pointwise on [a, bl (4.1.26), then J Ind>.o TJ Id>.o (4.4.3). Since I - In > 0, it follows that III - In II I -+ 0, therefore Tin -+Ti by the continuity of T. Also, pointwise on [a, bJ,

Ing -+ Ig and IIngl therefore

= Inlgl

:S Ilgl :S MI a.e.,

f Ing d>. f Igd>. -+

by Lebesgue's dominated convergence theorem (4.5.4). We know from (6) that

Tin =

f Ing d>.

for all n,

and passage to the limit yields the desired formula T

i = JIg d>..

0

Exercises 1. Give an example where IIIglioo < IIIlIoolIglioo ; that is, the inequality in (5) of Proposition 6.6.5 may be strict. {Hint: Consider a two-point set X = {a, b} , the q-a1gebra S = P(X) = {0, {a}, {b},X}, and the discrete measure J.L on S (§2.4, Exercise 3).}

2. The proof of Theorem 6.6.9 shows that

sup{ITil: I

E L:~(J.L), IIIII1 :S I} :S IIglioo.

Give an example for which the inequality is strict. {Hint: With (X, S) the measurable space in the hint for Exercise I, consider the measure J.L given by

J.L( 0) = 0, J.L({ a}) = 1, J.L( {b}) = +00, J.L(X) = +00.

§6.6. Loa

331

Then 'c1; consists of the scalar multiples of lO{a}, (/J is the only set of measure zero, and the linear forms on the one-dimensional vector space Lt = 'c1; are the scalar multiples of the linear form 1 >-> J 1dp. (in other words, the linear form CI"{a) >-> c). Let 9 = lO{b} and contemplate TI = J Igdp. (J E ,Ct).} 3. With notations as in Definition 6.6.6, lIuvll oo :S lIuli oo llvll oo for all u, v E Ll;" ; SO to speak, the norm on the algebra LC' is submultiplicative. {A Banach space with an associative and distributive multiplication that commutes with scalar multiplication (cf. §5.1, Exercise 4) and for which the norm is submultiplicative is called a Banach algebra.} 4. (i) The function 9 of the Riesz representation theorem (6.6.10) is a.e. unique. (ii) Extend the Riesz theorem to Lebesgue measure on IR, that is, to continuous linear forms on L~(IR, M, >.). {Hint: (i) Cf. 4.4.24. (ii) For each positive integer n, regard U [-n, n] as a linear subspace of LI (IR) by extending functions I: [-n, n] -+ IR to be 0 on IR - [-n,n]. If T is a continuous linear form on LI(IR) , apply Theorem 6.6.10 to each of the restrictions Tn = TIU[-n, n] to produce an essentially bounded function gn: [-n, nJ -+ IR, then argue that there is an essentially bounded function g: IR -+ IR such that gll-n, n] = gn a.e. for every n.} 5. Extend the Riesz theorem (6.6.10) to the complex case, that is, to continuous iC-linear forms T: L[;(>'o) -+ iC. {Hint: Regard L~ C L1; in the obvious way (remarks following Definition 6.4.4) and apply 6.6.10 to the IR-linear forms R,8: 4 -+ IR defined by Rj = Re(Tj) , 8j = Im(Tj) for all 1 E 'c~(>'o).} 6. Let 9 E 'cl;"(p.) and let T: Lt(p.) -+ iC be the linear form defined by T j = JIg dp.. We know from Theorem 6.6.9 that T is continuous; writing IITII = sup{ITjl: we see from the inequality IIlglll

1 E 4, 11/111 :S I},

< lIJ1hllglloo that IITII:S IIglloo· It can

happen that IITII < II gil 00 (Exercise 2). Prove: If p. is q-finite then IITII = IIglioo . {Hint: Write 9 = ulgl with u measurable and 1,:,1 = 1 (§6.4, Exercise 3), and let 8: L1; -+ iC be the linear form 81 = J Ilgldp.. From Igl = ug and II/ulh = II/lh = IlfUlh for all 1 E 'ct, conclude that IITII = 11811 ; we can suppose, therefore, that 9 ~ O. Given any E > 0, it will suffice to show that IITII ~ IIglioo - Eo If E = {x: g(x) > IIglioo - t}, necessarily p.(E) > 0 (the alternative is that O:S 9 :S IIglioo - E a.e., whence the absurdity IIglioo < II gil 00 - E). Let F E S with FeE and 0< p.(F) < +00; then 1 = (1/p.(F)}IOF is integrable, 111111 = I, and integration of the inequality Ig ~ (lIglioo - E}I leads to paydirt.}

6. Function Spaces

332

7. (i) With notations as in Exercise 6, if I-' is a-finite then 9 is 'essentially uniquely' determined by T, in the sense that if also h E .cC' (I-') and Tj = J Igdl-' = J Ihdl-' for all IE .cb(l-'), then 9 = h a.e. {Hint: Part (ii) of Theorem 4.7.2.} (ii) Give an example of a measure I-' (not a-finite) for which 'essential uniqueness' in the foregoing sense fails. {Hint: Exercise 2.}

6.7. V' (1 < p < +00)

Notations fixed throughout the section: (X, S, 1-') is a measure space, p a real number, 1 < p < +00, and q = ~ (the positive real number 'conjugate' to p); thus q> 1 and ~ + ~ = 1, in other words, p+ q = pq.

6.7.1. Definition. A complex-valued function I: X -+ C is said to be p-th power integrable if (i) I is measurable (with respect to S), and (ii) III" is integrable (with respect to 1-'), that is, III" E .c 1 • The set of all such functions I is denoted .c~(X, S, 1-'),

briefly

.c~,

and, for such a function

called the .c"-norm (or p-norm) of

I, we write

I.

The foregoing is a slight abuse of the term "norm"; as we shall see below (in 6.7.4), .c~ is a complex vector space for the pointwise linear operations, and I ...... 11111" is in fact a seminorm on .c~. The proof of the triangle inequality for this seminorm is based on the extension to integrals of an inequality proved in Chapter 3 (3.1.12): 6.7.2. Theorem. (Holder's inequality) ~), then 19 E.cb and

II I

E.c~

and 9

E.c~

(q

=

Proof At any rate, Ig is measurable (6.4.2); we have to prove that it is integrable. Let Ct = 11111", {3 = IIgll q . If Ct = 0 or {3 = 0, then I = 0 a.e. or 9 = 0 a.e., I 9 (= 0 a.e.) is integrable, and the asserted inequality is obvious (6.4.12).

§6.7. LP (1 < p < +oo)

333

Suppose 0 > 0 and {3 > O. For each x EX, application of Proposition 3.1.3 (with a = I/(x}l/o and b = Ig(x)I/{3} yields the inequality

lI(x}g(x} I < ~ . lI(x}IP + ~ . Ig(xW 0{3 thus the functions

- p

I and

oP

(3q

q

,

9 satisfy (identically) the inequality

...!-l/gl < ~ . I/IP + ~. Iglq . 0{3 - p oP q {3q

0

as m, n

-> 00,

then there exists a Iunction I E.c~ such that IIIn - Ill p -+ 0; moreover, any two such functions I are equal a.e.

Proof. (i) If I

E.c~ and c E lC, then

ci is measurable (6.4.2) and IcJIP = IclPIIIP is integrable, therefore ci E.c~ and IIcill p = Iclllill p ; the preceding theorem then completes the proof of (i). (ii) If lEN then I is measurable and IIIP = 0 a.e., therefore I E .c~ and f IIIPd~ = 0, thus IIIlI p = O. Conversely, if I E.c~ and IIIlI p = 0, then f IflPd~ = 0, whence IIIP = 0 a.e. (4.4.21), therefore lEN. (iii) The proof is similar to that of the analogous property of .ct (cf. 6.4.13). Let (In) be a sequence in .c~ such that 111m - In lip -+ 0 as m, n -+ 00. To simplify the notations, let us abbreviate IIIlI p to 11111, for I E .c~. By the triangle inequality (cf. 6.4.10) it clearly suffices to find an I E.c~ such that IIIn. - III -+ 0 for some subsequence (fn.) of (fn)

'F. Riesz (1880-1956) and E. Fischer (l875-1954).

§6.7. L" (1

< p < +00)

335

(cf. the proof of 6.1.26). Thus, passing to a subsequence, we can suppose that

IIfn+l - fnll :::; 2- n Write

0

for all n.

= E::'=I IIfn+l - fnll :::; E::'=I 2- n = 1. Let fo = 0

and define

n

9n=Llfk-fk-11

for n=1,2,3, ... ;

k=1 it follows from (i) that 9n E £P, therefore (9n)P E £1 for all n. Clearly o S 9n T, therefore also 0:::; (9n)P T; by Minkowski's inequality, n

n

k=l

k=2

119nll :::; L IIfk - fk-dl = 11M + L

IIfk -

fk-III :::; IIhll + 0,

thus

J

(9n)PdJL

= 119nll P :::; (11M + o)P < +00

for all n; by the monotone convergence theorem, there exists an h E £ 1 such that (9n)P Th a.e. Redefining the fn and h to be zero on a suitable null set (Le., on a suitable measurable set of measure 0), we can suppose that

o :::; (9n)P 1 h

pointwise on X.

Let 9 = h 1/p ; then 9 is measurable and 9P = h E £1, thus 9 E £P. Also,

by (*). For each x EX, n

L Ifk(x) -

fk-l(X)J

= 9n(X) T9(X) < +00,

k=1 therefore the series E~I[fk(x) - fk-I(X») is (absolutely) convergent and we may define f(x) to be its sum: n

f(x)

= n-oo lim ~[fk(x) L-, k=1

fk-l(X)]

= n-oo lim fn(x).

Thus fn .... f pointwise, therefore f is measurable (6.4.3). Also, for every

6. Function Spaces

336

x

e X, n

Ifn(x)1

= L[!k(x) - !k-I(X)) k=1 n

S;

L l!k(x) - !k-I(x)1 = gn(x) < g(x) k=1

for all n; passage to the limit yields III S; g, therefore IIIP S; gP = h e .c l , whence I/IP is integrable and so I e .ct· Next, we show that II/n - III -+ 0; the proof will make use of Fatou's lemma (4.5.5). At any rate, by the preceding paragraph, In - I e.ct for all n, so II/n - III makes sense. Let t > O. Choose an index N such that

m, n ~ N ~

Fix an index m

also, as n -+

00

111m - Inll

S; t.

N . Then

J11m -

(**)

=>

InlPdp S; t P

for all n

~ N;

t

11m - Inl P-+ 11m - liP e .c l

,

thus liminfl/mn

Inl P = n-oo lim 11m - Inl P = 11m -

liP

e .c l

,

and Fatou's lemma yields, in view of (**),

J11m -

IIPdp S; limninf

J11m -

whence 111m - III S; t (for every m ~ N). Finally, if also I. e.ct with II/n - 1.11 III -

/.II

S; III -

-+

InlPdp S; t P , 0, then

Inll + II/n - /.II

for all n; passage to the limit yields III by (i).

1.11 = 0, whence I - I.

= 0 a.e.

6.7.5. Definition. With notations as in 6.7.4, the quotient vector space is denoted

.ct/N

Lt(X,S,p) , briefly

Lt· For

u

e Lt ,say

u= j =

lI u li p =

I + N ,where I e .ct , one writes II/l1 p ,

called the nonn (or V-nonn) of u; if I,g e.ct and j = g, then I -g e N, I = 9 a.e., and II/l1 p = IIglip ,thus lIuli p depends only on the coset u, not on the particular function I e.ct selected to represent it.

§6.7. L" (1 < P < +00)

337

6.7.6. Corollary. With notations as in the preceding definition, L~ is a

(complex) Banach space with u ...... lIull" as norm. Proof. The proof is similar to that for L~ (6.4.18), with the requisite completeness supplied by Theorem 6.7.4.

A consequence of Holder's inequality is that each function in .c.~ induces a linear form on .c.~ (and, ultimately, on the Banach space L~), continuous in an appropriate sense: 6.7.7. Theorem. Let 9 E.c.~ (q = ~ ). (i) The lormula

=

L(/)

defines a linear lorm L on IL(/)I ~

J

(f E

IgdJt

.c.~,

~)

such that

11/11" IIgll q

lor all

I

E ~.

L(/n)

--+

L(/) .

(ii) L is continuous in the sense that

II/n - III"

--+

0

~

(iii) Moreover,

11911 q = sup{IL(/)I: I

E .c.~,

11/11" < I} .

Proof. (i) The indicated integrals exist by Theorem 6.7.2 and,

for I E .c.~ ,

IL(/)I =

J

Igdp.

~

J

I/gldp. = IIMI
LUn) - 0 LU); (b) there exists a constant M ~ 0 such that IL(J)I < Mllfll p for all fE.c~; (c) the set of complex numbers {L(J): f E .c~, IIfll p :S I} is bounded. When the foregoing conditions are verified, the correspondence j t-+ L(J) defines a continuous linear form on the Banach space L~ = .c~/N of 6.7.6. Proof· (a) => (b): Note first that IIfll p = 0 => L(J) = 0 (consider the sequence fn = 0 for all n). Assume to the contrary that no such M exists. Then, for every positive integer n, there exists a function 9n E .c~ 2 cr. H.L. Royden, Real analysis 13rd. edn., Macmillan, New York, 1988), p. 286, Theorem 30.

3 Op. cil, Chapter II, §7 (pp. 282-287).

§6.7. LP (1

< P < +00)

339

such that IL(gn)1 > nllgnllp (in particular, IIgnlip > 0 by the preceding remark); the functions fn = (nllgnllpflgn then satisfy

IIfn11p = lin

IL(Jn)1

and

= (nllgnllp)-lIL(gn)1 > I,

thus IIfnllp ..... 0 but L(Jn) f> 0, contrary to (a). (b) => (c): With M as in (b), IL(J)I ~ M whenever IIfll p ~ 1. (c) => (b): Let M be an upper bound for the numbers IL(J) I (f E £~, IIflip ~ 1). Given any 9 E £~, we assert that IL(g)1 ~ MlIgli p • If IIglip = o then, for every positive integer n, IInglip = nllgllp = 0 < 1, therefore IL(ng)1 ~ M by hypothesis; the validity of IL(g)1 ~ Min for all n means that L(g) = 0, thus the desired inequality holds trivially. On the other hand, if IIglip > 0 then the function f = (lIgllp)-lg satisfies IIflip = 1, therefore IL(J)I:5 M, whence IL(g)1 ~ MlIgli p by the linearity of L. (b) => (a): IL(Jn) - L(J)I = IL(Jn - J)I :5 Mllfn - flip· Finally, as noted in the proof of (a) => (b), such a linear form L satisfies IIflip = 0 => L(J) = 0; it follows that if U E Ll; = £tIN, say u = j = f + N, then the number L(J) depends only on u and not on the particular function f chosen from the coset. Thus, the correspondence u

t--+

L(J)

(u

=j

E

Lt)

is well-defined, it is clearly a linear form on Lt, and it follows from the condition (b) that this linear form is continuous for the metric topology on the Banach space Lt derived from its norm (see the remark following 6.4.15). 0 6.7.10. Lemma. Suppose the measure spoce (X,S,tL) is finite. If 9 : X ..... iC is a tL-integrable function such that the set of complex numbers

{J

fgdtL: f simple,

IIfllp :5 1 }

is bounded, then 9 E .c~ . Proof Recall that 1 < P < +00 and lip + I/q = I. Since tL is finite it is clear that, for every simple function f, f E £~ and fg E Let

.ct·

M = sudl

J

fgdtLl: f simple,

IIfllp :5 I};

by the arguments in 6.7.9, it is clear that

for every simple function f. Consider first the case that 9 is real-valued; we will prove that 9 E £~ assuming only that (*) holds for all real-valued simple functions f. Let E = {x: g(x);:: O} (a measurable set) and define u = 'PE - 'PX-E; then

6. Function Spaces

340

is a simple function, lui = 1 on X, and 9 sequence of simple functions such that

u

= ulgl·

Let (In) be a

TIglq ; 9 E £~ , we must show that Iglq E £~ , and for this it suffices 0::; In

to prove that to show that the sequence

f I ndJ.1.

hn = u/~/p

is bounded above (4.5.1). Let

(n = 1,2,3, ...);

since u and the In are real-valued simple functions, so are the h n , therefore

/ hngdJ.1. ::; Mllhnll p

Igl;::: 1~/q , we have 1~/p. ug = 1~/p ·Igi ;::: 1~/p . 1~/q

by the hypothesis on g. Since

hng = u/~/p . 9 =

= 1~/p+l/q = In,

therefore (i) but (ii)

and it follows from (i) and (ii) that

/ In dJ.1. ::; M ( / IndJ.1.) lip , whence (f IndJ.1.) 1-1/1' ::; M. Thus,

::;

M (even if f IndJ.1. = 0), that is, (f IndJ.1.) I/q for all n,

which completes the proof that 9 E £~ (when 9 is real-valued). Suppose now that 9 is complex-valued, and write 9 = gl + ig2 with gl, g2 E .c~ . For every real-valued simple function I,

/19dJ.1. = / IgldJ.1. + i / Ig2dJ.1. is the decomposition of the left member into its real and imaginary parts; since, by hypothesis, the left member remains bounded as I varies over all such functions satisfying 11/111'::; 1, the same is true of its real and imaginary parts, consequently gl, g2 E £~ by the first part of the proof, whence 9 E.c~ by Theorem 6.7.3. 0

§6.7. LP (1 < p < +00)

341

6.7.11. Theorem. (Riesz representation theorem) Let (X,S,p.) be a finite measure space, let 1 < p < +00, and let L:.c~ --+ C be a linear form satisfying the continuity conditions of 6.7.9. Then there exists a junction 9 E.ct (q = such that

#r)

LU)

=

1

f gdp.

for all f

E

.c~ .

Moreover,

IIgll q = sup{IL(f)I: f

E

.cl;, IIfllp :5 I} ,

and any two such functions g are equal to each other almost everywhere. Proof Define a set function v: S

--+

C by the formula

v(E) = L(IOE)

(E E S);

since 'PEuF = 'PE + 'PF when En F = 0 , it is clear from the linearity of L that v is finitely additive. In fact, v is countably additive. For, suppose E = U;;:I Ek , where (Ek) is a sequence of pairwise disjoint measurable sets; we are to show that the series E;;:I V(Ek) is convergent with sum v(E). Writing F n = U~=I Ek, we have v(F n ) = E~=I V(Ek) by finite additivity, so the problem is to show that v(F n) --+ v(E) . Since F n T E, we have E - F n 1 0, therefore p.(E - F n ) 1 0 by the finiteness of p. (2.6.3); then II'PE - 'PF.ll p = II'PE-F.ll p =

= thus II'PE - IOF. lip

--+

(I

(I )

l'PE-F.1PdP.) lip

I/P

'PE-F. dp.

= (p.(E -

»

II

Fn

p

10,

0, therefore

v(Fn )

= L('PF.) --+ L('PE) = v(E)

by the continuity assumption on L. Summarizing, v is a complex measure (6.5.1) on the finite measure space (X, S, p.). Moreover, v is absolutely continuous with respect to p.; for, if p.(E) = 0 then lI'PEli p = 0, therefore v(E) = L('PEl = 0, as shown at the beginning of the proof of 6.7.9. By the Radon-Nikodym theorem (cf. §6.5, Exercise 2), there exists a p.-integrable function 9 E.cb such that

llOEgdp.

= v(E) = L('PE)

It then follows from linearity that

1

f gdp.

= LU)

for all E E S.

6. Function Spaces

342

for all simple functions f; in view of the continuity condition on L, it follows from the preceding lemma (6.7.10) that 9 E It remains to show that the formula (.) is verified for every f E .ct· By linearity, it suffices to consider f (b): Obvious. (b) => (c): If the set is not bounded, then there exists a sequence (Yn) in E such that IIYnll ~ 1 and If(Yn)1 > n for all n. Let X n = (l/n)Yn and contemplate X n -+ 0 . 0 then IIxll- 1 x has norm 1. (c) => (d): If (d) => (a): If(x n ) - f(x)1 = If(x n - x)1 ~ Mllx n - xII.}

x'"

3. Let E be a normed space, f a linear form on E. (i) If f is continuous, then the number sup{lf(x)l: x E E, IIxll ~ 1} is the smallest number M ~ 0 satisfying condition (d) in Exercise 2. It is denoted IIfll and is called the nonn of f (the terminology is justified in Exercise 4). (ii) It can be shown that if a E E and a '" 0, then there exists a continuous linear form f on E such that IIfll = 1 and f(a) = lIall;4 in particular, every nonzero normed space admits nonzero continuous linear forms. 4. Let E be a normed space. (i) If f and 9 are continuous linear forms on E and if c is a scalar, then

the pointwise sum f (J

+9

and scalar multiple cf , defined by the formulas

+ g)(x) = f(x) + g(x) , (cf)(x) = cf(x)

for all x E E,

are also continuous, and

IIf + gil

~

IIfll + IIgll , IIcfll = Icillfll·

(ii) The set E' of all continuous linear forms on E is a vector space for the pointwise linear operations, and the correspondence f >-> IIfll defines a norm on E, thus E' is a normed space (over the same field of scalarsIR or iC-as E). In fact, E' is a Banach space (even if E is not complete), called the dual space of E. 5

5. Let (X,8, 1-') be a measure space, let 1 < p < +00, and let q = p/(P - 1). For each pair U E Lt and v E Ll:, define a complex number (u, v) as follows. Write u = j = f + N and v = !J = 9 + N (caution: these are cosets in different quotient spaces!) with f E £t and 9 E £l: '

Cf. the author, Lectures in functional analysis and operator theory [Springer-Verlag, New York. 1974), p. 169, 40.10. 5 Cf. the author, op. cit., p. 169, 40.9. 4

6. Function Spaces

344 and define (u,v)

=

J

fgdp.;

the definition is legitimate because the expression f f gdp. depends only on the cosets 1< and v, not on the particular functions f and 9 selected to represent them. (i) The mapping (u, v) ...... (u, v) is bilinear: (UI

+ U2, v) = (Ul> v) + (U2' v) (cu, v) = c(u, v)

(u, VI + U2) = (u, VI) + (u, V2) (u, cv) = c(u, v) for all U,UI,U2 E L~, V,Vl,V2 E L~ and cE C. (ii) I(u, v)l ~ lIull p IIvll•. (iii) For each v E L~ , the formula Lv(u)

= (u, v)

(u E 1{;)

defines a continuous linear form Lv on q;, that is, Lv E (L~)' . (iv) The mapping L~ -+ (L~)' defined by v ...... Lv is linear: L v1 +v,

= LVI + Lv"

Lev

= cL v ;

and isometric: II Lv II = IIvll.· (v) Theorem 6.7.11 shows that the mapping v ...... Lv of (iv) is surjective, assuming p. is finite. In fact, the same is true for an arbitrary measure p..6 The norm-preserving vector space isomorphism Ll: -+ (L~)' thus defined is usually expressed by writing (L:;)' = Ll: . In turn, (LV' = L:; , whence L~ = (q';)" , a property of 1{; (for 1 < P < +00 ) known as reflexivity. (vi) If one defines instead (u, v) = f f9dp. ,7 then the correspondence (u, v) becomes sesquilinear (linear in u, conjugate-linear in v), (u, v) and v Lv is a conjugate-linear mapping (Lev = cLv ) of Ll: onto (L:;)' . Linearity can be restored in two ways: (a) consider instead the mapping v ...... Lv where, for v = 9 + N E ~, one defines Ii = 9 + N, 9 being the complex-conjugate function g(x) = g(x) (xEX);or (b) stick to v ...... Lv but replace the natural (pointwise) scalar multiple (c, L) ...... cL on (1{;)' by the scalar multiple (c,L) ...... cL. If p = q = 2 then (u, v) is defined for all u, v E L~ (called the inner product, or scalar product, of u and v), in particular (u, u) = (lIuIl2)2; 6 cr. H.L. Royden, op. cit., p. 286, Theorem 30. 'E. Hewitt and K. Stromberg, Real and abstract analysis ISpringer-Verlag, New York, 1965], p. 223, (15.1).

§6.8. C(X)

345

L~ is an example of a (complex) Hilbert space, that is, a Banach space

whose norm satisfies the 'parallelogram law'

lIu + vll

2

+ lIu -

vll

2

= 211ull 2 + 211vll 2

for all u and v. 8 6. (i) If (X,S,I') is a finite measure space and 1 :5 p < r, then £i:(I') C £~(I'). (ii) If I' is not finite, the inclusion in (i) is in general false. (iii) If I: [0,1) -+ IR is the function defined by 1(0) = 0 and I(x) = x- I / 2 for 0 < x < 1 , then I is Lebesgue-integrable but its square is not. (iv) The measure space in the Hint for §6.6, Exercise 2 is not finite, but all of the spaces £1' (1:5 p < +00) coincide. {Hint: (i) If I E £i: and

E = {x: I/(x)l:5 I},

F

= {x:

I/(x)1 > I} ,

then 'PEl E £'t! c £~ and I'PF111' :5 !'PF Ilr E £1; contemplate 'PEl + 'PF I· (ii) Cf. §3.1, Exercise 4.}

I =

6.8. C(X)

In the preceding sections, the Banach spaces of functions on a set X have come from imposing a measure-theoretic structure on X and requiring the functions to be measurable. In the present section, the structure imposed on X is topological, the functions are required to be continuous, and the key results are obtained when X is assumed to be quasicompact; much of the ground has already been prepared in our discussion of uniform convergence (§6.2, especially 6.2.23). We begin with a discussion of algebraic operations in the space CIl(X) of continuous real-valued functions on X (6.2.20), then enlarge the discussion to accommodate complex-valued functions. The major applications come in the next section: Weierstrass's polynomial approximation theorem for continuous functions on a closed interval, and its spectacular generalization by M.H. Stone for continuous functions on a compact space.

6.8.1. Lemma. For a real-valued junction I: X space X, the lollowing conditions are equivalent: (a) I is continuous; (b) lor every real number c, the sets

-+

IR on a topological

{x EX: I(x) < c}, (x EX: I(x) > c} are open sets in X; 8 Hewitt and Stromberg, op. cit., p. 235, (16.8); or the author, op. cit., p. 164, (39.10) and p. 174, (41.1).

6. Function Spaces

346

(c) for every real number c, the sets {xEX:

f(x)~c},

{xEX:

f(x)~c}

are closed sets in X. Proof. (a) => (b): The sets {x: f(x) (a): For every open interval (a, b) in IR, its inverse image

r'(a,b»={x: a 0, then the following functions on X are also continuous:

f+g, cf, fg, Ref,

Imf, f, IW;

if, moreover, f(x) # 0 (Vx E X), then 1/ f is continuous. In particular, Cc(X) is an algebra Uor the pointwise operations f +g, cf, f g) containing the constant functions.

§6.8. C(X)

351

Proof Recall that the functions Re I and 1m I have final set lR (6.8.8); for the purposes of this proof, we assume that III'" also has final set lR. After the proof is done, we shall relax all this fussiness over final sets (6.8.11). We know from the preceding lemma that the real and imaginary parts of I and 9 are continuous (Le., belong to CUl(X)}, therefore so are the functions Re(l+g}=Re/+Reg and Im(l+g}=lm/+lmg (6.8.2), consequently 1+ 9 E Cc(X} (again by the preceding lemma). Similarly, ReI = ReI and Iml = -Iml are continuous, therefore so is I. The proofs for cl and I 9 are similar. In particular, I I is continuous, therefore so is III'" = (I f)"'/2 by Proposition 6.8.2 (note that, in view of Lemma 6.8.6, it does not matter whether we regard lR or C as the final set for Il and III"'). Finally, if I is never 0 then 1/1- 2 = (III 2 }-1 is continuous (6.8.2), therefore so 2 is III = 111- I· The foregoing results (particularly 6.8.6) amply justify the following simplification:

6.8.11. Scholium. With notations as in the preceding proposition, we regard CUl(X} as an lR-subaigebra of Cc(X} , by identifying 9 E CR(X} with to 9 E Cc(X} (see 6.8.6). In particular, for IE Cc(X} , the functions Re I, 1m I and III'" are henceforth regarded as continuous functions X -+ C (whose values happen to be in lR), and the formulas ReI = ! I + ! I, I = Re I + i 1m I , etc., may be regarded as linear combinations formed in the complex vector space Cc(X}. When X is quasicompact, Cc(X} has in addition a natural Banach space structure:

•

6.8.12. Theorem. Let X be a quasicompact topological space (6.1.3). Then: (i) Every continuous lunction I: X -+ C is bounded. (ii) Writing 1111100 = sUPxEX II(x}1 for IE Cc(X} , the correspondence I ...... 11/1100 defines a norm (6.4.9) on the complex vector space Cc(X}. (iii) Cc(X} is a complete metric space for the sup-metric (I, g) ...... III - glloo, thus is a Banach space (6.4.17). (iv) IIIglioo < IIIlIoolIglioo lor all I,g E Cc(X}. (v) IIfllloo = (1I/1100}2 lor all IE Cc(X) .

Proof (i) See Proposition 6.2.11. (il) The property III + glloo :s; 1111100 + IIglioo (called 'subadditivity') is verified in 3.1.10, and it is easy to verify that IIcilioo = lei 1111100 ('absolute homogeneity') and that 1111100 > 0 unless I is the function identically zero ('strict positivity'), thus I ...... 1111100 is a norm on Cc(X} Uustifying the term "sup-norm" already employed in 3.1.1O}.

6. Function Spaces

352

(iii) Completeness for the sup-norm metric is noted in 6.2.23, thus Cc(X) is a (complex) Banach space. (iv) For all x E X,

l(fg)(x)1 = I/(x)g(x)1 =

I/(x)I'lg(x)1 :5 IIIII 00 IIgII 00 ,

therefore 11191100:5 IIIII 00 IIglioo . (v) The asserted equality is a restatement of the equality

I/(xW = (SUPZEX I/(x)1)2, valid because r ...... r 2 is an order isomorphism [0, +00) sUPzEX

-+

[0, +00).

0

6.8.13. Definition. A Banach space (real or complex) with an associative bilinear multiplication (a, b) ...... ab satisfying lIabll :5l1allllbll is called a Banach algebra (real or complex).2 6.8.14. Theorem 6.8.12 shows that Cc(X) is a complex Banach algebra. The analogous theorem for real-valued functions holds, showing that CUI (X) is a real Banach algebra; property (v) then takes the form IIPlioo = (11/1100)2. The identity (v) (in both the real and complex cases) is characteristic of the so-called C·-algebras. 3

Exercises 1. The message of Lemma 6.8.4 is that the usual topology on IR is the relative topology induced on it by the usual topology on iC (cf. §3.3, Exercise 7). Extend this result to the relative topology on a subset of a metric space. 2. Let X and Y be nonempty topological spaces, Z = X x Y the product set. Call a subset W of Z open if, for each point z E W , there exist open sets U and V in X and Y, respectively, such that z E U x V C W , and let Oz be the set of all such subsets W. Prove: (i) Oz is the family of open sets for a topology on Z, that is, (Z,Oz) is a topological space in the sense of Definition 3.3.1; Oz is called the product topology on Z, and (Z,Oz) is called the product topological space (of the topological spaces X and Y, in that order). Assume for the rest of the exercise that Z is equipped with the product topology. (ii) Let prx: Z -+ X and pry: Z -+ Y be the projection mappings,

2C.E. Rickart. General theory of Banach algebras (Van Nostrand, Princeton. NJ.I960: reprinted by RE. Krieger, Huntington, NY. 1974J. 3Cf. J. Dixmier. C· -algebras [North.Holland. Amsterdam. 1977), R V. Kadison and J.R. Ringrose, FUndamentals of the theory of operator algebras. Vols. I-IV [Academic Press, New York, 1983-1992).

§6.9. Stone-Weierstrass Theorem

353

defined by prx(x,y) = x,

pry(x,y) = y.

For subsets A C X and BeY, prx -I(A) = A x Y and pry -I(B) = X x B; infer that prx and pry are continuous mappings. (iii) If W is an open set in Z, then prx(W) and pry(W) are open sets in X and Y, respectively. (iv) If I is a function defined on a topological space T and taking values in the product topological space Z = X x Y ,then I: T --+ Z is continuous if and only if both prx 0 I : T --+ X and pry 0 I : T --+ Y are continuous, concisely,

I :T

-+

X x Y continuous #

prx

0

I and pry 0 I continuous.

3. In contrast with Lemma 6.8.9, why was it not necessary in §6.4 to struggle to show that a complex-valued function on a measurable space is measurable if and only if its real and imaginary parts are measurable? 4. The set BVla, b] of all functions I: la, b) --+ IR of bounded variation, equipped with the pointwise linear operations and product, is a (real) Banach algebra for the norm 11111' = 1I/11ex> + V~I (§5.1, Exercise 4).4 5. Let T be a nonempty set, and let Bc(T) be the complex vector space of all bounded, complex-valued functions on T, equipped with the pointwise linear operations and the sup-norm (3.1.10). With products in Bc(T) defined pointwise, Bc(T) is a complex Banach algebra. Similarly, BR (T) is a real Banach algebra for the pointwise operations and the sup-norm. {Hint: Example 6.2.17 and the computation in (iv) of Theorem 6.8.12.}

6.9. Stone-Weierstrass Approximation Theorem If E:'o Cktk is a power series with real coefficients and radius of convergence R > 0 (§1.16, Exercise 3) and if la, b] is a nondegenerate closed subinterval of (- R, R) , then the series converges uniformly and absolutely on la, b] (Example 6.2.8); the formula

ex>

I(t)

=L

Ck tk

(a :5 t :5 b)

k=O

defines a function I: la, b] --+ IR that is continuous on la, b] and differentiable on (a, b) (Example 6.2.28). In particular, I is the uniform limit on la, b] of a sequence of polynomial functions (the sequence of functions defined by the partial sums). • cr. C.E. Rickart,

op. cit., p. 302, A.2.5.

354

6. Function Spaces

Not every continuous function g: [a, b] -+ IR has such a power series representation; for example, the continuous function g: [-1, 1] -+ IR defined by g(t) = ItI fails to be differentiable at the origin. Nevertheless, Weierstrass 1 proved that every continuous function 9 : [a, b] -+ IR is the uniform limit of a sequence of polynomial functions (we just can't expect the differences of successive terms of the sequence to be monomials of increasing degree). Amazingly, the crux of the matter is to prove that the function g(t) = It I on [-I,ll is such a uniform limit. Stated in topological terms, Weierstrass's theorem says that in the algebra Culla, bJ , equipped with the metric defined by the sup-norm (6.2.23), the subalgebra consisting of the polynomial functions is a dense subset. In a tour-de-foree of analysis, M.H. Stone2 isolated the key elements of the proof of Weierstrass' theorem and recast them in a vastly more general theorem about the approximation of continuous functions on a compact space X, the algebra of polynomial functions being replaced by a suitable subalgebra of CR(X) .3 The present section is devoted to an exposition of Stone's theorem.

Notations fixed for the rest of the section: X is a compact topological space (6.1.6); as in the preceding section, CUl(X) and Cc(X) are the algebras of real-valued and complex-valued continuous functions on X, equipped with the pointwise operations and the sup-metric. The core result is a theorem about linear subspaces of CUl(X): 6.9.1. Theorem. If I:- is a linear subspace of CUl(X) such that 1° I:. separates the points of X, 2° I:. annihilates no point of X, and 3° f E I:. ~ f n 1 E I:. , then I:. is dense in CR(X) for the sup-metric, that is, every f E CUl(X) is the uniform limit of a sequence of functions in I:.. Before embarking on the proof, which is divided into a series of five lemmas, some comments on the conditions 1°_3° are in order.

6.9.2. Remarks. 1. The meaning of 1°: If x, y E X with x oF y, then there exists a function f E I:. such that f(x) oF f(y). 2. The meaning of 2°: For each x E X there exists a function fEe such that f (x) oF O. (The condition is trivially satisfied if I:. contains the constant function 1.) 3. The meaning of 3°: If fEe then I:- also contains the function obtained by truncating the graph of f from above at 1, that is, the function (f n I)(x) = min{f(x), I}. 1 Karl Weierstrass (1815-1897). Harvey Stone (1903-1989). 3 M.H. Stone, liThe generalized Weierstrass approximation theorem" [Mathematics Magazine 21 (1948), 167-184,237-254]. 2 Marshall

§6.9. Stone-Weierstrass Theorem

355

4. None ofthe conditions 1°_3° can be omitted in Theorem 6.9.1 (Exercise 1). 6.9.3. Lemma. With .c as in 6.9.1, the uniform closure of .c is also a linear subspace of CIR(X) satisfying 1°-3°.

Proof. Write .c for the closure of .c in CR = CIl (X) for the sup-metric. It is obvious that .c satisfies 1° and 2°; our problem is to show that .c contains sums and scalar multiples and that it satisfies 3°. Let f, 9 E.c and choose sequences (fn), (gn) in.c such that fn --+ f, gn --+ 9 uniformly. Then f n +gn --+ f +9 uniformly and, for every c E IR , cfn --+ cf uniformly; since fn + gn and cfn belong to .c, their uniform limits f + 9 and cf belong to .c. Thus .c is a linear subspace of CR. Moreover, fn n 1 = !{In

+ 1 -Ifn -II},

since fn n 1 E .c and fn n 1 fnlE.c.

--+

f n1=

Hf + 1 -If - IlL

f n 1 uniformly, we conclude that

In view of the preceding lemma, the assertion of Theorem 6.9.1 is that if .c is a closed linear subspace of CR (for the sup-metric topology) satisfying 1°-3°, then .c = CIl; the next lemma is a small but crucial part of the assertion: 6.9.4. Lemma. If .c is a closed linear subspace of CR(X) satisfying 1°-3°, then

f E.c => If I E .c. Proof. Let f E .c. For every positive integer n, f n ~ .c and f n ~

= Hf + ~ -

If - ~I}

--+

= ~ ((n J) n

1»

E

~(f -If!)

uniformly, therefore f no E .c = .c. Then also so Ifl = (f U 0) - (f n 0) E .c.

=f n0 f U0 = - «(- J) n 0)

E .c ,

6.9.5. Lemma. If .c is a closed linear subspace of CR(X) satisfying 1°-3°, then

f, 9 E .c

=>

f U g, f n 9 E .c.

Proof. This is immediate from the preceding lemma and the formulas f U 9 = ~{J

+ 9 + If - gl},

f ng

= Hf + 9 -If -

gl}·

6.9.6. Lemma. If .c is a closed linear subspace of CIl(X) satisfying 1° -3°, then .c contains the constant functions.

Proof. We need only show that 1 E .c. For each x EX, choose f. E .c with f",(x) # 0 (possible by 2°). Replacing f", by If", I, we can suppose

6. Function Spaces

356

(Lemma 6.9.4) that f",(x) > 0, f", ~ 0 on X. Multiplying by a scalar, we can further suppose that

f",(x) > 1, f",

~

0 on X.

Let

then x E U", and, by the continuity of f"" U", is an open set in X. Thus (U",hex is an open covering of X. By compactness, there is a finite subcovering

X = U"'l U ... U U"'n , for suitable Xl>"" Xn in X. Then the function f = f"'l + ... + f"'n belongs to I:- and f > 1 on X, therefore 1 = f n 1 E I:- by the condition 3°. 6.9.7. Lemma. If I:- is a closed linear subspace of CR(X) satisfying 1° -3°, then I:- is f!-fold tmnsitive on X in the following sense:

x,y EX} x#y a,b E lR

=}

3gEI:- 3

g(x)=a and g(y)=b.

That is, for every pair of distinct points of X, there is a function in I:that takes on any specified values at the points. Proof. Let x, y EX, x # y, and let a, b E JR. Choose (by 1°) a function h E I:- such that h(x) # h(y) and let k = h - h(y)1 . Then k E I:- (by the preceding lemma) and k(y) = 0, k(x) = h(x) - h(y) Let gl

= (l/k(x})k; then

# o.

gl E I:- and

gl(x)=1 and gl(y) =0. Similarly, there exists a function g2 E I:- such that

g2(X) and the function 9 and g(y) = b.

=0

= agl + bg2

and g2(y)

= 1,

in I:- has the desired values g(x)

=a

Proof of Theorem 6. 9.1: Let I:- be as in the statement of the theorem. In view of Lemma 6.9.3, we can suppose that I:- is a closed linear subspace of CIR(X) satisfying 1°_3° and our problem is to show that I:- = CIR(X), Given f E CIR(X) and E > 0, it will suffice to show that there is a function gEl:- with IIg - flloo < E (this will show that f E I:- = 1:-).

§6.9. Stone-Weierstrass Theorem

357

The proof rests on the following two properties of £. (verified in Lemmas 6.9.5 and 6.9.7):

(A)

u,ve£. •

X}

x,ye x'!y a, be IR

(B)

uuv,unve£..

3 u e £. :l u(x)

•

=a

and u(y)

= b.

(In words, £. contains finite sups and infs, and is 2-fold transitive on X.) For each pair of points x, y eX, choose a function 9",y e £. such that

9",y(x)

= f(x),

9",y(y)

= f(y)·

{If x,! y, cite (B) with a = f(x) , b = f(y); if x = y, let 9",,,, = f(x)l, which belongs to £. by Lemma 6.9.6.} Trivially,

19",y(x) - f(x)1

(*)

< E and 19",y(y) - f(y)1 < E.

Let U",y = {z eX: 9",y(Z)

< f(z) +E} = (9"'y - f)-I«-oo,E»

V",y={zeX: 9",y(z»f(z)-E}=(9",y-J)-I«-E,+00». By the continuity of 9",y have

f , the sets

x,yeU",y

and

U",y, V",yare open, and by (*) we x,yeV",y.

Fix a point y eX. Construct a function 9y in £. as follows. The sets (U",y)",EX form an open covering of X; by compactness, X = U"'IY u ... U U"'nY for suitable points The function

Xl, .•• ,Xn

(more precisely, n = n(y) depends on y).

belongs to £. by (A).

claim 1: 9y fez) - e (by the definition of VZ'1/); therefore g1/(z) > fez) - e (by the definition of g1/)' whence the claim. The family (V1/)1/EX is an open covering of the compact space X, so X=V1/. U",UV1/_ for suitable points YI,"" Ym in X. The function

belongs to [. by (A), and g f - 11 on X. Let z EX. Say z E V 1/; . Then g(z)

~

g1/;(z) > fez) -e

(the first inequality by the definition of g, the second by claim 2), whence the claim. Thus f - e1 < 9 < f + 11 on X, therefore IIf - glloo :=:; e (in fact, the inequality is strict, since the range of f - 9 is a compact subset of 1R). 0 The motivation for what follows is the observation that

ItI = .Ji2

(t E 1R);

the form of the right-hand side shows that, to approximate the function t ...... ItI by polynomials in t, one need only approximate the square-root function by polynomial functions.

6.9.8. Lemma. In eR[O, 1) , the function t ...... ,fi is the uniform limit of a sequence of polynomial functions without constant term. Proof. Define a sequence of polynomial functions Po, PI , 112 , ••• recursively, as follows: Po(t) 0 and

=

(*)

Pn+l(t)

= Pn(t) + ~ [t -

(Pn(t»2J.

It is clear (by induction) that the Pn are all polynomial functions and that Pn (0) = 0 for all n. claim: O:=:; Po(t) :=:; PI(t) :=:; ... :=:; Pn(t) :=:; ,fi on [0,1). The proof is by induction on n. For n = 0 the assertion is trivial. Assuming all's well for n, for every t E [0,1) we have

(Pn(t»2 :=:; t,

Ht - (Pn(t»2J ~ 0,

§6.9. Stone-Weierstrass Theorem

359

therefore Pn+l(t);:: Pn(t) by (*). Also

,Ji - Pn+l(t)

= ,Ji -

Pn(t) - ~ [t - (Pn(t»2]

= (,Ji - Pn(t»)- M,Ji - Pn(t)][,Ji + Pn(t)J

(**)

= [,Ji -

Pn(t)]{l - ~(,Ji + Pn(t)]};

but ,Ji+Pn(t) ~ ,Ji+Vi ~ 2, therefore MVi+Pn(t)1 ~ 1; it follows that both factors in the rightmost member of (**) are ;:: 0 (the first factor, by the induction hypothesis), consequently Vt-Pn+l (t) ;:: 0, which completes the induction. Define f(t) = sUPnPn(t) for all t E [O,lJ; thus

o~

f(t)

~

1 and Pn(t) T f(t)

for all t E [0, 1). Passing to the limit in (*), we have

f(t)

= f(t) + ~ [t -

(/(t»2] ,

whence t - (/(t»2 = 0, f(t) = Vi, Thus Pn(t) T Vi for all t E [O,lJ. Since the Pn and the square-root function are continuous, it follows from Dini's theorem (6.2.24) that the Pn converge to the square-root function uniformly on [0, 11. 6.9.9. Theorem. (Stone-Weierstrass theorem, real case) Let A be a subalgebra of CR(X) such that lOA separates the points of X. and 20 A annihilates no point of X . Then A is uniformly dense in ClK (X) .

Proof By assumption, A is closed under pointwise sums, products and scalar multiples, in particular. A is a linear subspace of ClK. Let

B=A be the uniform closure of A in CR (i.e., the closure of A in ClK for the topology derived from the sup-metric). Since B:J A it is clear that B also satisfies 10 and 20 • claim: B is a subalgebra of CR. By the argument in Lemma 6.9.3, B is a linear subspace of ClK. If f.9 E B and if (fn). (9n) are sequences in A such that

fn then fn9n

--t

--t

f,

9n

--t

9

uniformly.

f9 uniformly. as one sees from the computation

fn9n - fg = (In - f)(gn - g) + f(gn - g) + (In - f)g. IIfngn - fglloo ~ IIfn - flloollgn - glloo + IIflloolign - glloo + IIfn - flloollglloo, therefore fg E A = B.

360

6. FUnction Spaces

Changing notation, we can suppose that A is closed for the uniform topology and our problem is to show that A = CR . claim: If f E A and f? 0 then Vl E A. By .J1 we mean the (continuous) function x ...... oJf(x) (x EX). Passing to a scalar multiple of f, we can suppose that 0:5 f :5 1. Let s : 10,lJ - t IR be the usual square-root function set) =...It (0:5 t :5 1), so that .J1 = so f . By Lemma 6.9.8, there exists a sequence (Pn) of real polynomial functions, without constant term, such that Pn - t S uniformly on [O,lJ. Since Pn has no constant term, Pn 0 f E A (for example, if Pn(t) == alt+a2t2+ ...+aNt N then Pnof = ad+ad 2 + ...+aNf N E A). Moreover, Pn 0 f - t so f uniformly on X, as one sees from the computation IIPn

0

f - s 0 flloo =

s) 0 flloo < IIPn - slloo

II (Pn -

(the first two sup-norIns are calculated as x varies over X, the third as t varies over [0, 1J), consequently s 0 f E A = A, whence the claim. (Alternatively, Pn 0 f - t so f uniformly on X because Pn - t S uniformly on f(X) C [0, 1J.) It follows from the preceding claim, and the formula If I = that f E A => IfI EA. Summarizing: A is a closed linear subspace of CIR, satisfying the conditions 10 and 2°, such that f E A => If I E A (hence A is also closed under finite sups and infs). Th complete the proof that A = CR , we need only show that A also satisfies condition 30 of Theorem 6.9.1; since f,g E A => f n g E A, it will suffice to show that 1 E A. claim: 1 EA. For each x E X there exists (by condition 2°) a function f" E A with f,,(x) ,;, O. Replacing f" by (f,,)2 (or by If"l) we can suppose that

VJ'i,

f,,? 0 on X and

f,,(x) > O.

Multiplying by a scalar, we can suppose further that f,,(x) > 1.

The set

U"

= {y EX:

f,,(y) > 1 }

is open and x E U", thus the family (U,,),.ex is an open covering of X; by compactness, X

for suitable points

= U". u ... U U"n

Xl, .•. , X n .

Define

f = f"l

+ ... + f"n

;

then f E A and f> 1 on X. Note that if r E IR, r > 1 ,then r l/n ! 1 (because log r > 0 therefore I ' logr II n = nlogr!O=logl). It follows that fl/n! 1 pointwise on X,

§6.9. Stone-Weierstrass Theorem

361

therefore

f lIn ! 1 uniformly on X R

by Dini's theorem (6.2.24). In particular, the subsequence fl / 2 converges to 1 uniformly; since fl / 2 E A (by induction: p/2 E A and f 1/2 + 1 = ,jf'/2R ), it follows that 1 E A = A as claimed, and the proof of the theorem is complete (by the discussion preceding the claim). R

R

6.9.10. Corollary. (Weierstrass approximation theorem) For every continuous real-valued function f: [a, bJ -+ IR on a closed interval [a, bj , there exists a sequence of real polynomial functions (Pn) such that Pn -+ f uniformly on [a, bj .

Proof. A real polynomial function on [a, b] is a function p: [a, bJ -+ IR such that p(t} = L:~=ocktk for all t E [a,b). where CO,C1>""Cn are suitable real numbers. Such functions are obviously continuous, and the set A of all such functions is a subalgebra of CR [a, b] that meets the requirements of the preceding theorem: for example, the monomial function p(t} t single-handedly separates all pairs of points of [a, bj, and the constant function p(t} 1 annihilates no point of [a, b].

=

=

For continuous complex-valued functions, it is necessary to assume that the subalgebra is closed under complex-conjugation of functions: 6.9.11. Corollary. (Stone-Weierstrass theorem, complex case) Let X be a compact space, B a (complex) subalgebra of Cc(X) such that (i) B separates the points of X, (ii) B annihilates no points of X, and (iii) fEB => fEB (where f is the complex-conjugate of Then B is uniformly dense in Cc(X).

n.

Proof. Recall that f(x} = f(x} for all x EX, where f(x} is the conjugate of the complex number f(x}. As in the preceding section (cl. 6.8.11), we regard CR(X} as the lR-subalgebra of Cc(X} consisting of all functions f E Cc(X} that are real-valued. Let

A= BnCR(X} be the set of real-valued functions in B; since both B and CR(X} are IR-sublgebras of Cc(X} (that is, subrings of Cc(X) that are also 1R-linear subspaces of Cc(X)}, the same is true of A. Moreover, it is clear from the assumption (iii) and the formulas

f=Ref+iImf, valid for every (*)

Ref=~(f+f), Imf=~i(f-f),

f E Cc(X} , that fEB

~

Ref, Imf E A.

6. Function Spaces

362

The idea of the proof is to apply the real case of the Stone-Weierstrass theorem to the subalgebra A of CIR(X); to this end, let us verify that A satisfies the conditions 1° and 2° of Theorem 6.9.9. (1°) If x,y E X, x -I y, choose IE B so that I(x) -I I(y); then one of ReI, Iml is a function 9 E A such that g(x) -I g(y). (2°) If x EX, choose I E B so that I(x) -I 0; then one of Re f , 1m I is a function 9 E A such that g(x) -10. It now follows from Theorem 6.9.9 that A is uniformly dense in CIR(X). To complete the proof, we need only show that every I E Cc(X) is the uniform limit of a sequence of functions In E B. Write I = 9 + ih with g, h E CIR(X) (the real and imaginary parts of I), and let (gn), (h n ) be sequences in A such that gn --+ 9 and hn --+ h uniformly. Then the functions In = gn + ih n belong to B by (*), and In --+ 9 + ih = I uniformly.

Exercises 1. In Theorem 6.9.1, none of the conditions 1°-3° can be omitted.

{Hint: Let X = {1,2} c JR be the discrete space with two points, so that CDl(X) = F(X,JR) can be identified with the set of all ordered pairs (Xl> X2) ofreal numbers, that is, with JR2. Consider, in turn, the following linear subspaces of JR2: [. = ({c, c) : c E JR}, the set of all constant functions; [. = {(O, c): c E JR} , the set of all functions that vanish at 1; [. = {(c,2c) : c E JR}, the set of all scalar multiples of the insertion mapping L: X --+ JR, L(X) = x.} 2. Let K be a nonempty compact subset of iC, u: K --+ iC the insertion mapping u(z) = z (z E K), u the conjugate function u(z) = Z (z E K), and let B be the subalgebra of Cc(K) generated by u and u; thus, a typical element of B is a linear combination of functions of the form

z ..... zmzn

(zEK),

where m, n are nonnegative integers. The functions p E B are called polynomials in z and z, a typical such function having the form 00

p(z)

=

L

Cm,n zmzn

(z

E K),

m,n=O

where all but finitely many of the coefficients Cm,n are equal to O. (The underlying algebraic concept: the algebra iC[s, t] of polynomials in two commuting indeterminates s and t; such a polynomial determines a function of z E K via the substitutions s ..... z, t ..... z.) (i) B is uniformly dense in Cc(K). (ii) If K = 1IJ = {z E iC: Izi = I} (the unit circle in the complex plane)

§6.9. Stone-Weierstrass Theorem

then

z = Z-l

363

for all z E K , and every p E B can be written in the form

p(z)

= 2>kZk

(z E lU)

kEZ where, in the notation of the earlier representation of p, Ck

=

:L:

Cm,n

m-n=k

( = 0 for all but finitely many integers k). If p E B then the function F: IR -> C defined by F(x) = p(e 2"iX) is continuous, periodic of period 1,

and has the representation F(x)

= I:: Ck e2"ikx = I:: cklcos 21l'kx + isin 21l'kx] kEZ

kEZ

(such functions are called trigonometric polynomials). (iii) With lU as in (ii), every continuous periodic function F: IR of period 1 has a representation F(x)

= g(e 2"ix)

-+

iC

(x E IR)

for a suitable function 9 E Cc(lU) , hence is the uniform limit of trigonometric polynomials. The proof of the existence of 9 entails a slight digression into "quotient topologies" .4

3. A topological space is said to be locally compact if it is separated (6.1.6) and if each point of the space has a compact neighborhood (in which case every neighborhood of a point contains a compact neighborhood of the point).5 Let X be a noncompact, locally compact space (for example, X = jRn with the usual topology). (i) A continuous function I: X -> IR (or iC) is said to vanish at infinity if, for every £ > 0, the set K. = {x EX: I/(x)1 ~ £} is compact, Such a function is necessarily bounded (/(K 1 ) is compact and I/(x)1 < 1 on X - K 1 ). The set Co(X) of all continuous functions vanishing at infinity is a Banach algebra for the pointwise operations and the sup-norm 11/1100 = SUPxEX I/(x)1 (cf. 6.8.14). (iii) The Stone-Weierstrass theorem extends to X, provided that C(X) is replaced by Co(X); that is, if A is a subalgebra of Co(X) that separates the points of X, annihilates no point of X, and is closed under complex conjugation, then A is dense in Co (X) for the norm topology."

• cr. J. Dixmier, General topology [Springer-Verlag, New York, 1984), p. 83, Corollary 7.5.6. 5 Op. cit., p. 46, Definition 4.5.2. • Op. cit., p. 84, Corollary 7.5.8.

CHAPTER 7

Product Measure

§7.1. §7.2. §7.3. §7.4.

Extension of measures Product measures Iterated integrals, the case of finite measures Fubini-Tonelli theorem for u-finite measures

If (X,S,/,) and (Y, T,v) are measure spaces, how can /' and v be combined to define a measure on a suitable u-algebra of subsets of X x Y ? The question is not academic: a satisfactory answer would open the door to constructing measures on 1R2 , 1R3 , •.• , starting with Lebesgue measure on IR. Lebesgue measure A on IR assigns to an interval [a, bl its length b - a; we expect the measure 11' on IR 2 derived from A to assign to a rectangle [a, bl x [c, dJ its area (b - a)(d - c). Thus, writing 1 = [a, b] 2 and J = [e, dl , our expectation is that the 'planar measure' 11' on 1R derived from A should satisfy the formula

11'(1 x J) = A(I)A( J) . More generally, we would like to define a measure 11' on a suitable u-algebra of subsets of X x Y that includes all 'measurable rectangles' Ex F (E E S, F E T) , such that the 'product rule' 1I'(E x F)

= /,(E)v(F)

holds for every measurable rectangle. Such a measure exists l but it is not necessarily unique (Exercise 2). However, when the given measures /' and v are u-finite, 11' is uniquely determined by the above 'product formula'; inasmuch as the uniqueness property is highly desirable (and ufiniteness is needed for other reasons as well), we shall limit the construction of 11' to the case of u-finite measures. The first step is to define a set function on the algebra of sets generated by the measurable rectangles E x F and satisfying the product rule, then extend it to the u-algebra generated 'CI. H.L. Royden, Real analysis (3rd edn., Macmillan, New York, 1988], p. 303ff; H.S. Bear, A primer of Lebesgue integration (Academic Press, New York. 1995J. Chapter 14. 364

§7.1. Extension of Measures

365

by the measurable rectangles; accordingly, we begin by considering a 'measure' defined on an algebra A of sets, and show how to extend it to a measure on the u-algebra S(A) generated by A (2.4.4).

7.1. Extension of Measures

Recall (2.4.1) that an algebm of subsets of a set X is a nonempty set A c 'P(X) that is closed under complementation and finite unions (thus a u-algebra is an algebra that is closed under countable unions). In Chapter 2, measures were defined to have domain au-algebra (2.4.12); it is useful to allow the domain to be merely an algebra, but a 'measure' is still required to be countably additive in the appropriate sense:

7.1.1. Definition. Let X be a set, A c 'P(X) an algebra of subsets of X. A function p.: A -+ [0, +ooJ is said to be a measure if 1° p.(u 1, the Fn are pairwise disjoint sets in .A with union E E .A, and Fn C En for all n, thus

by the countable additivity (7.1.1) and monotonocity (7.1.2) of /-I. claim: If E E.A then /-I. (E) = /-I(E) ; that is, the set function /-I. is an extension of the set function /-I. For, E C E E.A shows that /-I·(E) :5 /-I(E) by the definition of /-I•. On the other hand, if (En) is a sequence of sets in .A with E c U::'=l En , then E = U::'=l En En with En En E.A for all n, therefore

by the countable subadditivity and monotonicity of /-I; varying over all coverings of E by such sequences (En), we conclude that /-I(E):5 /-I. (E) . Thus /-I·(E) = /-I(E) as claimed. The proof that /-I. is an outer measure follows the format of the proof of Theorem 2.1.6. 7.1.4. Definition. Let p be an outer measure on a set X. Recall that a set B C X is said to be p-measurable if it splits every subset A of X additively, in the sense that p(A) =p(AnE)+p(AnE')

for all A M(p).

c

X (2.2.7). The set of all p-measurable sets will be denoted

7.1.5. Theorem. If p is an outer measure on a set X, then the set M(p) of all p-measurable sets is a u-algebra of subsets of X, and the restriction of p to M(p) is a measure. Proof. It is obvious that the empty set is p-measurable and that M(p) is closed under complementation. As noted in 2.2.6, M(p) is closed under countable unions and the restriction of p to M(p) is countably additive (d. 2.2.5).

A measure on an algebra can always be extended to a measure on a u-algebra:

§7.1. Extension of Measures

367

7.1.6. Theorem. (Hahn-Kolmogorov extension theorem)l Let A be an algebra of subsets of a set X and let J.L be a measure on A (7.1.1). Then the set M(J.L°) of all J.L0 -measurable sets is a u-algebra containing A, and the restriction of J.L0 to M(J.L°) is a measure that extends J.L.

Proof. We know that 1'0 is an outer measure (7.1.3), that M(J.L°) is a u-algebra and that the restriction of J.L0 to M(J.L°) is a measure (7.1.5). From Theorem 7.1.3 we also know that J.L°IA = J.L, thus it remains only to show that A c M(J.L°). Assuming E E A and A c X, we need only show that J.L°(A)

~

1'0 (A () E)

+ J.L°(A () E')

(cf. 2.2.3, (i». We can suppose that J.L°(A) < +00; given any will then suffice to show that

J.L0 (A)

+f

~

J.L°(A () E)

f

> 0, it

+ J.L°(A () E') .

By the definition of J.L0 , there exists a sequence (En) of sets in A such that A c UEn and

L J.L(En) :5 J.L°(A) Then A () E c

U En () E ,where

+ f.

En () E E A for all n, therefore

J.L°(A n E) < LJ.L(En n E);

(1) similarly,

J.L°(A n E') < LJ.L(En n E').

(2)

By the finite additivity of J.L,

J.L(En () E) + J.L(En n E') = J.L(En ) for all n, thus the addition of (1) and (2) yields

J.L°(A n E) + J.L°(A n E') < LJ.L(En) :5 J.L°(A) + f, which verifies (*). 0 With notations as in the preceding theorem, since M(J.L°) is au-algebra containing A. it contains the u-algebra S(A) generated by A (2.4.4), thus

A

c S(A) c Mi(Fn n E)

(i=1,2);

n=l

it will thus suffice to show that for all n. Thus, assuming F E A and J.ii (F) < +00, it will suffice to show that for all E E S.

7. Product Measure

370 For i == 1, 2 define

Vi:

S

Vi (E)

-+

==

IR by the formula

lLi{F

n E)

(' iC is a simple function with respect to S x T (hence is integrable with respect to p. x v), then every section of h is integrable, both iterated integrals of h exist, and

Proof. By linearity (Remark 7.3.4, (ii» we can suppose that h is the characteristic function ofa set M E S x T , a case that is covered by Remark 7.3.4, (i). Explicitly, if h = L:~=l Ck'PM. then, for all x E X and y E Y, n

n

h%

= L Ck'P(M.l. ,

hY =

k=l

and

L

Ck'P(M.l· ,

k=l

J

hYdp.

= f:Ckg M • (y), k=l

thus the requirements of Definition 7.3.3 are met with E = rz:" L:~=l Ck/M., F = rz:" 9 = L:~=l CkgM •. 0

f =

The following theorem will be generalized to a-finite measures in the next section: 7.3.6. Theorem. (Fubini-Tonelli theorem! finite case) Let (X,S,p.) and (Y, T, v) be finite measure spaces, h : X x Y -> 1R. If h ~ 0 and h is measurable with respect to S x T , then the following conditions are equivalent: (a) h is integrable with respect to JL x v;

'Guido Fubini (1879-1943), Leonida Tonelli (1885-1946).

7. Product Measure

386

JJ JJ

(b) hdvdp. exists; (c) hdp.dv exists. When this is the case, J hd(p. x v)

= JJ hdvdp. = JJ hdp.dv.

Proof. Let (hn) be a sequence of simple functions (with respect to S x T) such that 0::; h n Th. Every section of hn is simple (with respect to S or T, as the case may be), hence integrable (with respect to p. or v), and 0::; (hn)x Th x , 0::; (hn)U Th U for all x EX, Y E Y. Define fn(x)

=J

(hn)x dv ,

9n(Y)

=J

(hn)Udp.

for x EX, Y E Y. As noted in the proof of the lemma, In E £.I(p.) , 9n E £1(£1) and (*)

J Indp.

=J

9n dv

=J

hnd(p. x v)

for all n.

Since (hn)x T for each x, and (hn)U T for each y, we have fn

T and

9n

T

by the monotonicity of integration. (a) => (b): Suppose hE £1(p. X v). Then J hnd(p. x v) T J hd(p. x v)

(by Definition 4.4.3, or by the monotone convergence theorem, 4.5.3), thus J Indp.

T J hd(p. x v) < +00

by (*). Since (In) is an increasing sequence, it follows from the monotone convergence theorem (for p.) that there exists f E £1(p.) such that In T I a.e. (for p.) and J fdp.

Choose E E S, p.(E) x EX - E,

= s~p J

=

=J

hd(p. x v).

0 so that In T I

sup J(hn)xdV n

fndp.

on X - E; then, for all

= sup In (x) = I(x) < +00, n

and since (hn)x T h x and h x is measurable (with respect to T), it follows from the monotone convergence theorem (for v) that h x E £ 1 (v) and

§7.3. Iterated Integrals

387

The pair E, f meets the requirements of Definition 7.3.3, thus the iterated hdlldlL exists and integral

JJ

II

hdlldlL

=

I

fdlL

=

JJ

I

hd(lL x II).

J

(a) ~ (c): Similarly, with hdlLdll = hd(lL x II). (b) ~ (a): Suppose hdlldlL exists, and let E,f be as in Definition 7.3.3. For each x E X - E we have 0:5 (h,.lz Th z and

JJ

s~p I(h,.)zdll =

I

hzdll

< +00

by the monotone convergence theorem, in other words fn(x) T f(x).

Thus fn T f a.e. (for IL) and J fndlL TJ fdlL by the monotone convergence theorem; in view of (*), we see that the sequence

I

h,.d(1L x II) =

I

fndlL

is bounded, whence h E £1 (IL x II) by the definition ofintegrability (4.4.3). (c) ~ (a): Similarly. 7.3.7. Corollary. (Fubini's theorem for finite measures) Let (X,S,IL) and (Y, T, II) be finite measure spaces. If h E £~(JL x II) then both iterated integrals of h exist and are equal to J hd(lL XII) .

Proof. By linearity (7.3.4, (Ii)) we can suppose that h;::: 0, then cite the theorem. The following corollary will be useful in the next section for generalizing to the u-finite case: 7.3.8. Corollary. Let (X,S,IL) and (Y, T,II) be finite measure spaces. If h : X x Y -+ C is bounded and measurable (with respect to S x T), then every section of h is integrable and, defining f(x)

=

I

hzdll,

g(y)

=

I

h!ldlL

for all x E X and y E Y, we have f E £I(IL) , g E £1(11) and

II

hdlldlL

=

II

hdlLdll

=

I

f dlL

=

I

gdll

=

I

hd(lL XII) .

Proof We can suppose by linearity that h;::: 0 (7.3.4, (Ii)). At any rate, we know that h is integrable with respect to the finite measure IL x II (4.4.5). For every x EX, hz is bounded and measurable with respect

7. Product Measure

388

to T (7.3.2, (iii)), therefore integrable with respect to v; similarly for ysections. Thus I and 9 are defined everywhere on X and Y, respectively. Note that I and 9 are bounded. For, if 0 < h :5 K < +00, then f h x dv:5 Kv(Y) < +00 for all x E X, thus

o :5 I

< K v(Y) < +00

and similarly 0:5 9 :5 K Jl(X) < +00. The case that h is simple is covered by Lemma 7.3.5. In general, let (h,,) be a sequence of simple functions (with respect to S x T) such that 0:5 hn Th. Define

In(x)

=

J

(h,,)xdv

(x E X);

we know that In E £l(p) , in particular In is measurable with respect to S.Foreach XEX, (h,,)x

Th x

E £l(V) ,

and integration with respect to v yields In(x) T I(x). Thus In T I pointwise. Since the In are measurable, so is their pointwise (real-valued) limit I; also, I is bounded, so IE £l(p) . Similarly, 9 E £l(V) , and the asserted formulas are immediate from the theorem.

Exercises

I :X

Y, 'PB : Y -+ {O, I} is the characteristic function of a set BeY, and 'PB 0 I : X -+ {O, I} is the composite function, then 1. If

'PB 0

-+

1= 'PI-I(B) .

2. Let Ao be Lebesgue measure on the closed interval [0,1) (Example 4.7.6), defined on the u-algebra Mo of Lebesgue-measurable sets contained in [O,IJ, and let 11: = Ao x AO' Let h: [0,1] x [0,11 -+ IR be the function such that h(x,O) = l/x for 0 < x :5 1 and h(x, Y) = 0 for all other (x, y) . (i) h is measurable with respect to Mo x Mo (indeed, with respect to 8 0 x 80, where 8 0 is the u-algebra of Borel sets contained in [O,lJ). (ii) h is 1I:-integrable. (iii) The section hO of h is not Ao-integrable. {Hint: (i) Apply criterion (b) of 4.1.8; for example, if c > 0 then ((x,y): h(x,y»c}=(O,I/c) x(O,I). (ii) hO(x)=I/x for x>O.}

7.4. Fubini-Tonelli Theorem for u-Finite Measures Throughout this section, (X,S,p) and (Y, T,v) are u-finite measure spaces, that is, p and v are u-finite measures on the u-algebras Sand T

§7.4. Fubini-Tonelli Theorem for a-Finite Measures

389

(7.1.9). Let (X x Y,S x T,J1. x v) be the product measure space (7.2.13); as observed in the course of its construction, the measure J1. x v is also a-finite (7.2.12). Our objective is to extend the results of the preceding section (for finite measures) to the a-finite case. The strategy is the same as in the construction of J1. x v: express each of J1. and v as the supremum of a suitable increasing sequence of finite measures, let the finite case do all of the hard work, then 'pass to the limit' via some easy technical lemmas. As in the proof of 7.2.12, let (Pn ) be a sequence of sets in S such that P n TX and J1.(P n ) < +00 for all n, and let (Qn) be a sequence of sets in T such that Qn TY and v(Qn) < +00 for aU n. Then

(J1. x v)(P n x Qn) = J1.{Pn)v(Qn) < +00 for all n. The formulas

= J1.(PnnE) vn(F) = v(Qn n F)

J1.n (E)

(EE S) (FE T)

define finite measures J1.n on S and V n on T such that J1.n V n Tv. The measures J1.n x V n are finite (7.2.11). 7.4.1. Lemma. J1.n x V n

T J1.

and

TJ1. x v.

Proof In view of the definition of J1. x v (7.2.13), the argument for this is given in the proof of Corollary 7.2.12. 0 The following notation, already employed in 4.8.4 in connection with finite signed measures, will also be useful in the present context:

7.4.2. Definition. For each measurable set A E S, define a set function J1.A : S ..... [0, +00) by the formula (E E S).

(One calls J1.A the contraction of J1. to A.)

7.4.3. Remarks. (i) The set function J1.A introduced in the foregoing definition is a measure, by a straightforward argument that is evidently valid when J1. is replaced by an arbitrary (not necessarily a-finite) measure. (ii) Every contraction of the a-finite measure J1. is also a-finite. In order that J1.A be a finite measure, it is necessary and sufficient that J1.(A) < +00. (iii) With the notations introduced at the beginning of the section, J1.n = J1.p. and V n = vQ•. 7.4.4. Lemma. Suppose A E Sand respect to S (6.4.1). Then

f

E

.cHJ1.A)

~

f : X .....

ce

f'PA E 4(J1.),

is measurable with

7. Product Measure

390

and in this case

J

Idp.A

=

J

l'PAdp..

Prool (valid for any measure space). By linearity, we can suppose without loss of generality that I ~ o. {The crux of the matter is that Re(f'PA) = (Re/)'PA and Im(f'PA) = (1m f)'PA; and, when I is realvalued, that (f'PA)+ = !+'PA and (f'PA)- = I-'PA.} case 1: I = 'PE for some E E S. Then

IE .c1(P.A)

¢}

p.A(E) < +00 p.(A n E) < +00

¢}

'PAnE E .c1(p.)

¢}

'PA'PE E .c1(p.)

¢}

!'PA E .c1(p.)

¢}

and in this case we have

J

Idp.A

= p.A(E) = p.(A n E) =

J

'f'AnEdp.

=

J

!'PAdp..

case 2: I simple (and ~ 0). Immediate from case 1 and linearity. case 9: the general case (I ~ 0) . Let (fn) be a sequence of simple functions such that 0 pointwise on X (4.1.26). Then

< In T I

0::5 In'PA T I'PA' where the In'PA are simple functions. Citing the definition of integral (4.4.1,4.4.3) and case 2, we have IE .c1(P.A)

¢}

In E .c1(P.A) (\f n) and

¢}

In'PA E .c1(p.) (\f n) and

¢}

!'PA E .c1(p.) ,

J J

Indp.A is bounded In'PAdp. is bounded

and in this case,

J

Idp.A

= s~p

J

Indp.A

= s~p

J

In'PAdp.

=

J

!'PAdp..

0

7.4.5. Lemma. Let (fn) be a sequence 01 functions on X such that, lor every n, 0::5 In E .c1(p.n) and In ::5 In+! p.-a.e. The lollowing

§7.4. Fubini-Tonelli Theorem for ,,-Finite Measures

391

conditions are equivalent: (a) the sequence J fndJ.Ln is bounded; (b) there exists a function Ie £1(J.L) such that In TI J.L-a.e. When this is the case,

J

IdJ.L =

s~p

J

fndJ.Ln.

Proof. Note that In(x) is an increasing sequence for J.L-almost every

= J.LPn , by the preceding lemma we

x eX, briefly In T J.L-a.e. Since J.Ln have In'PP n e £1(J.L) and

J

fn'PP n dJ.L

=

J

fndJ.LP n

=

J

IndJ.Ln;

also In'PP.. Tj.L-a.e. By the monotone convergence theorem for J.L, in order thatthereexistafunction Ie £1(J.L) such that In'PPn TI J.L-a.e. (equivalently In T I J.L-a.e., since, for each x eX, 'PP.. (x) = 1 ultimately), it is necessary and sufficient that

J s~p J

s~p in other words

In'PPndJ.L

< +00,

IndJ.Ln < +00;

when this is the case, we have

s~p

J

IndJ.Ln

= s~p

J

In'PpndJ.L

=

J

IdJ.L.

The analogue of the preceding lemma of course holds for the sequence Vn Tv. 7.4.6. Lemma. II A e Sand BeT then (J.L x V)AxB

= J.LA

X

VB .

Proof. All measures in sight are a-finite and, for every measurable rectangle ExFeSxT,

(J.L x V)AxB(E x F) = (JL x v)«A x B) n (E x F»

= (J.L x v)(A n E) x (B x F») = J.L(A n E)v(B n F)

= J.LA(E)VB(F) = (J.LA x VB)(E x F), therefore (J.L x V)AxB = J.LA 7.2.12.

X

VB by the uniqueness part of Corollary

In particular, (J.L x v)PnxQn = J.LP n X vQn = J.Ln x V n , therefore the analogue of Lemma 7.4.5 also holds for the sequence J.Ln x V n TJ.L x v.

7. Product Measure

392

7.4.7. Theorem. (Fubini-Tonelli) If (X, S, J.l) and (Y, T, II) are u-finite measure spaces, and if h is a measurable function with respect to S x T such that h;:: 0 on X x Y, then the following conditions are equivalent: (a) hE £1(J.l X /I); (b) JJ d/ldJ.l exists; (c) JJ hdl-'dll exists. When this is the case,

J

hd (J.l x /I) =

JJ

JJ

hdlldl-' =

hdJ.ld/l.

Proof. The foregoing notations for the measures I-'n and /In are in force. Let (hi) be a sequence of simple functions on X x Y (with respect to S x T) such that 0 ~ hi t h. The hi are of course integrable with respect to the finite measures J.ln x /In ; replacing hi by hi'Pp; x Q, , we can suppose further that It;. is integrable with respect to I-' x II. For each i and n, define

(x E X); by Lemma 7.3.5 (or Corollary 7.3.8) we know that fi(n) E £1(J.ln) and (1)

Note also that for each n. fi(n)

(2)

(because, for each x EX, (h i )%

as i

t

T pointwise on Y). and that

for each i, ft>

(3)

t t

as n

t

(because (lin) is an increasing sequence of measures). (a) ~ (b): Suppose hE £1(J.l X /I). Then, citing (1), we have

dJ.ln

=

J

hid(J.ln x /In)

~

J

hid(J.l x /I)

~

J

hd(J.l x /I)

< +00

for all i and n. Fix an index i. By (3), (*) and Lemma 7.4.5, there exists /; E £1 (1-') such that f!n) t /; J.l-a.e., and

(4)

J

/;dJ.l

= s~p

J

fi(n> dJ.ln

= s~p =

J

J

hid(J.ln x lin)

hid(J.l x II)

(the first equality. by 7.4.5; the second, by (1); the third. by 7.4.5 applied

7. Product Measure

392

7.4.7. Theorem. (Fubini-Tonelli) If (X, S, J.l) and (Y, T, II) are u-finite measure spaces, and if h is a measurable function with respect to S x T such that h;:: 0 on X x Y, then the following conditions are equivalent: (a) hE £1(J.l X /I); (b) JJ d/ldJ.l exists; (c) JJ hdl-'dll exists. When this is the case,

J

hd (J.l x /I) =

JJ

JJ

hdlldl-' =

hdJ.ld/l.

Proof. The foregoing notations for the measures I-'n and /In are in force. Let (hi) be a sequence of simple functions on X x Y (with respect to S x T) such that 0 ~ hi t h. The hi are of course integrable with respect to the finite measures J.ln x /In ; replacing hi by hi'Pp; x Q, , we can suppose further that It;. is integrable with respect to I-' x II. For each i and n, define

(x E X); by Lemma 7.3.5 (or Corollary 7.3.8) we know that fi(n) E £1(J.ln) and (1)

Note also that for each n. fi(n)

(2)

(because, for each x EX, (h i )%

as i

t

T pointwise on Y). and that

for each i, ft>

(3)

t t

as n

t

(because (lin) is an increasing sequence of measures). (a) ~ (b): Suppose hE £1(J.l X /I). Then, citing (1), we have

dJ.ln

=

J

hid(J.ln x /In)

~

J

hid(J.l x /I)

~

J

hd(J.l x /I)

< +00

for all i and n. Fix an index i. By (3), (*) and Lemma 7.4.5, there exists /; E £1 (1-') such that f!n) t /; J.l-a.e., and

(4)

J

/;dJ.l

= s~p

J

fi(n> dJ.ln

= s~p =

J

J

hid(J.ln x lin)

hid(J.l x II)

(the first equality. by 7.4.5; the second, by (1); the third. by 7.4.5 applied

§7.4. Fubini-Tonelli Theorem for ,,-Finite Measures to the sequence of mueasures JLn

393

x Vn TJ1. x v and the 'constant sequence'

of functions h;, h; , h; ... ) claim: 1; T J1.-a.e. Let us show, for example, that h:::; 12 a.e. (with respect to J1.). Since I~n) T It a.e. and IJn) T 12 a.e., there exists a set E E S such that J1.(E) = 0 and such that, for all x E X - E,

It (x)

= sup I~n)(x),

12(x) = sup IJn) (x) .

n

n

Let x EX - E; by (2), we have

It)(x):::; IJn)(x)

for all n,

and It(x):::; 12(x) results on takingsuprememover n. Thus h:::; 12 a.e. Similarly I; < Ii+ 1 a.e. for each i. Since a countable union of null sets is null, we conclude that 1; T a.e. Summarizing, we have 1; E .c 1 (J1.) for all i, I; T J1.-a.e., and, citing (4) and the definition of integral (4.4.3), we have

J l;dJ1. = J h;d(J1. x

(5)

v) T J

hd(J1. x

v)

by the monotone convergence theorem, there exists Ii TI J1.-a.e., and

< +00;

I

E

.c 1 (J1.)

such that

J IdJ1. = s~p J l;dJ1. = J hd(J1. x v).

(6)

Let E E S with J1.(E) = 0 and 1; T I pointwise on X - E. For each i, I;(n) T I; J1.-a.e.; choose E; E S with J1.(E;) = 0 and I;(n) T I; on X-E;. Let 00

A=EUUE;; i=l

then J1.( A) = 0 and (7)

xEX-A

=>

+OO>/(X)=S~P/;(X)=sUP(suP/;(n)(X») , ,n =

in particular, (8)

x E X - A =>

J s~p J

(h;)zdvn

:::;

s~p (s~p J(h;hdVn)

,

I(x) < +00 for all i and n.

Fix xEX-A.Fix n.Then (h;)zdvn :::; I(x)

< +00

7. Product Measure

394 by (8); since (h;)% .c1 (vn ) and

T h% and h%

is measurable, we conclude that h% E

(9)

Now vary n; since and

V

n Tv, by Lemma 7.4.5 we conclude that h%

(10)

j h%dv

= s~p j

Gathering up all the strands, we have all XEX-A,

j

~dv = s~p j

h%dvn

I

E £l(v)

•

E £1(1'), p.(A)

=0

and, for

h%dvn

= s~p (s~p j(hi)%dvn ) = s~p (s~p j

(hi)%dvn)

= I(x) (the first equality, by (10); the second, by (9); the third, by the 'associativity of sups,' cf. §1.l5, Exercise 7; the fourth, by (7», thus II hdvdp. exists and is equal to I Idp.-hence, by (6), to I hd(p. x v). (a) => (c): The proof is similar, leading to the equality If hdp.dv =

I hd(p. x

v).

(b) => (a): Suppose If hdvdp. exists. Let E E S and such that p.(E) = 0 and

x EX - E

=>

h%

E

.c 1 (v)

and j h%dv

I

E

.c1(p.)

be

= I(x).

Let x E X - E; then 0 ::; (h i )% T h% E .c 1 (v) , where the (h,)% are measurable (7.3.2, (iii» hence v-integrable (4.4.20), and

I(x) = j h%dv =

=

s~p j(h;hdv

s~p (s~p j(h,)%dVn)

= s~p (s~p I,(n) (x)) (the first equality, by the choice of E and I; the second, by the monotone convergence theorem; the third, by Lemma 7.4.5; the fourth by the definin n tion of If ». In particular, 0::; If ) ::; I p.-a.e. Since I is p.-integrable and It) is measurable with respect to S, it follows (by 4.4.20) that

§7.4. Fubini-Tonelli Theorem for u-Finite Measures

ft)

395

E £1 (Il) and

J

(11)

fi(n) dll


0)

nr1(u.(J(t») isaneighborhoodoft. lEe

In particular, it is clear that if £ is equicontinuous at t, then every is continuous at t (3.4.2).

1E£

8.1.4. Definition. With notations £ C F(T, Y) as in 8.1.2, we say that £ is equicontinuous (on T) if it is equicontinuous at every point t E T. 8.1.5. Remark. If £ C F(T, Y) is equicontinuous on T, then every 1 E £ is continuous on T (cf. 8.1.3). 8.1.6. Definition. Let £ C F(X, Y) be a set of functions on the metric space (X, d) with values in the metric space (Y, p). We say that £ is equi-uniformly continuous (on X) if, for every £ > 0, there exists a 6 > 0 such that (**)

d(x,x') $ 6

~

p(J(x),f(x'» $

£

for all

1 E £.

(The term 'uniformly equicontinuous' is also used.) 8.1.7. Remark. With notations as in 8.1.6, to say that a function 1 : X -+ Y is uniformly continuous means (6.3.2) that, for every £ > 0, there exists a 6 > 0 such that

d(x, x') $ 6 ~ p(J(x),/(x'» $ £; thus, to say that £ is equi-uniformly continuous on X means that, for every £ > 0, there exists a 6 > 0 that 'works' simultaneously for all functions 1 E £ . Thus, for a set £ C F(X, Y) , we have the diagram of implications

£ equi-uniformly continuous

/ every 1 E £ is uniformly continuous

£ equicontinuous

/ every 1 E £ is continuous

§8.1. Equicontinuity

401

or, more succinctly,

E: equi-UC ,/

'\.

E: equi-C

every feE: is UC

'\.

,/ every feE: isC

The central result to be proved is as follows: 8.1.8. Theorem. (Ascoli's theorem)! Let fn : X

->

Y (n

= 1,2,3, ...)

be a sequence of functions, where (X, d) is a compact metric space and (Y, p) is a metric space. If the set

E: =

Un:

n = 1,2,3, ... } C C(X, Y)

is equicontinuous and pointwise totally bounded, then there exists a subsequence Un.) that is uniformly Cauchy. The proof of Ascoli's theorem is arranged in a series of lemmas. Although the concept of equi-uniform continuity does not appear explicitly in the statement of the theorem, it arises in the proof and is, in fact, equivalent to equicontinuity in the presence of the compactness of X (Lemma 8.1.10 below). 8.1.9. Lemma. If (X,d) is a compact metric space and U is any open covering of X , then there exists a real number or > 0 such that AcX } diamA:5or

'*

AcU forsome UeU,

that is, every subset of X of diameter < or is entirely contained in at least one of the sets of the covering. (Such a number or is called a Lebesgue number for the covering.)

Proof Assume to the contrary that every or > 0 fails to have the desired property. Then, for every positive integer n, the number lin fails to have the property, so there exists a subset An of X such that diam An < lin but no one set of U contains An, that is, (lfUeu)

1 Giulio

AnItU.

ABeoli (1843-1896). For an interesting historical note, see N. Bourbaki, Gen· eral tcpology. VoL II (Chs. 5.10) (Springer-Verlag, New York, !9881, p. 347.

8. The Differential Equation y'

402

= f(x, y)

In particular, every An is nonempty; for each n, choose a point X n E An such that X n 1. V. Since X is compact, the sequence (x n ) has a convergent subsequence (6.1.11), say X n • -> x. Since U covers X, there exists a set V E U such that x E V . A contradiction to (*) will be reached by showing that An. C V for some index k. Since V is open, there exists an E > 0 such that V,(x) C V. Choose an index k such that

d(xn.,x) < E/2 and

link < E/2;

it will suffice to show that An. C V,(x). Let yEAn•. By the triangle inequality,

d(y,x) ::; d(y,x n.) + d(x n., x) , where dey, x n.) ::; link (because diam An.

::; link ), therefore d(y,x) ::; link + d(xn., x) < E/2 + E/2 = E by the choice of k, whence y E V,(x). 8.1.10. Lemma. Let [ c .r(X, y) be a set of junctions from

X to Y, where (X,d) and (Y,p) are the given metric spaces and X is compact. Then

[ is equicontinuous

¢}

[

is equi-uniformly continuous.

Proof. ~: Noted in 8.1.7. =>: Let E > O. For each x EX, choose 6", > 0 so that f(U6.(X))

c U'/2(1(x»)

for all fEE.

The family (U6.(x))"'EX is an open covering of X; let 6 > 0 be a Lebesgue number for the covering (8.1.9). If X,I EX, then

d(x, x') < 6

=> => => =>

diam{x, x'} < 6 {x,x'} C U6.(Z)

for some z E X

f({x,x'}) c f(U6.(Z») c U'f2(1(z)) P(l(x),j(x')) < E for all fEE,

thus [ is equi-uniformly continuous (8.1.6).

for all fEE

8.1.11. Lemma. Let (X, d) and (Y, p) be the given metric spaces and suppose that (X, d) is totally bounded. Let Un) be an equi-unifonnly continuous sequence in C(X, Y) and suppose there exists a dense subset A of X such that (In(X)) is Cauchy for each x EA. Then Un) is uniformly Cauchy on X.

Proof We observe first that if f: X -> Y is uniformly continuous, then f(X) is a totally bounded subset of Y. For, let (Yn) be any sequence in f(X) and, for each n, choose Xn E X so that f(x n ) = Yn' Since X

§8.1. Equicontinuity

403

is totally bounded, the sequence (x n ) has a Cauchy subsequence (x n .)

(6.1.24), and since I is uniformly continuous it follows that (J(xn.» = (Yn.) is Cauchy in Y (6.3.4). Thus, every sequence in I(X) has a Cauchy subsequence, consequently I(X) is totally bounded (6.1.24). In particular, diamI(X) < +00,

I is bounded (6.2.9). Thus, I : X -> Y uniformly continuous

that is,

'*

I E C(X, Y) n B(X, Y) ,

so that the metric

D(f,g)

= sup p(J(x),g(x») :>:EX

on B(X, Y) is defined for all pairs I, 9 of uniformly continuous functions (X, Y) of all uniX -> Y; in other words, D defines a metric on the set formly continuous functions X -> Y , convergence relative to D signifying uniform convergence (6.2.15). With (fn) as in the statement of the lemma, weare to show that D(fm, In) -+ 0 as m, n -+ 00. Given any f > 0, we seek an index N such that

c..

By the assumed equi-uniform continuity of the sequence, there exists a 6 > 0 such that

(1)

d(x, x') < 6

'*

p(Jn(x), In(x'» < f/3 for all n.

Let Xl, ... , X r be a 6/2-net in the totally bounded space X; by the density of A, there exist points al, ... , ar in A such that

d(a;,x,) N

'*

p(Jm(ai),In(a;» < f/3 for i

= 1" ... ,r.

Fix a pair of indices m, n ~ N and fix a point X EX; we need only show that p(Jm(x),In(x») < f. By (2), there exists an index j (1 < j ::; r) such that (4) Then

d(x,ai) j then 9~ is strictly to the right of 9] in the sequence (9~). hence (by the preceding paragraph) in the sequence (9n). In particular, 9~ is strictly to the right of 9l in the sequence (9~) (because h ;:: 2 > 1). hence in the sequence (9n). Also. (9~(al» is a subsequence of (9~ (al») • therefore (9~(al») is a Cauchy sequence in Y.

Continuing recursively, we construct a tableau of functions 1 1 1 91' 92' 93' 222 91' 92' 93' 333 91' 92' 93' such that each row is a subsequence of the preceding row (hence of all preceding rows. and of the original sequence 91. 92 •93 •... ). and such that for each k.

(9~(aj») is a Cauchy sequence in Y, for j

= 1•...• k,

that is, if the kth row of the tableau is applied to any of the elements aI.' ..• ak , the result is a Cauchy sequence in Y. Moreover, the element 9Z of the kth row is strictly to the right of ~ in the (k - 1)th row (hence in the original sequence 91,92.93 •... ). We now look at the sequence

9Z:

1

91

2

I

92

3

I

93'

...

I

the dia90nal of the tableau. From the foregoing remarks. it is clear that this is a subsequence of the original sequence (gn). Moreover. for each k, the 'truncated diagonal' k k+1 k+2 gk' 9k+I' 9k+2' is a subsequence of the kth row. thus the sequence

k+l() k+2() 9kk() ak , 9k+1 ak, 9k+2 ak , is Cauchy in Y, therefore so is the sequence (9;:(ak»). In other words, the diagonal sequence (g;:) is a subsequence of (9n) that is pointwise

Cauchy on the set A.

0

8. The Differential Equation y'

406

= f(x, y)

We are now ready to prove Ascoli's theorem.

Proof of Theorem 8.1.8: By assumption X is compact, therefore totally bounded (6.1.12, 6.1.13), therefore separable (6.1.19); let A be a countable dense subset of X. The sequence Un), assumed to be pointwise totally bounded on X, is in particular pointwise totally bounded on A; applying the preceding lemma to the sequence of restrictions 9n = fnlA, we obtain a subsequence Un.) that is pointwise Cauchy on A. Since the fn are equicontinuous by hypothesis, so are the fn.; by Lemma 8.1.10, the fn. are equi-uniformly continuous, so it follows from Lemma 8.1.11 that the sequence f is uniformly Cauchy on X.

n.

8.1.13. Corollary. As in Ascoli's theorem, let (X,d) be a compact metric space and (Y, p) a metric space. If E c C(X, Y) is equicontinuous and pointwise totally bounded, then E is totally bounded for the sup-metric.

Proof. By Ascoli's theorem, every sequence in E has a uniformly Cauchy subsequence, therefore E is totally bounded for the sup-metric by Theorem 6.1.24. 8.1.14. Corollary. Let (X,d) be a compact metric space and (Y,p) a complete metric space. The following conditions on a set E C C(X, Y) are equivalent: (a) E is totally bounded (for the sup-metric); (b) E is compact (for the topology of uniform convergence, where the bar denotes closure relative to that topology); (c) E is pointwise totally bounded and equicontinuous (hence equi-uniformly continuous).

Proof. (c) =* (a): This is Corollary 8.1.13. (a) =* (b): Given any E > 0, let ft, ... ,fr be an E/2-net for E. If f E [. then f is within E/2 of some point of E, hence within E of some /;, thus ft, ... , fr is an E-net for E. Thus, the total boundedness of E implies that of [. (and the argument is valid in any metric space). Since C(X, Y) is complete for the sup-metric (6.2.22), its closed subset E is also complete (6.1.29). Thus, [; is complete and totally bounded, hence compact (6.1.26). (b) =* (c): Proof that E is pointwise totally bounded. Let x EX. The mapping C(X, Y) --+ Y defined by f >-+ f(x) is continuous (because uniform convergence implies pointwise convergence), consequently the image E(x) = {/(x): f E E} of the compact set E is compact in Y (6.1.27), therefore is totally bounded; it follows easily that its subset E(x) is also totally bounded (cf. the proof of (2) of 8.1.11). Proof that E is equicontinuous: In view of Lemma 8.1.10, it is the same to show that E is equi-uniformly continuous. Let E > O. We seek a {, > 0 such that d(x, x') < {,

=*

p(J(x), f(x'» < E for all f E [; .

§8.1. Equicontinuity

407

Since £ is compact, hence totally bounded, it follows that £ is totally bounded; let ft, ... , fr be an l/3-net for £. For each index i, f; is continuous, therefore uniformly continuous (6.3.7), so there exists a 6; > 0 such that d(x,x') < 6;

writing 6

= min{6 b

d(x, x')

=>

p(/;{x), f;(x'») < l/3;

we have

p(f;(x),f;(x'») O. Let C = CR(A) be the real Banach algebra of continuous real-valued functions on the compact rectangle A (6.8.14), let u, v E C be the functions defined by u(x, y) = x and v(x, y) = Y, and let A be the subalgebra of C generated by u, v and the constant function 1; as noted in the preceding lemma, A is the set of all functions on A that are polynomials in x and Y «x,y) E A) with real coefficients. The algebra A separates the points of A; indeed, if (Xl> Yl) # (X2, Y2) then either Xl # X2 or Yl # Y2, in other words, U(Xl,yr) # U(X2,Y2) or V(Xl,yr) # V(X2,Y2). It follows from the Stone-Weierstrass theorem (6.9.9) that A is uniformly dense in C. Choose a sequence (fn) in A such that In ..... I uniformly on A, that is, II/n - 11100 ..... O. Then II/nlloo ..... 11/1100 > 0; for, if doo denotes the metric doo(g, h) = IIg - hll oo derived from the sup-norm, then

II/nlloo = doo(fn, 0) ..... doo(f, 0) = 11/1100 by Corollary 3.2.4. Suppressing at most finitely many terms, we can suppose that II/nlloo > 0 for all n. Writing Cn = 1l/1100/llfnil00 , we have Cn ..... 1, therefore

IICn/n - 11100 = II(Cn - I)/n + (fn - moo ~ ICn - IllI/nlloo + II/n - 11100 ..... o· 11/1100 + 0 = 0, that is, Cn/n ..... I uniformly; moreover,

IICnfnlloo = Cnll/niloo = 1111100

for all n.

Thus, replacing In by Cn/n, we can suppose that (1)

II In 1100 = 11/1100 = M

for all n.

By the Lemma, every In satisfies a Lipschitz condition in the second variable Y, hence satisfies the hypotheses of Picard's theorem (8.2.1). The significance of the normalization (1) is that the interval I defined for I

§8.3. Peano's Theorem

419

in the present theorem is the same as the interval 1 defined in Picard's theorem for every fn. Thus, for each n, Picard's theorem provides a continuously differentiable function 'Pn: 1 -+ IR such that

'Pn(XO)

= Yo,

'Pn(I)

C

J

and

'P~(x)

= fn(x,'Pn(X»

for all x

E

I

(with one-sided derivatives at the endpoints of I). Preparatory to obtaining a uniformly convergent subsequence of ('Pn) (via Ascoli's theorem) we show that the sequence is equicontinuous (even equi-uniformly continuous-but see 8.1.10). Let e > 0; we seek a 6 > 0 such that

(2)

Xi> X2 E I, IXI - x21 < 6

'*

l'Pn(XI) - 'Pn(x2)1 < e for all n.

In considering pairs Xi> X2 in I, it will suffice to suppose that Xl < X2. When Xl < X2 , by the mean-value theorem there exists, for each n, a point t n E (Xi> X2) such that

'Pn(XI) - 'Pn(X2)

= 'P~(tn)' (Xl

- X2)

= fn(tno'Pn(t n »

. (Xl - X2),

whence l'Pn(xt} - 'Pn(x2)1 ~ Mixi - x21. If 6 > 0 is chosen so that M6 < e, then 6 meets the requirements of (2). Since 'Pn (I) C J for all n, and since J is compact hence totally bounded (6.1.26), it is trivial that the sequence ('Pn) is pointwise totally bounded (in the sense of Definition 8.1.1). By Ascoli's theorem (8.1.8), ('Pn) has a subsequence ('Pn.) that is uniformly Cauchy. Passing to the subsequence and changing notations, we can suppose that ('Pn) is uniformly Cauchy in CR(I) = C(I, IR) . Since IR is complete, ('Pn) converges uniformly to a function 'I' E CR(I) (6.2.22). It follows from the properties of the 'Pn that 'P(xo) = Yo and '1'(1) C J. It remains only to show that 'I' is continuously differentiable on I and that 'P'(x) = f(x,'P(x» for all X E I. We know that for each n, 'Pn is continuously differentiable on I, 'Pn(XO) = Yo and 'P~(x) = fn(x,'Pn(x» for all X E I, thus, by the fundamental theorem of calculus, (3)

'Pn(x)

= Yo +

1% fn(t,'Pn(t»dt

for all X E I.

%0

Our problem is to "pass to the limit under the integral sign" , so as to obtain in the limit (4)

'P(X)

= Yo +

1% f(t, 'P(t»dt

for all X E I;

%0

it will then follow from (4) that 'I' is continuously differentiable on I and that it has the properties listed in (**) of the statement of the theorem. Let g: 1 -+ IR and gn: I -+ IR (n = 1,2,3, ...) be the functions defined by the formulas

g(x)

= f(x,'P(x»

, gn(x)

= fn(x,'Pn(x»

(x

E

I).

420

8. The Differential Equation y' = f(x, y)

Then 9 and the gn belong to CR (I) (by the continuity of the functions that figure in their definition); to deduce (4) from (3), we need only show that

1 %

gn (t)dt

-+

1% g(t)dt

%'0

for all X E I.

%0

We know that IIfn - fll"" -+ 0 and IICPn - cplI"" -+ 0; it will suffice to infer that IIgn - gil"" -+ 0, for then it will follow that

1 %

gn(t)dt

:to

-1% g(t)dtl ~ IIgn - gil"" Ix - xol

-+

0

%'0

for each x E I. Now comes a very clever move: for every pair of indices m, n, consider the identity

gn(x) - g(x)

= fn (x, CPn(X» - f(x,cp(x») = fn(x,CPn(X» - fm(x,CPn(X» + fm(x'CPn(X» - fm (x, cp(x») + fm(x,cp(X») - f(x,cp(x».

We are interested in the left-most member, but we have introduced a parameter m so that in each of the three differences making up the telescoping sum on the right side, just one index changes: the index on f, the index on cP, and the index on f, respectively. It then follows from the triangle inequality that

(5)

Ign(x) - g(x)1 ~ IIfn - fmll"" + Ifm(x,CPn(X» - fm(x,CP(X»)j + IIfm - fll""

for all x E I. Given any (> 0, it will suffice to show that IIgn - gil"" ultimately. Choose an index N such that

n?N

'*

~

4(

IIfn-fll",,~(;

then (by the triangle inequality)

m, n ? N

'*

IIfm - fnll"" ~ 2(

and it follows from (5) that (6)

m,n? N

'*

for all x E I. Setting m (7)

n? N

'*

Ign(x)-g(x)1 ~ 3Hlfm(x,CPn(x»)-fm(x,cp(X»\

=N

in (6), we have

Ign(x) - g(x)1 ~ 3£+ IfN (x, CPn(x») - fN(x,cp(x»

thus, if K N > 0 is a constant such that

I;

§8.3. Peano's Theorem

421

for all x, Yl, Y2 , it follows from (7) that n ~

N

'*

19n(X) - 9(x)1 :5 3t + KNICPn(x) - cp(x)1

for all x E I , whence (8)

119n - 91100 :5 3t + KNIICPn - '1'1100

for all n ~ N.

Since IICPn -'1'1100 :5 t/ K N ultimately, it follows from (8) that 119n -91100 :5 4t ultimately; we have shown that 119n - 91100 --> 0, which completes the verification of (4) and hence of (**). To appreciate the cunning of the three-term telescoping sum (with a parameter m) in the foregoing proof, contemplate trying to reach the same goal with the following two-term telescoping sum (without the parameter m):

9n(X) - g(x) = fn(x,CPn(x» - f(x,cp(x»

= fn (x, CPn(x» - f(x,CPn(X»

+ f(x'CPn(X»

- f(x,cp(x»;

in the last difference on the right, we have no Lipschitz condition on f to push the proof along. Impasse. Try again: if K n is a Lipschitz constant for fn then, from the formula

9n(X) - 9(x) = fn(x,CPn(x» - f(x,cp(x»

= fn(x,CPn(X» - fn{x,cp(x» + fn(x,cp(x» - f(x,cp(x» , we infer that

19n(X) - 9(x)1 :5 KnICPn(x) - cp(x)1 + Ifn{x, cp(x» - f(x, cp(x» I IL2 are not subtractible.}

=

9.1.3. Theorem. If ILl and JL2 are measures on S, at least one of which is finite, then the function v = ILl - J.L2 has the following properties: 1° v(0) =0. 2° If (En) is a sequence of pairwise disjoint sets in S, and if E = U~lEn, then 00

veE) =

L v(E

n) ,

n=l

in the sense that L~=l v(Ek ) is defined for every nand n

L v(Ek)

-t

veE)

in lR

k=l

asn--+oo.

3° v does not take on both of the values +00 and -00. Proof Suppose, for example, that ILl is finite. Then v does not take on the value +00, whence 3°. Property lOis obvious. We know from the

9. Topics in Measure and Integration

424

countable additivity of /11 and /12 that n

in JR,

L/1l(Ek}-+/1l(E} k=l n

L /12(Ek) -+ /12(E)

in JR

k=l

(convergence in JR is defined in 1.16.8), so it clearly suffices to prove the following lemma: If On -+ 0 in JR and f3n -+ f3 in JR, then On + f3n -+ o + f3 in JR. At any rate, all the sums in question are defined, and the sequence (on) is bounded. If f3 E JR then f3n is ultimately bounded (cf. 1.16.6) and, since convergence is undisturbed by amputating a finite number of terms, it is clear that on+f3n -+o+f3.If f3= +00 then (1.16.9) r E JR, r

>0

=>

ultimately f3n

> 2r -

0

and On

therefore On + f3n > r ultimately, thus On + f3n -+ +00 = if f3 = -00 then -On -+

-0

0

>0

-

r,

+ f3. Finally,

in JR and - f3n -+ +00 in JR,

therefore -On + (-f3n) -+ +00 by the preceding case, whence On + f3n -00 = 0 + f3.

-+

9.1.4. Definition. A signed measure on S is a function II: S -+ JR satisfying the conditions 1° and 2° of the above theorem; the property 2° is expressed by saying that II is countably additive. (The property 3° is automatically verified, as we shall see in 9.1.7 below.) 9.1.5. Example. The real measures on S considered in §6.5 are precisely the signed measures all of whose values are finite; in other words, they are the finite signed measures discussed in §4.8. For the rest of the section, II denotes a signed measure on S. Our main objective is to show that II can be expressed as a difference of two measures (at least one of them finite) as in Theorem 9.1.3, a result known as the Jordan-Hahn decomposition theorem. We prepare the way for this by establishing some basic properties of II in a series of six propositions. 9.1.6. Proposition. II is finitely additive, that is, if E l , ... , E r are paIrwise disjoint sets in S, then r

II(E l U ... U E r )

=L

II(Ek}

k=l

(in particular, the sum on the right side exists in JR).

§9.1. Jordan-Hahn Decomposition

425

Proof. Define Ek = 0 for every positive integer k > r and apply property 2° of v to the sequence (En). 0

9.1.7. Proposition. v_ satisfies 3° of 9.1.3. In particular, the sum v(E) + v(F) exists in IR for every pair of sets E, F in S . and v(F) = -00 for suitable sets E, F in S. Consider the following decompositions of E UF : proof.l Assume to the contrary that v(E)

Eu F

= +00

= Eu (E' n F) = (EnF') u F.

By the finite additivity of v (9.1.6), v(E U F)

= v(E) + v(E' n F) = v(E n F') + v(F) = +00 + v(E' n F) = v(E n F') + (-00)

(in particular, the indicated sums in IR exist), whence the absurdity v(E U F) = +00 = -00. 0 9.1.8. Proposition. If E E Sand v(E) E IR. then FE S, Fe E =} v(F) E IR,

hence also F ES

=}

v(E n F) E IR .

Proof. Assuming F E S and FeE, so that E v(E)

= v(E -

= (E -

F) U F, then

F) + v(F)

by the additivity of v; in particular. the sum is defined in IR and is equal to the real number v(E) , consequently both terms of the sum must be in IR. The second implication is immediate from the first. 0 9.1.9. Proposition. Let (En) be a sequence of pairwise disjoint sets in S and let E = U::"=l En, so that 00

v(E)

=L

v(En)

n=1

by the condition 2° of 9.1.4. If v(E) E IR then v(En) E IR for all n. and the series is absolutely convergent. Proof. The terms v(E n ) of the series are real numbers by 9.1.8, and the

convergence is absolute by the same argument as for 4.8.3, (i).2

0

1 E. Hewitt and K. Stromberg, Real and abstract analysis[Springer, New York, 1965J, p. 304, (19.2). 2 As remarked in the proof of 4.8.3, (i), a 'commutatively convergent' infinite series of real numbers (i.e., a series which is convergent for every permutation of its terms) is absolutely convergent; for a straightforward elementary proof, see E. Landau, Differential and integral calculus (Chelsea, New York, 1951J, p. 158, Theorem 217.

9. Topics in Measure and Integration

426

9.1.10. Proposition. Let (En) be a sequence of sets in S and let E E S. (i) If En T E then /.I(En ) -> /.I(E) in 1R. (ii) If En ! E and /.I(Ed E 1R, then /.I(En) -> /.I(E) in 1R.

Proof. (i) Define Eo = 0 and F n = En - En-I for every positive integer n. Then (Fn) is a sequence of pairwise disjoint sets in S with union E, consequently 00

/.I(E)

= I:: /.I(F n) ; n=l

since, for each n,

this means that /.I(En) -> /.I(E) as n -> 00 (9.1.4). (ii) By 9.1.8, /.I(E) and the /.I(E n) are real numbers. In particular, from E I = (E I - E) U E and the additivity of /.I, we infer that /.I(E I - E) = /.I(Ed - /.I(E). Writing G n = E I - En , we have E I = G n U En and

consequently /.I(G n ) = /.I(Ed - /.I(En); since G n T E I - E, it follows from (i) that /.I (G n)

->

/.I(E I - E)

in 1R,

that is,

/.I(E I ) - /.I(En ) whence /.I(En)

->

->

/.I(Ed - /.I(E) ,

/.I(E).

9.1.11. Proposition. Suppose E is a set in S such that /.I(E) is a real number. If (Ei)iEI is a pairwise disjoint family of sets in S such that Ei C E for all i E I, then /.I(E i ) = 0 for all but countably many indices i, that is, the set

J = {i E I: /.I(E i )

i

0}

is a countable subset of I. Proof. As noted in 9.1.8, /.I(E i ) E IR for all i E I. For each positive integer n, let

In

= {i E I:

1/.I(Ei)I ~ lIn} ;

clearly I n T J, so it will suffice to show that every I n is finite.

§9.1. Jordan-Hahn Decomposition

427

Assume to the contrary that I n is infinite for some index n. Choose a sequence il.i2.ia •... of distinct indices in I n (1.9.4) and let F=UEi.; k=l

then F E S and FeE, therefore II(F) E JR. Since the Ei. are pairwise disjoint. we infer from 9.1.9 that the series

is convergent, contrary to the fact that III(Ei.)1 ik E I n ).

~

lin for all k (because

The following definition extends Definition 4.8.4 (the special case of a finite signed measure):

9.1.12. Definition. Let A E S. Define a set function IIA: S the formula IIA(E) = II(A n E)

--+

JR by

(E E S).

The basic formal properties of this notation are gathered in the following proposition: 9.1.13. Proposition. Let A. B. C E S. (i) II(/> = O. (ii) IIA is a signed measure on S. (iii) IIA is a finite signed measure ¢? II(A) E IR. (iv) (IIA)B = 11MB. (v) An B = (/> ~ IIAUB = IIA + liB. (vi) If II(A) E IR then lIX-A = II - IIA . (vii) If either II(A) E IR or II(B) E JR then

IIAUB

+ IIAnB = IIA + liB·

Proof (i) Obvious. (ii) The countable additivity of IIA follows from that of II and the identity An

(QEk)

= Q/nEk'

(iii) If IIA is finite then in particular II(A) = II(A n X) = IIA(X) E JR; conversely, if II(A) E IR then II(A n E) E IR for all E E S by 9.1.8. thus IIA is finite. (iv) IIMB(E) = II(A n B n E) = IIA(B n E) = (IIA)B(E) for all E E S. (v) Clear from the additivity of II.

9. Topics in Measure and Integration

428

(vi) By (v), v = VX = VX-A + vA, and the term VA is transposable by (iii). (vii) Suppose v(A) E R, so that VA and VAnB are finite signed measures. From AuB - A = B - AnB, we have VAuB-A = VB-AnB . But Au B = (AUB - A) U A yields VAuB = VAuB-A

+ VA ,

whence VAuB-A = VAuB - VA , and similarly B = (B - A n B) U (A n B) yields VB = VB-AnB + VAnB ,

whence VB-AnB = VB - VAnB; substituting these equations into (*), we have VAUB - VA = VB - VAnB , and the finite signed measures VA and

V

AnB are transposable.

9.1.14. Remarks. Let A E S. Writing AnS={AnE: EES}={EES: EcA}, it is easy to see that An S is a u-algebra of subsets of A and that the restriction V A n S of V to A n S is a signed measure in the context of the measurable space (A, A n S) .

I

9.1.15. Definition. With the preceding notations, the signed measure

viA n S is abbreviated viA. (Abuse of notation: viA is not a function on A.)

9.1.16. Definition. A set A E S is said to be purely positive with respect to v if VA ?: 0, that is, if VA is a measure on S (equivalently, viA is a measure on An S), and we then write A?: 0 (with respect to v). Similarly, a set A E S is said to be purely negative with respect to v, written A:5 0, if VA :5 0 (equivalently, -VA is a measure on S). 9.1.17. Remarks. The preceding notations have the following properties: (i) VA = 0 A?: 0 & A :5 O. (ii) v ?: 0 VA?: 0 for all A E S. (iii) A ?: 0 An B ?: 0 for all B E S. (iv) A?: 0 & B ?: 0 Au B ?: o. (v) An ?: 0 (n = 1,2,3, ...) U::='=l An ?: O.

'* '*

'*

'*

{Proof: (i), (ii) are obvious. (iii) This follows from (ii) and the formula VAnB -:- (VA)B . (iv) When AnB = 0 this follows from VAuB = VA +VB , and the general case then follows from (iii) and the formula A U B = A U (B n A') .

§9.1. Jordan-Hahn Decomposition

429

(v) Let A=U~=IAn.Replacing An by A1u...uA n onecansuppose, in view of (iv), that An T A; it then follows from 9.1.10 that

vA(E) = v(A n E) = lim v(A n n E) = lim VA. (E) :::: 0 for all E E S, thus VA:::: O.} The key "existence theorem" of this section is as follows: 9.1.18. Lemma. If A E S and 0

< v(A) < +00, then there exists a set

Ao E S such that

Ao c A, Ao :::: 0 (with respect

to v) and v(Ao)

>0.

Proof· If A:::: 0, that is, if VA :::: 0, then Ao = A meets the requirements. Otherwise, there exists a set B E S with B c A and v(B) Let (Bi)iEI be a maximal family of pairwise disjoint sets such that B i E S, B i C A and v(B i )

< 0 for all

< O.

i EI

(such a family exists by Zorn's lemma). By 9.1.11, the index set I is countable, therefore the set B = UiEI B i belongs to S. Of course B C A, and v(B) = Lv(Bi )

< O.

iEI

Let Ao = A - B . Since v(A) E IR, we have v(A o) = v(A) - v(B) by 9.1.8 and the additivity of v, and since v(A) > 0 and v(B) < 0 we conclude that v(A o) > O. It remains only to show that Ao:::: o. Thus, if E E S and E C Ao , we need only show that v(E) :::: O. The alternative, v(E) < 0, would contradict the maximality of the family (Bi)iEI. 0

9.1.19. Remark. Application of the lemma to -v yields the dual result: If -00 < v(A) < 0 then there exists a set Ao E S such that Ao C A, Ao :::; 0 (with respect to v) and v(Ao) < O. All the needed tools are in hand for an efficient proof of the desired decomposition theorem: 9.1.20. Theorem. (Jordan-Hahn decomposition)3 Let (X, S) be a measurable space. If v is any signed measure on the u-algebra S, then there

exists a set A E S such that A :::: 0

and X - A :::; 0

Defining J1.1 = VA and J1.2 such that v = J1.1 - J1.2 •

3

= -VX-A,

(with respect to v).

J1.1 and J1.2 are measures on S

Camille Jordan (1838-1922), Hans Hahn (1879-1934).

9. Topics in Measure and Integration

430

Proof. We know that v does not take on both of the values +00 and -00 (9.1.7). Suppose, for example, that the value +00 is not taken on, so that

< veE) < +00 for all E E S.

-00

Let

l' = {A E S: A ~ 0 with respect to v} (for example, (/) E 1'). We know from 9.1.17, (v) that l' is closed under countable unions. Moreover, the values of v on l' are real numbers ~ o. We assert that v takes on a largest value on 1'. For, let Ct

= sup{v(B):

BE 1'}

and let (An) be a sequence in l' such that v(A n ) -+ Ct in lR. Replacing An by Al U ... u An , we can suppose that An T. Then, writing A = U::":I An, we have An TA, therefore v(An) -+ veAl by 9.1.10, thus

veAl

= limv(An ) = Ct = sup{v(B):

BE 1'};

since A E 1', we conclude that v takes on its largest value at A. In particular, 0:5 veAl < +00. Since A E l' , we know that A ~ o. We need only show that X - A :5 o. Assuming to the contrary that there exists a set E E S such that E C X - A and veE) > O. we then have 0 < veE) < 00; by the lemma, there exists a set ~ E S such that ~ C E, ~ ~ 0 and v(Ao) > o. Thus Ao E l' and Ao is disjoint from A (because ~ C E c X - A), therefore

v(A U Ao) = veAl

+ v(~) > veAl =

but AU~ E l' by 9.1.17, (iv), therefore of Ct. a contradiction.

Ct;

v(AU~):5 Ct

by the definition

9.1.21. Remark. The measures J.ll> J.l2 constructed by the method of the preceding theorem are unique. That is, if also B E S, B > 0 and X - B :5 0 (with respect to v) then VA

= lIB

and

!IX-A

= !IX-B.

For. An B' > 0 (because A ~ 0) and An B' :5 0 (because B':5 0). consequently VAnB' = o. Similarly VA'nB = o. From A = (A n B) U (A n B') we infer that

= VAnB + VAnB' = VAnB lIB = VAnB ,thus VA = lIB· Similarly VA

and similarly

!IX-A

= '"'X-B.

9.1.22. Definition. With notations as in Theorem 9.1.20, one writes v+

= VA

and

v-

= -!lX-A

§9.1. Jordan-Hahn Decomposition

431

(the measures v+ and v- depend only on v by the preceding remark), and the formula

is called the Jordan-Hahn decomposition of the signed measure v. The measure v+ + v- is called the total variation of v and is denoted

Ivl =

v+ +v-.

Inspecting the proof of Theorem 9.1.20, we see that if v does not take on the value +00 then v+ is finite. If v does not take on the value -00, then v- is finite. The following proposition will be useful on several occasions in the next two sections: 9.1.23. Proposition. Let v be a signed measure on S. Then: (i) v = 0 ¢} Ivl = o. (ii) For every measumble set E E S,

(VE)+ = (V+)E,

(VE)- = (V-)E,

IVEI = IvIE.

Proof (i) If v = 0 then, with notations as in Definition 9.1.22, v+ = VA = OA = 0 and similarly v- = 0, therefore Ivl = v+ + v- = O. Conversely, if Ivl = 0 then v+ = v- = 0, therefore v = v+ - v- = 0 . (ii) With notations as in 9.1.22, we have (VE)A = VEnA = (VA)E = (V+)E ?: 0 (VE)X-A = VEn(X-A) = (VX-A)E = (-V-)E :5 0; thus, at least one of the measures (VE)A, -(VE)X-A is finite, and A defines a Jordan-Hahn decomposition of the signed measure VE, with

(VE)+

= (VE)A = (V+)E,

(VE)- = -(VE)X-A = -(VX-A)E = -(-V-)E = (V-)E, IVEI = (VE)+ + (VE)- = (V+)E + (V-)E = (v+ + V-)E = IvIE. 0

Exercises 1. Let (X, S) be a measurable space. (i) The u-algebra S is a commutative ring with unity for the operations of sum and product defined, respectively, by the formulas

E Ell F = (E - F) U (F - E) , E0F=EnF,

having rlJ as zero element, X as unity element, and satisfying E 0 E = E for all E E S . (The same is true for every algebra of sets.)

9. Topics in Measure and Integration

432

(ii) If

V

is a signed measure On S, and if

So = {E e S: v(E) e IR }, then So is an ideal in S (for the ring structure just described). {Hint: 9.1.8.} 2 (iii) (Theorem of M.H. Stone}4 If R is a ring with unity such that x = x for all x e R (such rings are called Boolean) then R is commutative (xy = yx for all x, y in R) and R may be regarded as (i.e., is isomorphic to) an algebra of subsets of a suitable set, with operations as described in (i). 2. Let (X, S) be a measurable space, v a signed measure on S. Define S+={AeS: VA>O}

S_ = {AeS: VA ::;O} = {AeS: (-V}A ;:::O} So={AeS: VA=o}=s+nS_. Then S+ is a a-ring of subsets of X (therefore so are S_ and So), and S+ is a a-algebra if and only if v ;::: O. {A ring of subsets of X is a set n of subsets of X, with (/) en, such that if A, Ben then also A - B, Au Ben; a ring that is closed under countable unions is called a a-ring. }

9.2. Radon-Nikodym Theorem

Throughout this section, (X, S, It) is a measure space. (Later in the section, it will be assumed that It is a-finite.) If f e .c1(1t) , that is, if f : X -+ IR is It-integrable (4.4.7), we know that the indefinite integral f· It : S -+ 1R, defined by (J. It)(E) =

k

fdlt

(E e S) ,

is a real measure on S (4. 7.3) such that E e S, It(E}

=0

~

(J . 1t}(E)

=0

(see 4.7.2, (vii». Thus, for every It-integrable function f: X -+ 1R, the set function v = f . It is a real measure On S that is absolutely continuous with respect to It in the sense of Definition 4.8.6 (written v« It). It was shown in Corollary 4.8.12 that if It is a finite measure, then every real measure v on S such that v« It has the form v = f . It for some It-integrable function f. Our objective in this section is to generalize this result so as to permit It to be a-finite and v to be a signed measure (with

4

cr. P.R. Halmos, Measure theory (Van Nostrand, New York, 1950; reprinted Springer-

Verlag, New York, 1974), p. 170, Exercise (lSa).

§9.2. Radon-Nikodym Theorem

433

possibly infinite values). The passage to a-finite p. is straightforward, but admitting signed measures poses two technical problems: when v takes on infinite values, (1) the function f can no longer be required to be p.integrable-we will need to define f· p. for certain measurable functions f that are not p.-integrable, and (2) the condition v« p. will no longer suffice, but must be augmented with the assumption that the measure Ivl (defined in 9.1.22) is also u-finite. We commence by laying the groundwork for item (1). 9.2.1. Definition. If f ~ 0 is a nonnegative measurable function (with respect to S), we define

+oo

J { fdp.

=

f ¢ .c 1 (p.) if f E .c 1 (p.). if

as usual

The first properties of this notation are gathered in the following proposition: 9.2.2. Proposition. Let f, 9 and fn (n = 1,2,3, ...) be measumble functions ~ 0, and let c be a real number 2: o. Then: (i) J cfdp. = c J fdp.. (ii) J(f + g)dp. = J fdp. + J gdp.. (iii) f :5 9 p.-a.e. => J fdp. :5 J gdp.. (iv) fn T f p.-a.e. => J fndp. TJ fdp.. Proof. (i) When J fdp.

day (1.15.4). (ii) Since 0:5 J(f+9)dP.

= +00, the convention o· (+00) = 0

saves the

f, 9 < f + g,

< +00

~

f+g E.c

1

~

f,g E.c

1

~

J fdp.+ J gdp.

< +00,

in which case the asserted equality is true by the additivity of integration (4.4.6). Otherwise, the equality reduces to +00 = +00. (iii) Assuming f:5 9 p.-a.e., we are to show that J fdp. :5 J gdp.. This is trivial if J gdp. = +00. Otherwise, 9 E .c 1 (p.) , therefore f E .c 1 (p.) and J fdp. :5 J gdp. by 4.4.19. (iv) Assuming fn T f /ira.e., we are to show that J fndp. TJ fdp.. At any rate, J fndp. T by (iii). If J fdp. < +00 then f and the fn are p.-integrable and the assertion follows from the monotone convergence theorem (4.5.3). Otherwise f ¢ .c 1 (p.); it then follows from the monotone convergence theorem that either some fn fails to be integrable, or every fn is integrable but the sequence! fndp. is unbounded, and in either case the assertion that sup J fndp. = fdp. reduces to +00 = +00. The concept of indefinite integral extends to measurable functions 2: 0 (and, with trivial modifications, to functions that are > 0 p.-a.e., an extension for which we have no need):

9. Topics in Measure and Integration

434

~

9.2.3. Definition. If f is a measurable function f . J.L : S -+ [0, +00] is defined by the formula (f . J.L)(E) =

J

'PEfdJ.L

0, a set function

(E E S) ,

where the symbol on the right side, also written assigned to it by Definition 9.2.1.

IE fdJ.L,

has the value

The properties of this notation are readily derived from Proposition 9.2.2: 9.2.4. Proposition. Let f,g and fn (n = 1,2,3, ...) be measurable functions ~ 0, and let e be a real number ~ O. Then: (1) f . J.L is a measure on S such that E

E S,

J.L(E) = 0

~

(f . JL)(E) = O.

(2) (ef) . J.L = e(f . J.L) . (3) (f + g) . J.L = f . J.L + g. JL. (4) f ~ 9 J.L-a.e. ~ f· J.L ~ 9 . J.L. (5) fn T f J.L-a.e. ~ fn' JL T f . J.L on S. (6) 'PF . JL = J.LF for all F E S. (7) f . J.L is a finite measure $} fEl:. l (J.L) . (8) (fg) . IJ. = f· (g. J.L) .

Prool. (1) We verify the criteria of Definition 2.4.12, by showing that the nonnegative function f· J.L vanishes at the empty set and is countably additive. Since 'P0f = 0, we have (f ·1J.)(0) = IOdJ.L = o. If E, F E Sand En F = 0 then 'PEuF f = 'PEl + 'PF I , whence (f. IJ.)(E U F)

= (f . J.L)(E) + (f . J.L)(F)

by (ii) of 9.2.2, thus f· IJ. is finitely additive. If (En) is a sequence of pairwise disjoint sets in S with union E then, writing F n = U~=1 Ek , we have 'PF.! T'PEf, consequently (f. J.L)(Fn ) T (f. IJ.)(E) by (iv) of 9.2.2; since f· IJ. is finitely additive, this means that n

(f. J.L)(E)

= }.!..~(f. J.L)(Fn ) = }.!..~ L(f ·1J.)(Ek) , k=1

thus I is countably additive. Finally, if J.L(E) = 0 then 'PEf = 0 J.L-a.e., thus 'PEf is J.L-integrable with integral 0, that is, (f. J.L )(E) = 0 . (2) For all E E S, citing (i) of 9.2.2 at the appropriate step we have, for . all EES,

[(ef) . J.L](E)

=

whence (ef)· J.L

J

O. at least one of which is JL-integrable, consequently 'PEh E V and

J

'PEhdJL

=

J

'PEfdJL -

J

'PEgdJL

by the definition of the left side (9.2.8), that is, I 'PEhdJL = (h· JL)(E) . This expression may also be written hdJL. extending the notation in 9.2.3.}

IE

The functions in V are conveniently characterized as follows: 9.2.10. Proposition. The following conditions on a function h: X -+ lR are equivalent: (a) hE V; (b) h is measurable and at least one of h+, h- is JL-integrable. Proof. (b) => (a): Immediate from h = h+ - h- . (a) => (b): Write h = f - 9 as in 9.2.6. At any rate, h is measurable and we know that either f or 9 is JL-integrable. case 1: f E .c1(JL) . Let A = {x: h(x) ~ O} = {x: f(x) ~ g(x) } . Then 'PAh = h+ and

o ~ h+ = 'PAh = therefore h + E .c 1 (JL) .

'PAU - g) = 'PAf - 'PAg ~ 'PAf E .c1(JL) ,

case 2: 9 E .c1(JL). Then -h = 9 - f, so by the proof of case I, (-h)+ E .c1(JL) , that is,

h- E .c1(tt). For the signed measures h'JL (h E V) of the present section, the HahnJordan decomposition and total variation defined in the preceding section are readily expressed in terms of h: 9.2.11. Corollary. For every hE V,

(h'JL)+=h+'JL, (h·JL)-=h-·JL. Ih·JL!=lhl·JL· Proof. We recall that Ih· JLI = (h· JL)+ + (h· JL)- (9.1.22); on the other hand, Ihl is a measurable function ~ 0, thus Ihl· JL is the measure defined in 9.2.3. If

A={x:

h(x)~O},

§9.2. Radon-Nikodym Theorem

439

then 'PAh = h+ . Let

B = X - A = {x: h(x) < O} , C = {x: h(x):::; O} = {x: (-h)(x)

~

O};

then 'Pc ( -h) = (-h)+ = h- , thus h-

= -'Pch = -'PBh

(note that the latter equality holds trivially at the points where h is 0). Summarizing, 'PAh = h+, 'PBh = -h- .

Now, h = h+ - h- , where one of h+, h- is JL-integrable by the preceding proposition, therefore h· I' = h+ . I' - h- . I'

by Definition 9.2.8. If (h· JL)A is the signed measure defined in 9.1.12 then, forall EeS,

(h· JL)A(E)

= (h· JL)(An E)

(by 9.1.12)

=

(by 9.2.9)

f =f f =

'P AnEhdJL 'PE'PAhdJL lpEh+dJL

= (h+ . JL)(E) thus (h· JL)A = h+ . I' (h· JL)B

~

O. Similarly, for all E e S,

f =f

= (h· JL)(B n E) = = =

(by 9.2.3)

f f

'PE'PBhdJL

'PBnEhdJL

lpE(-h-)dJL

(0 - lpEh-)dJL

f

= JOdI' - 'PEh-dJL = -(h- . JL)(E) ,

(by 9.2.8)

thus (h· JL)B = -(h- .1') :::; O. In other words, A ~ 0 and X - A < 0 with respect to the signed measure h· I' (9.1.16); thus A and X - A define a Jordan-Hahn decomposition (9.1.22) of h· 1', with (h.JL)+ = (h· JL)A = h+ . 1', (h·JL)- = -(h· JL)X-A = h- . 1',

440

9. Topics in Measure and Integration

consequently, citing (3) of 9.2.4 at the appropriate step, Ih ·/-11 = (h ./-1)+ + (h ./-1)- = h+ ./-1 + h- ./-1 =(h++h-)·/-I=lhl·/-I·

9.2.12. Definition. A signed measure v on S is said to be absolutely continuous with respect to the measure /-I, written

v « /-I , if it satisfies the condition E E S, /-I(E) = 0 =*

v(E) =

o.

(The case that v is a finite signed measur~that is, a real measur~was considered in 4.8.6.) For example, if I is a measurable function > 0, then I· /-1« /-I by (I) of 9.2.4. 9.2.13. Proposition. For a signed measure v on S, the lollowing conditions are equivalent:

(a)

V«/-I; (a' ) E E S

/-I(E) = 0 =* VE = 0 ; (b) Ivl «/-I; (c) v+ «/-I and v-« /-I . I

Proof. (a) =* (a' ): If /-I(E) = 0 then, for all F E S, /-I(E n F) = 0, therefore v(E n F) = 0 by (a); thus VE = o. (a' ) =* (c): Let A, X - A be a Hahn decomposition for v (9.1.22); thus A E S and

If /-I(E) = 0 then VE = 0 by (a' ), therefore by 9.1.23 we have

(V+)E = (lIE)+ = 0+ = 0

and similarly (V-)E = 0; thus v+ (E) = 0 and v- (E) = 0, and we have shown that v+« /-I and v-« /-I . (c) =* (a): This is clear from the formula v(E) = v+(E) - v-(E) (E E S). Summarizing: (a) ¢} (a/ ) ¢} (c). Finally, (b) ¢} (c): Since Ivl = v+ + v- (9.1.22) we have Ivl(E) = v+(E) + v-(E)

for all E E S.

By the positivity of lvi, v+ and v- , it follows that Ivl(E) = 0

¢}

v+(E) = 0 & v- (E) = 0,

whence the equivalence of (b) and (c).

The signed measures h· /-I (h E V) are all absolutely continuous with respect to /-I:

§9.2. Radon-Nikodym Theorem

441

9.2.14. Theorem. Let hE'D. Then: (i) h· J1.« J1.. (ii) If J1. is u-finite, then so is Ih· J1.1.

Proof. (i) Since Ih· J1.1 = Ihl . J1. (9.2.11) and Ihl· J1.« J1. by (1) of 9.2.4, we have h· J1. «J1. by the preceding proposition. (ii) Since Ih· J1.1 = Ihl . J1. we can suppose that h ;:: 0, in which case h . J1. is a measure. Assuming J1. u-finite, there exists a sequence (Pn) of sets in S such that Pn T X and J1.(Pn) < +00 for all n. Let Qn

= {x:

h(x) ~ n};

then Qn T X, therefore P n nQn T X ,where J1.(PnnQn) < +00 for all n. It will suffice to show that (h· J1.)(Pn n Qn) < +00 for all n; indeed,

o ~ h'PP.nQ.

~ ncpP.nQ.. E £1(J1.) ,

therefore hCPP•• nQ. E £1(J1.) , thus (h·J1.)(P n nQn) =

f

'PP.. nQ.hdJ1.
·2 c1 /J. Write hi = II -gl and h2 = h -g2

444

9. Topics in Measure and Integration

as in Definition 9.2.6. Argue that if gl E £I(p.) then also g2 E £I(p.) ,

rt

£I(p.) then the left member of the asserted inequality whereas if gl is -oo.} (iii) Suppose p. is a-finite and let hI. h2 E 1). Then hI . P. < h2 . P.

$}

= h2 . P.

$}

hI . P.

hi S h2 hI = h2

p.-a.e. p.-a.e.

{Hint: It clearly suffices to prove the first eqUivalence. In view of (ii), it suffices to show that hI' p. h2' p. => hi h2 p.-a.e. By (i) there exists a sequence (En) of measurable sets such that En T X and 'PEn hI, 'PEn h 2 are p.-integrable for all n. Note that ('PEn hI)' p. ('PEn h2 )· p. and cite (iv) of 4.7.2.}

:s

:s

:s

4. (Polar decomposition) If v is any signed measure on S, then there exists a function I E 1)(lvl) such that v = I . Ivl and III = 1. {Hint: Choose A E S such that v+ = VA and v- = -VX-A (9.1.22)

and contemplate

1= 'PA

- 'Px-A.}

5. In the proof of the Radon-Nikodym theorem (9.2.15) the general case can be derived from the special case of a measure by means of Exercise 4. {Hint: Write v = I 'Ivl as in Exercise 4 and apply the first part of the proof of9.2.15 to write Ivl = g.p. with 9 ~ O. Look at v = I 'Ivl = I·(g·p.) in the light of Proposition 9.2.16.}

9.3. Lebesgue Decomposition of Measures

Throughout this section, (XS) is a fixed measumble space; all measures and signed measures under considemtion are defined on S. The topic taken up in this section is an analogue, for measures, of the decomposition of a function of bounded variation as the sum of an absolutely continuous function and a singular function (§5.12).

9.3.1. Definition. A pair p., v of signed measures on S are said to be mutually singular, written p. .L v, if there exists a set A E S such that P.A = p. and

VX-A

= v.

The first properties of this notation are as follows: 9.3.2. Proposition. Let p. and v be signed measures on S, and let EES. (i) P. .L v $} v.L p..

§9.3. Lebesgue Decomposition of Measures

445

(ii) J.LE = J.L ~ J.LX-E = O. (iii) J.L 1. v ~ 3 A E S " J.LA = J.L & VA = O. (iv) J.L 1. v ~ 3 A, B E S :3 J.LA = J.L, va = v, A n B (v) J.LE = 0 ~ IJ.LIE = O. (vi) J.LE = J.L ~ IJ.LIE = IJ.LI· (vii) J1. 1. v ~ 1J1.I1. Ivl. (viii) J.L 1. J.L ~ J.L = 0 .

= (1) •

Proof For brevity we write CE = X - E. (i) Clear from C(CE) = E. (ii) If J.LE = J.L then

J.LCE

= (J.LE)cE = J.LEnCE = J.LQ) = O.

Conversely, if J.LCE = 0 then, citing (v) of 9.1.13, we have J.L

= J.LX = J.LEuCE = J.LE + J.LCE = J.LE + 0 = J.LE . VA = 0 ~ VX-A = v, thus (iii) is equivalent

(iii) By (ii), to the condition defining J.L 1. v (9.3.1). (iv) If J.L 1. v and if A E S is chosen as in Definition 9.3.1, then the pair A, B = X - A meets the requirements of the condition on the right. Conversely, if A and B meet the requirements of that condition, so that in particular B c X - A, then VX_A

= (Va)x-A = Van(X-A) = va = v,

thus A satisfies the condition defining J.L 1. v (9.3.1). (v) By Proposition 9.1.23, IJ.LIE = IJ.LEI and IJ.LEI = 0 ~ J.LE = O. (vi) If J.LE = J.L then; by 9.1.23, IJ.LIE = IJ.LEI = IJ.LI. Conversely, if IJ.LIE = IJ.LI then 1J.LIX-E = 0 by (ii), that is, lJ.Lx-Ei = 0, whence J.LX-E = 0, and finally J.LE = J.L by (ii). (vii) If A E S then, by (vi), J.LA

= J.L &

VX-A

=V

~

IJ.LIA

= IJ.LI &

Ivlx-A

= lvi,

thus J.L 1. v ~ 1J.LI1. Ivl by Definition 9.3.1. (viii) If J.L 1. J.L and if A E S is chosen as in 9.3.1, so that J.LA = J.L and J.LX-A = J.L, then, by (ii), 0 = J.LX-A = J.L. Conversely, if J.L = 0 then every set in S can play the role of A in 9.3.1 in showing that J.L 1. J.L. If J1. is a measure and v is a signed measure, the relation v «J.L is expressible in terms of null sets (9.2.12): J.L(E) = 0 => veE) = O. When J.L and v are both signed measures, the useful concept is as follows: 9.3.3. Definition. Let J.L and v be signed measures on S. We say that v is absolutely continuous with respect to J.L, written

446

9. Topics in Measure and Integration

if v« IILI in the sense of 9.2.12, that is, if E E S, IILI (E) = 0

=>

v(E) = O.

9.3.4. Remarks. With notations as in 9.3.3, it follows from 9.2.13 that v« IL

$>

Ivl «IILI

$>

v+« IILI & v-« IILI.

therefore v«1L

v+ «IL & v- «IL·

$>

Also, since IILI is a measure, IILI(E) = 0 is equivalent to IILIE = O. in other words (9.1.23), to IILEI = 0 and hence to ILE = O. Thus the relation v« IL-equivalently Ivl«llLl---{;an also be expressed by the implication E E S, ILE

= 0 =>

VE

= O.

9.3.5 CAUTION. if v and IL are signed measures on S such that v« IL, it does not follow that IL(E) = 0 => v(E) = 0 (Exercise 1).

9.3.6. Definition. Let IL and v be signed measures on S. A Lebesgue decomposition of v with respect to IL is a pair VI, V2 of signed measures on S such that (In particular, the sum VI + V2 is defined. and IVII« IILI, IV21.l IILI by 9.3.4 and (vii) of 9.3.2.) One calls VI the absolutely continuous part of v. and V2 the singular part of v, with respect to IL. Our goal in this section is to show that if IL and v are signed measures on S such that IILI and Ivl are u-finite, then there exists a Lebesgue decomposition of v with respect to IL (and, of course, of IL with respect to v, by the symmetry of the hypotheses). First we look at the question of uniqueness of such a decomposition: 9.3.7. Proposition. Let IL and v be signed measures on S and suppose

v

= VI + V2 = PI + P2

are two Lebesgue decompositions of v with respect to IL (9.3.6). Then: (1) V2 = P2 . (2) If. moreover, IV21 is u-finite, then VI = PI . Proof· By assumption VI« IL, V2 .l IL and PI« IL, P2 .l IL. From Definition 9.3.3 we know that VI« IILI and PI« IILI ; also, by (vii) of 9.3.2, V2 .l IILI and P2.i IILI . Replacing IL by IILI. we can therefore suppose that IL is a measure. (1) Since v2.iIL, by (iii) of 9.3.2 there exists a set A E S such that (V2)A

= V2,

ILA

=0

(that is, IL(A)

= 0) ,

§9.3. Lebesgue Decomposition of Measures

447

and similarly there exists a set B E S such that

(P2)B = P2, P.B = 0 (that is, p.(B) = 0) . Then p.(A u B) :5 p.(A) + p.(B) = 0, thus p.(A u B) = 0; since follows from (a / ) of 9.2.13 that (VI)AUB = 0, therefore (9.1.13)

VI« p.,

it

VAuB = (VI + V2)AUB = (VI)AUB + (V2)AUB =O+(V2)AUB = «V2)A)AUB = (V2)An(AUB) = (V2)A = Similarly VAUB = P2 ,thus (2) By (1), we can write

V2

V2'

= VAuB = P2 .

assuming IV21 is u-finite, our task is to 'cancel' V2 in the second equation. Let (En) be a sequence of pairwise disjoint sets in S such that 00

X

= U En,

IV21(En) < +00 for all n.

n_1

Then, for each n, the measure !(V2)EnI = IV21E" is finite, therefore so are the measures (V2)E.. +, (V2)E.. - ,consequently (V2)E.. is finite-valued (Le., is a real measure). Now,

and similarly

thus

for all n; since (V2)E.. is finite-valued, it can be canceled in the preceding equation, so that

Then, for each F E S, the sets En nF are pairwise disjoint with union F, consequently 00

00

VI (F) = :~::>I (En n F) = n=l

L (VI )E,.(F) n=l 00

= L(PI)E.. (F) = n=l

thus VI = PI.

00

L PI (En n F) = PI (F) , n=l

9. Topics in Measure and Integration

448

9.3.8. Definition. Signed measures written

if both

p.« /I

p., /I

on S are said to be equivalent,

and /1« p. .

9.3.9. Remarks. In view of Definition 9.3.3 and the remarks following it, we have

p. ;: /I

~

~

p.« Ivl & v« 1p.1 1p.1 « Ivl & Ivl « 1p.1

the latter condition says that the measures null sets, that is, for sets E E S,

~

1p.1

1p.1 and

1p.I(E) = 0

~

Ivl(E) = 0,

=0

~

IlIEl = 0,

=Ivl ; Ivl

have the same

equivalently

that is (9.1.23), IP.EI

equivalently (9.1.23 again)

We approach the main decomposition theorem (Corollary 9.3.13) through three special cases (a lemma, a theorem and a corollary): 9.3.10. Lemma. If p. and v are q-finite measures on S such that v« p., then there exists a set A E S such that v P.A and v = VA'

=

Proof. By the Radon-Nikodym theorem, there exists a measurable function 9 ~ 0 such that v = 9 . P. (see the proof of 9.2.15). Let

A = {x: g(x) > O} . Then 'P Ag

= g, therefore 'PA . V = 'PA . (g.

p.) = ('PAg) . p. = g. p. = v

(the second equality by (8) of 9.2.4), thus v

= 'PA . V = VA •

§9.3. Lebesgue Decomposition of Measures It will now suffice to show that I/(E) = 0

$} $} $}

$} $}

thus

1/

itA . Indeed,

lIE = 0 $} 'PE·I/=O 'PE' (g. It) = 0 $} ('PEg) . It = 0 [('PEg) . It](X) = 0 (cf. (1) of 9.2.4)

J

'PEgdlt = 0

$}

$}

1/;:

449

(4.4.21) 'PEg = 0 It-a.e. 'PE'PA = 0 It-a.e. $} 'l'EnA = 0 It-a.e. It(EnA) = 0 $} /LA(E) = 0,

and itA have the same null sets.

The Lemma is sufficient to yield the Lebesgue decomposition of a O"-finite measure 1/ with respect to a finite measure It: 1/

9.3.11. Theorem. If It and 1/ are measures on S with It finite and 0" -finite, then there exist pairwise disjoint sets A, B, C E S such that X=AUBUC, I/A=ltA, lIJl=O, Itc=O,

whence 1/ = I/A + I/C is a (the!) Lebesgue decomposition of to /L (unique by 9.3.7).

1/

with respect

Proof. The measure It + 1/ is also O"-finite and 1/ $ It + 1/; obviously 1/ « It + 1/ , so by the Lemma there exists a set E E S such that (0)

1/

=(It + I/)E,

1/

= lIE.

Writing B = X - E, we have I/B = 0 by (ii) of 9.3.2. From (0) we see that 1/

=(It + I/)E = ItE + lIE = ItE +

1/,

whence it is clear that /LE« 1/; applying the Lemma to the pair ItE, there exists a set F E S such that (00)

ItE

=I/f',

1/,

ItE = (/LE)F = ItEnF·

Writing A = E n F and citing (0) and (*0) at the appropriate steps, we have I/A = lIEnF = (lIE)F = I/f' ;: ItE = ItEnF = itA , thus I/A = itA . Since AcE and B = X-E, we have AnB = (/). Writing C = E-A, the measurable sets A, B, C are pairwise disjoint with Au B U C = X. Moreover, since E = Au C and itA = ItE (noted in ("», we have ItE = ItAUC = itA

+ J.Lc =

ItE + J.Lc ,

and Itc = 0 results on cancelling the finite measure ItE.

9. Topics in Measure and Integration

450

Finally, we see from (*) and the disjointness of A and C that II = lIE = IIAue = IIA

+ lie;

the relations (lIe)e = lie and 1J.e = 0, that is, (llek = lie, 1J.x-e = 1J.,

=

show that lie 1. 1J., whereas IIA 1J.A« 1J., thus II = IIA +lIe is a Lebesgue decomposition of II with respect to 1J.. 0 The extension to a pair of u-finite measures is straightforward: 9.3.12. Corollary. (Lebesgue decomposition) Same conclusion as the Theorem, assuming that both 1J. and II are u-finite measures. Proof. Let (En) be a sequence of pairwise disjoint sets in S such that X = U::"=l En and 1J.(En ) < +00 for all n. Then, for each n, 1J.E. is finite,

lIE. is u-finite,

so by the preceding theorem there exist pairwise disjoint sets Un, Vn , W n in S with X=UnUVnUWn,suchthat (IIE.)U.

= (1J.E.)U

n

,

(IIE.)V. = 0, (1J.En )W. = 0,

that is,

(*)

lIE.nu..

=1J.E..nU.. ,

lIE.nv. = 0, 1J.E.nw.. =

o.

For each n, the sets ~=~n~,~=~n~,~=~n~

are pairwise disjoint sets with AnUBnUCn=En , therefore the sets 00

A=

U An,

00

B=

n=1

U Bn ,

00

U Cn

C=

n=1

n=1

are pairwise disjoint sets in S with Au B U C = we have IIA ..

=

1J.A.. , lIB.

= 0,

1J.e.

U::"=l En

= X. By (*),

=0

for every n, whence it is clear from the countable additivity of measures that IIA

=

1J.A,

lIB = 0, 1J.e = 0,

thus the conclusions of the preceding theorem also hold for 1J. and II. Finally, the extension to signed measures is a triviality:

0

§9.4. Convolution in LI(IR)

451

9.3.13. Corollary. Same conclusion as the Theorem, assuming that Il and v are signed measures such that the measures IIlI and Ivl are u-finite. Proof. Apply the preceding corollary to the u-finite measures IIlI and Ivl: there exist pairwise disjoint sets A, B, C in S with union X, such that

IvlA

=liliA,

IvlB = O. III Ie = O.

in other words (9.1.23)

IVAI - lilAI, IIIBI = 0, IJLCI = 0, that is,

and finally (9.3.9) VA

=

IlA,

lIB =0, JLC =0,

thus the conclusions of the Theorem also hold for J.1 and v.

0

Exercises 1. Let A be Lebesgue measure on the closed interval [-1,1), let h: [-1, 1J -IR be the function h = -'1'(-1,01 + '1'(0,1) that is -Ion [-l,OJ and + 1 on (O, 1) , and let Il = h . A be the indefinite integral associated with the A-integrable function h (4.7.1). (i) IIlI = A, therefore A 4: J.1 in the sense of Definition 9.3.3. {Hint: 9.2.11.} (ii) J.1([-l,l]} = 0 but A{[-l,l]} = 2.

2. If Il, v are signed measures on S such that the measures IIlI and Ivl are u-finite. then the following conditions are equivalent:

(a) Il (b) v

= v; = I ·11l1

for some I E V(l1l1) such that I(x) # 0 11l1-a.e. {Hint: (a) ~ (b): Write v = I· IIlI by the Radon-Nikodym theorem (9.2.15) and apply the equality Ivl = 1/1·11l1 to the set E = {x: I(x) =

O} . (b) ~ (a): For every E E S, Ivl{E) ~ 'l'E = 0 11l1-a.e. (cf. 4.4.21).}

= J'l'El/ldllll; argue that

Ivl(E)

=0

9.4. Convolution in L I (1R)

We know that, in the context of Lebesgue measure A on IR (4.4.7), L' = 4;(IR, M. A) is a Banach space (6.4.18). In fact. there is a natural

452

9. Topics in Measure and Integration

way of introducing a product in L' in such a way that l' becomes a Banach algebra (in the sense of Definition 6.8.13) of capital importance in the theory of the Fourier transform.' The crux of the matter is this: given 11.,vEL',say 11.=j, v=g with f,gE£'=£HIR,M,A} (see6.4.14 for the dot notation), the problem is to create a function hE £' suitable for defining a product 11.V = h. There is a trivial solution: always take h = 0; the result is a trivial algebra in which 11.V = 0 for all elements 11. and v. A useful solution must make h dependent on f and 9 in a subtler way: we shall show that there exists a function h E £' such that

(*)

h(x}

=

J

f(x - y}g(y}dy

for almost every x (here dy stands for Lebesgue measure A in the context of y variable while x is fixed, and "almost every" refers to an exceptional set of Lebesgue measure zero). More precisely, we wiII show that the integral on the right side of (*) exists for almost every x, and is equal to a Lebesgue-integrable function on the complement of a Lebesgue-measurable set (or of a Borel set) of measure zero; the proof is a deft application of the Fubini-Tonelli theorem (§7.4). For the proof that this leads to an associative (and commutative) algebra structure on L', we refer the reader to more specialized texts. 2 For the rest of the section, A denotes Lebesgue measure, defined on the a-algebm M of Lebesgue-measumble sets (2.2.9); we write J.L = AlB for the restriction of A to the a-algebm B = B(IR} of Borel sets of JR (2.4.5). In the context of either A or J.L, we also use the notations J f(x, y}dx and J f(x, y}dy to indicate the variable of integmtion when the other variable in a function of two variables is held fixed.

Thus, there are two measure spaces in the picture: (JR, M, A) and (IR, B, J.L). A complex-valued function f : IR ..... C on JR is called Lebesgue-measumble if it is measurable with respect to M, and Borel if it is measurable with respect to B (cf. 9.4.3 below). On the one hand, every J.L-integrable (Borel) function f: IR ..... C is obviously A-integrable (i.e., Lebesgue-integrable), with J fdA = J fdJ.L. On the other hand, if f : IR ..... C is A-integrable then there exists a Borel function g: IR ..... C such that f = 9 on the complement of a Borel set of measure zero (see Proposition 9.4.7 below); in particular, f = 9 A-a.e. (in fact J.L-a.e.) and 9 is J.L-integrable with J gdJ.L = J fdA. Thus, although £t(J.L} is a proper linear subspace of £t(>·} , the Banach spaces Lt(J.L} and LHA} may be identified. For technical reasons explained below, it will be convenient to

'cr. E. Hewitt and K. Stromberg, Real and abstract analysis [Springer, New York, 1965), p. 250, (16.36). 2 Cf. L.H. Loomis, An introduction to abstroct hannonic analysis (Van Nostrand, New York, 19531, E. Hewitt and K. Stromberg, op. cit., p. 399. (21.34), or the author, Measure and integration [Macmillan, New York, 1965; reprinted Chelsea, New York. 19701. §86.

§9.4. Convolution in L1(lR)

453

discuss Borel functions first, then adapt the results to Lebesgue-measurable functions. Some preparatory material on Borel functions is in order. Recall that if X is any topological space, the class B(X) of Borel sets of X is the a-algebra generated by the open sets of X (4.1.2,(ii)). In particular, we abbreviate B = B(lR) for the a-algebra of Borel sets of lR (cf. 2.4.11). A function f: lR -lR is called a Borel function if f-I(B) C B (4.1.10); in other words f, regarded as a function defined on the measurable space (R,B), is measurable with respect to B (4.1.3)-equivalently (4.1.6), U open in lR

'*

rl(U) E B.

This suggests the following generalization: 9.4.1. Definition. Let X and Y be topological spaces. A function

f: X - Y is said to be a Borel function if B Borel in Y

'*

equivalently (cf. the proof of 4.1.6), open set U in Y.

r

1

(B(Y)) C B(X) , that is,

rl(B) Borel in X;

r

I

(U) is a Borel set in X for every

9.4.2. Remarks. (i) Every continuous function is a Borel function. {The inverse image of an open set is open, hence is a Borel set.}

(ii) If f: X - Y and g: Y - Z are Borel functions (X, Y, Z topological spaces), then the composite function go f : X -+ Z is Borel. {If B is a Borel set in Z then (gof)-I(B) = f-I(g-I(B)) is the inverse image under f of a Borel set in Y.} (iii) If X and Yare topological spaces, then

B(X) x B(Y) C B(X x Y) .

{If U, V are open sets in X, Y , respectively, then U x V is open in X x Y hence is a Borel set.} (iv) If X and Yare topological spaces having a countable base for the topology (6.1.20)-for example if X and Y are separable metric spaces (6.1.21)-then

B(X) x B(Y)

= B(X x Y) .

{Every open set in X x Y is the union of a sequence of sets Un xVn with Un, Vn open in X, Y, respectively, whence B(X x Y) c B(X) x B(Y).} In particular,

B(lR) x B(lR)

= B(lR x

lR) .

We are particularly interested in Borel functions in the case that Y (and, ultimately, X = lR):

=C

9.4.3. Proposition. Let X be a topological space. The following conditions on a function f: X - C are equivalent:

9. Topics in Measure and Integration

454

(a) I is a Borel function (in the sense of Definition 9.4.1); (b) I is measurable with respect to the u-algebra SeX) (in the sense of Definition 6.4.1); (c) the Iunctions Re I, 1m I : X -+ IR are measurable with respect to SeX) (in the sense of Definition 4.1.3); (d) the Iunctions Re I, 1m I: X -+ IR are Borel (in the sense of 9.4.1).

Prool. (b)

(c) by the Definition of (b) (6.4.1). (c) ¢} (d) by the remarks in the paragraph preceding 9.4.1. Recall that C and 1R2 may be identified as metric spaces (cf. 3.1.9, 3.1.15), hence as topological spaces. (a) =? (d): If U is an open set in IR, then the 'rectangle' U + ilR (in the Gaussian plane C = 1R2 ) is open in C, so by the hypothesis (a), I-I(U + ilR) is a Borel set in X; since ¢}

rl(U

+ ilR) = (Ref)-I(U) n (Imf)-I(IR) = (Ref)-I(U),

we see that (Ref)-I(U) E SeX) for all open sets U in IR, thus ReI is a Borel function (in the sense of 9.4.1). Similarly 1m I is a Borel function (consider IR + iU). (d) =? (a): Every open set W in C is the union of open rectangles with Gaussian-rational vertices, hence can be expressed as a countable union 00

W

= U(Un

+iVn ),

n=1

where Un and Vn are open intervals in 1R; by the hypothesis (d), the set rl(W)

=

00

00

n=1

n=1

Url(U n + iVn) = U(Ref)-I(Un ) n (Imf)-I(Vn)

is clearly a Borel set in X. Thus 9.4.1). 0

I is a Borel function (in the sense of

A bonus of criterion (b) is that the set of all Borel functions X a complex algebra for the pointwise operations:

-+

C is

9.4.4. Corollary. Let X be a topological space. II I, 9 : X -+ C are Borel functions, c E IR and c< > 0, then the functions I + g, cI, I 9 and IIIQ are also Borel functions. Proof This is immediate from 6.4.2 and criterion (b) of Proposition 9.4.3. 0

9.4.5. Corollary. II X is a topological space and In : X -+ C is a pointwise convergent sequence 01 Borel functions, then the limit function lim In is also a Borel Iunction. Proof Immediate from 6.4.3 and criterion (b) of 9.4.3.

0

§9.4. Convolution in L'{IR)

455

9.4.6. Corollary. II X and Y are topological spaces and I : X --+ ce, 9 : Y --+ ce are Borel Iunctions, then the junction I 0 9 : X X Y -+ ce defined by

(J 0 g)(x, y)

= I(x)g(y)

(x EX. Y E Y)

is measurable with respect to B(X) x B(Y) ; in particular, 109 is a Borel junction. Proof By criterion (b) of 9.4.3, our hypothesis is that I. 9 are measurable with respect to B(X). B(Y) , respectively. If I and 9 are characteristic functions, say I = 'PE • 9 = 'PF with E. F Borel sets in X. Y. respectively, then I0g = 't'ExF is measurable with respect to B(X) x B(Y) because Ex F E B(X) x B(Y). The case that I and 9 are simple (with respect to B(X) and B(Y) , respectively) then follows from the bilinearity of the operation 0. In the general case, I and 9 are the pointwise limits of sequences In, gn of simple Borel functions (because their real and imaginary parts are. by criterion (b) of 9.4.3 and Theorem 4.1.26), therefore

I0g

= limIn 0gn

is measurable with respect to B(X) x B(Y) by Proposition 6.4.3. Finally, (J 0 g)-1 (B(CC)) c B(X) x B(Y) c B(X x Y) by (iii) of 9.4.2, thus I 0 9 is a Borel function on X x Y.

Our last generality before we get down to business (convolution): 9.4.7. Proposition. II I : IR --+ ce is Lebesgue-measurable, then there exists a Borellunction g: IR --+ ce such that 1= 9 on the complement 01 a Borel set 01 measure 0 (thus 1= 9 j1.-a.e. and hence >.-a.e.).

Proof The assumption on I is that its real and imaginary parts are measurable with respect to the a-algebra M of Lebesgue-measurable subsets of IR (4.1.5, (iii) and 6.4.1). In view of Proposition 9.4.3. we can suppose that I: lR --+ IR. and by Theorem 4.1.15 we can suppose further that I > O. Contemplating Theorem 4.1.26 and Corollary 4.1.20. we can suppose in addition that I is simple; finally, taking into account Theorem 4.1.9, we can suppose that I is the characteristic function of a Lebesgue-measurable set. (At every step of the reduction, we note that a union of a finite or denumerable set of Borel sets of measure 0 is also such a set.) Say I = 'PE, E EM. By Corollary 2.4.15. there exist Borel sets F and G such that FeE c G and >'(G - F) = O. The function 9 = 'PF is Borel. I = 9 = 1 on F and I = 9 = 0 on CG. thus I = 9 on the complement C(G - F) = F u CG of the Borel set G - F of measure O. (The argument shows that every Lebesgue-measurable set of measure 0

9. Topics in Measure and Integration

456

is contained in a Borel set of measure 0, thus the concepts "A-a.e." and "JJ.-a.e." coincide-and may indifferently be written "a.e.")

9.4.8. Definition. If f, 9 : IR by the formula (fVg)(x, y)

-+

C we define a function fV 9 : 1R

= f(x -

2

-+

C

(x, Y E IR)

y)g(y)

(note that the arguments of f and 9 on the right side add up to x); writing -rr: 1R2 -+ IR for the function -rr(x, y) = x - y, we have

(fVg)(x, y)

= f(-rr(x, y»)g(y) = (f

0

-rr)(x, y). (I 0 g)(x, y),

thus fVg = (f 0 -rr)(10 g). 9.4.9. Proposition. If f, 9 : IR -+ C are Borel functions, then the function fV 9 : 1R2 -+ C defined above is measurable with respect to B(1R2 ) = B(IR) x B(IR) , thus is a Borel function on 1R2 . IR is continuous, f 0 -rr : 1R2 -+ C is a Borel function on 1R2 by (i) and (ii) of 9.4.2; in other words, by (iv) of 9.4.2, f 0 -rr is measurable with respect to B(IR) x B(IR). (This is the step where Borel sets are more convenient than Lebesgue-measurable sets.) On the other hand, 10g is measurable with respect to B(IR) x B(IR) by Corollary 9.4.6, therefore 50 is the pointwise product (f 0 -rr)(10 g) = fVg.

Proof. Since -rr: 1R2

-+

Recall that JJ. = AlB denotes the restriction of Lebesgue measure A to the u-algebra of Borel sets B = B(IR) . Prior to calculating the iterated integrals of fV 9 (with respect to JJ. x JJ.), we describe the sections (cr. 7.3.1) of fV 9 : 9.4.10. Lemma. If f, 9 : IR

-+

C then, for all x, y E IR ,

(fVg)% = (f 0 -rr)%. g,

(fVg)Y = g(y)f- y ,

where f- y denotes the translate of f by -y, that is, f-y(x) for all x E IR .

= f(x -

y)

Proof. For all x, y E IR,

= (fVg)(x, y) = (f -rr)(x, y)g(y) = (f -rrMy) . g(y) , (fVg)Y(x) = (fVg)(x,y) = f(x - y)g(y) = g(y)f-y(x) (fVgMy)

0

by the formulas of Definition 9.4.8.

0

9.4.11. Lemma. Let f,g: IR -+ C. (i) If f E.cb (Jor either A or JJ.) then, for every y and

J

(JVg)Y =

J

f(x - y)g(y)dx =

(f

E IR,

f)g(y).

(fVg)Y

E.et

§9.4. Convolution in L1 (1R)

457

(ii) If f. 9 E C~ (for either>. or p.) then, for every x E IR.

(J and

0

f

-rr)%

E C~,

(J'\1g)% = (J 0 -rr)z- 9

f

f(x - y)g(y)dy =

(f

0

E

ct

-rr)%(y)g(y)dy.

Proof. (i) By the second formula of the preceding lemma, (J'\1g)Y = g(y)f-Y for every y E IR. From the translation-invariance of Lebesgue measure (2.1.10) it is straightforward to show that f- y E Cb and f f- y = f f (first consider the case that f is a characteristic function, then a simple function 2': 0, etc.). (ii) Let x E IR. By the first formula of the preceding lemma, (J'\1g)% = (J 0 -rr)% . g; since (f

0

-rr)%(y)

= (f 0

-rr)(x. y)

= f(x -

y)

and the function Ifl 2 is by assumption integrable (6.7.1), it is straightforward to show, using the invariance of Lebesgue measure under the transformations y ...... -y ...... x - y (2.1.11, 2.1.10), that (J 0 -rr)% E C~. It follows that (f'\1g)z = (f o-rr)%' 9 is integrable by Holder's inequality (6.7.2) and

f

f(x - y)g(y)dy =

f

(J'\1g)%(y)dy =

f

(f

0

-rr)%(y)g(y)dy.

{Incidentally, the right member of the equality can be written as the inner product in L~ of the equivalence classes of (f 0 -rr)% and 9 (see §6.7, Exercise 5).} 9.4.12. Theorem. If f,g E ct(p.) then f'\1g E CH/-L x p.) and

Proof. Since the operation f'\19 (hence its integral) is bilinear. as is the expression on the right side, we can suppose that f 2': 0 and 9 2': 0 (express the real and imaginary parts of f, 9 as a difference of positive integrable Borel functions); then also f'\1g 2': O. From 9.4.9 we know that f'\1 9 is measurable with respect to the domain B(IR) x B(IR) of p. x p.. By (i) of the preceding lemma, (J'\1g)Y is p.-integrable for every y E IR , and

f

(f'\1g)Ydp. =

(J fdp.)g(y).

The function (f fdp.)g being /-L-integrable, we see (Definition 7.3.3) that the iterated integral

ff

(J'\1g)(x, y)dxdy

458

9. Topics in Measure and Integration

exists and is equal to / (/

Id~) gd~ = (/ Id~)

(/

By the Fubini-Tonelli theorem (7.4.7), l"ilg is integral (J I d~)( f gd~).

9d~) .

(~

x

~)-integrable

with

9.4.13. Corollary. II I,g E .ct(~) then the iterated integral

/ / (f"ilg)(x, y)dydx exists and is equal to (J I d~)( f gd~) . Proal. This is immediate from the preceding theorem and Fubini's theorem (7.4.8). 9.4.14. Remarks. The preceding corollary can be expressed as follows: If I and 9 are Lebesgue-integrable Borel functions, then there exist a

Lebesgue-integrable Borel function h and a Borel set E of measure 0 such that x E IR - E

and

=>

3 / I(x - y)g(y)dy

= h(x)

f hd>' = (J Id>.){ f gd>') .

9.4.15. Definition. Let I,g: lR -+ C. The convolution of I and 9 (in that order-but see Exercise 2) is the function I * 9 whose domain is the

set D'*g

= {x E lR: = {x E lR:

(f"ilgh E 4(>.)} the function y 0-+ I(x - y)g(y) is >'-integrable}

and whose values are given by the formula (f

* g) (x) = /

I(x - y)g(y)dy

(x E D,*g).

9.4.16. Remarks. The message of 9.4.14: If I

and 9 are Lebesgueintegrable Borel functions, then there exist a Lebesgue-integrable Borel function h and a Borel set E of measure 0 such that D'*g ::> IR - E

and

I *9 =

h on lR - E.

{It can be shown3 that D,*g is itself a Borel set (whose complement has measure O).} Moreover, f hd>' = (J Id>.)( f gd>') . Abusing the notations slightly, f (f * 9 )d>' = (J I d>.)( J gd>') . 3 cr. the author, op. cit., p. 293, Theorem 1.

§9.4. Convolution in L1 (IR)

459

Convolution is an operation that 'smoothes away' behavior on negligible sets: 9.4.17. Lemma. If on IR such that

It, h. 91> 92 It

=

: IR -+ C are complex-valued functions

h a.e.

and

91 = 92 a.e.

(with respe£t to Lebes9ue measure) then It * 91 = of the domains).

h *92 (includin9 equality

Proof. For each x E IR, the functions

Y ...... !I(x - Y)91(Y) ,

Y ...... h(x - Y)92(Y)

are equal a.e.; since Lebesgue measure is complete (§2.4, Exercise 8), each of these functions is Lebesgue-integrable (or Lebesgue-measurable) if and only if the other is. and their integrals are then equal. The extension of convolution to Lebesgue-integrable (not necessarily Borel) functions is now effortless: 9.4.18. Theorem. If f,9: IR -+ C are Lebes9ue-integmble, then there exist a Lebes9ue-inte9mble Borel function h and a Borel set E of measure 0 such that

DJ.g:>IR-E

and

f*9=h onlR-E.

Proof Let fo,90 : IR -+ C be Borel functions such that f = fo and 9 = 90 on the complement of a Borel set of measure 0 (9.4.7). Then fo, 90 are also Lebesgue-integrable and. as noted in 9.4.16, there exist an integrable Borel function h and a Borel set E of measure 0 such that

DJo'Yo :>IR-E and

fo*90=h on IR-E.

By the lemma, f * 9 = fo * 90 , thus the function h and the Borel set E of measure 0 meet the requirements of the theorem.

9.4.19. Remark. With notations as in the theorem, we have shown that the iterated integral

!

!UV9)(X, y)dydx

exists (note that in Definition 7.3.3. the function of two variables is not required to be measurable), without knowing whether fV9 is measurable with respect to M x M. The Borel functions fO.90 in the foregoing proof have done the heavy lifting (via 9.4.12 and 9.4.13).

460

9. Topics in Measure and Integration Exercises

1. There exist compact spaces X and Y for which the inclusion B(X) X B(Y) c B(X X Y) of 9.4.2, (iii) is proper. 4

2. (i) (Commutative law) If I, 9 : IR -+ C are Lebesgue-integrable, then l*g=g*1 a.e. 5 (ii) (Associative law) If I,g,h: IR -+ C are Lebesgue-integrable, and if p, q : IR -+ C are Lebesgue-integrable functions such that 1* 9 = P a.e. and g* h = q a.e., then p*h= I*q a.e. 6 3. With notations as in Lemma 9.4.11, if (with respect to either A or J1.) one of I,g is in L1; and the other is in L'C', then (f'ilg)% E L1; for allxEIR. {Hint: Lemma 6.6.8 and the formula (f'ilg)% = (f 0 11')%' g.} 4. If u, v E Lt = L1;(IR, M, A), write u = j, v = 9 as in 6.4.14, choose h E L1; so that h = 1* 9 a.e. (9.4.18), and define a product in Lt by the formula uv = h. Then the function iC. We are going to use K to define a linear mapping T: £P -> C, of which K is called the kernel function. Since I x I is compact for the Euclidean metric topology (see 6.1.23 or §6.1, Exercise 10) and K is continuous, K is bounded (6.8.12); as in 6.8.12, we write

IIKlloo = sup{IK(x, y)l:

x, y E I},

the sup-norm of K in Cc(1 x I). 9.5.8. Lemma. Let K be the function 01 the preceding definition and let IE £P. (i) For each x E I, the function y ...... K(x, y)/(y)

= Kx(y)/(y)

(y E I)

is Lebesgue-integrable, and (1)

where q is the exponent complementary to p (as in 9.5.6, (iii».

9. Topics in Measure and Integration

464

(ii) The function g: I

->

g(x) =

(2)

C defined by

J

K(x,y)f(y)dy

(x E I)

satisfies the inequality Ig(x) - g(x')1 :5 IIKz - Kz'ilqllfll"

(3)

for all x,x' E I.

(iii) 9 is continuous. (iva) If p> 1 then IIglloo:5 IIKlloo(b - a)l/qllflh, . (ivb) If p = I, q = +00 then IIglloo:5 IIKlloollflil . (v) Regarding 9 as an element of J:.q, and K as an element of

J:.l:(I x I,S x S,>. x >.), IIgllq :5 IIKll q llfll,,· Proof Recall that if p = 1 then q = +00; if 1 q =p/(P-l); and if p= +00 then q = 1.


1 or p = 1, whence pointwise (in fact, uniform) total boundedness. (b) Given any E > 0, choose {j > 0 as in the proof of (iii) of the lemma. Then, by (4) and (5) of the proof, either

Ix - x'i :5 {j

=?

I(TJ)(x) - (TJ)(x')1 :5 E(b - a)l/ q for all fEB

or

Ix - x'i

~ {j

=?

I(Tf)(x) - (TJ)(x')1 ~

E

for all fEB,

§9.5. Integral Operators

467

according as p > 1 or p = 1, whence the equicontinuity of T(B). That the equicontinuity is unifonn is not news (8.1.10). 0 9.5.13. Corollary. Let 1 $ r < +00. The linear mappings (i) T : C ..... C, (ii) T: C ..... C and (iii) T: £P ..... £r defined by the lormula in 9.5.11 are also compact. Proof If lEe then III :5 11/11",,· 1, therefore II/l1p :5 1I/11",,1I111p (where 1I111p = (b-a)l/ p when p < +00, and 1I111p = 1 when p = +00); it follows that the insertion mapping C ..... £P is continuous. Consider the diagram T

C

C

ir

_-'--+1

r

L,;r

where T is the linear mapping of the Theorem and i p , i r are the insertion mappings. The mappings contemplated in the present corollary are (i) To i p = TiC, (ii) i r 0 To i p , and (iii) i r 0 T; since it is clear that the composite of a continuous linear mapping and a compact linear mapping (in either order) is compact, the corollary is immediate from the Theorem. 0 The case p = q = 2 is especially transparent: T: £2 ..... £2 leads to a compact operator t in the Hilbert space L 2 = L 2 (I, S,.x) (see Exercise 2) with IItll $ IK(x, YWdxdy) 1/2 by (v) of 9.5.8. The theory of operators in Hilbert space is especially well-developed (cf. Exercise 5).

(II

Exercises 1. Let (E, II II) be a seminormed space and let N = {x E E: IIxll = O} .

As noted in § 6.4, Exercise 1 , N is a linear subspace of E, and the mapping x + N ...... IIxll on the quotient vector space E/N (well-defined because IIx + zll = IIxll for all zEN) is a norm. We abbreviate E = E/N and x = x+N; thus E is a normed space with norm IIxll = IIxll for all x E E. (i) A sequence (x n ) in E is Cauchy in the sense of (9.5.2) if and only if (:i: n ) is Cauchy in E, and x n ..... x in E if and only if :i:n -+:i: in E. (ii) E is complete in the sense of 9.5.2 if and only if E is a Banach space. 2. With notations as in Definition 9.5.4, let T: E -+ F be a linear mapping, and form the quotient normed spaces E, F by the method of Exercise 1. (i) If, for x E E, IIxll = 0 => IITxll = 0 (cf. 9.5.5, (4)), then there exists a linear mapping t: E -+ F such that tx = (Tx)' for all x E E. (ii) With T as in (i), T is continuous in the sense of Definition 9.5.4 if and only if t is continuous. (iii) With T as in (i), T is compact in the sense of Definition 9.5.4 if and only if t is compact. (iv) In particular, if T : E -+ F is any continuous (compact) linear

468

9. Topics in Measure and Integration

mapping then there exists a continuous (compact) linear mapping T: E -+ F such that T± = (Tx)' for all x e E. 3. (i) Let (An) be a sequence of pairwise disjoint Lebesgue-measurable subsets of 1= la,b] such that >.(An ) > 0 for all n, and let fn = 'PAn be the characteristic function of An. Then IIfnlloo = 1 for all nand IIfm - fnlloo = 1 when m 'f n. Infer that the identity mapping £.00-+ £.00 , though continuous, is not compact. (ii) Adapt (i) to £.p (1 :5 P < +00) by SUitably modifying the functions fn. (iii) (Theorem of F. Riesz) In order that the identity mapping on a normed space be compact, it is necessary and sufficient that the space be finite-dimensional. 2 4. (i) When 1 :5 p < +00, every f e £.p is the limit in mean (of order p) of a sequence (fn) in C.3 (ii) The corresponding statement is false for p = +00, since a sequence (fn) in C with IIfm - fnlloo -+ 0 is uniformly convergent to a function in C-and there exist functions in £.00 that are not equal a.e. to a continuous function (for example, the characteristic function of an interval la, c] with a < c < b). 5. With notations as in Corollary 9.5.13, let p = q = 2. The following conditions on the compact linear mapping T: £.2 -+ £.2 are equivalent: (a) J(Tf)gd>. = J f (Tg)d>' for all f,g e £.2; (b) J(Tf)gd>. = J f(Tg)d>. for all f,g e C; (c) f(Tf)gd>. = f f(Tg)d>. for all polynomial functions f,g on I; (d) K(x, y) = K(y, x) for all x, y e 1. Such a mapping T is said to be self-adjoint, and the mapping T it induces in the Hilbert space L2 has an explicit representation ("Spectral Theorem") in terms of its eigenvalues. 4

2Cf. the author, Lectures in functional analysis and operator theory [Springer, New

York. 19741. p. 91, (23.10).

cr. the author, Measure and integration [Macmillan, New York, 1965j reprinted Chelsea. New York. 1970J. p. 220. Theorem 2. 4 cr. the author, Introduction to Hilbert space [Oxford University Press, New York, 1961; reprinted Chelsea, New York. 1976], p. 186. Theorem 6. 3

Bibliography

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HARTMAN, P., Ordinary differential equations, 2nd edn., Birkhiiuser, Boston, 1982. HAUSDORFF, F., Set theory, 3rd edn., Chelsea, New York, 1957. HEWITT, E. AND STROMBERG, K., Real and abstract analysis, Springer-Verlag, New York, 1965. HILDEBRANDT, T.H., Introduction to the theory of integration, Academic, New York,1963. HOBSON, E. W., The theory of functions of a real variable and the theory of Fourier's series. Vol. I, Dover, New York, 1957. KADISON, R.V. AND RINGROSE, J.R., Fundamentals of the theory of operator algebras. Vols. I·IV, Academic, New York, 1983-1992. KAPLANSKY, I., Set theory and metric spaces, 2nd edn., Chelsea, New York, 1977. KESTELMAN, H., Modern theories of integration, Oxford, 1937; 2nd revised edn., Dover, New York, 1960. KURATOWSKI, K., Topalogie. I., Monografie Matematiczne, 2nd edn., Warsaw, 1948. LANDAU, E., Foundations of analysis, Chelsea, New York, 1951. LANDAU, E., Differential and integral calculus, Chelsea, New York, 1951. LOOMIS, L.H., An introduction to abstract harmonic analysis, Van Nostrand, New York, 1953. MCSHANE, E.J., Integration, Princeton, 1944. OXTOBY, J., Category and measure, Springer-Verlag, New York, 1971. RICKART, C.E., General theory of Banach algebras, Van Nostrand, Princeton, 1960; reprinted R.E. Krieger, Huntington, 1974. ROYDEN, H., Real analysis, 3rd edn., Macmillan, New York, 1988. RUDIN, W., Principles of mathematical analysis, 3rd edn., McGraw-Hill, New York,1976. SUPPES, P., Axiomatic set theory, Van Nostrand, Princeton, 1960; reprinted Dover, 1972. SZ.·NAGY, B., Introduction to real functions and orthogonal expansions, Oxford, New York, 1965.

Index of Notations

SYMBOL

PAGE

SYMBOL

PAGE

xEA

2 2 3 4 4 8 9 10

lim sup Iln liminf Iln limn_oo Iln )'0 (A) M()'O) ).i(A) , )..(A)

79 80 82 87 96 97 99 101 103 107 119 119 119 122 124 124 126 127 129 134 137 137 137 142 142

IP',N,Z,Q,IR,C AcB AUB,AnB A' , A-B

{a} f:X-+Y

'PA gof P(X)

X/~

X""Y E~F

E-