Measure, Integration and Functional Analysis

If(a)g(a) I <- (z If(a) I

I g(a) I

q)

1/q

If f, g e P(O), 1 < p < oo, then f + g e lp(Q) and p) 1 /P

If(a) + g(a) I

<

P) 1 /P (Y_

+

"IP.

I g(a) I p)

I f(a) I

As in 2.4.5, we obtain the Cauchy-Schwarz inequality for sums from the Holder inequality. If f, g e l2(S2), then fg a 11(52) and

Y_ f(a) g(a) s (E I f(a) 12)

"' (Y

I g(a) 12)

If in the above discussion we replace S by {1, 2, ..., n}, all convergence difficulties are eliminated, and all the spaces Ip(S2) coincide with C". If 0 < p < 1, II p is not a seminorm on Lp(Q, F, µ). For let A and B be disjoint sets with a = u(A) and b = p(B) assumed finite and positive. If f = IA, II

g ='B, then

88

2

FURTHER RESULTS IN MEASURE AND INTEGRATION THEORY

1/p

Ilf+gllp=(f If+gl"du) Ill lip = al/p,

1/P

=(fn(IA+10)dic) Ilgllp =

=(a+b)1/P,

bl/P

But (a+b)1/P>a'1"+b1/Pif a, b > 0, 0 1, and has the value 0 when x = 0. Thus the triangle inequality fails. We can, however, describe convergence in LP, 0

(a+b)P
a,b>0, 0
which is proved by considering (a + x)P - aP - xP. It follows that

f If +gIPdµ<_ fn IfvPdp+ fn IgIPdP,

f,geL",

(2)

and therefore d(f, g) = In I f - g I P dp defines a pseudometric on M. In fact the pseudometric is complete (every Cauchy sequence converges); for Eq. (2) implies that if f, g e L", then f + ge LP, so that the proof of 2.4.11 goes through.

If fl is an interval of reals,

is the class of Borel sets of Q, and p is

Lebesgue measure, the space L"(0, .F, µ) will be denoted by Lp(c2). Thus, for example, L°[a, b] is the set of all complex-valued Borel measurable functions f on [a, b] such that 6 lifil = f lf(x)IP dx < co.

If f is a complex valued Bore] measurable function on (S2, p) and fl, f2, ... e LP(Q, F, p), we say that the sequence f f,) converges to f in LP if II f, - f II p - 0, that is, if in I f" - f I P dµ -- 0 as n -+ oD. We use the notation f" i 4 f. In Section 2.5, we shall compare various types of convergence of sequences of measurable functions. We show now that any f e LP is an LP-limit of simple functions.

Theorem. Let f e LP, 0 < p < oo. If E > 0, there is a simple function g e LP such that Ilf - gilp < a; g can be chosen to be finite-valued and to satisfy I g 1 5 I f 1. Thus the finite-valued simple functions are dense in R. 2.4.13

PROOF. This follows from 1.5.5(b) and 1.6.10. 1

If we specialize to functions on R" and Lebesgue-Stieltjes measures, we may obtain another basic approximation theorem. Theorem. Let f e LP(S2, F, p), 0 < p < oo, where n = R", .F R(R"), and it is a Lebesgue-Stieltjes measure. If e > 0, there is a continuous function g e LP(S2, F, µ) such that II f - gll p < s; furthermore, g can be chosen so that sup I g 1 S sup If 1. Thus the continuous functions are dense in Y.

2.4.14

2.4

LP SPACES

89

PROOF. By 2.4.13, it suffices to show that an indicator IA in L" can be approximated in the LP sense by a continuous function with absolute value at most 1. Now IA c L" means that u(A) < oo; hence by Problem 12, Section 1.4, there is

a closed set C e A and an open set V

A such that p(V - C) < e"2-P.

Let g be a continuous map of f) into [0, 11 with g = 1 on C and g = 0 on V` (g exists by Urysohn's lemma). Then

fn IIA -gIPdu=

f{IA#g} IIA

- gl"du.

But {IA $ g} e V- C and IIA - g I _< 2; hence

IIIA-gIIpS2Pg(V- C) <eP.

Since g=g-IA+IA, we have geL". Theorem 2.4.14 shows that the continuous functions in V do not form a closed subset, for if they did, every function in LP would be continuous. Equivalently, the continuous functions in LP are not complete, in other words,

there are Cauchy sequences of continuous functions in L" that do not converge in LP to a continuous limit. Explicit examples may be given without making use of 2.4.14; see Problem 2.

The Space L. If we wish to define LP spaces for p = co, we must proceed differently. We define the essential supremum of the real-valued 2.4.15

Bore] measurable function g on (S2, .F, u) as

esssup g=inf{cER:p{(O:g(cu)>c}=0) that is, the smallest number c such that g < c a.e. [p]. If f is a complex-valued Borel measurable function on (0, F, p), we define IIfII

=esssup If1.

The space L°°(Q, F, u) is the collection of all f such that (If IIA < oo. Thus f e L°° iff f is essentially bounded, that is, bounded outside a set of measure 0. Now I f + g j < I f I + g -< If II. + (I g l(,° a.e.; hence

Ilf+A. <- 11f 11. + Ilgll. In particular, f, g e L°° implies f + g e U. The other properties of a seminorm are easily checked. Thus Loo is a vector space over the complex field,

II

11

,0

is a

seminorm on L°', and becomes a norm if we pass to equivalence classes as before. If f, f,, f2 ,

... a L°° and 1,f, -f II,° -> 0, we write f II,° -+ 0 if there is a set A e F with p(A) = 0 such that f -+f

uniformly on A`.

2

90

FURTHER RESULTS IN MEASURE AND INTEGRATION THEORY

For, assume 11f. - f II - 0. Given a positive integer m, Il f" -f II < 1/m for sufficiently large n; hence I f"(w) - f(w)I < 1/m for almost every ('0, say for w 0 Am, where u(Am) = 0. If A = Um=, Am, then p(A) = 0 and f" -+f uniformly on A`. Conversely, assume p(A) = 0 and f" -+f uniformly on A`. Given e > 0, 1 f,, - f I < s on A` for sufficiently large n, so that I f" - f 1 5 E a.e. Thus ll f" - f Ii - E for large enough n, and the result follows.

An identical argument shows that f f.) is a Cauchy sequence in L°

(llf"-fmll.-+0asn,m->oo)iffthereisasetAe.Fwithp(A)=0andf"fm - 0 uniformly on A`. It is immediate that the Holder inequality still holds when p = 1, q = oo, and we have shown above that the Minkowski inequality holds when p = oo. To show that L°° is complete, let {f.) be a Cauchy sequence in L°°, and let A be a set of measure 0 such that f"(w) - fm(w) - 0 uniformly for w e A`. But then "(w) converges to a limit f (w) for each co E A`, and the convergence is uniform on A`. If we define f (co) = 0 for w E A, we have f e L°° and Theorem 2.4.13 holds also when p = oo. For if f is a function in L°°, the standard approximating sequence {f,,) of simple functions (see 1.5.5) converges to f uniformly, outside a set of measure 0. However, Theorem 2.4.14 fails when p = oo (see Problem 12). If C is an arbitrary set, F consists of all subsets of 11, and p is counting measure, then L°°(fl, ,F, p) is the set of all bounded complex-valued functions f = (f (a), a e fl), denoted by The essential supremum is simply the supremum; in other words, Ilf II = sup( I f (a) I : a e fl). If f) is the set of positive integers, l°°(Q) is the space of bounded sequences of complex numbers, denoted simply by 1°°. Problems

If f = {a,,, n = 1, 2, ...}, the a are real or complex numbers, and p is counting measure on subsets of the positive integers, show that la f dp = Y', a", where the sum is interpreted as in 2.4.12. (b) I f f = (f (a), a e n) is a real- or complex-valued function on the arbitrary set 0, and u is counting measure on subsets of 0, show that lo f du = Ea f (a), where the sum is interpreted as in 2.4.12. 2. Give an example of functions f, fl, f2, ... from R to [0, 1] such that (a) each f,, is continuous on R, (b) ,,(x) converges to f (x) for all x, f f (x) I p dx - 0 for every p e (0, oo), and (c) f is discontinuous at some point of R. {a(,"), a2"), ..} be a sequence of complex 3. For each n = 1, 2, ..., let 1.

(a)

I

numbers.

2.4 L° SPACES (a)

91

If the a(n) are real and 0 5 ak") < ak"+1) for all k and n, show that 00

00

lim E a(n) n-g00 k=1

lim ak").

k=1n-'oo

Show that the same conclusion holds if the ak") are complex and I ak"> I < bk for all k and n, where _k= 1 b, < oo. (b) If the ak") are real and nonnegative, show that

k=1 n=1

(c)

Y_ akn'' ak")= n=1 Y, k=1

If the a(k") are complex and Y- Yk I ak" I < oo, show that ['n 1 Yk00 i a(kn) and Yk 1 E ak ) both converge to the same 1

1

1

4.

finite number. Show that there is equality in the Holder inequality if I f I ° and I g I ° are linearly dependent, that is, iff A I f I ° = B I g I a a.e. for some constants

5.

6. 7.

A and B, not both 0. If f is a complex-valued p-integrable function, show that I f of d.I = fn I f I dp iff arg f is a.e. constant on {w: f (w) # 0}. Show that equality holds in the Cauchy-Schwarz inequality iff f and g are linearly dependent. (a) If 1 < p < oo, show that equality holds in the Minkowski inequality if Af = Bg a.e. for some nonnegative constants A and B, not both 0.

(b) What are the conditions for equality if p = I ? 8.

If 1 < r < s < oo, and f e LS(S2, .F, u), u finite, show that Ilf 11, < kJI f II, for some finite positive constant k. Thus LS e L" and LS convergence implies L' convergence. (We may take k = 1 if u is a probability measure.) Note that finiteness of u is essential here; if p is Lebesgue measure on

R(R) and f (x) = I /x for x

1, f (x) = 0 for x < 1, then f e LZ but

f0L'. 9.

If It is finite, show that Ilf IIp -i Ilf III as p -* oo. Give an example to

10.

show that this fails if 14(S)) = oo. (Radon-Nikodym theorem, complex case) If p is a or-finite (nonnegative, real) measure, A a complex measure on (S2, and A < p, show that

there is a complex-valued u-integrable function g such that 2(A) _ JA g dp for all A e F. If h is another such function, g = h a.e. Show also that the Lebesgue decomposition theorem holds if A..

is a complex measure and u is a a-finite measure. (See Problem 6, Section 2.2, for properties of complex measures.) 11.

(a)

Let f be a complex-valued p-integrable function, where u is a nonnegative real measure. If S is a closed set of complex numbers

92

2

FURTHER RESULTS IN MEASURE AND INTEGRATION THEORY

and [l /µ(E)] f E f dµ e S for all measurable sets E such that µ(E) > 0, show that f (w) e S for almost every a). [If D is a closed disk with

center at z and radius r, and D e S`, take E = f -'(D). Show that I J E (f - z) dp I < rp(E), and conclude that µ(E) = 0.] (b) If 2 is a complex measure, then A < I A I by definition of 121 ;

hence by the Radon-Nikodym theorem, there is a I A I -integrable complex-valued function h such that A(E)= 5 E h d I A I for all

E e F. Show that I h I= 1 a.e.

[ I A I ]. [Let A, = {co: I h(w) I< r},

0 < r < 1, and use the definition of 12I to show that I h I >_ 1 a.e. Use part (a) to show I h I < I a.e.] (c) Let p be a nonnegative real measure, g a complex-valued p-integrable function, and 2(E) = fE g dµ, E e F. If h = d.1/dl A I as in part (b), show that 12 I (E) = f E lig dµ. (Intuitively, Jig dp = h d l - hh dI l I = I h 12 dill = dI l I . Formally, show that f a fh dI ).I = f a fg dp if f is a bounded, complex-valued, Borel measurable function, and set f = RE.) (d) Under the hypothesis of (c), show that

IAI(E)=J IgI dµ 12.

for all Ee..

Give an example of a bounded real-valued function f on R such that there is no sequence of continuous functions f such that if -f IL -> 0. Thus the continuous functions are not dense in L°° (R).

2.5 Convergence of Sequences of Measurable Functions

In the previous section we introduced the notion of LP convergence; we are also familiar with convergence almost everywhere. We now consider other types of convergence and make comparisons.

Let f, f f2, ... be complex-valued Bore] measurable functions on (S2, .F, µ). We say that f -> f in measure (or in µ-measure if we wish to

emphasize the dependence on p) if for every e > 0, p(w: I f,(w) - f (w) I >- e} When µ is a probability measure, the con- 0 as n -+ oo. (Notation: vergence is called convergence in probability. The first result shows that LP convergence is stronger than convergence in measure. 2.5.1

Theorem. If f, f1, f2, ... e LP (0 < p < oo), then f

PROOF. Apply Chebyshev's inequality (2.4.9) to If, - f I. 1

f implies f

f.

2.5

93

CONVERGENCE OF SEQUENCES OF MEASURABLE FUNCTIONS

is a Cauchy sequence in L°, then The same argument shows that if is Cauchy in measure, that is, given e > 0, p{w: I fm(m) I >_ E) - 0

{

as

If f, f,, f2 .... are complex-valued Borel measurable functions on (Q, .9, p),

we say that f - f almost uniformly if, given e > 0, there is a set A e F such that p(A) < E and f -> f uniformly on A`. Almost uniform convergence is stronger than both a.e. convergence and convergence in measure, as we now prove. 2.5.2 Theorem. If f, - f almost uniformly, then f -f in measure and almost everywhere.

PROOF. If s > 0, let f, -+f uniformly on Ac, with p(A) < e. If 6 > 0, then eventually If, -f I< bon A`, so fl f, f -f I > S} c A. Thus p{ I f - f > S} < p(A) < E, proving convergence in measure. To prove almost everywhere convergence, choose, for each positive integer

k, a set Ak with p(Ak) < 1/k and f --+f uniformly on Ak`. If B = Uk

Akc, 1

then f -+f on B and p(B`) = p(f lk 1 A,):!:-:: p(Ak) - 0 as k -+ oo. Thus p(Bc) = 0 and the result follows. I The converse to 2.5.2 does not hold in general, as we shall see in 2.5.6(c), but we do have the following result.

Theorem. If {f} is convergent in measure, there is a subsequence converging almost uniformly (in particular, a.e. and in measure) to the same limit function. 2.5.3

is Cauchy in measure, because if I f - fm I PROOF. First note that { then either If, - f I e/2 or I f - fm I >- e/2. Thus

Pflfn-fmI

>E}
E,

as n,m -> oo.

Now for each positive integer k, choose a positive integer Nk such that Nk+, > Nk for all k and

p{w: If,((o) -f.(w)I > 2-k} <

2_k

for n, m > Nk.

Pick integers nk Z Nk, k =/1, 2, ...; then if gk = /lw)I

: 2-k} < 2-k. p{w. I9k(w) - 9k+I Let Ak = { I9k - 9k+1 I >- 2-k}, A = lim sup* Ak. Then p(A) = 0 by 2.2.4; but if co 0 A, then co e Ak for only finitely many k; hence I9k(w) - 9k+, (w) I < 2-k

94

2 FURTHER RESULTS IN MEASURE AND INTEGRATION THEORY

for large k, and it follows that gk(w) converges to a limit g(w). Since µ(A) = 0 we have gk -+ g a.e.

If B,. = Uk At, then µ(Br) < k r µ(A k) < s for large r. If co 0 Br, then gk(w) - gk1(w)I < 2-k, k = r, r + 1, r + 2, .... By the Weierstrass M-test, 9k -+ g uniformly on Br , which proves almost uniform convergence. Now by hypothesis, we have f "+ f for some f, hence --+f. But by 2.5.2,

f = g a.e. (see Problem 1). Thus f, converges almost uniformly to f, completing the proof. I There is a partial converse to 2.5.2, but before discussing this it will be convenient to look at a condition equivalent to a.e. convergence:

2.5.4 Lemma. If µ is finite, then f -> f a.e. if for every S > 0, 0

U {w: Ifk(w) -f(w)I >- S})

k=n

as n - oo.

PROOF. Let B, = {w : Ifn(w) - f (w) I >- S}, B,5 = lim sup Bna = f, 1 Uk n Bka Now Uk n Bka I Ba ; hence µ(U n Bka) -+ µ(Ba) as n - oo by 1.2.7(b). Now {w : fn(w)

U Ba

a>o 00

U B1/m

since Ba, c Bat for

S1 > S2.

M=1

Therefore,

f. -f

a.e.

if µ(Ba)=0 if

for all 6>0

µ UBka -,0

for all

S>0. 1

G='On

2.5.5

Egoroff's Theorem. If µ is finite and f -+f a.e., then f -+f almosi

uniformly. Hence by 2.5.2, if µ is finite, then almost everywhere convergence implies convergence in measure.

PROOF. It follows from 2.5.4 that given c > 0 and a positive integer j, for sufficiently large n = n(j), the set Aj = Uk n(j) {Ifk -f I ? 1/f) has measure less than s/2j. If A = U; A, then µ(A) < YT I µ(A j) < s. Also, if S > 0 and j is chosen so that 1/j < S, we have, for any k >- n(j) and w e A` (hence 1

co 0 Aj), I fk(w) - f (co) I < 1 /j < S. Thus f -+ f uniformly on K.

I

2.5

CONVERGENCE OF SEQUENCES OF MEASURABLE FUNCTIONS

95

We now give some examples to illustrate the relations between the various types of convergence. In all cases, we assume that F is the class of Borel sets and p is Lebesgue measure. 2.5.6 Examples. (a)

Let .0 = [0, 1] and define Le

f(x) =

if

0<-x
elsewhere.

Then f" -+ 0 a.e., hence in measure by 2.5.5. But for each p E (0, 00],f,, fails to

converge in R. For if p < oo, IIf"IID= f 1If"(x)IIdx= o

n

and

(b)

Let SZ = R, and define 1

if 0<x<en,

0

elsewhere.

f(x) = n

Then f" -+ 0 uniformly on R, so that f.----+ 0. It follows quickly that f" -+ 0

a.e. and in measure. But for each p e (0, oo), f" fails to converge in L°, since Ilf"IIp = npe" -+ Co. (c)

Let fl = [0, oo) and define

n<x
1

if

0

elsewhere.

f"(x) =

Then f" -o. 0 a.e. and in measure (as well as in L°, 0 < p < co), but does not converge almost uniformly. For, if f" - 0 uniformly on A and µ(A`) < e, then eventually f" < I on A; hence if A" = [n, n + (1/n)] we have A n UkZ" Ak = 0 for sufficiently large n. Therefore, A` Ukz" Ak, and consequently µ(A`) z k " p(Ak) = oo, a contradiction. (d) Let I = [0, 1], and define

m-I

1

if

0

elsewhere.

fnm(x) =

n

<x< m

m=1,...,n, n=1,2,...,

96

2

FURTHER RESULTS IN MEASURE AND INTEGRATION THEORY

Then Ilfnml!P =I In --+0, so for each p e (0, oo), the sequence fL1,f21,f22,f31, f32 , f33, ... converges to 0 in L° (hence converges in measure by 2.5.1). But

the sequence does not converge a.e., hence by 2.5.2, does not converge almost uniformly. To see this, observe that for any x 0, the sequence {fnm(x)} has infinitely many zeros and infinitely many ones. Thus the set on which fnm

converges has measure 0. Also, fnm does not converge in L', for if fnm f, then fnm e-, f, hence f = 0 a.e. (see Problem 1). But Ilfnmll, = 1, a contradiction.

Problems 1.

If fn converges to both f and gin measure, show that f =g a.e.

2.

Show that a sequence is Cauchy in measure if it is convergent in measure. (a) If p is finite, show that U convergence implies L° convergence for all p e (0, oo). (b) Show that any real-valued function in L°[a, b], - oo < a < b < oo,

3.

0 < p < oo, can be approximated in L° by a polynomial, in fact by a polynomial with rational coefficients. 4.

If it is finite, show that { fn} is Cauchy a.e. (for almost every co, { fn(w)} is

a Cauchy sequence) if for every S > 0, U {co: I ff{w) -fk(w)I >- S}

J,k=n 5.

0

as

n -, oo.

(Extension of the dominated convergence theorem) If I fn I < g for all n = 1, 2, ..., where g is Ep-integrable, and f,2->f, show that f is it-integrable

and faf,,d;t-'fnfdp. 2.6

Product Measures and Fubini's Theorem

Lebesgue measure on Rn is in a sense the product of n copies of onedimensional Lebesgue measure, since the volume of an n-dimensional rect-

angular box is the product of the lengths of the sides. In this section we develop this idea in a general setting. We shall be interested in two constructions. First, suppose that (Q3 , F j, A) is a measure space for j = 1, 2, ... , n. x Qn such that We wish to construct a measure on subsets of .0, x C12 x x A. [with each A3 E .f';] is the measure of the "rectangle" A, x A2 x p,(A,)p2(A2) µn(A,). The second construction involves compound experiments in probability. Suppose that two observations are made, with the first observation resulting in a point co, a Q,, the second in a point w2 a C12. The probability that the first observation falls into the set A is, say, µ,(A). Furthermore, if the first observation is co, the probability that the second observation

2.6

PRODUCT MEASURES AND FUBINI'S THEOREM

97

falls into B is, say, p(w B), where p(w,, ) is a probability measure defined on F2 for each co, a 52,. The probability that the first observation will belong to A and the second will belong to B should be given by p(A x B) =

JA

p(wi, B)pi(dwl),

and we would like to construct a probability measure on subsets of f21 x E12

such that p(A x B) is given by this formula for each A e.F, and B e . 2 . [Intuitively, the probability that the first observation will fall near w, is p,(dw,); given that the first observation is w,, the second observation will fall in B with probability p(w B). Thus p(a),, B)p,(dco,) represents the prob-

ability of one possible favorable outcome of the experiment. The total probability is found by adding the probabilities of favorable outcomes, in other words, by integrating over A. Reasoning of this type may not appear natural at this point, since we have not yet talked in detail about probability theory. However, it may serve to indicate the motivation behind the theorems of this section.] Definitions. Let F be a a-field of subsets of 52;, j = 1, 2, ... , n, and let fl = S2, X 02 x x fl,,. A measurable rectangle in 0 is a set A = Al x A2 x x A. where A; e F; for each j = 1, 2, ..., n. The smallest a-field containing the measurable rectangles is called the product a-field, written F, X F2 x x If all .F; coincide with a fixed a-field .F, the 2.6.1

product a-field is denoted by 39'". Note that in spite of the notation, .F, X F2 x x F" is not the Cartesian product of the #j; the Cartesian product is the set of measurable rectangles, while the product a-field is the minimal a-field over the measurable rectangles. Note also that the collection of finite

disjoint unions of measurable rectangles forms a field (see Problem 1). The next theorem is stated in such a way that both constructions described above become special cases.

Product Measure Theorem. Let (S2 'I, p,) be a measure space, with p, a-finite on F and let Q2 be a set with a-field F 2. Assume that for each

2.6.2

co, a 0, we are given a measure p(w,, ) on F2. Assume that p(w,, B), besides being a measure in B for each fixed w, a f2,, is Borel measurable in co, for each fixed B e .F2.. Assume that the p(w, ) are uniformly a-finite; that is, S 2 can be written as UM, B", where for some positive (finite) con-

stants k" we have p(w B") < k" for all to, e f2,. [The case in which the p(w -) are uniformly bounded, that is, p(w,, f22) < k < oo for all w is of course included.]

98

2 FURTHER RESULTS IN MEASURE AND INTEGRATION THEORY

Then there is a unique measure p on F = F, x F2 such that for all A e .F,, B E F2,

µ(A x B) = fA µ(w1, B)µ1 (dwl) namely,

p(F) = f µ(w1, F(wl))pl (dwl), n,

F e .°F,

where F(wl) denotes the section of Fat w, : F(wl) = {w2 a 522: (w,, (02) a F}.

Furthermore, p is or-finite on F; if µl and all the p(w,, ) are probability measures, so is µ. PROOF. First assume that the µ(w,, ) are finite. If C e .F, then C((01) E ;F2 2 for each w, c52,. To prove this, let 8 = {C e F: C(w,) E .F2}. Then W is a o-field since (1)

(Jc)(W1 ) = U Cn(wl), n1

Cc(wl) = (C(wl))c.

n=1

If A e -w,, B E then (A x B)(w,) = B if co, e A and 0 if w, 0 A. Thus f contains all measurable rectangles; hence W = F.

If C e .F, then u(w,, C(w,)) is Borel measurable in to, To prove this, let 9 be the class of sets in .F for which the conclusion of (2) (2)

holds. If C = A x B, A e .°r1, B e 92, then µ(w1, C(w,)) _

u(w,, B)

µ(w1, 0) = 0

if co, a A, if w1 0 A.

Thus µ(w,, C(wt)) = µ(w B)IA(w,), and is Borel measurable by hypothesis.

Therefore measurable rectangles belong to W. If C,, ..., C. are disjoint measurable rectangles,

'U(wl, (Cic)1) = µ(w1,

U C1(wl))

_ Y_ µ(wl, Ci(w,)) i=1

is a finite sum of Borel measurable functions, and hence is Borel measurable in co, Thus ' contains the field of finite disjoint unions of measurable rect-

2.6

PRODUCT MEASURES AND FUBINI'S THEOREM

99

angles. But f is a monotone class, for if C. E', n = 1, 2, ..., and C. T C, hence p(w,, C,,(co,)) -- p(w,, C(w,)). Thus F(col, C(co,)), then C(w,) a limit of measurable functions, is measurable in w,. If C, . C, the same conclusion holds since the p(w,, ) are finite. Thus W = .F. (3)

Define F e IF

p(F) = f p(w1, F(wl))pl(dwl), n

[the integral exists by (2)]. Then p is a measure on .F, and

p(A x B) = f p(w,, B)µ, (dw,)

Ac-.F-,, B E F2.

for all

A

To prove this, let F,, F2, ... be disjoint sets in F. Then

p n1 U F = fSl, p wl, n=1 U Fn(wl))µ1(dwl) 00

Y_ p(w1, Fn((o1))p1(dw1)

-ffl,n=1

_ n=1 f 11, p(w1, Fn(w1))p1(dw1) w

n=1

proving that p is a measure. Now p(A

x B) = f p(w1, (A =

x B)((o1))p1(dw1)

fn, p(wl, B)IA(w1)p1(dw1)

[see (2))

= fA p(wl, B)p1(dw1) as desired. Now assume the p(w1, ) uniformly a-finite. Let 522 = Un 1 B, where the

Bn are disjoint sets in Pre and p(w B.)< kn < oo for all co, E 521. If we set

B) =µ(w,, B n Be),

B E .F2 ,

100

2

FURTHER RESULTS IN MEASURE AND INTEGRATION THEORY

-) are finite, and the above construction gives a measure

the

on

F such that un'(A x B) = f

A e. ,,

B)u,(dwl),

B E F2

A

= f u((ol, B n Bn)ul(dw1), A

namely,

un'(F) = f un'(w,, F(wl))p,(dw)

u(w,, F(w,) n Bn)ul(dw,).

Let it = En , pn'; p has the desired properties. For the uniqueness proof, assume the p((o,, ) to be uniformly a-finite. If A is a measure on F such that A(A x B) = JA p(w,, B)p1tdco,) for all Ae B e F2, then A = p on the field F U of finite disjoint unions of measurable rectangles. Now It is a-finite on .moo, for if S22 = Un , B. with

B. C_ F2 and p(c) ,, Bn) < k < oo for all co, and fl, = Um=, Am, where the and Am belong to .y, and p,(Am) < oo, then S2, X K22 = U.n-, (Am x u(Am x Bn) = f

(w,, Bn)p1(dw1) < knu,(Am) < oo.

Am

Thus A = it on Jr by the Caratheodory extension theorem. We have just seen that it is a-finite on .moo, hence on F. If it, and all the p(w,, ) are probability measures, it is immediate that it is also. Corollary : Classical Product Measure Theorem. Let (fl,, F; , p) be a measure space for j = 1, 2, with pj a-finite on Ft,. If S2 = S2, X S22, .F _ .F, X F21 the set function given by

2.6.3

u(F) = f u2(F(w,)) du,(w,) = f u1(F(w2)) du2(w2) n,

nZ

is the unique measure on F such that p(A x B) = p,(A)p2(B) for all A E .F B e .F2. Furthermore, p is a-finite on F, and is a probability measure if it, and p2 are. The measure it is called the product of it, and p2, written

u=u, x P2 PROOF. In 2.6.2, take p(w1, ) = p2 for all co, The second formula for p(F) is obtained by interchanging it, and u2 . I

As a special case, Let 0, = f12 = R, .F, = F 2 = M(R), it, = p2 = Lebesgue measure. Then F, x F 2 = R(R2) (Problem 2), and p = p, x P2 agrees with Lebesgue measure on intervals (a, b) = (a,, b,] x (a2, b2]. By the Carathbodory extension theorem, It is Lebesgue measure on R(R2), so we have another method of constructing two-dimensional Lebesgue measure. We shall generalize to n dimensions later in the section.

2.6 PRODUCT MEASURES AND FUBINI'S THEOREM

101

The integration theory we have developed thus far includes the notion of a multiple integral on R"; this is simply an integral with respect to n-dimensional Lebesgue measure. However, in calculus, integrals of this type are evaluated by computing iterated integrals. The general theorem which justifies this process is Fubini's theorem, which is a direct consequence of the product measure theorem. 2.6.4 Fubini's Theorem. Assume the hypothesis of the product measure theorem 2.6.2. Let f: (S2, .F) -+ (R, R(R)). (a) If f is nonnegative, then Jn2 f(w,, w2)p(w1, dw2) exists and defines a Borel measurable function of wl. Also

fna (b)

dp = f ` f f(a) I, w2)µ(w1, dwi))p1(dwi). n,

n2

If So f dp exists (respectively, is finite), then in, f (w1, (02)µ(wl, d(02)

exists (respectively, is finite) for µl-almost every w1, and defines a Borel measurable function of (o, if it is taken as 0 (or as any Borel measurable function of w,) on the exceptional set. Also, fn f dp =

f,sz (L2 f(wl,

dw2))il(dwl)

[The notation In, f (w1, w2)p(w1, dw2) indicates that for a fixed w1, the function given by g(w2) = f (wl, (02) is to be integrated with respect to the measure µ(w1, PROOF. (a)

First note that:

(1) For each fixed wl we have f (w ): (02 , .F2) -- (R, E? (R)). In other words if f is jointly measurable, that is, measurable relative to the product a-field .7, x 2 , it is measurable in each variable separately. For if B e R(R), _1(B)} {02 : f (w,, w2) a B) = {w2: (W11 (02) of =f -'(B)(wl) E 2 by part (1) of the proof of 2.6.2. Thus SO, A(011 w2)p(wl, dw2) exists. Now let IF, FE .F, be an indicator. Then

412 IF(wl, W2)u(wl, d(02) = f IF((,,)((02)µ(wl, dw2) = p(wl, F(wl)),

and this is Borel measurable in w, by part (2) of the proof of 2.6.2. Also

f nIF du = p(F) =

f

n,

p(wl, F(w1))p1(dwl)

=f f J

n2

by 2.6.2

'F(w1, w2)µ(wl, dw2)µl(dwl)

2

102

FURTHER RESULTS IN MEASURE AND INTEGRATION THEORY

Now if f = Yj=, xj!FJ, the Fj disjoint sets in .f, is a nonnegative simple function, then

f f (O)

11

n2

(02)p(w1, dw2) _ Y xjp(w1, Fj(w1)), j=1

Borel measurable in w1, and

f f2f d,u

j=

xjfn IFj du = j= 1 xj f f IFf(w1, w2)p(w1, dw2)p1(dw1) f2,

n2

by what we have proved for indicators

fTl, f

(w1, w2)pwl, dw2)p1(dw1). 2

Finally, if f: (0, f) - (R, R(R)), f >_ 0, let 0
The product measure theorem and Fubini's theorem may be extended to n factors, as follows: 2.6.7

Theorem. Let .F; be a a-field of subsets of 52,, j = 1, ... , n. Let µ, be

x52; , let a a-finite measure on F1, and, for each (co, ..., w;) c-01 x µ(w,, ... , cop B), B e Fjt1i be a measure on .F;+t (j = 1, 2, .. ., n - 1). Assume the µ(w,, ... , (o;, ) to be uniformly a-finite, and assume that x .FJ) - (R, R(R)) µ(w ... , wJ, C) is measurable: (521 x x S2; , F, x

for each fixed C e .F;+,. Let 52 = )1 x

x .F..

x Stn , .F _ .F, x

(a) There is a unique measure µ on .°F such that for each measurable rectangle A, x x An e .I", x An)

µ(A 1 x

= f µt(dwl) f µ(w1, dw2) Al

...

Az

... , wn- 1, do),). A f-1 *0 ... , wn-2 , dwn- I) f µ(w1,

[Note that the last factor on the right is µ(w1, ..., co,, -,An).] The measure µ is a-finite on .F, and is a probability measure if µ1 and all the µ(w1, ..., w;, ) are probability measures. (b)

Let f: (52, .`F) -* (R, R(R)). If f > 0, then

f µ(w1, d(02) ... lfl(wt, f nf dµ = fStµl(dwl) n in, nz fn .

... , on-2 , dwn- 1)

w1, ... , wn)µ(wt, ... , wn- t, dwn),

(1)

where, after the integration with respect to µ(w1, ... , w;, ) is performed (j = n - 1, n - 2, ... , 1), the result is a Borel measurable function of

(w

,

w;).

If 1, f dµ exists (respectively, is finite), then Eq. (1) holds in the sense that for each j = n - 1, n - 2, ... , 1, the integral with respect to µ((o 1, ... , wj, ) exists (respectively, is finite) except for ((o ... , wi) in a set of 0,

where Aj is the measure determined [see (a)) by µ, and the measures u(w1, ), ..., µ(w1, ..., ('0j_ I, ). If the integral is defined on the exceptional

2.6

105

PRODUCT MEASURES AND FUBINI'S THEOREM

set as 0 [or any Borel measurable function on the space (S2, x ... x f1;,

Jr, x

x ,Fj)j, it becomes Borel measurable in (wl, ... co).

PROOF. By 2.6.2 and 2.6.4, the result holds for n = 2. Assuming that (a) and (b) hold up to n - I factors, we consider the n-dimensional case. By the

induction hypothesis, there is a unique measure An-, on F, x 12 x such that for all A, E.F,, ..., E

)..-,(A, x ... x

f A,µl(dwl) ... J

f

.X

dw2)

A2

µ(w,, ... , (On-2 , dwn- 1)

and An_, is a-finite. By the n = 2 case, there is a unique measure p on x F (which equals F, x (.F, x x x Fn; see Problem 3) such x

that for each A x

p(wl, ..., (on-1, An) dAn-1(w,,

I

..., wn-1)

A

IA(w1,

...

wn-1, An)

S2, x ... x S2 _ ,

dAn-1(w1,

... , wn-1).

(2)

x If A is a measurable rectangle A, x then IA(wl, ...) (On-1) = IA,(w,) thus (2) becomes, with the aid of the induction

hypothesis on (b),

,u(A, x ... x An) = f µl(dwl) A,

µ(w

..., wn-1+ An)p(w1, ..., wn-2, dwn-1)

which proves the existence of the desired measure p on F. To show that µ is a-finite on .F o, the field of finite disjoint unions of measurable rectangles, and consequently p is unique, let f2, = U , AJr, j = 1, ..., n, where µ(w ..., w;_1, Air) < k;r < co for all co,, ..., wj_1, j = 2, ..., n, and P,(A,r) = krr < co. Then co

fl =

U

(A,,, x A2i2 x ... X Ani),

with

p(A,i, x A2,2 x ... x This proves (a).

k,i,k2i2 ... kni < co.

2

106

FURTHER RESULTS IN MEASURE AND INTEGRATION THEORY

To prove (b), note that the measure p constructed in (a) is determined by An-1 and the measures µ(W1, ... , on_1, ). Thus by the n = 2 case.

J f dp = f

J

f (W 1, ..., Wn)µ(W1, ..., Wn-1, dWn)

W,1) where the inner integral is Borel measurable in (w1, ..., Wn-I), or becomes so after adjustment on a set of fin-I -measure 0. The desired result now follows by the induction hypothesis. Comments. (a) If we take f = IF in formula (1) of 2.6.7(b), we obtain an explicit formula for p(F), F e °F, namely,

2.6.8

.

p(F) = Jf1µ1(d01) J µ(0I, d(02) ,

J

s12

f1I

F(W 1,

i.!(01, ... , on-2, dwn- 1) ... , 0n)µ(01, ... , on - 1, don)

(b) We obtain the classical product measure and Fubini theorems by taking µ(W1, ..., cop ) = p;+1, J = 1, 2, ..., n - 1 (with pj+l a-finite). We obtain a unique measure p on .F such that on measurable rectangles, µ(A1 x ... X A") = 1z1(A1)µ2(A2) ... µ"(A")

If f: (S2, F) - (R, V(R)) and f

- 0 or In f dµ exists, then

f f dp = $d, J n

a,

dµ2 ... nz

J nnf

dun

and by symmetry, the integration may be performed in any order. The measure p is called the product of p1. ..., p", written p = p, x x p,,. In particular, if each p; is Lebesgue measure on R(R), then p, x Lebesgue measure on R(R"), just as in the discussion after 2.6.3.

x pn is

Problems

2.

Show that the collection of finite disjoint unions of measurable rectangles in S21 x x Stn forms a field. x 4(R) (n times). Show that -V(R") = R(R) x

3.

If.gf1, ... , F,, are arbitrary a-fields, show that

1.

(F I x ... X ,F"-1) x -n = F1 x'2 X ... X F'n.

2.6

PRODUCT MEASURES AND FUBINI'S THEOREM

107

4.

Let u be the product of the a-finite measures ut and 112 . If C E 'I X .F2, show that the following are equivalent: (a) u(C) = 0, (b) u2(C(w1)) = 0 for u,-almost all co, a fl,, (c) u1(C(w2)) =0 for 102-almost all w2 "12-

5.

In Problem 4, let (Q',

be the completion of (f2, .F, u), and

assume It,, P2 complete. If B E .y', show that B(w1) a .F2 for p,-almost all co, e f2, [and B(w2) e .F, for u2-almost all w2 a (12). Give an example in which B((o,) 0 F2 for some w1 E Q. 6. (a) Let f2, = f22 = the set of positive integers, .F, = .''2 = all subsets, u1 = 92 = counting measure, f (n, n) = n, f(n, n + 1) = -n, n = 1,

2, ..., f(i,j)=0 ifj#i or i+1. Show that fn, fn2f due du,=0, f n2 f fl, f du, dµ2 = oo. (Fubini's theorem fails since the integral off with respect to p, x P2 does not exist.) (b) Let f2, = 112 = R, .F, _ 2 = R(R), u1 = Lebesgue measure, 92 = counting measure. Let A = {(w1, (02): w, = (02} E X F' 2 . Show that J S1, J S12

'A due edit 1 =

f u2(A(w1)) dui(w1) = oo, 11i

but fn2 fn 'A du1 due = f u1(A(w2)) du2(w2) = 0. ,

n2

[Fubini's theorem fails since u2 is not a-finite; the product measure theorem fails also since in, u2(F(w,)) dp,(w,) and

fn2 p1(F(w2)) du2(w2) do not agree on F, x .F2.] 7.f Let n, = f22 = the first uncountable ordinal, .F, = .'2 = all subsets, f2 = f2, x'12 , .F = .;z, x F2 . Assume the continuum hypothesis, which identifies f2, and f22 with [0, 1].

If f is any function from f2, (or from a subset of f21) to [0, 1] and G = {(x, y): x c- Q1, y = f (x)} is the graph of f, show that G e .F. (b) Let C, = {(x, y) a f2: y < x}, C2 = {(x, y) E f2: y > x). If B a Cl or B - C2, show that B e .F. (The relation y < x refers to the ordering of y and x as ordinals, not as real numbers.) (c) Show that ,F consists of all subsets of 0. Show that a measurable function of one variable is jointly measurable. Specifically, if g: (Q .F,) -+ (f2', F') and we define f: 1, x 122 - f2' by f (w1, (02) = g(w,), then f is measurable relative to Pr, x .qf2 and .F', regardless of the nature of F2 . (a)

8.

t Rao, B. V., Bull. Amer. Math. Soc. 75, 614 (1969).

108 9.

2

FURTHER RESULTS IN MEASURE AND INTEGRATION THEORY

Give an example of a function f: [0. 1] x [0, 1] ->' [0, 1] such that (a) f (x, y) is Bore) measurable in y for each fixed x and Borel measurable in x for each fixed y, (b) f is not jointly measurable, that is, f is not measurable relative to the product a-field -4[0, 1] x 9[0, 1], and (c) f u (J f (x, y) dy) dx and f o (jo f (x, y) dx) dy exist but are unequal. (One example is suggested by Problem 7.)

2.7 Measures on Infinite Product Spaces

The n-dimensional product measure theorem formalizes the notion of an n-stage random experiment, where the probability of an event associated with the nth stage depends on the result of the first n - I trials. It will be convenient later to have a single probability space which is adequate to handle n-stage experiments for n arbitrarily large (not fixed in advance). Such a space can be

constructed if the product measure theorem can be extended to infinitely many dimensions. Our first task is to construct the product of infinitely many a-fields.

Definitions. For each j = 1, 2, ..., let (Q;, F) be a measurable space. Let f = 1100, 12;, the set of all sequences (w,, (02 , ...) such that w; a 0j, 2.7.1

j=1,2,.... If B"crlj_,Qj,we define B"={wED:(w,,...,cw")eB"}. The set B. is called the cylinder with base B"; the cylinder is said to be measurable if B" a fl F F. If B" = A, x x A,,, where A; c f1i for each i, B" is called a rectangle, a measurable rectangle if A i e .°F, for each i. A cylinder with an n-dimensional base may always be regarded as having a higher dimensional base. For example, if

B={wE(I:(w,,w2,w3)EB3), then

B = {w6f):(w1,w2,w3)EB3, (04Ei24) = {w E fl: (w1 . 0)2, w3 , (04) a B3 X f14).

It follows that the measurable cylinders form a field. It is also true that finite disjoint unions of measurable rectangles form a field; the argument is the same as in Problem 1 of Section 2.6. The minimal a-field over the measurable cylinders is called the product of the a-fields .F;, written flj 1 F ; rl , F; is also the minimal a-field over

2.7

109

MEASURES ON INFINITE PRODUCT SPACES

the measurable rectangles (see Problem 1). If all .F, coincide with a fixed a-field F, then H, .`, is denoted by F', and if all S2, coincide with a fixed set S, f 1 S2, is denoted by Sr'. The infinite-dimensional version of the product measure theorem will be used only for probability measures, and is therefore stated in that context. (In fact the construction to be described below runs into trouble for nonprobability measures.)

2.7.2

Theorem. Let (S2, , .F), j = 1, 2, ..., be arbitrary measurable spaces;

let i2 = f S2,' .F = fly 1 ., 1

.

Suppose that we are given an arbitrary prob-

ability measure P, on .F,, and for each j = 1, 2, ... and each (w...... co) e 92, x x S2, we are given a probability measure P(w,, ..., w, , ) on .°F,+,. Assume that P(w1, ..., w,, C) is measurable: (R, R(R)) for each fixed C e .v,+ If B" e j % F; , define P"(B") = fn PI(dwi) Jn2 P(wi, do),.) ... , J

an

Gl_1 52,,

f_1-F,)-+

N(01' ... , (0"-2 , dui"-1)

w")P(w1, ..., w"-1, dw").

Note that P,, is a probability measure on f;=1 ,F, by 2.6.7 and 2.6.8(a). There is a unique probability measure P on .F such that for all n, P agrees with P on n-dimensional cylinders, that is, P{w e S2: (w1, ..., w") e B") _

P"(B") for all n=1,2,...and all B"efj=1,F;. PROOF. Any measurable cylinder can be represented in the form B. = w,;) a B"} fot some n and some B" a f;=1.F,; define P(B") = P"(B"). We must show that P is well-defined on measurable cylinders. For sup{w c -0 :

pose that B can also be expressed as {co e Q: (w,, ..., (9 ) e C'j where Cm e j=1 .F,; we must show that P"(B") = Pm(Cm). Say m < n; then (co,, ... , co,,,) x K2,,. It follows from E C' iff (w 1, ... , co,,) e B", hence B" = C' x S2,"+, x P(Cm). (The fact that the P((O ..., w, , ) the definition of P,, that are probability measures is used here.)

Since P. is a measure on fl ., F,, it is immediate that P is finitely additive on the field .Fo of measurable cylinders. If we can show that P is continuous from above at the empty set, 1.2.8(b) implies that P is countably additive on .F0, and the Caratheodory extension theorem extends P to a

probability measure on fl. t, F,; by construction, P agrees with P,, on n-dimensional cylinders.

110

2

FURTHER RESULTS IN MEASURE AND INTEGRATION THEORY

Let (B., n = n, , n2,

... } be a sequence of measurable cylinders decreasing

to 0 (we may assume n, < n2 <

, and in fact nothing is lost if we take n; = i for all i). Assume lim"-,, P(B") > 0. Then for each n > 1, P(B") = fn 9n1)(w1)P1(dwl), ,

where

9(1)(w1) = f P(w1, dw2) ... nx

4

w"-1, dwn)

an fan

Since B"+1 = B", it follows that Brt+1 = B" x IB.,+,(w,, ..., wn+1)

hence

IB.,(w1, ..., w").

Therefore gn1 (co,) decreases as n increases (co, fixed); say g,(,1) (w1)-h,((o1). By the extended monotone convergence theorem (or the dominated convergence theorem), P(B") -+ In, h,(w,)P,(dw1). If lim"-,, P(B") > 0, then h1((O1') > 0 for some w,' a 52,. In fact co,' E B', for if not, IB"(w1', (02, ..., Co.) = 0 for all n; hence g,(,1)(w,') = 0 for all n, and h,(w1') = 0, a contradiction.

Now for each n > 2,

9n"(wl') =

J lZ

9n2)(w2)P((9 ,', dw2),

where P(w1', w2 , dw3) 9(w2) = L3 n2) J

5 'B-(C4)1" w2, ..., w")P(w,...... wn-,, dww) an

As above,

j h2(w2); hence

9(1)(w,') -' Lh2(w2)P(wt', dw2) Since gn1)(w1') -> h1(w1') > 0, we have h2(w2') > 0 for some (02' E 522, and as above we have (w1', w2') e B2.

The process may be repeated inductively to obtain points co,', w2', ... such that for each n, (w,', ..., co.') e B. But then (w1', w2', ...) a n"'=1 B. _

0, a contradiction. This proves the existence of the desired probability measure P. If Q is another such probability measure, then P = Q on measurable cylinders, hence P = Q on .F by the uniqueness part of the Caratheodory extension theorem. I

2.7

ill

MEASURES ON INFINITE PRODUCT SPACES

The classical product measure theorem extends as follows:

Corollary. For each j = 1, 2, ... , let

2.7.3

P,) be an arbitrary prob-

52; , .F _ f 1 ,F j . There is a unique probability space. Let 0 = ability measure P on F such that

P{w a 52: co, e A1, ..., co E

f j=1P;(A;)

... and all A; e F j , j = 1, 2, .... We call P the product of the Pj, and write P = f 1 P. for all n = 1, 2,

take P(cvli ..., (oj, B) = Pj+1(B), B e j+1. Then f;=1 P;(A j), and thus the probability measure P of 2.7.2 has the desired properties. If Q is another such probability measure, then P = Q on the field of finite disjoint unions of measurable rectangles; hence P = Q on W by the Caratheodory extension theorem. I PROOF.

P (A1 x

In

2.7.2,

x

Problems 1.

Show that f; ,F j is the minimal a-field over the measurable rectangles.

2.

Let ,F = R(R); show that the following sets belong to F°° :

1

(a)

{x a R°°: sup,, x

< a},

(b) {xeR°°:Yn IIxxl
(d) {x a R': lim sup,,.. , x < a}, (e) {x a R'°: Y"= 1 xk = 0 for at least one n > 0}.

3.

Let F be a c-field of subsets of a set S, and assume F is countably

generated, that is, there is a sequence of sets A1, A2, ... in F such that the smallest a-field containing the A; is 9. Show that 9' is also countably generated. In particular, .(R)°° is countably generated; take the A; as intervals with rational endpoints. 4. How many sets are there in R(R)'? 5.

Define f: R°° -+ R as follows:

Ax 1, x2 ,

the smallest positive integer n such that x1 + - +x,2: 1, ) = if such an n exists, I foralln. 0o if

Show that f: (R00, .4(R)°°) -+ (R, 9(R)).

112

2

FURTHER RESULTS IN MEASURE AND INTEGRATION THEORY

2.8

References

The presentation in Chapters 1 and 2 has been strongly influenced by several sources. The first systematic presentation of measure theory appeared in Halmos (1950). Halmos achieves slightly greater generality at the expense

of technical complications by replacing a-fields by a-rings. (A a-ring is a class of sets closed under differences and countable unions.) However, a-fields will be completely adequate for our purposes. The first account-of measure theory specifically oriented toward probability was given by Loeve (1955). Several useful refinements were made by Royden (1963), Neveu (1965), and

Rudin (1966). Neveu's book emphasizes probability while Rudin's book is particularly helpful as a preparation for work in harmonic analysis. For further properties of finitely additive set functions, and a development of integration theory for functions with values in a Banach space, see Dunford and Schwartz (1958).

3 Introduction to Functional Analysis

3.1

Introduction

An important part of analysis consists of the study of vector spaces endowed with an additional structure of some kind. In Chapter 2, for example, we studied the vector space L"(fl, .F, p). If I < p < oo, the seminorm II II, allowed us to talk about such notions as distance, convergence, and completeness.

In this chapter, we look at various structures that can be defined on vector spaces. The most general concept studied is that of a topological vector space, which is a vector space endowed with a topology compatible with the algebraic operations, that is, the topology makes vector addition and scalar multi-

plication continuous. Special cases are Banach and Hilbert spaces. In a Banach space there is a notion of length of a vector, and in a Hilbert space, length is in turn determined by a "dot product" of vectors. Hilbert spaces are a natural generalization of finite-dimensional Euclidean spaces. We now list the spaces we are going to study. The term "vector space" will always mean vector space over the complex field C; "real vector space" indicates that the scalar field is R; no other fields will be considered.

Definitions. Let L be a vector space. A senrinorin on L is a function II from L to the nonnegative reals satisfying for all aeC, xcL, Ilaxll = jai Ilxlj for all x, y c L. !x+yll-IlxlI+IYII

3.1.1 II

113

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INTRODUCTION TO FUNCTIONAL ANALYSIS

The first property is called absolute homogeneity, the second subadditivity. Note that absolute homogeneity implies that 11011 = 0. (We use the same

symbol for the zero vector and the zero scalar.) If, in addition, llxll = 0 implies that x = 0, the seminorm is called a norm on L and L is said to be a normed linear space. If II

II

is a seminorm on L, and d(x, y) = Ilx - Yll, x, y E L, d is a pseudo-

metric on L; a metric if 11 II is a norm. A Banach.space is a complete normed

linear space, that is, relative to the metric d induced by the norm, every Cauchy sequence converges. An inner product on L is a function from L x L to C, denoted by (x, y) <x, y>, satisfying = a<x, z> + b

for all

a,beC, x,y,zeL,

<x, y> _ for all x, y e L <x,x>0 for all xEL,

<x,x>=0

if and only if x=0

(the over-bar indicates complex conjugation). A vector space endowed with an inner product is called an inner product space or pre-Hilbert space. If L is an inner product space, Ilxll = (<x, x>)112 defines a norm on L; this is a consequence of the Cauchy-Schwarz inequality, to be proved in Section 3.2. If, with this norm, L is complete, L is said to be a Hilbert space. Thus a Hilbert space is a Banach space whose norm is determined by an inner product. Finally, a topological vector space is a vector space L with a topology such that addition and scalar multiplication are continuous, in other words, the mappings

(x, y)-->x+y

of L x L into L

(a, x) -. ax

of C x L into L

and

are continuous, with the product topology on L x L and C x L. In many books, the topology is required to be Hausdorfl, but we find it more convenient not to make this assumption. A Banach space is a topological vector space with the topology induced by the metric d(x, y) = 11x - yll. For if xn -+ x and y y, then

+Y )II s Ilx. - xll + IIY

if a,, - a and x,, - x, then 11a. x. - ax it < 11a. x. - a xll + Il a x - axll

= Ia.I Ilx,, - xll + Ia - aI IlxII-0.

3.1

INTRODUCTION

115

The above definitions remain unchanged if L is a real vector space, except of course that C is replaced by R. Also, we may drop the complex con-

jugate in the symmetry requirement for inner product and simply write (x, y> = (y, x> for all x, y E L. If (0, .,c, p) is a measure space and I < p:!5; oo, II v is a seminorm on the vector space L°(S2, F, µ). If we pass to equivalence 3.1.2 Examples. (a)

II

classes by identifying functions that agree a.e. [p], we obtain L°(S2, °F, P), a Banach space (see 2.4). When p = 2, the norm II IIP is determined by an inner product (.f, g> _ Jnf4 dp

Hence L2(S2, .F, p) is a Hilbert space. If .F consists of all subsets of f and p is counting measure, then f = g a.e.

[p] implies f =_ g. Thus it is not necessary to pass to equivalence classes; L°(S2, .F, p) is a Banach space, denoted for simplicity by I°(S2). By 2.4.12, if I < p < co, then P(S2) consists of all functions f =

(f (a), a E S2) from f to C such that f(a) = 0 for all but countably many a, I f(c) ° < oo. When p = 2, the norm on 12(52) is induced by and If IIP = the inner product = Y-f(a)9(«) a

When p = oo, the situation is slightly different. The space 1°°(f2) is the collection of all bounded complex-valued functions on 0, with the sup norm If II = sup{1 f(x) I: X C -Q).

Similarly, if 0 is a topological space and L is the class of all bounded continuous complex-valued functions on S2, then L is a Banach space under the sup norm, for we may verify directly that the sup norm is actually a norm, or

equally well we may use the fact that L c Thus we need only check completeness, and this follows because a uniform limit of continuous functions is continuous. (b) Let c be the set of all convergent sequences of complex numbers, and

put the sup norm on c; if f = If 11 =sup(

n >_ 1) E c, then

n = 1, 2, ...}.

Again, to show that c is a Banach space we need only establish completeness. Let be a Cauchy sequence in c; if ff = k > 1}, then limk., a,,,, exists from each n since f E c, and bk = ank exists,

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INTRODUCTION TO FUNCTIONAL ANALYSIS

uniformly in k, since I ank -a k I

Ilfn -fmll - 0 as n, nm -* oo. By the

standard double limit theorem,

lim lim ank = lim lim an,.

n-oo k-oo

k-co n-oo

In particular, limk-a, bk exists, so if f = {bk, k > 11, then f e c. But Ilfn -f II = supk I ank - by

0 as n - oo since ank --> bk uniformly in k. This proves

completeness. (c) Let L be the collection of all complex valued functions on S, where S

is an arbitrary set. Put the topology of pointwise convergence on L, so that a sequence or net { fn} of functions in L converges to the function f e L if and only if fn(x) --+f(x) for each x e S. (See the appendix on general topology,

Section Al, for properties of nets.) With this topology, L is a topological vector space. To show that addition is continuous, observe that if fn -).f and gn -+ g pointwise, then f,, + gn --),f + g pointwise. Similarly if an a C, n = 1, 2, ..., a" - a, and J,, f pointwise, then an fn of pointwise. 3.2

Basic Properties of Hilbert Spaces

Hilbert spaces are a natural generalization of finite-dimensional Euclidean

spaces in the sense that many of the familiar geometric results in R" carry over. First recall the definition of the inner product (or "dot product ")

on R": If x = (x...... x") and y = (y,, ..., yn), then <x, y> = y;=, x; y; . (This becomes Y;_, x y; in the space C" of all n-tuples of complex numbers.) The length of a vector in R" is given by IIxII = (<x, x>)1"2 = , x2)'2, and the distance between two points of R" is d(x, y) = Ilx - YII In order to show

that d is a metric, the triangle inequality must be established; this in turn <_ IIxII IIYII In fact the Cauchy-Schwarz inequality holds in any inner product space, as we now

follows from the Cauchy-Schwarz inequality I <x, y> I prove:

3.2.1 Cauchy-Schwarz Inequality. If L is an inner product space, and IIxII = (<x, x>)112, x e L, then I <x, Y> I < Ilx!I

IIYI!

for all x, y e L.

Equality holds if x and y are linearly dependent. PROOF. For any a e C,

0<<x+ay,x+ay>_<x+ay,x>+<x+ay,ay> _ <x, x> + a + i<x, y> + I a 12.

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BASIC PROPERTIES OF HILBERT SPACES

Set a = - <x, y>l (y, y> (if = 0, then y = 0 and the result is trivial). Since _ <x, y>, we have

0<<x,x>-2

I<x,y>12+ I<x,Y>I2



proving the inequality.

Since <x + ay, x + ay> = 0 if x + ay = 0, equality holds it x and y are linearly dependent. I

Corollary. If L is an inner product space and

3.2.2

x c- L, then 11

II

11.x1! = (<x,

x>)"2,

is a norm on L.

PROOF. It is immediate that Ilxli ? 0, Ilaxll = I a I Ilxll, and Ilxll = 0 iff x = 0. Now

I1x+Y112 = <x+y,X+Y> = ilx112+ IIY112+<x,y>+ = IIx112+ IIY112+2Re<x,y> <- IIxI12 + 1ly112 + 211x11 IIYII

by 3.2.1.

Therefore Ilx + y112 <- (11x11 + IIYIl)2. 1

Corollary. An inner product is (jointly) continuous in both variables,

3.2.3

that is, x,, -+ x, y y implies <x,

y>

can be a net as

well as a sequence). PROOF.

<X.,

<x, Y> I = I <x,, , Y. - Y> + <x. - x, Y> I -<

llyn - YII + iix - xli IIYII

by 3.2.1.

But by subadditivity of the norm, 111x.11 - llxii 1 < llx. - x11- 0; hence

11xii. It follows that <x,,, y.>

<x, y>. I

The computation of 3.2.2 establishes the following result, which says geometrically that the sum of the squares of the lengths of the diagonals of a parallelogram is twice the sum of the squares of the lengths of the sides:

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3.2.4 Parallelogram Law. In an inner product space, IIx + y112 + IIx - YII2 = 2(11X112 + 11y112). PROOF.

IIx + YII2 = IIXII2 + IIYII2 + 2 Re<x, y>, and

IIx-YII2=IIXI12+IIYII2-2Re<x,y>. I Now suppose that x,, ..., x are mutually perpendicular unit vectors in Rk, k > n. If x is an arbitrary vector in Rk, we try to approximate x by a linear

combination Y;=, ajxj. The reader may recall that E;_, ajxj will be closest to x in the sense of Euclidean distance when aj = <x, xj>. This result holds in an arbitrary inner product space.

Definition. Two elements x and y in an inner product space L are

3.2.5

said to be orthogonal or perpendicular if <x, y> = 0. If B c L, B is said to be orthogonal iff <x, y> = 0 for all x, y E B such that x:0 y; B is orthonormal if it is orthogonal and IIXII = 1 for all x c- B. The computation of 3.2.2 shows that if x,, x2 , ..., x,, are orthogonal, the Pythagorean relation holds: x+`12 = D°_' IIx;112. 3.2.6 Theorem. If {x ..., x,j is an orthonormal set in the inner product space L, and x c- L, n

is minimized when a j = <x, xj>, j = 1, ..., n.

x- Y-ajxj II

j=1

11

PROOF. IIx

a xjII2 = ` x - JY la. x j , x - kI' ak xk

\

n`

IIX112

k=1

n

ak<X, Xk> -

akxk ). + K j =1 ajxj, E k=1 /

aj<xj, x> j=1

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3.2 BASIC PROPERTIES OF IIILBERT SPACES

The last term on the right is >. I a, 12 since the x; are orthonormal. Furthermore, -a,<x, x,> - a,<xl , x> + I a;1 2 = _I <x, x,> 12 + I a,; - <x, x,> 12. Thus 1

0<

2

x-

a,x,

n

n

11X112- YI<x,x;>I2+ Yla;-<X,x;>12, (1) J=1

.1=1

.1=1

so that we can do no better than to take a, = <x, x,>. I The above computation establishes the following important inequality. 3.2.7 Bessel's Inequality. If B is an arbitrary orthonormal subset of the inner

product space L and x is an element of L, then 11X112>-

yeB

I<x,y>12.

In other words, <x, y> = 0 for all but countably many y e B, say y = x1,

x...... and IIX112>_E I<x,x,>12.

Equality holds if El= <x, x,>x, -> x as n -+ oo. 1

PROOF. If x1, ..., xn E B, set

<x, x,> in Eq. (1) of 3.2.6 to obtain

Il x - Yi = 1 <x, xl>xi II 2 = 11 X 112 - Li = 1 I <x, X j> 12 > 0. I

We now consider another basic geometric idea, that of projection. If M is a subspace of Rn and x is any vector in Rn, x can be resolved into a compo-

nent in M and a component perpendicular to M. In other words, x = y + z where y e M and z is orthogonal to every vector in M. Before generalizing to an arbitrary space, we indicate some terminology. 3.2.8

Definitions. A subspace or linear manifold of a vector space L is a subset M of L that is also a vector space; that is, M is closed under addition and

scalar multiplication. If L is a topological vector space, M is said to be a closed subspace of L if M is a subspace and is also a closed set in the topology of L. A subset M of the vector space L is said to be convex if for all x, y e M,

we have ax + (I - a)y e M for all real a e [0, 1]. The key fact that we need is that if M is a closed convex subset of the Hilbert space H and x is an arbitrary point of H, there is a unique point of M closest to x.

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Theorem. Let M he a nonempty closed convex subset of the Hilbert space H. If x e H, there is a unique element yo e M such that 3.2.9

IIx - yoll = inf{IIx-Y1l:yEM}. PROOF. Let d = inf{llx - Y1l : y e M}, and pick points yl, Y21 ... C_ M with is a Cauchy sequence. IIx - y.11 - d as n ce ; we show that Since Jlu + vlJ 2 + flu - x112 = 211u112 + 211vIl2 for all u, v e H by the parallel-

ogram law 3.2.4, we may set u = yR - x, v = ym - x to obtain IIYn+Ym-2x112+ IIY.-Yn112

=2IIY,,-xII2+211ym-x112

or

IIYn-Ym112=211Y,,-xII2+211Ym-xII2-4111(y,,+Ym)-xII2.

Since z(y,, + y,,) e M by convexity, 11 (y + ym) - xII2 >- d2, and it follows 0 as n, m -> oo. that IIy,, - Y,,11

By completeness of H, y,, approaches a limit yo, hence IIx fix - Yo11 . But then JJx - yo J1 = d, and yo e M since M is closed; this finishes the existence part of the proof. To prove uniqueness, let yo, zo e M, with IIx - Yoli = IIx - zoll = d. In the parallelogram law, take u = yo - x, v = zo - x, to obtain

IIYo+zo-2x112+llyo-z0112=211Yo-x112+211zo-x1f2=4d2. But IIYo + zo - 2x112 = 411R(Yo + zo) - x112 >- 4d2; hence IIYo - zoll = 0, so

Yo=zo I If M is a closed subspace of H, the element yo found in 3.2.9 is called the projection of x on M. The following result helps to justify this terminology.

Theorem. Let M be a closed subspace of the Hilbert space H, and yo an element of M. Then 3.2.10

lix -- Yo 11 = inf{ IIx - Yil : y e M}

iff

x - Ye 1 M,

that is, <x - yo, y> = 0 for all y e M. PROOF. Assume x - yo 1 M. If y e M, then I1x-Y112 = IIx - Yo - (y-Yo)112

= IIx-Yo112 + IIY-Yo112-2Re<x-Yo,Y-Yo> since y - yoeM = IIx-YoII2+ IIY-YoII2 IIx - YoII2

Therefore, JJx - Yoll = inf{Ilx - ylj : y e M}.

3.2

BASIC PROPERTIES OF HILBERT SPACES

121

Conversely, assume Ilx - Yol! = inf{Ilx - yll : y e M}. Let y e M and let c be an arbitrary complex number. Then yo + cy e M since M is a subspace, hence Ilx - Yo - cyll Z llx - Yoll. But I1x-Yo-cY112 = llx-Yo112+ IcI211Y112-2Re<x-yo,cy>; hence

IcII11Y112 -2Re<x-Yo,cy>>0.

Take c = b<x - yo, y>, b real. Then <x - yo, cy> = b I <x - yo, y> 12. Thus I <x - Yo' Y> 12[b211Y112 - 2b] > 0. But the expression in square brackets is negative if b is positive and sufficiently close to 0; hence <x - yo, y> = 0. 1

We may give still another way of characterizing the projection of x on M.

Projection Theorem. Let M be a closed subspace of the Hilbert space H. If x e H, then x has a unique representation x = y + z where 3.2.11

y e M and z 1 M. Furthermore, y is the projection of x on M.

PROOF. Let yo be the projection of x on M, and take y = yo, z = x - yo. By 3.2.10, z 1 M, proving the existence of the desired representation. To prove uniqueness, let x = y + z = y' + z' where y, y' e M, z, z' 1 M. Then

y - y' e M since M is a subspace, and y - y' 1 M since y - y' = z' - z. Thus y - y' is orthogonal to itself, hence y = y'. But then z = z', proving uniqueness. I If M is any subset of H, the set M1 = {x e H: x I M} is a closed subspace by definition of the inner product and 3.2.3. If M is a closed subspace, Ml is called the orthogonal complement of H, and the projection theorem is expressed by saying that H is the orthogonal direct sum of M and Ml, written

H=M(B M1. In R", it is possible to construct an orthonormal basis, that is, a set {x1, .... , x"} of n mutually perpendicular unit vectors. Any vector x in R"

<x, x.>x;, so that <x, x.> is the commay then be represented as x ponent of x in the direction of x; . We are now able to generalize this idea to an arbitrary Hilbert space. The following terminology will be used.

Definitions. If B is a subset of the topological vector space L, the space spanned by B, denoted by S(B), is the smallest closed subspace of L containing all elements of B. If L(B) is the linear manifold generated by B, 3.2.12

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INTRODUCTION TO FUNCTIONAL ANALYSIS

that is, L(B) consists of all elements F"., a; xi, a E C, Xt c B, i = 1, ... , n, n = 1, 2, ... , then S(B) = L(B). If B is a subset of the Hilbert space H, B is said to be an orthonormal basis for H if B is a maximal orthonormal subset of H, in other words, B is not a proper subset of any other orthonormal subset of H. An orthonormal

set B c H is maximal if S(B) = H, and there are several other conditions equivalent to this, as we now prove. Theorem. Let B = {x8, a e 1} be an orthonormal subset of the Hilbert space H. The following conditions are equivalent: 3.2.13

(a) B is an orthonormal basis. (b) B is a "complete orthonormal set," that is, the only x c- H such that

xlBisx=0.

B spans H, that is, S(B) = H. For all x e H, x = & <x, xa)xa. (Let us explain this notation. By 3.2.7, <x, xa) = 0 for all but countably many xa , say for x,, x2 , ... ; the assertion is that Y j=, <x, x;)x; -1, x, and this holds regardless of the order in which the xj are listed.) <x, xa)<xa, y). (e) For all x, y e H, <x, y) (f) For all x e H, JJxJJ 2 = 1]a J <x, xa) 1 2. (c)

(d)

Condition (f) [and sometimes (e) as well] is referred to as the Parseval relation.

If x 1 B, x :A 0, let y = x/JlxIl. Then B v {y} is an orthonormal set, contradicting the maximality of B. (b) implies (c): If x e H, write x = y + z where y e S(B) and z 1 S(B) (see 3.2.11). By (b), z = 0; hence x e S(B). (c) implies (d): Since S(B) = L(B), given x e H and e > 0 there is a finite set F c I and complex numbers a8, a e F, such that PROOF. (a) implies (b):

x - Y as x,, < e. aeF

By 3.2.6, if G is any finite subset of I such that F c G,

x - E <x, xa)xa

11

-<

I x - aY as xa

II

y as xQ l < s. - x - aeF

where

as = 0 for a 0 F

3.2

BASIC PROPERTIES OF HILBERT SPACES

123

Thus if x,, x2 , ... is any ordering of the points xQ E B for which <x, x8> # 0, lix - Ei=, <x, x;>x; < e for sufficiently large n, as desired. (d) implies (e): This is immediate from 3.2.3. (e) implies (f): Set x = y in (e). (f) implies (a): Let C be an orthonormal set with B c C, B 0 C. If x e C, x 1 B, we have IIxI12 = a I <x, x2> 2 = 0 since by orthonormality of C, x is orthogonal to everything in B. This is a contradiction because llxll = 1 for all

3.2.14 Corollary. Let B = {x,,, a e I} be an orthonormal subset of H, not necessarily a basis. (a) B is an orthonormal basis for S(B). [Note that S(B) is a closed subspace of H, hence is itself a Hilbert space with the same inner product.] (b) If x e H and y is the projection of x on S(B), then

y=

<x, x.>xa

[see 3.2.13(d) for the interpretation of the series]. PROOF. (a)

Let x e S(B), x 1 B; then x I L(B). If y e S(B), let y,, y2, ...

e L(B) with y,, -+ y. Since <x,

0 for all n, we have <x, y> = 0 by 3.2.3.

Thus x 1 S(B), so that (x, x> = 0, hence x = 0. The result follows from 3.2.13(b). (b) By part (a) and 3.2.13(d), y = Y,,(y, x >x8 .

But x - y 1 S(B) by

3.2.11, hence <x, x8> = for all a. I

A standard application of Zorn's lemma shows that every Hilbert space

has an orthonormal basis; an additional argument shows that any two orthonormal bases have the same cardinality (see Problem 5). This fact may be used to classify all possible Hilbert spaces, as follows: Theorem. Let S be an arbitrary set, and let H be a Hilbert space with an orthonormal basis B having the same cardinality as S. Then there is an isometric isomorphism (a one-to-one-onto, linear, norm-preserving map) 3.2.15

between H and 12(S).

PROOF. We may write B = {x., a e S}. If x e H, 3.2.13(d) then gives x = 1. <x, xz>xa, where Y I <x, xa> 12 = 11x112 < oo by 3.2.13(f). The map x - (<x, xe>, a E S) of H into 12(S) is therefore norm-preserving; since it is also linear, it must be one-to-one. To show that the map is onto, consider any

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INTRODUCTION TO FUNCTIONAL ANALYSIS

collection of complex numbers as , a e S, with I as 12 < 00. Say as = 0 except for a = a,, a, ..., and let x = a a,,,xa,. [The series converges to an element of H because of the following fact, which occurs often enough to be stated separately: If {y,, Y2, ...} is an orthonormal subset of H, the series >; c, y, converges to some element of H if y; I c1I 2 < oo. To see this, observe that 1117" c; ZIIY;I12 = ;_" c3 2; thus the partial sums form a Cauchy sequence iff yj I c; 12 < co.] Since the x,, are orthonormal, it follows that <x, xa> = aQ for all a, so that

xmaps onto (a,,,aeS).I We may also characterize Hilbert spaces that are separable, that is, have a countable dense set. Theorem. A Hilbert space H is separable if it has a countable orthonormal basis. If the orthonormal basis has n elements, H is isometrically isomorphic to C"; if the orthonormal basis is infinite, H is isometrically iso3.2.16

morphic to 12, that is, 12(S) with S = {1, 2, ...}. PROOF. Let B be an orthonormal basis for H. Now IIx - Yll2 = 11x112 + Ily112 = 2 for all x, y e B, x 5e y, hence the balls A,, = {y: Ily - x11 < j;}, x e B, are

disjoint. If D is dense in H, D must contain a point in each A,, so that if B is uncountable, D must be also, and therefore H cannot be separable. Now assume B is a countable set {x,, X2, ...). If U is a nonempty open subset of H (= S(B) = L(B)], U contains an element of the form Y_;_, a; x; with the a; a C; in fact the a; may be assumed to be rational, in other words, to have rational numbers as real and imaginary parts. Thus D= (jY,,-1 a; x; : n = 1, 2, ... ,

the a3 rational}

is a countable dense set, so that H is separable. The remaining statements of the theorem follow from 3.2.15. 1

A linear norm-preserving map from one Hilbert space to another automatically preserves inner products; this is a consequence of the following proposition: 3.2.17 Polarization Identity. In any inner product space,

4<x, y> = IIx + YII2 - IIx - YII2 + illx + iYI12 - illx - iy112.

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BASIC PROPERTIES OF HILBERT SPACES

PROOF.

IIx+y112=I1Xll2+11y112+2Re<x,y> IIx-y112=IIXII2+I1Y112-2Re<x,y> Ilx + iy112 = IIxii2 + IIYI12 + 2 Re<x, iy> iyl[2

= IIXI12 + 1ly112 - 2 Re<x, iy> IIx But Re<x, iy> = Re[-i<x, y>] = Im<x, y>, and the result follows.

Problems 1.

In the Hilbert space 12(S), show that the elements e,,, a e S, form an orthonormal basis, where s 96 a,

e2(s) = r0,

s=a. If A is an arbitrary subset of the Hilbert space H, show that l1,

2.

(a)

A1' = S(A). (b)

If M is a linear manifold of H, show that M is dense in H if Ml = {0}.

3.

Let x,, ... , x" be elements of a Hilbert space. Show that the xi are

linearly dependent if the Gramian (the determinant of the inner products <x;, x;>, i, j = 1, ..., n) is 0. 4. (Grain-Schmidt process) Let B = {x,, x2 , ...} be a countable linearly independent subset of the Hilbert space H. Define e, = x,/IIx, II ; having chosen orthonormal elements e,, ..., en, let yn+, be the projection of xn+, on the space spanned by e,, ..., en: n

Y.+, _ Y <xn+,, ei>e, i=, Define en+, =

X'1+1 iIxn+,

(a)

_Y..,,,* Yn+,

Show that L{e,, ..., en} = L{x ..., xn} for all n, hence xn+, #

Yn+, and the process is well defined. (b) Show that the e" form an orthonormal basis for S(B). Comments.

Consider the space H = L2(-1, 1); if we take xn(t) = t",

n = 0, 1, ..., the Gram-Schmidt process yields the Legendre polynomials en(t) = an d"[(t2 - 1)"]/dt", where a" is chosen so that Ilenll = 1. Similarly, t"e-121'2, if in L2(- oo, oo) we take xn(t) = n = 0, 1, ..., we obtain the Hermite polynomials en(t) = an(-1)"e`2 d"(e-`2)/dt".

126 5. 6.

3

INTRODUCTION TO FUNCTIONAL ANALYSIS

Show that every Hilbert space has an orthonormal basis. Show that any two orthonormal bases have the same cardinality. Let U be an open subset of the complex plane, and let H(U) be the collection of all functions f analytic on U such that

(a) (b)

11f1I2= f f If(x+iy)I2dxdy
If we define

= f f f(x +.iy)g(x + iy) dx dy,

f, g e H(U),

U

H(U) becomes an inner product space. (a) If K is a compact subset of U and f e H(U), show that sup{I f(z) I : z e K} < 11f 111,[n_ (lo

where do is the Euclidean distance from K to the complement of U. Therefore convergence in H(U) implies uniform convergence on

compact subsets of U. (If z e K, the Cauchy integral formula yields

f(z) = (2n)-`

f

zn

f(z + re") d8,

r < do.

0

Integrate this equation with respect to r, 0 < r <_ d < do. Note also that if U is the entire plane, we may take do = oo, and it follows that H(C) = {0}.) (b) Show that H(U) is complete, and hence is a Hilbert space. 7.

(a)

If f is analytic on the unit disk D = {z: I zI < 1) with Taylor o a" z", show that

expansion J(z)

-

fz"If(reie)I2dO= sup os.
00

"o

It follows that if H2 is the collection of all functions f analytic on D such that I

za

os.<12n fo

N2(f) = sup

(b)

If(reie)I2 dO < oo,

then H2, with norm N, is a pre-Hilbert space. If f e H2, show that

f f I f(x+iy)I2dxdy
3.3

127

LINEAR OPERATORS ON NORMED LINEAR SPACES

hence H2 c H(D) and convergence in HZ implies convergence in H(D). (c)

If f"(z) = z", n = 0, 1, ..., show that f" - 0 in H(D) but not in

H2. (d) Show that H2 is complete, and hence is a Hilbert space. [By (a),

H2 is isometrically isomorphic to a subspace of 12.] (e)

If en(z) = z", n = 0, 1, ..., show that the e" form an orthonormal

basis for H2 If en(z) = [(2n + 2)/27r]'/2z", n = 0, 1, ..., show that the e" form an orthonormal basis for H(D). Let M be a closed convex subset of the Hilbert space H, and yo an element of M. If x e H, show that (f)

8.

IIx - YoII = inf{jjx - YII : y e M} if

Re<x - yo, y - yo> < 0 9.

(a)

for all y c- M.

If g is a continuous complex-valued function on [0, 2n1 with g(0) = g(2n), use the Stone-Weierstrass theorem to show that g can be uniformly approximated by trigonometric polynomials Yk= _n ck e`kt. Conclude that the trigonometric polynomials are

dense in L2[0, 2n]. (b) if f e L2[0, 21r], show that the Fourier series Z'

an e'"t, an =

(1 /2n) Jo" f (t)e-'"t dt, converges to f in L2, that is, f2" 0

2

J(t)- I akeik' dt-s0 k= -n

as n ->oo.

(c)

Show that {e`"t/J2n, n = 0, ± 1, ±2, ...} yields an orthonormal

(a)

Give an example to show that if M is a nonempty, closed, but not convex subset of a Hilbert space H, there need not be an element of minimum norm in M. Thus the convexity hypothesis cannot be

basis for L2[0, 2n]. 10.

dropped from Theorem 3.2.9, even if we restrict ourselves to existence and forget about uniqueness. (b) Show that convexity is not necessary in the existence part of 3.2.9 if H is finite-dimensional. 3.3 Linear Operators on Normed Linear Spaces

The idea of a linear transformation from one Euclidean space to another is familiar. If A is a linear map from R" to R'", then A is completely specified

by giving its values on a basis e1, ..., en :

A(Y-°=1 cr e;) _

°=1 c; A(e;);

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INTRODUCTION TO FUNCTIONAL ANALYSIS

furthermore, A is always continuous. If elements of R" and R' are represented by column vectors, A is represented by an m x n matrix. If n = m, so that A

is a linear transformation on R", A is one-to-one if it is onto, and if A-1 exists, it is always continuous (as well as linear). Linear transformations on infinite-dimensional spaces have many features not found on the finite-dimensional case, as we shall see.

In this section, we study mappings A from one normed linear space L to another.such space M. The mapping A will be a linear operator, that is, A(ax + by) = aA(x) + bA(y) for all x, y e L, a, b e C. We use the symbol 11 for the norm on both spaces; no confusion should result. Linear operators can of course be defined on arbitrary vector spaces, but in this section, it is always understood that the domain and range are normed. Linearity does not imply continuity; to study this idea, we introduce a 11

new concept.

Definitions and Comments. If A is a linear operator, the norm of A is defined by: 3.3.1

(a) 11A 11 = sup{IlAxll : x e L, Ilxll 5 1}. We may express IIAII in two other ways. (b) IIAII = sup{IlAxll : x e L, Ilxll = 1}. (c) IIAII = sup(IlAxll/llxll : x e L, x # 0).

To see this, note that (b) <_ (a) is clear; if x 96 0, then IlAxll/Ilxll = IIA(x/Ilxll)ll, and xlllxll has norm 1; hence (c):5 (b). Finally if Ilxll <_ 1, x:96 0, then IlAxll
The linear operator A is said to be bounded if 11 All < oo. Boundedness is often easy to check, a very fortunate circumstance because we can show that boundedness is equivalent to continuity. 3.3.2

Theorem. A linear operator A is continuous if it is bounded.

PROOF. If A is bounded and {x"} is a sequence in L converging to 0, then 0. (We use here the fact that the mapping x--* Nxll of L IIAx"II S IIAII llx"II into the nonnegative reals is continuous; this follows because I llxll - llYll 15 11x

- YII.) Thus A is continuous at 0, and therefore, by linearity, is con-

tinuous everywhere. On the other hand, if A is unbounded, we can find elements x" e L with Ilx"II < 1 and IIAx,,ll - oo. Let y,, = x"/IIAx"II ; then y" 0, but II Ay" II = 1 for all n, hence Ay" does not converge to 0. Consequently, A

is discontinuous. I

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LINEAR OPERATORS ON NORMED LINEAR SPACES

We are going to show that the set of all bounded linear operators from L to M is itself a normed linear space, but first we consider some examples: Examples. (a) Let L = C[a, b], the set of all continuous complexvalued functions on the closed bounded interval [a, b] of reals. Put the sup norm on L: 3.3.3

IIx11 = sup{ I x(t) I

: a < t < b},

x e C[a, b].

With this norm, L is a Banach space [see 3.1.2(a)]. Let K = K(s, t) be a continuous complex-valued function on [a, b] x [a, b], and define a linear operator on L by (Ax)(s) =

Ja

K(s, t)x(t) dt,

a < s < b.

(By the dominated convergence theorem, Ax actually belongs to L.) We show that A is bounded: II A x 11 = sup

as
f K(s, t)x(t) dt b

l

a

< sup lx(t) I.:5' sup f b I K(s, t) I d t a
:5b

a

Max

=

IIxll

f IK(s, b t) I dt a5s5b a

(note that fa I K(s, t) I dt is a continuous function of s, by the dominated convergence theorem). Thus IIAII <_ max fbIK(s, t) I dt < oo. a<s5b a

In fact IIAII = max. 5.,5b fa I K(s, t) I dt (see Problem 3). (b) Let L = 1°(S), where S is the set of all integers and I 5 p < oo.

Define a linear operator T on L by (Tf)a =f., 1, n e S; T is called the sided shift or the bilateral shift. It follows from the definition that T is one-to-

one onto, and (T -'f )a = fa _,, n e S. Also, 11 Tf 11 = 11.f11 for all f e L; hence 11 T -.111 = II f 11 for all f e L, so that T is an isometric isomorphism of L with itself. ([n particular, 11 T11 = 11 T-'11 = 1.) If we replace S by the positive integers and define T as above, the resulting operator is called the one-sided or unilateral shift. The one-sided shift is onto

with norm 1, but is not one-to-one; T(f1, f2, ...)= (f2, f3 .

)

The shift operators we have defined are shifts to the left. We may also define shifts to the right; in the two-sided case we take (Af)a = n e S (so n >_ 2; (Af), =0; that A = T-'). In the one-sided case, we set (Af)

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INTRODUCTION TO FUNCTIONAL ANALYSIS

3

thus A(f1,f2 , ...) = (0,f1,f2 , ...).The operator A is one-to-one but not onto; A(L) is the closed subspace of L consisting of those sequences whose first coordinate is 0. (c) Assume that the Banach space L is the direct sum of the two closed subspaces M and N, in other words each x e L can be represented in a unique way as y + z for some y e M, z e N. (We have already encountered this situation with L a Hilbert space, M a closed subspace, N = M'.) We define a linear operator P on L by

Px=y. P is called a projection; specifically, P is the projection of L on M. P has the following properties:

(1) P is idempotent; that is, P2 = P, where P2 is the composition of P with itself.

(2) P is continuous. Property (1) follows from the definition of P; property (2) will be proved later, as a consequence of the closed graph theorem (see after 3.4.16).

Conversely, let P be a continuous idempotent linear operator on L. Define

M={xeL:Px=x},

N={xEL:Px=0}.

Then we can prove that M and N are closed subspaces, L is the direct sum of M and N, and P is the projection of L on M. By continuity of P, M and N are closed subspaces. If x e L, then x = Px

+ (I - P)x = y + z where y e M, z e N. Since M n N = {0}, L is the direct

sum of M and N. Furthermore, Px = y by definition of y, so that P is the projection of L on M. If f is a linear operator from a vector space L to the scalar field, f is called

a linear functional. (The norm of a scalar b is taken as I b 1.) Considerable insight is gained about normed linear spaces by studying ways of representing continuous linear functionals on such spaces. We give some examples:

Let f be a continuous linear functional on the Hilbert space H. We show that there is a unique element y e H such that 3.3.4 Representations of Continuous Linear Functionals. (a)

f(x)=<x,y)

for all

xeH.

This is one of several results called the Riesz representation theorem.

If the desired y exists, it must be unique, for if <x, y) = <x, z> for all x, then y - z is orthogonal to everything in H (including itself), so y = z.

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LINEAR OPERATORS ON NORMED LINEAR SPACES

To prove existence, let N be the null space of f, that is, N = {x e H: f(x) = 0}. If N' = {0}, then N = H by the projection theorem; hence f = 0, and we may take y = 0. Thus assume we have an element u e N' with u 0. Then u 0 N, and if we define z = u/f (u), we have z e Nl and f (z) = 1. If x e H and f (x) = a, then with

x = (x - az) + az,

x - az E N,

az 1 N.

If y = z/Ilzll2, then <x, y> = <x - az, y> + a

since y I N

= a(z, y> = a =.f (x) as desired.

The above argument shows that if f is not identically 0, then N' is one-

dimensional. For if x e N' and f(x) = a, then x - az e N n N', hence x = az. Therefore N' = {az: a e Q. Notice also that if 11fII is the norm off, considered as a linear operator, then 111'11 = IIyII

For I f(x) I = I<x; y>I <- Ilxll Ilyll for all x, hence 11f 11 < IIYII; but I f(y)I = I = Ilyll 11Y11, so 11f 11 t 11Y11.

Now consider the space H* of all continuous linear functionals on H;

H* is a vector space under the usual operations of addition and scalar multiplication. If to each f E H* we associate the element of H given by the Riesz representation theorem, we obtain a map qi: H* -,. H that is one-to-one onto, norm-preserving, and conjugate linear; that is,

f,gEH*, a,bEC.

0(af+bg) =do(f)+b0(g),

[Note that if f(x) = <x, y> for all x, then af(x) = <x, ay>.j Such a map is called a conjugate isometry. (b) Let f be a continuous linear functional on l (= l°(S), where S is the set of positive integers), I < p < oo. We show that if q is defined by (1/p) + (1/q) = 1, there is a unique element y = (yl, y2, ...) a 14 such that AX) = Y_ xk yk

for all x e !°.

k=1

Furthermore, ao

111

l1=IIY11=(E lYklq) k= 1

1/q

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INTRODUCTION TO FUNCTIONAL ANALYSIS

3

To prove this, let en be the sequence in 1P defined by en(j) = 0, j 96 n; If x e 1P, then llx - Yk=, xkekliP = Ek n+, IxkIP->0; hence x = J:k 1 xk ek where the series converges in 1P. By continuity off, en(n) = 1.

where yk =f(ek)

f(x) = k xkyk, 00

k=1

(1)

Now write yk in polar form, that is, yk = rk eiek, rk > 0. Let

by (1), n

n

Al.) _ k=1 rk-1e-£Bkrkei0"= k=1 Y_ IA

(2)

But If(Zn)I C 11f11 I1Zn11 n

= IIJ

1/P

II k=1 Y_ IYkI1q-11P)

llfll( k-1 IYkI9)1/P By Eq. (2), n

1/4

Ek1IYk19)

G Ilfll;

hence y e 19 and IIYII <_ If II. To prove that If Il < Ilyll, observe that Eq. (1) is of the form f (x) = J .Q xy du where Q is the positive integers and µ is counting measure. By Holder's inequality, If(x)I <- Ilxll IIYII, so that 11f 11 <_ IIYII

Finally, we prove uniqueness. If y, z e 14 and f(x) = Y-k

1 xkYk =

for all x e 1P, then g(x) = Yk 1 xk(yk - Zk) = 0 for all x e P. The above argument with f replaced by g shows that Ilgll = Ily - zll; hence liY - zll = 0, and thus y = z. (c) Let f be a continuous linear functional on I. We show that there is a unique element y e F such that k

1 xkZk

AX) = E xk yk

for all x e 11.

k=1

The argument of (b) may be repeated up to Eq. (1). In this case, however,

if yk=rke'Bk,k= 1, 2, ..., we take zn = (0, ..., 0, e - ie°, 0, ...),

with

a-'e in position n.

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LINEAR OPERATORS ON NORMED LINEAR SPACES

Thus by (1),

f(zn) = e `

e'e

=IY

But If(zn)I s Ilfil Ilz.ll = Of 11.

Therefore I y I <_ llf II for all n, so that y e F and IIYII <_ IIf II. But by (1),

I f(x)I < (suPlYki) k

k=l

Ixkl = 11A Ilxii;

hence ilfll _< IIYII Uniqueness is proved as in (b). In 3.3.4(b) and (c), the map f -, y of (1")* to 14 [q = c in 3.3.4(c)] is linear,

and is one-to-one by uniqueness of y. To show that it is onto, observe that any linear functional of the form f(x) = YA , xk yk , x e 1", y e 1", satisfies IIYII by the analysis of 3.3.4(b) and (c). Therefore f is continuous, so IIIII that every y e 19 is the image of some f e (1")*. Since If II = Ilyll, we have an isometric isomorphism of (I")* and 14.

If we replace yk by h, we obtain the result that there is a unique y e 1q such that f (x) = Yk 1 xk yk for all x e 1". This makes the map f -+ y a conjugate isometry rather than an isometric isomorphism. If p = 2, then q = 2 also, and thus we have another proof of the Riesz representation theorem for separable Hilbert spaces (see 3.2.16). In fact, essentially the same argument may be used in an arbitrary Hilbert space if the e are replaced by an arbitrary orthonormal basis. (Other examples of representation of continuous linear functionals are given in Problems 10 and 11.) We now show that the set of bounded linear operators from one normed linear space to another can be made into a normed linear space.

Theorem. Let L and M be normed linear spaces, and let [L, M] be the collection of all bounded linear operators from L to M. The operator 3.3.5

norm defined by 3.3.1 is a norm on [L, MI, and if M is complete, then [L, M] is complete. In particular, the set L* of all continuous linear functionals on L is a Banach space (whether or not L is complete).

PROOF. It follows from 3.3.1 that IIA ll > 0 and ilaA II = Pal IlA ll for all A e [L, M] and a e C. Also by 3.3.1, if 11 All = 0, then Ax = 0 for all x e L, hence A = 0. If A, B e [L, M], then again by 3.3.1, IIA + BII = sup{II(A + B)xll : x e L,

Since II(A + B)xll :5 IlAxll + IlBxll for all x,

iiA + BII s IIAil + IIBll

Ilxll < 1}.

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INTRODUCTION TO FUNCTIONAL ANALYSIS

and it follows that [L, M] is a vector space and the operator norm is in fact a norm on [L, M]. Now let A1, A2 , ... be a Cauchy sequence in [L, M]. Then

II(A"-Am)xli S IIA"-A,"II Ilxll-r0

as

n,m->oo.

(1)

Therefore {A" x} is a Cauchy sequence in M for each x e L, hence A. converges

pointwise on L to an operator A. Since the A. are linear, so is A (observe that Ajax + by) = aA" x + bA" y, and let n -+ oo). Now given e > 0, choose N such that 11 A,, - Amll < s for n, m >- N. Fix n >- N and let m - co in Eq. (1) to conclude that II (A" - A)xll <- e IlxiI for n >- N; therefore IIA" - A II - 0 as n -a oo. Since 11A 11 IIA - A"II + IIA"II, we have A e [L, M] and A. -> A in

the operator norm.

In the above proof we have talked about two different types of convergence of sequences of operators. 3.3.6

Definitions and Comments. Let A, A1i A2, ... e [L, M]. We say that

A" converges uniformly to A if IIA" - A II - 0 (notation: A"-" . A). Since II(A" - A)xll < IIA" - All Ilxll, uniform operator convergence means that A"x-. Ax, uniformly for llxll <_ 1 (or equally well for llxll <- k, k any positive real number). We say that A. converges strongly to A (notation: A"-4 A) if A" x -+ Ax for each x e L. Thus strong operator convergence is pointwise convergence on all of L.

Uniform convergence implies strong convergence, but not conversely. For example, let {e1, e2 , ...) be an orthonormal set in a Hilbert space, and let A. x = <x, e">, n = 1, 2, .... Then A"+ 0 by Bessel's inequality, but A. does not converge uniformly to 0. In fact 11 A. e"II = 1 for all n, hence IIA"ll = 1.

There is an important property of finite-dimensional spaces that we are now in a position to discuss. In the previous section, we regarded C" as a Hilbert space, so that if x e C", the norm of x was taken as the Euclidean norm Ilxll = (Ei=1 f x, z)l"z. The metric associated with this norm yields the standard topology on C". However, we may put various other norms on C", for example, the LP norm Ilxll, = (E,=1 I xi I P)1"P, 1 < p < oo, or the sup norm llxll = max(I xl 1, ..., I x" 1). Since the space is finite-dimensional, there are no convergence difficulties and all elements have finite norm. Fortunately, the proliferation of norms causes no confusion because all norms on a given finite-dimensional space induce the same topology. The proof of this result is outlined in Problems 6 and 7.

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LINEAR OPERATORS ON NORMED LINEAR SPACES

Problems 1.

Let f be a linear functional on the normed linear space L. If f is not identically 0, show that the following are equivalent:

(a) f is continuous. (b) The null space N =f -'{0} is closed. N is not dense in L. (d) f is bounded on some neighborhood of 0. [To prove that (c) implies (d), show that if B(x, e) r N = 0, and f is unbounded on B(0, s), then f(B(O, e)) = C. In particular, there is a (c)

point z e B(0, r) such that f(z) = -f(x), and this leads to a contradiction.] 2.

Show that any infinite-dimensional normed linear space had a discontinuous linear functional. (Let e1, e2, ... be an infinite sequence of linearly independent elements such that 1 for all n. Define jr

appropriately on the e and extend f to the whole space using linearity.) In Example 3.3.3(a), show that IIAII = maxa5ssb fb I K(s, t)1 dt. [If J ,b I K(s, t) J dt assumes a maximum at the point u, and K(u, t) = r(t)e'etr0, Let x1, x2, ... be continuous functions such that let z(t) = lx,,(t) 1 < 1 for all n and t, and f o I xa(t) - z(t) I dt - 0 as n - co. Since K(s, t)z(t) di as n - oo, it follows that II Ax,, II < II A II and (Axa)(s) fQ K(u,')I dt:5 IIAII.] The same argument, with integrals replaced by sums, shows that if A is a matrix operator on Ca, with sup norm, IIAII = max1 s ts Y;=' I a;; I . 4. Let M be a linear manifold in the normed linear space L. Denote by [x] the coset of x modulo M, that is, {x + y: y e M}. Define

3.

r(t)e-t90).

II [x] II = inf{lLvll : y e [x]);

note that II [x] II = inf{ Ilx - z11 : z e M} = dist(x, M). Show that the 5. 6.

above formula defines a seminorm on the quotient space L/M, a norm if M is closed. If L is a Banach space and M is a closed subspace, show that L/M is a Banach space. (a) Let A be a linear operator from L to M. Show that A is one-to-one with A-' continuous on its domain A(L), if there is a finite number m > 0 such that IlAxll ? m Ilxll for all x e L. (b) Let 111), and 11 JJ2 be norms on the linear space L. Show that the

norms induce the same topology if there are finite numbers m, M > 0 such that mllxll1:5 11x112 <Ml141

for all

xeL.

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INTRODUCTION TO FUNCTIONAL ANALYSIS

(This may be done using part (a), or it may be shown directly that, for

7.

example, if 11x111 <_ (l/m)IIx112 for all x, then the topology induced by 1111 , is weaker than (that is, included in) the topology induced by 11112.) Let L be a finite-dimensional normed linear space, with basis e1, ... , e". Let 11 111 be any norm on L, and define n

11x112 = ( lxil2}

1/2

n

,

=1

(a)

where

x = Y_ xiei. i=1

Show that for some positive real number k, 11x111 <_ k 11x112 for all

xeL. (b)

Show that for some positive real number in,

11x111 ? in on

{x: hIxll2 = 1}. [By (a), the map x -+ 11x!11 is continuous in the topology induced by II 112 ] (c) Show that 11x111 ? in lIx211 for all x e L, and conclude that all norms on a finite-dimensional space induce the same topology. (d) Let M be an arbitrary normed linear space, and let L be a finite8.

dimensional subspace of M. Show that L is closed in M. (Riesz lemma) Let M he a closed proper subspace of the normed linear space L. Show that if 0 < S < 1, there is an element x6 e L such that 11x611 = 1 and Ilx - x611 ? S for all x e M. [Choose x1 0 M, and

let d = dist(x1, M) > 0 since M is closed. Choose x0 e M such that 9.

IIx1 - x011 <- d/S, and set x6 = (x1 - xo)/11x1 Let L be a normed linear space. Show that the following are equivalent: (a) L is finite-dimensional.

(b) L is topologically isomorphic to a Euclidean space C" (or R" if L is a real vector space), that is, there is a one-to-one, onto, linear, bicontinuous map between the two spaces. (c) L is locally compact (every point of L has a neighborhood whose closure is compact). (d) Every closed bounded subset of L is compact. The set {x e L: llxll = 1} is compact. The set {x e L: IIx11 = 1} is totally bounded, that is, can be covered by a finite number of open balls of any preassigned radius. [Problem 7 shows that (a) implies (b); (b) implies (c), (d) implies (e),

(e)

(f)

and (e) implies (f) are obvious, and (c) implies (d) is easy. To prove 10.

that (f) implies (a), use Problem 8.] Let c be the space of convergent sequences of complex numbers [see 3.1.2(b)]. If f is a continuous linear functional on c, show that there is a unique element y = (yo , Y1, . . .) a l1 such that for all x e c,

AX) = lI m

)(0

k1 xk A

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LINEAR OPERATORS ON NORMED LINEAR SPACES

Furthermore, 11111 = I YO - Y_k = I Yk I +Y-k'=1

I Yk I

If co is the closed

subspace of c consisting of those sequences converging to 0, the repre-

sentation of a continuous linear functional on co is simpler: f(x) = j Yk . Thus co* is isometrically isomorphic to 1'. Let (f', .F,, p) be a measure space.

Y-k I xk Yk 1 11111 = Y_k I 11.

(a)

Assume it finite, I

AX) = 5 xy tip n

Furthermore, IIfII = IIYIIq

for all

x e B'.

If y1 is another such function, show , and apply the

that y = YI a.e. [p]. [Define ).(A) = f (IA), A e , Radon-Nikodym theorem.]

(b) Drop the finiteness assumption on It. If A e .F and p(A) < ce,

part (a) applied to (A, ,F n A, p) provides an essentially unique YA e L4 such that y,, = 0 on A` and f (xIA) =

Ja

xyA dp

for all

x e L°;

also, IIYAIIq is the norm of the restriction off to L°(A), so IIYAIIq < Ilfll (i)

If p(A) < oo and p(B) < oo, show that YA = yB a.e. [p] on A n B. Thus yA B may be obtained by piecing together YA and yB.

(ii)

k = sup{IIYAIIq Let n = 1, 2, ... , be sets with A e .F, p(A) < oo} < II! 11. Show that YA converges in Lq to

a limit function y, and y is essentially independent of the (iii)

particular sequence Furthermore, IIYIIq 5 II! I1 Show that f(x) = fn xy dp for all x e L°. Since IIYIIq 5 I!.f 11

by (ii) and IIf1 S IIYIIq by Holder's inequality, we have (c) (d)

If 11 = IIYIIq Thus the result of (a) holds for arbitrary it. Prove (a) with it finite, p = 1, q = oo. Prove (a) with p o-finite, p = 1, q = co. [For an extension of this result, see Kelley and Namioka (1963, Problem 14M).1

It follows that there is an isometric isomorphism of (LP)* and Lq if I

also forp=1,q=co.

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Basic Theorems of Functional Analysis

Almost every area of functional analysis leans heavily on at least one of the three basic results of this section : the Hahn-Banach theorem, the uniform

boundedness principle, and the open mapping theorem. We are going to establish these results and discuss applications. We first consider an extension problem. If f is a linear functional defined on a subspace M of a vector space L, there is no difficulty in extending f to a linear functional on all of L; simply extend a Hamel basis of M to a Hamel

basis for L, define f arbitrarily on the basis vectors not belonging to M, and extend by linearity. However, if L is normed and we require that the extension of f have the same norm as the original functional, the problem becomes more difficult. We first prove a preliminary result.

Lemma. Let L be a real vector space, not necessarily normed, and let p be a map from L to R satisfying 3.4.1

p(x+y)_
for all for all

x,yeL x e L and all a > 0.

[The first property is called subadditivity, the second positive-homogeneity. Note that positive-homogeneity implies that p(O) = 0 [set x = 0 to obtain

p(O) = ap(0) for all a > 0]. A subadditive, positive-homogeneous map is sometimes called a sublinear functional.] Let M be a subspace of L and g a linear functional defined on M such that g(x) < p(x) for all x e M. Let x0 be a fixed element of L. For any real number c, the following are equivalent: (1)

g(x) + Ac < p(x + Axo) for all x e M and all .1 a R.

(2)

-p(-x - xo) - g(x) < c< p(x + xo) - g(x) for all x e M.

Furthermore, there is a real number c satisfying (2), and hence (1).

PROOF. To prove that (1) implies (2), first set ) = 1, and then set A = -1 and replace x by -x [note g(-x) = -g(x) by linearity]. Conversely, if (2) holds and 2 > 0, replace x by x/.l in the right-hand inequality of (2); if A < 0, replace x by x/i in the left-hand inequality. In either case, the positive homogeneity of p yields (1). If ). = 0, (1) is true by hypothesis. To produce the desired c, let x and y be arbitrary elements of M. Then

g(x) - g(y) = g(x - y) <_ p(x - y)

< p(x + x0) + p(-y - x0)

by subadditivity.

3.4 BASIC THEOREMS OF FUNCTIONAL ANALYSIS

139

It follows that

sup[ -P(-y - x0) - g(y)] < inf [P(x + x0) - g(x)].

yEM

xEM

Any c between the sup and the inf will work. I We may now prove the main extension theorem. 3.4.2

Hahn-Banach Theorem. Let p be a subadditive, positive-homogeneous

functional on the real linear space L, and g a linear functional on the subspace M, with g < p on M. There is a linear functional f on L such that f = g on M and f < p on all of L. M, consider the subspace M1 = L(M u {xo}), consisting of all elements x + Axo, x E M, A e R. We may extend g to a linear functional on M1 by defining gl(x + Axo) = g(x) + Ac, where c is any real number. If PROOF. If x0

we choose c to satisfy (1) of 3.4.1, then gl :!5-: p on M1. Now let ' be the collection of all pairs (h, H) where h is an extension of

g to the subspace H z) M, and h

There is a version of the Hahn-Banach theorem for complex spaces. First, we observe that if L is a vector space over C, L is automatically a vector space over R, since we may restrict scalar multiplication to real scalars. For example, C" is an n-dimensional space over C, with basis vectors (1, 0,..., 0),

..., (0, ... , 0, 1). If C" is regarded as a vector space over R, it becomes 2n-dimensional, with basis vectors

(1,0,...,0),...,(0,...0, 1),

(i,0,...,0),...,(0,...,0,i).

Now iff is a linear functional on L, with ft = Ref, f2 = Imf, then ft and f2 are linear functionals on L', where L' is L regarded as a vector space over R. Also, for all x e L,

f(ix) =fi(ix) + if2(ix). But

f(ix) = if(x) _ -f2(x) + ifl(x)

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Thus f1(ix) _ -f2(x), JZ(ix) =fl(x); consequently f(X) =f1(X) - 1f1(IX) =f2(IX) + i 2(X)

Therefore f is determined by f1 (or by f2). Conversely, let f1 be a linear functional on L'. Then f1 is a map from L to R such that f1(ax + by) = af1(x) + bf1(y) for all x, y e L and all a, b e R. Define f(x) = f1(x) - if1(ix), x e L. It follows that f is a linear functional on L (and f1 = Ref). For f1 is additive, hence so is f, and if a, b e R, we have f ((a + ib)x) = f1 (ax + ibx) - if, (- bx + iax) = af1(x) + bf1(ix) + ibf, (x) - iaf1(ix) = (a -i ib)[f1(x) - if1(ix)] = (a + ib) f (x).

Note that homogeneity of f1 does not immediately imply homogeneity of f. For f1(..x) = Af1(x) for real .1 but not in general for complex A. [For example, let L = C, f (x) = x, f1(x) = Re x.] We now prove the complex version of the Hahn-Banach theorem. 3.4.3 Theorem. Let L be a vector space over C, and p a seminorm on L.

If g is a linear functional on the subspace M, and IgI < p on M, there is a linear functional f on L such that f = g on M and If I < p on L. PROOF. Since p is a seminorm, it is subadditive and absolutely homogeneous,

and hence positive-homogeneous. If g1 = Reg, then g1 < I g I
functional on L and f = g on M. Fix x e L, and let f(x) = re", r > 0. Then

If(x)j = r =f(e-'Bx) = f1(e-iex) < p(e-i°x)

since

r is real

since

f1 < p on L

= p(x)

by absolute homogeneity.

3.4.4 Corollary. Let g be a continuous linear functional on the subspace M of the normed linear space L. There is an extension of g to a continuous linear

functional f on L such that Ilfll = Ilgll. PROOF. Let p(x) = Ilgll Ilxhl; then p is a seminorm on L and IgI

by definition of Ilgll. The result follows from 3.4.3.

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A direct application of the Hahn-Banach theorem is the result that in a normed linear space, there are enough continuous linear functionals to distinguish points; in other words, if x y, there is a continuous linear functional f such that f(x) 96f(y). We now prove this, along with other related results. Theorem. Let M be a subspace of the normed linear space L, and let L* be the collection of all continuous linear functionals on L. 3.4.5

(a)

If x0 0 M, there is an f e L* such that f = 0 on M, f (xo) = 1, and 11111

= 1/d, where d is the distance from x0 to M. (b) xo e M iff every f e L* that vanishes on M also vanishes at xo . (c) If x0 0, there is an f e L* such that Of 11 = I and f(xo) = Ilxoll ; thus the maximum value of I f (x) I 111x11, x 0, is achieved at x0. In particular, if x 0 y, there is an f e L* such that f (x) sE fly). PROOF. (a) First note that L(M u {xo}) is the set of all elements y = x + axo, x e M, a e C, and since xo 0 M, a is uniquely determined by y. Define f on N = L(M V {x0}) by f(x + axo) = a; f is linear, and furthermore, 11111 = 1/d, as we now prove. By 3.3.1 we have

11111

=sup{I1(Y)l .YEN, Y*O) IIYII

sup{II

=sup(

111

lal

x+axo lal

IIx + axoll

ll.xEM, aeC, x&0 or

a

0

xeM, ac-C, a960} 111

since f(y) = 0 when a = 0. Now Ial

Ilx + axoll

1

xIl

[lx' + -u

1

Ilxo - zll

for some

z e M;

hence 11111 = (inf{Ilxo - zll : z e M})-' = 1/d < oo. The result now follows

from 3.4.4. (b) This is immediate from (a). (c) Apply (a) with M = {0}, to obtain g e L* with g(xo) = I and 11911 = 1/Ilxoll; set f = Ilxollg 1

The Hahn-Banach theorem is basic in the study of the concept of reflexivity, which we now discuss. Let L be a normed linear space, and L* the

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set of continuous linear functionals on L; L* is sometimes called the conjugate space of L. By 3.3.5, L* is a Banach space, so that we may talk about L**, the conjugate space of L*, or the second conjugate space of L. We may identify L with a subspace of L** as follows: If x e L, we define x** e L** by

x**(f) =f(x),

f e L*.

If 11f, - f II --+ 0, then f .(x) -+ f (x); hence x** is in fact a continuous linear functional on L*. Let us examine the map x - x** of L into L**. Theorem. Define h: L -+ L** by h(x) = x**. Then h is an isometric isomorphism of L and the subspace h(L) of L**; therefore, if x e L, we have, by 3.4.6

3.3.1, IIxII = Ilx**II = sup{If(x)I : fe L*, 11f 11 s 1}.

PROOF: To show that h is linear, we write

(h(ax + by)](f) =f(ax + by) = af(x) + bf(y)

[ah(x)](f) + [bh(y)](f).

We now prove that h is norm-preserving (Ilh(x)II = Ilxll for all x e L); conse-

quently, h is one-to-one. If x e L, I [h(x)](f) I = I f(x) I < JJxJJ

11f 11, and

hence Jlh(x)JJ < JJxJJ. On the other hand, by 3.4.5(c), there is an f e L* such that II f ll =1 and I f(x) I = Ilxil. Thus sup{I [h(x)](f) I : f e L*,

11f 11 = I} z IIxII

so that Jlh(x)IJ >- IIxII, and consequently 11h(x)II = IIxII. 1

If h(L) = L**, L is said to be reflexive. Note that L** is complete by 3.3.5 so by 3.4.6, a reflexive normed linear space is necessarily complete. We shall now consider some examples. 3.4.7 Examples. (a) Every Hilbert space is reflexive. For if >G is the conjugate isometry of 3.3.4(a), H* becomes a Hilbert space if we take =

_ = f (x), where x = di(g). Therefore q = h(x). (b) If 1 < p < oo, 1° is reflexive. For by 3.3.4(b), (IP)* is isometrically isomorphic to 19, where (1/p) + (1/q) = 1. Thus if t e (1P)** we have t(y) _ _k , yk Zk , y e 14, where z is an element of (14)* =1p. But then t = h(z). Essentially the same argument, with the aid of Problem 11 of Section 3.3, shows that if (0, F, p) is an arbitrary measure space L°(i2, .F, p) is reflexive

fort
The space 11 is not reflexive. This depends on the following result:

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BASIC THEOREMS OF FUNCTIONAL ANALYSIS

143

Theorem. If L is a normed linear space and L* is separable, so is L. Thus if L is reflexive and separable (so that L** is separable), then so is L*. 3.4.8

PROOF. Let fl, f2, ... form a countable dense subset of (f e L*: 11f II = 1 } (note that any subset of a separable metric space is separable). Since Ilf"II = 1, we can find points x" e L with IIx"II = land I f"(x")I >_ I for all n. Let M be

the space spanned by the x" ; we claim that M = L. If not, 3.4.5(a) yields an f e L* with f = 0 on M and 11f ll = 1. But then I < f"(x") I = I f"(xn) .f(x") I :5 Il.f" -.f II for all n, contradicting the assumption that { f1, fZ , ...} is

dense. I To return to 3.4.7(c), we note that l' is separable since {x ell : xk = 0 for all but finitely many k} is dense. [If x e !1 and P) _ (x1, ... , x", 0, 0, ...) then x(") -+ x in I1.] But (11)* = I°° by 3.3.4(c), and this space is not separable. For if S = {x e I': xk = 0 or I for all k}, then Sis uncountable and lix - yll = I

for all x, y e S, x y. Thus the sets Bx = {y c- 1': Ily - xII < i}, x e S, form an uncountable family of disjoint open sets. If there were a countable dense set D, there would be at least one point of D in each Bx, a contradiction. Thus 11 is not reflexive.

We now consider the second basic result of this section. Suppose that the A,, i belonging to the arbitrary index set I, are bounded linear operators from L to M, where L and M are normed linear spaces. The uniform boundedness principle asserts that if L is complete and the A; are pointwise bounded, that is, sup{IIA, x11 :

i e I} < oo for each x e L, then the Ai are uniformly bounded,

that is, sup{IIA,II : I e I} < oo. Completeness of L is essential; to see this, let

L be the set of all sequences x = (x1, x2 , ...) of complex numbers such that xk = 0 for all but finitely many k, with the IP norm, I

PROOF. Let C" _ {x e L: sups IIA,xII <- n}, n = 1, 2..... Since the A, are pointwise bounded, UM 1 C" = L, and since the A, are continuous, each C. is

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closed. By the Baire category theorem, for some n there is a closed ball B = {x E L: IIx - x011 <_ r} c C.. Now if Ilyll <_ 1 and i E I, we have IIA1yll = 1 IIA1zli

r

where

I

z = ry

1

- IIA1(x0 + z)II + - IIA1x0Il

r

2n

r

Thus II A, 11

r

since

x0 + z and

x0

belong to

<_ 2n/r. I

The uniform boundedness principle is used in an important way in the study of weak convergence, a concept that we now describe. 3.4.10

Definitions and Comments. The sequence (or net) in the normed linear space L is said to converge weakly to x e L iff f (x.) --).f(x) for every f c- L* (notation: x). Convergence in the metric of L (IIx - xll -> 0) will be called strong convergence and will be written simply as x,, -+ x. [In this terminology, strong convergence of the sequence of linear operators A,, to the linear operator A means strong convergence of A,,x to Ax for each x (see 3.3.6).]

For each x0 E L, consider the collection of sets U(xo) = {x e L : I f,(x) - f,(xo) I < e1,

n},

where

We form a topology by taking the U(xo) as a base for the neighborhood system at x0. (There is no difficulty checking that the required axioms for a neighborhood system are satisfied.) The resulting (Hausdorff) topology is called the weak topology on L; it follows from the definition that convergence relative to the weak topology coincides with weak convergence. (The topology induced by the norm of L will be called the strong topology.) It follows from the definitions that strong convergence implies weak convergence, hence every set that is open in the weak topology is open in the strong topology (and similarly for closed sets). To see that the converse does not hold, let {e1, e2 , ...} be an infinite orthonormal sequence in a Hilbert x> -+ 0 as n -> oo by Bessel's inequality, hence space H. If x e H, then But 11e,,11 = 1, so e,, does not converge strongly to 0.

In a Euclidean space, strong and weak convergence coincide. For if x _ + akfl ek converges weakly to x = ate, + + ak ek (where the e1 a,,,e, +

3.4

145

BASIC THEOREMS OF FUNCTIONAL ANALYSIS

are basis vectors for Ck), let f be a continuous linear functional that is 1 at that e;and Oat

Thus x - x strongly. The weak topology is an important example of a product topology (see the appendix on general topology, Section A3). For L may be identified (see 3.4.6) with a subset h(L) of the set C` of all complex-valued functions on L*, and the weak topology is the product topology, also called the "topology of pointwise convergence," on CL*, relativized to h(L). We give a few properties of weak convergence. is bounded, that is, Theorem. (a) A weakly convergent sequence sup. Ilxnll < oo. In fact if x."'+ x0, then Ilxoll <_ lim infs. IIx.N (b) If A is a bounded linear operator from L to M and the net {x,) converges weakly to x0 in L, then Ax, converges weakly to Axo in M. (c) If the net {x,} converges weakly to x0, then x0 belongs to the subspace spanned by the x; . (d) If M is a linear manifold of L, the closures of M relative to the weak

3.4.11

and strong topologies coincide. Consequently if M is closed in the strong topology, it is closed in the weak topology. PROOF. (a)

The xn may be regarded as continuous linear functionals on L*

with x (f) =f(x) (see 3.4.6). The x are pointwise bounded on L* since f(xn) -f(xo); hence by the uniform boundedness principle, sup. Ilxnll < oo. Also, if f s L*, If(xo)I = lim l n- 00

lim inf I f(xn)I <_ 11f 11 lim inf IIx,,II n_ OD

n-oo

Since 1x011 = sup{lf(xo)l :.1E L*, II! II < l}, we may conclude that 11x011 -<

lim inf.

, Ilxnll

If g e M*, define f= g o A; since A is continuous, we have f e L*, so that f (x,) -+ f(xo), that is, g(Ax,) -+ g(Axo). But g is arbitrary ; hence (b)

Ax, "'+ Axo. (c) If this is false, 3.4.5(a) yields an f e L* with f (xo) = I and f (x,) = 0, contradicting x,-!4 x0. (d) The strong closure is always a subset of the weak closure, so assume

x belongs to the weak closure of M. Let {x,) be a net in M with x; "'-* x. By (c),

x is a strong limit of a net of finite linear combinations of the x,. But these finite linear combinations belong to M since M is a subspace; hence x belongs

to the strong closure. I In order to characterize weak convergence in specific spaces, the following result is useful.

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3.4.12

Theorem. Let E be a subset of L* such that S(E) = L*, and let be a sequence in L. If x0aL,then x0iffsup, lIxnII

for allfe E. PROOF. The "only if" part follows from 3.4.11(a) and the definition of weak convergence. For the "if" part, let f e L*, and choose elements fk a L(E) with

If -fk1I - 0. Then If(xn) -f(x0)I

I f(xn) -fk(xn)I + I fk(xn) -fk(xo)I + I fk(x0) -f(x0)I -< Ilf-All IIX-II + Ifk(xn)-fk(X0)I + Ilfk-.fII 11X011-

Since the x are bounded in norm, given s > 0, we may choose k such that th right-hand side is at most I fk(XO) I + s; but fk(xo) as n -. o0 since fk e L(E), and since a is arbitrary, the result follows. I We now describe weak convergence in IP and L'. 3.4.13 (a)

Theorem. Assume I

x II < oo and Xnk -24 Zk as n -+ oo for each k.

(b)

Let x,, e LP(fl, Pte, p), n = 1, 2, ..., where µ is assumed finite. If

z e LP(11, F, Ic), then x,, -- z if sup,,

oo and JA x du -* JA z dp for each

A e F. (It will often be convenient to blur the distinction between L' and L', and treat the elements of L' as functions rather than equivalence classes.) Define fk e (lP)* by fk(x) = xk ; then fk corresponds to the sequence in l4 with a one in position k and zeros elsewhere [see 3.3.4(b)]. Take E = {fl, f2, ...} and apply 3.4.12. (b) For each A e F, definef4 e (LP)* by f,(x) = f A x dp; fA corresponds to the indicator function 1A a L9 (see Problem 11, Section 3.3). Take E as the set of all fA, A e F; E spans (LP)* because the simple functions are dense in L", and the result follows from 3.4.12. 1 PROOF. (a)

We now consider the third basic result, the open mapping theorem. This will allow us to conclude that under certain conditions, the inverse of a one-toone continuous linear operator is continuous. We cannot make this assertion in general, as the following example shows: Let L be the set of all continuous complex-valued functions x on [0, 1] such that x(0) = 0, M = {x e L: x has a continuous derivative on [0, fl); put the sup norm on L and M. If A is defined

by (Ax)(t) = Jo x(s) ds, 0 < t < 1, then A is a one-to-one, bounded, linear

3.4 BASIC THEOREMS OF FUNCTIONAL ANALYSIS

147

operator from L onto M. But A-' is discontinuous; for example, if x(t) = sin nt, then (1 - cos nt)/n, so that y -+ 0 in M, but x has no limit in L. A hypothesis under which continuity of the inverse holds is the

completeness of both spaces L and M. In the above example, M is not complete; for example, a sequence of polynomials may converge uniformly to a continuous function without a continuous derivative (in fact to a continuous nowhere differentiable function). We now state the third basic theorem. 3.4.14 Open Mapping Theorem. Let A be a bounded linear operator from the Banach space L onto the Banach space M. Then A is an open map, that is, if D is an open subset of L, then A(D) is an open subset of M. Consequently if A is also one-to-one, then A-' is bounded. PROOF. We defer this until the next section (see 3.5.11) since it will not be much

more difficult to prove a more general statement about mappings of topological vector spaces.

The open mapping theorem allows us to prove the closed graph theorem, which is often useful in proving that a particular linear operator is bounded.

First we observe that if L and M are normed linear spaces, we may define a norm on the product space L x M by II(x, y)II = (11x11' + IIYIIP)"P, x e= L, y e M, where p is any fixed real number in [1, oo). For if x, x' e L, Y, Y e M, (Ilx + x'I1P + IIY +Y'NP)!'P < [(Ilxll + Ilx'll)P + (IIYII + IIY'll)P]11P <- (11x11' + IIYIIP)1'P + 0141P + IIY'11P)1 '

by Minkowski's inequality applied to R2.

Therefore the triangle inequality is satisfied and we have defined a norm on

L x M (the other requirements for a norm are immediate). Furthermore, (x., (x, y) if x,, -+ x and y -+ y; thus regardless of the value of p, the norm on L x M induces the product topology; also, if L and M are complete, so is L x M. [The same result is obtained using the analog of the L°° norm, that is, 11(x, y)Ii = max(IIxII,

3.4.15

Definition. Let A be a linear operator from L to M, where L and M

are normed linear spaces. We say that A is closed if the graph G(A) = {(x, Ax): x e L} is a closed subset of L x M. Equivalently, A is closed if the following condition holds:

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If x e L, x -+ x, and Ax,, -+ y, then (x, y) c G(A); in other words, y = Ax. This formulation shows that every bounded linear operator is closed. The converse holds if L and M are Banach spaces.

Closed Graph Theorem. If A is a closed linear operator from the Banach space L to the Banach space M, then A is bounded.

3.4.16

PROOF. Since G(A) is a closed subspace of L x M, it is a Banach space. Define P: G(A) -+ L by P(x, Ax) = x. Then P is linear and maps onto L, and IIP(x, Ax) II = II x II

tively, if (x,,

11(x, Ax) II

; hence Il Pll <_ 1 so that P is bounded. [Alterna-

(x, y), then x -+ x, proving continuity of P.] Similarly,

the linear operator Q : G(A) -+ M given by Q(x, Ax) = Axis bounded. If P(x, Ax)

= 0, then x = Ax = 0, so P is one-to-one. By 3.4.14, P-1 is bounded, and since A = Q o P-1, A is bounded. As an application of the closed graph theorem, we show that if P is the projection of a Banach space L on a closed subspace M, then P is continuous [see 3.3.3(c)]. Let {x} be a sequence of points in L with x -+ x, and assume Px,, converges to the element y e M. Recall that in defining a projection operator it is assumed that L is the direct sum of closed subspaces M and N; thus

x,,

x y P is continuous.

y y

z x that y = Px, proving P closed. By 3.4.16,

Problems 1.

Show that a subadditive, absolutely homogeneous functional on a

vector space must be nonnegative, and hence a seminorm. Give an example of a subadditive, positive-homogeneous functional that fails to be nonnegative. 2. Let (S), F, µ) be a measure space, and assume .F is countably generated, that is, there is a countable set W (-- F such that aff) = F. (Note that the minimal field Fo over f is also countable; see Problem 9, Section 1.2.) If y is a-finite on Pro and 1 < p < co, show that L°(S2, .F, p) is separable. If in addition there is an infinite collection of disjoint sets A e F with µ(A) > 0, show that L'(0, .9, p) is not reflexive.

3.4

BASIC THEOREMS OF FUNCTIONAL ANALYSIS

149

If L and M are normed linear spaces and [L, M] is complete, show that M must be complete. 4. Let A e [L, M], where L and M are normed linear spaces. The adjoint of A is an operator A* : M* L*, defined as follows: If f e M* we take (A*f)(x) = f(Ax), x e L. Establish the following results: 3.

IIA*II = IIAII (b) (aA + bB)* = aA* + bB* for all a, b e C, A, B e [L, M]. (a)

(c)

If A e [L, M], B E [M, N], then (BA)* = A*B*, where BA is the composition of A and B.

(d) If A e [L, M], A maps onto M, and A-' exists and belongs to 5.

[M, L], then (A-')* = (A*)-'. Define the annihilator of the subset K of the normed linear space L as Kl = f f e L*: f (x) = 0 for all x E K}. Similarly, if J c L*, define J' = {x e L: f (x) = 0 for all f e J}. If A e [L, M], we denote by N(A) the null space {x e L: Ax = 0}, and by R(A) the closure of the range of A, that is, the closure of {Ax: x e L}. Establish the following: (a) (b) (c) (d) (e) (f) (g)

For any K c L, K'1 = S(K), the space spanned by K. R(A)' = N(A*) and R(A) = N(A*)l. R(A) = M if A* is one-to-one. R(A*)1 = N(A).

For any J c L*, S(J) e J- - ; S(J) = J11 if L is reflexive. R(A*) c N(A)'; R(A*) = N(A)' if L is reflexive. If R(A*) = L*, then A is one-to-one; the converse holds if L is reflexive.

6.

7.

8.

Consider the Hahn-Banach theorem 3.4.2, with the additional assumption that L is a normed linear space (or more generally, a topological vector space) and p is continuous at 0; hence continuous on all of L since I p(x) - p(y) I < p(x - y). Show that if L is separable, the theorem may be proved without Zorn's lemma. It follows that 3.4.3 and 3.4.4 do not require Zorn's lemma under the above hypothesis. If po is a finitely additive, nonnegative real-valued set function on a field .moo of subsets of a set Q, use the Hahn-Banach theorem to show that po has an extension to a finitely additive, nonnegative real-valued set function on the class of all subsets of fl. Thus in one respect, at least, finite additivity is superior to countable additivity. If the sequence (or net) of bounded linear operators A,, on a Banach space converges strongly to the (necessarily linear) operator A, show that A is bounded; in fact IIAII <_ lim inf IIAnll < sup IIAII < oo

n- rn

150

9.

10.

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INTRODUCTION TO FUNCTIONAL ANALYSIS

Let {Ai, i e I} be a family of continuous linear operators from the Banach space L to the normed linear space M. Assume the A, are weakly bounded, that is, supi I f (Ai x) I < co for all x e L and all f e M*. Show that the Ai are uniformly bounded, that is, sup; IIAiII < co. (a) I f the elements x, , i e 1, belong to the normed linear space L, and sup, Jf (x;) I < oo for each f e L*, show that sup, Ilxill < oo. (b) If A is a linear operator from the normed linear space L to the

normed linear space M, and foA is continuous for each fe M*, 11.

show that A is continuous. Let L, M, and N be normed linear spaces, with L or M complete, and let B: L x M -+ N be a bilinear form, that is, B(x, y) is linear in x for each fixed y, and linear in y for each fixed x. If for each f e N*, f (B(x, y)) is continuous in x for each fixed y, and continuous in y for each fixed x,

show that B is bounded, that is, sup{ B(x, y) I : Ilxll,

IIYII < 1} < oo.

Equivalently, for some positive constant k we have I B(x, y) < klixll IIYII for all x, y. 12. 13.

Give an example of a closed unbounded operator from one normed linear space to another. Let A be a bounded linear operator from the Banach space L onto the Banach space M. Stow that there is a positive number k such that for each y e M there is an x e L with y = Ax and Ilxll <_ kllyll This result is sometimes called the solvability theorem.

3.5 Some Properties of Topological Vector Spaces

Recall that a topological vector space is a vector space L with a topology that makes addition and scalar multiplication continuous; that is, the maps (x, y) -+ x + y of L x L -> L and (a, x) ax of C x L -> L are continuous, with the product topology on L x L and C x L. In this section we look at some of the important properties of such spaces. One cannot put an arbitrary topology on a vector space L and obtain a topological vector space. For example, consider the discrete topology (all subsets of L are open); with this topology, L is not a topological vector space (exclude the trivial case L = {0}). To see this, pick a nonzero x e L, and let a = 1/n, n = 1, 2, .... If scalar multiplication were continuous, we would have a x -> 0 x = 0; since {0} is open, this implies that a x = 0 for sufficiently large n, a contradiction.

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151

The following result is helpful in checking whether a given topology makes L into a topological vector space. {Throughout this section, we use the following notation: If A and B are subsets of L,

A+B={x+y:xEA, yEB},

A-B={x-y:xeA, yEB). IfxeL, x+B={x +y:yeB}

x-B={x-y:yeB}.] Theorem. Let L be a topological vector space, and let all be a base of overneighborhoods at 0 (the sets in * are overneighborhoods of 0, and for each neighborhood V of 0 there is a set U e all with U e V). Then 3.5.1

(a) if U, V e all, there is a set W e W such that W c U n V; (b) if U e* then for each x e L there is a S > 0 such that ax c- U for

all complex numbers a with I al < S; (c) for each U e all, there is a set V e ?l with V + V e U.

Conversely, if all is a collection of subsets of L satisfying (a)-(c), and in addition, (d) each U e all is circled, in other words, if x c- U, then ax r= U whenever

aI < 1, there is a unique topology that makes L a topological vector space with all a base of overneighborhoods at 0.

This follows since the intersection of two neighborhoods of 0 is a neighborhood of 0. (b) By continuity of scalar multiplication, ax - 0 as a -> 0, with x fixed; hence ax e U for sufficiently small I al. (c) By continuity of the map (x, y) - x + y at x = y = 0, there are sets V V2 e all such that if x e V, and y e V2, then x + y e U; in other words, V, + V2 e U. By (a), there is a set V E all with V c VI n V2 ; then V + V c PROOF. (a)

V1 + V2 C-_ U.

Conversely, let all satisfy (a)-(d). Take all as an overneighborhood base at 0, and, for each x e L, x + all = {x + U: U e all} as an overneighborhhood

base at x. Thus

'(x) = { V c L: V c x + U for some U e all} is the pro-

posed class of overneighborhoods of x. Now: (i)

If V e 1'(x), then x e V. This follows because 0 belongs to each U e all, by (b).

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(ii)

If V1, V2 E *(x), then V1 n V2 e *^(x). For if x + U. e Vi, i = 1, 2, by (a) there is a set W e all with W c U1 n U2. Then x + W c (x + U,) n (x + U2) c V1 n V2.

(iii)

If V E *(x) and V c W, then W e 'V(.x). This is immediate from the definition of V(x).

(iv)

If V e 'fi(x), there is a set W e *'(x) such that W c V and V e 'V(y) for each y e W. For if x + U c V, by (c) there is a set

Teall with T+TcU. If W=x+T and t +Tcx+T+Tcx+ Uc V, hence Ve'r(y).

y

Thus the axioms of a system of overneighborhoods are satisfied, so there is a topology on L with all an overneighborhood base at 0. By definition of

the *(x), we have x,, - x if x - x -* 0; hence if the addition operation (x, y) -* x + y is continuous at (0, 0), it is continuous everywhere. But continuity of addition at (0, 0) is immediate from (c). Finally, we look at continuity of the multiplication operation (a, x) ---I' ax.

By (b), the map is continuous at a = 0 for each fixed x, hence, as above, continuous at any a for each fixed x. If U e all and a e C, choose a positive integer n Z I a I and apply (c) inductively to obtain V e au such that V + .. + V (n times) is a subset of U. (Note, for example, that V + V + V c V + V+ V+ V since 0 e V.) If n V = {ny: y e V}, then n V e V+ + V c U; hence by (d), aV c U, so that multiplication is continuous at x = 0 for each fixed a, and thus continuous at any x for each fixed a. Now if U e all, then by (d), if a 1 < 1 and x e U, then ax a U, so that multiplication is jointly continuous at (0, 0). Thus if x --* x and a -* a, then (a - a)(x - x) 0, and by the above

analysis, a x -* ax, ax,, - ax. It follows that a x - ax. Finally, to prove uniqueness of the topology, observe that if a& is a base of overneighborhoods at 0, continuity of addition and subtraction makes the map y -+ x + y, y e L, a homeomorphism, and therefore *'(x) is a system of overneighborhoods of x.

We have seen that a seminorm on a vector space L induces a topology under which L becomes a topological vector space. We now show that a similar result holds if instead of a single seminorm we use an entire family.

3.5.2

Theorem. Let {pi, i e I} be a family of seminorms on the vector space

L. Let all be the class of all finite intersections of sets {x e L: pi(x) < bi} where i e 1, bi > 0. Then all is a neighborhood base for a topology that makes L a topological vector space. This topology is the weakest, making all in L, x -+ x if pi(x - x) - 0 for each the p, continuous, and for a net

ieI.

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153

PROOF. We show that the conditions of 3.5.1 are satisfied. Condition (a) follows because an intersection of two finite intersections of sets {x: pi(x) < S,} is a finite intersection of such sets. If x e L, then p,(ax) = I a I pi(x), which is less than S, if I a I is sufficiently small; this proves (b). To prove (c), suppose

that U = n,= I{x: pi(x) < S;}, and let V = ni= I{x: pi(x) < S} where 0 < S < mini S; . If y, z e V, then p,(y + z) 5 pi(y) + p,(z) < S,, hence y + z e U. Finally, each U e all is circled since pi(ax) = la Ip,(x), proving (d). If x e U = n,.,{z: p,(z) < S,} and y e V =n" I{z: pi(z) < minj[S; - pi(x)ii, then p,(x + y) <- pi(x) + pi(y) < p,(x) + S, - pi(x) = S; ; hence x + V c U. Thus U is an overneighborhood of each of its points, so the sets in a& are open and all is a neighborhood base, not merely an overneighborhood base.

Now continuity of p, at 0 is equivalent to continuity everywhere [ lp,(x) - pi(y) 1 5 pi(x - y)] and the sets U e* are open in any topology for which the pi are continuous at 0. This proves that the given topology is the

weakest, making the pi continuous. Finally, x - x if x - x - 0 (see the proof of 3.5.1), and x - x -+ 0 if for each i, p,(x - x) 0, by definition of all.

The topology induced by a family of seminorms is locally convex; that is, for each x e L, the family of convex overneighborhoods forms a base of overneighborhoods at x. This follows because the sets {x: pi(x) < S,} are convex. We shall prove later (see 3.5.7) that every locally convex topology is generated by a family of seminorms.

If for each x 96 0 there is an i e I such that pi(x) # 0, the topology is Hausdorff. For if x # y and p,(x - y) = S > 0, then {z: p,(z - x) < 6/2) and {z: p,(z - y) < 6/2) are disjoint neighborhoods of x and y. Examples. (a) Let L be a vector space of complex-valued functions on a topological space Q. For each compact subset K of f), define PK(x) = sup{ I x(t) I t e K). In the (Hausdorff) topology induced by the seminorms pK, convergence means uniform convergence on all compact subsets of .0. If we restrict the K to finite subsets of f), we obtain the topology of pointwise convergence. In general, if the K are restricted to a class W of subsets of fl, we obtain the topology of uniform convergence on sets in W. (b) Let L = C°° [a, b], the collection of all infinitely differentiable complex-valued functions on the closed bounded interval [a, b] a R. For each n, 3.5.3

:

let p (x) = sup{ I x("'(t) I : a < t < b} where P) is the nth derivative of x. In

the topology induced by the p,,, convergence means uniform convergence of all derivatives. We now examine convex sets in more detail.

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Definitions. Let K be a subset of the vector space L. Then K is said to

be radial at x if K contains a line segment through x in each direction, in other words, if y e L, there is a b > 0 such that x + Ay e K for 0:5 A < 6. (If K is radial at x, x is sometimes called an internal point of K.) If K is convex and radial at 0, the Minkowski functional of K is defined as p(x) = inf (r > O: x e K). 111

Intuitively, p(x) is the factor by which x has to be shrunk in order to reach the boundary of K. The Minkowski functional has the following properties.

3.5.5 Lemma. Let K be a convex subset of L, radial at 0, and let p be the Minkowski functional of K. (a) The functional p is sublinear, that is, subadditive and positivehomogeneous. (b) {x e L: p(x) < l} is the radial kernel of K, defined by rad ker K = {x e K: K is radial at x}; also, K c {x e L: p(x) < 1}. (c) If K is circled, then p is a seminorm. (d) If L is a topological vector space and 0 belongs to the interior K° of K, then p is continuous, K = {x e L: p(x) < I}, and K° = {x e L: p(x) < 1); hence {x e L: p(x) = 1) is the boundary of K. PROOF. (a)

If x/r e K and y/s e K, then

x+ y __ r

r+s

x+

y

r+sr r+ss e K S

by convexity.

Thus p(x + y) < r + s; take the inf over r, then over s, to obtain p(x + y) 1; hence x e K by convexity. [Write x = (1/s)(sx) + [1 - (1/s)]0.] Now if p(x) < I and y e L, then p(x + Ay) < p(x) + )p(y) < I for A sufficiently small and nonnegative; hence K is radial at x. Conversely, if K is radial at x, then x + Ax e K for some A > 0; hence p(x + Ax) < 1 by definition of p. Thus p(x) < (I + A)-1 < 1. The last statement follows from the definition of p. (c) if x/r e K and a e C, a 96 0, then ax/ I a I r e K since K is circled ; consequently p(ax) < I a I r. Take the inf over r to obtain p(ax) < I a Ip(x). Replace x by x/a to obtainp(x) < j a f p(x/a) or, with b = 1 la, p(bx) f b I p(x). Now p(0) = 0 since 0 e K, and the result follows. (d) Since 0 e K° there is a neighborhood U of 0 such that U e K. If

A > 0 and y e AU, that is, y = Ax for some x e U, then p(y) = Ap(x) < A.

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155

[Note x e K implies p(x) < 1, by (b).] Thus p is continuous at the origin, and therefore continuous everywhere by subadditivity. Since p is continuous, {x: p(x) < 1} is closed, so by (b), K c {x: &) :5 1}. But if 0 < A < 1 and p(x) < 1, then p(Ax) < 1; hence AX E K. If A -* 1, then x; therefore p(x) < 1 implies x e K. Ax Again, by continuity of p, {x: p(x) < 1} is open, and hence is a subset of K°. But if p(x) >_ 1, then by considering {x with A > I we see that x is a limit of a sequence of points not in K, hence x 0 K°. I

Before characterizing locally convex spaces, we need the following result.

3.5.6 Lemma. If U is a neighborhood of 0 in the topological vector space L, there is a circled neighborhood V of 0 with V c U, and a closed circled overneighborhood W of 0 with W c U. If L is locally convex, V and W can be taken as convex. PROOF. Choose T e 'W and S > 0 such that aT (-- U for I a l < 6, and take V = U{aT: I a 1< b}. Now if A c L, we claim that

A= n{A+N: Ne.'},

(1)

where .K is the family of all neighborhoods of 0. [This may be written as

A = n{A - N: N e X); note that N e .' if - N e .N' since the map y - -y is a homeomorphism.] For if x e A and N e X, then x + N is a neighbor-

hood of x; hence (x+N)nA 00. Ifye(x+N)nA, then xey-N. If x0A,then (x+N)r A=0forsomeNe.,V;hencexoA-N. Now if U is a neighborhood of 0, let V, be a circled neighborhood of 0 with V, + V, c U. By (1), V, e V, + V,, and since V, is circled, so is V,. Thus we may take W = V, V.

In the locally convex case, we may as well assume U convex [the interior

of a convex set is convex, by 3.5.5(d) and the fact that a translation of a convex set is convex]. If V2 is a circled neighborhood included in U, the convex hull r

n

.I,>0,

YA,=1, n=1,2,...

111+=1

the smallest convex overset of V2, is also included in U. Since V2 is circled so is V2 , and therefore so is (V2)°. (If x + N c P2, N e X, and I a l < 1, then ax + aN c V2 .) Thus we may take V = (V2)°. Finally, if V3 is a circled convex neighborhood of 0 with V3 + V3 e U, then W = V3 is closed, circled, and convex, and by (1), W c V3 + V3 c U.

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3.5.7 Theorem. If L is a locally convex topological vector space, the topology of L is generated, in the sense of 3.5.2, by a family of seminorms. Specifically, if 3?1 is the collection of all circled convex neighborhoods of 0, the Minkowski functionals pu of the sets U e W are the desired seminorms.

PROOF. By 3.5.5(c), the Pu are seminorms, and 3.5.6, iW is a base at 0 for the topology of L. By 3.5.5(d), for each U e ?l we have U = {x: pu(x) < 1}, and

it follows that the topology of L is the same as the topology induced by the

Pu I The fact that the Minkowski functional is sublinear suggests that the Hahn-Banach theorem may provide useful information. The next result illustrates this idea. Theorem. Let K be a convex subset of the real vector space L; assume K is radial at 0 and has Minkowski functional p. 3.5.8

(a) If f is a linear functional on L, then f < 1 on K iff f < p on L. (b) If g is a linear functional on the subspace M, and g< 1 on K n M,

then g may be extended to a linear functional f on L such that f < 1 on K. (c) If in addition K is circled and L is a complex vector space, and g is a linear functional on the subspace M with I g I < 1 on K n M, then g may be extended to a linear functional f on L such that I f I < 1 on K. (d) A continuous linear functional g on a subspace M of a locally convex topological vector space L may be extended to a continuous linear functional on L. PROOF. (a)

If f < p on L, then f < 1 on K by 3.5.5(b). Conversely, assume

f < 1 on K. If x/r a K, then f(x/r) < 1, so f(x) <- r. Take the inf over r to obtain f (x) < p(x). (b) By (a), g < p on M, so by the Hahn-Banach theorem, g extends to a linear functional f with f < p on L. By (a), f <- 1 on K. (c) Let L' be L regarded as a real vector space; by (b), g, = Re g may be extended to a linear function f, on L' with f, :5 1 on K. If fix) = f,(x) - if,(ix), then (as in 3.4.3) f is a linear functional on L, and on M we havef(x) = g,(x) - ig, (ix) = g(x). Finally, if f (x) = re19, r >- 0, then I f (x) I = r = f (e-!9x) _ f,(e-19x) since risreal. IfxeK,thene-1Axe KsinceKiscircled,so I f(x)1 <- 1. (d) By continuity, g must be bounded on some neighborhood of 0 in M, so that by 3.5.6 there is a convex circled neighborhood K of 0 in L such that gI is bounded on K n M. By (c), g has an extension to a linear functional f on L with I f I bounded on K. But if If 1 <- r on K, then, given S > 0, we have If I < 6 on the neighborhood (b/r)K, proving continuity.

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SOME PROPERTIES OF TOPOLOGICAL VECTOR SPACES

157

In 3.4.14 we promised a general version of the open mapping theorem, and we now consider this question. The following technical lemma is the key step. 3.5.9 Lemma. Let A be a continuous linear operator from L to M, where L is a complete, metrizable topological vector space and M is a Hausdorff topological vector space. Assume that for every neighborhood W of 0 in L, A(W) is an overneighborhood of 0 in M. Then in fact A(W) is an overneighborhood of 0 in M. PROOF. Denote by °?l the collection of overneighborhoods of 0 in L, and by 'V the corresponding collection in M. It suffices to show that if U and V are neighborhoods of 0 in L, then A(U) c A(U + V). For then given the neighborhood W we may find a closed circled neighborhood U with U + U + U e

W; then [see 3.5.6, Eq. (1)] U + U c U + U + U c W. Now A(U), which belongs to Y," by hypothesis, is a subset of A(U + U) by the result to be estab-

lished, and this in turn is a subset of A(W), as desired. Thus let U and V be neighborhoods of 0 in L. Since L is metrizable it has a countable base of neighborhoods at 0, say U1, U2,.... It may be assumed that Un+1 + Un+, c Un for all n [3.5.1(c)], and that U1 = U, U2 + U2 c U r) V. Let y e A(U); we try to find elements xn e Un such that if yn = Axn, then,

for all n= 1,2,..., Yl + ... +y,, - Y e A(Un+1)

(1)

Since y e A(U1) (=A(U)) and y + A(U2) is an overneighborhood of y, there is an element y1 e A(U1) n (y + A(U2)). Thus y1 = Ax, for some x, e U1, and y1 - y e A(U2). If x1, ... , xn have been found such that y1 + + yn - y e A(Un+1), then since y - (y1 + + yn) + A(Un+2) is an overneighborhood of y - (y, + ' + yn), there is an element yn+, e A(Un+1) n (y - (y1 + + yn) + A(Un+2)), so that we have Xn+1 a Un+,, Y.+]= Axn+1, and y1 + + yn+, - y e A(U,,+2), completing the induction. then Let z , = '+ +** ' Un+k C U. (Note Un+k-1 + Un+k C Un+k-1 + Un+k-1 C Un+k-2, and proceed backward.) Since the Un form a base at 0, (zn) is a Cauchy sequence in L; hence z,, converges to some x e L by completeness. [By the appendix on general

topology, Theorem A 10.9, the metric d of L may be assumed invariant, that is, d(x, y) = d(x + z, y + z) for all z. We know that zn,-z,, -,, 0 as n,m-+co;

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hence by invariance, d(zm , zn) = d(zm - z., 0) -* 0, so that {zn} is Cauchy.] Since

it follows that x e U + V. If we can show that y = Ax, then y e A(U + V) and we have finished. If y 0 Ax, the Hausdorff hypothesis for M yields a closed circled W' E 'fi' such that Ax + W' and y + W' are disjoint; let W be a closed set in 11 such that W + W c W'. By continuity of A there is a set W, E 0l such that A(W,)

c W. Since the Un form a base at 0, Un c W, for large n; hence A(U,,) c-A(WI) c W; since W is closed, we have

A(U,,) c W

for sufficiently large n.

(2)

Ifn<m, m

n

m

Aj=1 Exj -y= j-1 Eyj-y+ j=n+l y- yj E A(Un+1) + A(Un+l + ... + Um)

by (1).

By (2), A(Y; ,x) Let m -- oo ; by continuity of A we find Ax - y E W + W C W' = W' C W' + W', contradicting the disjointness of Ax + W' and y + W'. Under the hypothesis of 3.5.9, if W e K, then by continuity of A there is a set U,, in the countable base at 0 in L such that A(U,) c V. By 3.5.9, A(U,)e Yl^ for all n; hence M also has a countable base at 0. It is shown in the appendix

on general topology (Theorem A10.8) that the existence of a countable base at 0 in a topological vector space is equivalent to pseudo-metrizability. Thus under the hypothesis of 3.5.9, M, being Hausdorff, is metrizable. One more preliminary result is needed. 3.5.10 Lemma. Let A be a linear operator from the topological vector space L to the topological vector space M. Assume A(L) is of the second category in M; that is, A(L) cannot be expressed as a countable union of nowhere dense sets. If U is an overneighborhood of 0 in L, then A(U) is an overneighborhood of 0 in M. PROOF. Let 'W and 'V be the overneighborhoods of 0 in L and M, respectively.

If U e O?1, choose a circled V e /1 with V+ V c U. By 3.5.1(b) applied to V

3.5

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SOME PROPERTIES OF TOPOLOGICAL VECTOR SPACES

we have L = Un i nV, hence A(L) = Un=, nA(V). Since A(L) is of the second category, nA(V) [= nA(V)] has a nonempty interior for some n, hence A(V) has a nonempty interior. If B is a nonempty open subset of A(V), then

OeB-BcA(V)-A(V)cA(V)-A(V) = A(V) + A(V)

since

V,

hence

A(V),

is circled

c A(U). But B - B = U{x - B: x e B}, an open set; hence A(U) e 3.5.11

Open Mapping Theorem. Let Land M be metrizable topological vector

spaces, with L complete. If A is a continuous linear operator from L to M and A(L) is of the second category in M, then A(L) = M and A is an open map. PROOF. Take 0ll and Y'' as in 3.5.9 and 3.5.10. Let U be any set in ?l ; by 3.5.9

and 3.5.10, A(U) a 'Y'. If y c M, then y e nA(U) for some n by 3.5.1(b); hence y = A(nx) for some x e U, proving that A(L) = M. If G is open in L and x0 a G, choose V e °l1 such that xo + V c G. Then A(V) e 1 and Ax. + A(V) e A(G), proving that A(G) is open. 3.5.12

Corollary. If A is a continuous linear operator from L onto M, where

L and M are complete metrizable topological vector spaces, then A is an open map.

PROOF. By completeness of M, A(L) (=M) is of category 2 in M, and the result follows from 3.5.11. 1 We conclude this section with a discussion of separation theorems and their applications. If K, and K2 are disjoint convex subsets of R3, there is a plane P such that K, is on one side of P and K2 is on the other side. Now P can be described as {x e R3 : f (X) = c} for some linear functional f and real number c; hence we have f(x) < c for all x in one of the two convex sets, and f(x) >- c for all x in the other set. We are going to consider generalizations of this idea. The following theorem is the fundamental result of this type. 3.5.13

Basic Separation Theorem. Let K, and K2 be disjoint, nonempty

convex subsets of the real vector space L, and assume that K, has at least one

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internal point. There is a linear functional f on L separating K, and K2, that is, f 0 0 and f(x) < f(y) for all x e K, and y e K2 [for short, f(K,) < f(K2)]. PROOF. First assume that 0 is an internal point of K,. Pick an element z e K2;

then -z is an internal point of -z + K, c K, - K2 ; hence 0 is an internal point of the convex set K = z + K, - K2. If z e K, then K, n K2 Qf , contradicting the hypothesis; therefore z 0 K; so if p is the Minkowski functional of K, we have p(z) > I by 3.5.5(b). Define a linear functional g on the subspace M = (),z: A e R} by g(Az) _ A. If a > 0, then g(az) = a = a(l) < ap(z) = p(az), and if a < 0, then g(az) = a < 0 <- p(az). By the Hahn-Banach theorem, g extends to a (nontrivial) linear functional f with f < p on L. By 3.5.8(a), f S I on K. Since f(z) = g(z) = 1, f(K,) < f(K2). If x is an internal point of K,, then 0 is an internal point of -x + K,, and

-x + K, is disjoint from -x + K2. If f(-x + Kl) < f(-x + K2), then f(KI) -
is a nontrivial linear functional f on L such that Ref(x) < Re f(y) for all x e K y e K2 ; this is what we shall mean by "f separates K, and K2 " in the complex case. 3.5.14 Corollaries. Assume the hypothesis of 3.5.13. (a) If L is a topological vector space and K, has nonempty interior, the

linear functional f constructed in 3.5.13 is continuous. (b) If L is locally convex, and K, and K2 disjoint convex subsets of L, with K, closed and K2 compact, there is a continuous linear functional f on L

strongly separating K, and K2, that is,.f(x) < c, < c2 < f(y) for some real numbers c, and c2 and all x e K1, y e K2. In particular if L is Hausdorll' and x, y e L, x y, there is a continuous linear functional f on L with f (x) 96f(y). (c) If L is locally convex, Ma closed subspace of L, and x0 e M, there is a continuous linear functional f on L such that f = 0 on M and f(x°) 96 0. Since K,° 96 0, the interior and internal points of K, coincide, by 3.5.5(b) and (d). Thus in the proof of 3.5.13, 0 is an interior point of K, so that K° = {x: p(x) < l} by 3.5.5(d). Since f < p on Lit follows that f < 1 on some neighborhood U of 0 (for example, U = K°); since U may be assumed PROOF. (a)

circled, f(-x) < 1 for all x e U; hence I f 1 < 1 on U. But then If < 6 on the neighborhood 5U, proving continuity. (b) Let K = K2 - K,; K is convex, and is also closed since the sum of a compact set A and a closed set B in a topological vector space is closed. (If

3.5

SOME PROPERTIES OF TOPOLOGICAL VECTOR SPACES

161

then y,,,-+z- x, hence z - x e B since B is closed.) Now 0 0 K by disjointness of K, and K2 ; so by local convexity, there is a convex circled neighborhood V of 0 such that V n K = 0. By 3.5.13 and part (a) above there is an f e L* separating V and K, say f( V) < c < f(K). Now f cannot be identically 0 on V, for if so f- 0 on L by 3.5.1(b). Thus there is an x e V with f(x) # 0, and since V is circled,

we may assume f (x) > 0. Thus c must be greater than 0; hence f(K2) f(K1) + c, c > 0, as desired. (c) Let K, = M, K2 = {x0}, and apply (b) to obtain fe L* with f(M) < C' < c2 < f(xo). Since f is bounded above on the subspace M, f must be identically 0 on M. [If f(x) 96 0, consider f(Ax) for arbitrarily large A.] But then .f (x0) >- c2 > 0. 1

If we adopt the above definition of separation in complex vector spaces, all parts of 3.5.14 extend immediately to the complex case.

Separation theorems may be applied effectively in the study of weak topologies. In 3.4.10 we defined the weak topology on a normed linear space; the definition is identical for an arbitrary topological vector space L. Specifically, for each f e L*, p1(x) = f (x) I defines a seminorm on L. The locally

convex topology induced by the seminorms pr, f e L*, is called the weak topology on L. By 3.5.1 and 3.5.2, a base at x0 for the weak topology consists of finite intersections of sets of the form {x: p f(x - x0) < s}, so in the case of a normed linear space we obtain the topology defined in 3.4.10.

There is a dual topology defined on L*; if x e L, then px(f) = I f(x) I defines a seminorm on V. The locally convex topology induced by the seminorms px is called the weak* topology on V. By 3.5.2, the weak topology is the weakest topology on L making each f e L* continuous, so the weak topology is weaker than the original topology

of L. Convergence of x to x in the weak topology means f (x,,) -+f(x) for each f e V. The weak* topology on L* is the weakest topology making all evaluation maps f-+f(x) continuous. Convergence of f to fin the weak* topology means f (x) -+f(x) for all x e L; thus weak* convergence is simply pointwise convergence, so if L is a normed linear space, the weak* topology is weaker than the norm topology on V. We have observed in 3.4.10 that the weak topology is an example of a product topology. Since weak* convergence is pointwise convergence, the weak* topology is the product topology on the set C` of all complex-valued functions on L, relativized to L*. In distinguishing between the weak topology and the original topology on L, it will be convenient to call the original topology the strong topology. By the above discussion, a weakly closed subset of L is closed in the original

topology. Under certain conditions there is a converse statement:

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INTRODUCTION TO FUNCTIONAL ANALYSIS

3.5.15 Theorem. Let L be a locally convex topological vector space. If K is a convex subset of L, then K is strongly closed in L iff it is weakly closed.

PROOF. Assume K strongly closed. If y 0 K, then by 3.5.14(b) there are real

numbers c, and c2 and an f e L* with Re f(x) < c, < c2 < Re f(y) for all x e K. But then W = {x e L: I f (x) - f (y) I < c2 - c, j is a weak neighborhood

of y, and if xeK, we have If(x - y)I > - IRe f(y) - Re f(x)I

c2-c1;

therefore W n K = 0, proving K` weakly open. For the remainder of this section we consider normed linear spaces. The closed unit ball f f: If II <- 1} of L* is never compact (in the strong topology) unless L (hence L*) is finite-dimensional; see Problem 9, Section 3.3. However, it is always compact in the weak* topology, as we now prove:

3.5.16

Banach-Alaoglu Theorem. If L is a normed linear space and B =

{ f e L*: if II < 11, then B is compact in the weak* topology. PROOF. If f e B, then f(x) I < if II Ilxll _< Ilxjj for all x e L. If I(x) is the set (z e C: I zI <_ Ilxlj), then B c fl{1(x): x e L), the set of all functions defined on L such that f(x) e 1(x) for each x e L. By the Tychonoff theorem (see the appendix on general topology, Theorem A5.4), fl{I(x): x e L) is compact in the product topology (the topology of pointwise convergence). be a net in B; by the above discussion there is a subnet converging Let pointwise to a complex-valued function f on L, and since the f, are linear so and it is f. Now I f(x)I I

f

is

L

the weak topology

topology on the to the is weakly compact. Conversely, weak compactness of the closed unit ball implies reflexivity. To prove this we need the following auxiliary

on

results:

3.5.17 Lemma. Let L be a topological vector space. (a) If f is a linear functional on L, f is continuous relative to the weak topology iff f is continuous relative to the strong topology, that is, iff f e V. (b) If g is a linear functional on L*, g is continuous relative to the weak* topology iff there is an x e L such that g(f) =f(x) for all f e L*.

3.5

SOME PROPERTIES OF TOPOLOGICAL VECTOR SPACES

PROOF. (a)

163

If f e L* and x converges weakly to x, then f(x)

definition of the weak topology, so f is weakly continuous. Conversely, if f is continuous relative to the weak topology and x converges strongly to x, then x,, converges weakly to x, hence f(x) Thus f e V. (b) The "if" part follows from the definition of the weak* topology, so assume g weak* continuous. A basic neighborhood of 0 in the weak* topology

is of the form U=ni=1{feL*: If(x;)I 0, n = 1, 2, ... ; thus there is a U of this type such that g(f) I < I forallfe U. By linearity, I f(x;)I
Al is arbitrary, g(f) = 0. But this implies, by a standard algebraic result (see Problem 15) that g(f) = Yl=1 c;f(x;) for some complex numbers c1, ..., c,,.

Setx=Y;_,clxi.

3.5.18 Lemma. Let L be a normed linear space, and let h be the isometric isomorphism of L into L** (see 3.4.6). Let B be the closed unit ball in L, B** the closed unit ball in L**. Then h(B) is dense in B** relative to the weak* topology. Furthermore, h(L) is dense in L**.

PROOF. Let M be the weak* closure of h(B), and let x** E B**, x** 0 M. By 3.5.16, B** is weak* compact (and convex); so by 3.5.14(b), there is a linear functional g on L** such that g is continuous relative to the weak* topology

and Re g(Al) 5 c, < c2 < Re g(x**). Since M is a subspace (in particular is circled), g(y**) I < cl for all y** e M. [If y** e M and g(y**) = rei°, then I g(y**) = g(e-Fey**) = r = Re g(e-rOy**) < c,.] By 3.5.17(b), g is of the form g(y**) = y**(f) for some f e V. If y E B, and y** = h(y) is the corresponding element of h(B) c M, then If(y)I = Iy**(f)I c,; hence

-

IlfII s c, < c2 < Reg(x**) < Ix**(f)1 < Ilx**II 11f 11 :g 1181,

a contradiction. Consequently M = B**. Finally, if y** e L**, y** 96 0, then y**/Ily**II e B**, which we have just seen is the weak* closure of h(B). Thus y**/Ily**II belongs to the weak* closure of h(L), which is a subspace. The result follows. I We now characterize reflexive normed linear spaces.

Theorem. A normed linear space L is reflexive if its closed unit ball B is weakly compact. 3.5.19

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INTRODUCTION TO FUNCTIONAL ANALYSIS

PROOF. The "only if" part has been pointed out after 3.5.16, so assume B weakly compact. The isometric isomorphism h of L into L** is a homeomorphism of the weak topology of L and the weak* topology of h(L) [note that x"- + x iff f (x,,) --+f(x) for all f E L*; in other words, if h(x")-"' . h(x)J. Thus h(B) is a weak* compact (hence closed) subset of B**, and is dense in B** by 3.5.18. It follows that h(B) = B**, so as in 3.5.18, h(L) = L**, proving reflexivity.

3.5.20 Corollaries. (a) If M is a closed subspace of the reflexive normed linear space L, then M is reflexive. (b) If L is a Banach space, L is reflexive if L* is reflexive.

If B is the closed unit ball in L, then B n M is the closed unit ball in M. By hypothesis, B n M is strongly closed, hence weakly closed by 3.5.15. But B is weakly compact by 3.5.19, so B n M is weakly compact. Again by 3.5.19, M is reflexive. (b) Assume L reflexive. The weak* topology on L* then coincides with its weak topology, hence by 3.5.16, the closed unit ball of L* is weakly coinpact, and consequently L* is reflexive by 3.5.19. (Completeness of L is not used in this part.) PROOF. (a)

If L* is reflexive, so is L** by the first part of the proof. Since L is complete,

so is h(L); thus h(L) is a closed subspace of L**, and the result follows from

(a). I Problems 1.

2.

3.

Let W be the family of overneighborhoods of 0 in a topological vector space L. Show that L is Hausdorff iff (}{U: U E °ll} = {0}. If f is a linear functional on the topological vector space L, with f not identically 0, show that the four conditions of Problem 1, Section 3.3, are still equivalent. If L is a finite-dimensional Hausdorff topological vector space, show that L is topologically isomorphic to C" for some n. Show also that a finitedimensional subspace of an arbitrary Hausdorff topological vector space

is always closed. [It can be shown that for a Hausdorff topological vector space, finite-dimensionality is equivalent to local compactness; see Kelley and Namioka (1963, Theorem 7.8).] 4. (a) Let K be a convex set, radial at 0, in the vector space L. If g is any nonnegative sublinear functional on L such that {x c- L: g(x) < 1} c K c {x e L: g(x) < 1}, show that g is the Minkowski functional of K. [Thus by 3.5.5(d), if 0 e K°, then the Minkowski functionals of K°, K, and K coincide.)

3.5

SOME PROPERTIES OF TOPOLOGICAL VECTOR SPACES

(b)

165

Use part (a) to find the Minkowski functionals of each of the following subsets of R2: (i) {(x, y): IxI + I Y1 (ii) (iii)

(c)

<_ 1}.

{(X, Y): -1<x<1, -1
If K, and K2 are convex and radial at 0, with Minkowski functionals

and p2, show that the Minkowski functional of K, n K2 is 5.

6.

max(p1, P2) Let L be the set of all complex-valued functions on [0, 1], with the topol-

ogy of pointwise convergence. Then L is a locally convex topological vector space, with topology induced by the seminorms p,,(f) = I f(x)I, f e L, where x ranges over [0, 1]; see 3.5.3(a). Show that L is not metrizable. (The same result holds even if we restrict L to the continuous complex-valued functions on [0, 1].) Let L be a locally convex topological vector space. Show (without using the theory of uniform spaces given in the appendix on general topology, Section A10) that the following conditions are equivalent:

(a) L is pseudometrizable, that is, there is a pseudometric inducing the topology of L. (b) L has a countable base at 0 (hence at every point of L). (c) The topology of L is generated by a countable family of seminorms. 7.

Let L be the collection of analytic functions on the unit disk D = {z: IzI < 1}, with the topology of uniform convergence on compact subsets [see 3.5.3(a)]. Show that L is metrizable but not normable.

8.

Let L = LP[0, 1) where 0 < p < 1; with the pseudometric d(x, y) _ x(t) - y(t) I P dt (see 2.4.12), L is a topological vector space. [Apply 3.5.1, using the sets U = {x: f o x(t) I P dt < e}, e > 0.] ,t o I

Let f be a continuous linear functional on L. If fix) = 1, show that x can be written as y + z where d(y, 0) = d(z, 0) = # d(x, 0). (b) With f as in part (a), show that there are functions x,, X2, ... e L 0) -- 0 as n - oo, but I f (x.) I > I for all n. Conclude such that (a)

that the only continuous linear functional on L is the zero functional.

[In part (a), either I f(Y) I ? I or I f(z) I > 1; if, say, I f(Y) I #, take x, = 2y and proceed inductively.] 9. Let L be a topological vector space. Show that there is a nontrivial continuous linear functional on L iff there is a proper convex subset of L with nonempty interior. 10. Let B1 and B2 be (not necessarily convex) nonempty subsets of the topological vector space L, with B1° 96 0. If f is a linear functional on

166

11.

3

INTRODUCTION TO FUNCTIONAL ANALYSIS

L that separates B, and B2, show that f is continuous. [This gives another proof of 3.5.14(a).] Let K be a convex subset of a locally convex topological vector space L.

Show that the closure of K in the weak topology coincides with the 12.

13.

closure in the strong topology. Let K be a convex set, radial at 0, in the real vector space L. Let x be a point of L and M a subspace of L such that x + M does not contain any internal points of K. Show that there is a hyperplane H (that is, H is the null space of a nontrivial linear functional on L) such that M c H and x + H contains no internal points of K. [Define g on L(M u {x}) by g(y + ax) = a, and extend using the Hahn-Banach theorem.] (a) If L is a topological vector space and 91, 2 are topologies on L such that ,% , c 9 2 and L is complete and metrizable under both

, and 9-,2, show that J, (b) If L is a Banach space under two norms 11 11, and II 112, and, for

some k > 0, llxll S klfxll2 for all x e L, show that for some r > 0,

14.

IIx112 < rllxll, for all x e L; in other words (Problem 6, Section 3.3) the norms induce the same topology. (a) (Closed graph theorem) Let A be a linear map from L to M, where L and M are complete metrizable topological vector spaces.

Assume A is closed; in other words, the graph of A is a closed subset of L x M, with the product topology. Show that A is continuous. (b) If A is a continuous linear operator from L to M where L and M

are topological vector spaces and M is Hausdorff, show that A 15.

16.

is closed. Let g, f1, ... , f" be linear functionals on the vector space L. If N denotes null space and n;= I N(f;) e N(g), show that g is a linear combination of the f, . Let L be a normed linear space. If the weak and strong topologies coincide on L, show that L is finite-dimensional, as follows: (a) The unit ball B = {x: IIxli < 1) is strongly open, hence weakly e L* and b1, ..., open by hypothesis. Thus we can find .

S" > 0 such that {x: l f;(# < S,, i = 1, ..., n} c B. Show that (with N(f) denoting the null space off) (),!=I N(f;) = (0). (b)

Define T(x) = (f, (x), ... , f"(x)) ; by (a), T is a one-to-one map of L

onto a subspace M of C". If {y...... yk) is a basis for M and x; =

T - ' (y;), j = 1, ... , k, show that {x,, ... , xk} is a basis for L, 17.

hence L is finite-dimensional. Let L be a separable normed linear space. If B is the closed unit ball in L*, show that B is metrizable in the weak* topology. Thus since B is compact by 3.5.16, every sequence in B (hence every norm-bounded

3.6

REFERENCES

167

sequence in L*) has a weak* convergent subsequence. Similarly, if L is reflexive, the closed unit ball of L is metrizable in the weak topology. (Let {x,, X21 ...} be a countable dense subset of and set w {{

1

If(xn) - g(Xn)1

d(f, g) - =1 2" 1 + ((((x.) g(xn)I ,,JJ

18.

19.

f, g E L*. f,

Show that weak* convergence implies d-convergence.) Show that a reflexive Banach space is weakly complete; in other words every weak Cauchy sequence [{ f (x )} is Cauchy for each f e L*] converges weakly. A subset B of a topological vector space L is said to be absorbed by a

subset A if B c aA for sufficiently large I a I ; B is said to be bounded if it is absorbed by every neighborhood of 0. (a) Show that B is bounded if for each sequence of points x e B and each sequence of complex numbers .1 -+ 0, we have A. x -+ 0. (b) A bornivore in a topological vector space L is a convex circled set that absorbs every bounded set; L is bornological if L is locally convex and every bornivore is an overneighborhood of 0. Show that every metrizable locally convex space is bornological. (c) Let T be a linear operator from L to M, where L is bornological and M is locally convex. If T maps bounded sets into bounded sets, show that T is continuous. (The converse is true for arbitrary L and M.) (d) If L is a Hausdorff topological vector space, show that there is no bounded subspace of L except {0}.

3.6

References

There is a vast literature on functional analysis, and we only give a few representative titles. Readable introductory treatments are given in Liusternik and Sobolev (1961), Taylor (1958), Bachman and Narici (1966), and Halmos (1951); the last deals exclusively with Hilbert spaces. Among the more ad-

vanced treatments, Dunford and Schwartz (1958, 1963, 1970) emphasize normed spaces, Kelley and Namioka (1963) and Schaefer (1966) emphasize topological vector spaces. Yosida (1968) gives a broad survey of applications to differential equations, semigroup theory, and other areas of analysis.

4 The Interplay between Measure Theory and Topology

4.1

Introduction

A connection between measure theory and topology is established when a a-field .F is defined in terms of topological properties. In the most common situation, we have a topological space fl, and F is taken as the smallest a-field

containing all open sets of Q. If this is done, there is a natural connection between measure-theoretic and topological questions. For example, if µ is a measure on , and A e .F, we may ask whether A can be approximated by a compact subset. In other words, we wish to know if u(A) = sup{µ(K): K a compact subset of A}. As another example, we may ask whether a function in L°(S2, .07, p) can be approximated by a continuous function. One formulation of this is to ask whether the continuous functions are dense in

P. In this chapter we investigate questions of this type. The results in the first two sections are not topological, but they serve as basic tools in the later development. We first consider a result that is a companion to the monotone class theorem: Definition. Let -9 be a class of subsets of a set 0. Then -9 is said to be a Dynkin system (D-system for short) if the following conditions hold.

4.1.1

168

4.1

INTRODUCTION

169

(a) QE -'. (b) If A, B e 2, B c A, then A - B e 9. Thus -9 is closed under proper differences. (c)

If A1,A2,...e9andA,, TA, then AeO.

Note that by (a) and (b), -9 is closed under complementation; hence by (c), -9 is a monotone class. If -9 is closed under finite union (or closed under finite intersection), then g is a field, and hence a a-field (see 1.2.1). 4.1.2

Dynkin System Theorem. Let .9' be a class of subsets of 0, and assume 9' closed under finite intersection. If -9 is a Dynkin system and -9 .9', then -9 includes the minimal a-field F = a(9'). PROOF. Let -9o be the smallest D-system including 9. We show that -9Q = F,

in other words the smallest D-system and the smallest a-field over a class closed under finite intersection coincide. Since -9o c -9, the result will follow.

Now -9o (-- .;t since F is a D-system. To show that .F c .9o , let ' = {A e -9o : A r B e -90 for alI B e .9'}. Then 91' c ' since .9' is closed under finite intersection, and since -9o is a D-system, so is W. Thus .90 a' , hence -9o=W. Now let " _ {C ego : C n D c-90 for all D e 3o). The result .90 = W implies that 51 T, and since 'e' is a D-system we have .9o c W', hence 0o = V. It follows that -90 is closed under finite intersection; by 4.1.1, -90 is

a a-field, so that F c go. 1 If Y is a field and .,# is the smallest monotone class including So, then Af = a(.9') (see 1.3.9). In the Dynkin system theorem, we have a weaker hypothesis on .9' (it is closed under finite intersection but need not be a field)

but a stronger hypothesis on the class of sets including . 9' (9 is a Dynkin system, not merely a monotone class). Corollary. Let 9 be a class of subsets of 0, and let µl and P2 be finite measures on a(.9'). Assume 0 E . 9' and . 9' is closed under finite intersection. If µ, = µ2 on So, then µl = µ2 on a(.9'). 4.1.3

PRoor. Let g be the collection of sets A e a(.9') such that µ,(A) = 102(A). Then

2 is a D-system and 59' c .9, hence g = a(.9') by 4.1.2. 1 4.1.4 Corollary. Let .9' be a class of subsets of 0; assume that S2 e .9' and . 9'

is closed under finite intersection. Let H be a vector space of real-valued functions on fl, such that IA E H for each A E .9'.

4

170

THE INTERPLAY BETWEEN MEASURE THEORY AND TOPOLOGY

Suppose that whenever f f2 , ... are nonnegative functions in H, I f 1 < M < co for all n, and f, T f, the limit function f belongs to H. Then IA e H for all A c a(.9'). PROOF. Let _q = {A c fl: IA a H}; then ,9'c -9 and 2 is a D-system. For if

A,Bc ,BcA,then IA_B=IA-JBeH,sothat A-Be -9. If A. e-9 and A T A, then IA T IA, hence I. e H by hypothesis; thus A e -9. The result now follows from 4.1.2. 1 4.2

The Daniell Integral

One of the basic properties of the integral is linearity; if f and g are pintegrable functions and a and b are real or complex numbers, then

fn (af+bg)dp=a$n fdp+b fn g dp. Thus the integral may be regarded as a linear functional on the vector space of integrable functions. This idea may be used as the basis for a different approach to integration theory. Instead of beginning with a measure and constructing the corresponding integral, we start with a given linear functional E on a vector space. Under appropriate hypotheses, we extend E to a larger space, and finally we show that there is a measure u such that E is in fact the integral with respect to p. We first fix the notation to be used.

Notation. In this section, L will denote a vector space of real-valued functions on a set f ; L is assumed closed under the lattice operations, in other words if f, g e L and f vg = max(f, g), f A g= min(f, g), then f v g, f A g e L. 4.2.1

There are several familiar examples of such spaces; L can be the class of continuous real-valued functions on a given topological space, or equally well, the bounded continuous real-valued functions. Another possibility is to take L as the collection of all real-valued functions on a given set. The letter E will denote a positive linear functional on L, that is, a linear map E from L to R such that f'>_ 0 implies E(f) >- 0. This implies that E is monotone, that is, f < g implies E(f) <- E(g).

If H is any class of functions from 0 to R (or R), H+ will denote f f e H : f > 0). The collection of functions f: S2 -+ R of the form lim f where the f form an increasing sequence of functions in L+, will be denoted by L';

if the f are allowed to form an increasing net in L+, the resulting class is denoted by L. (See the appendix on general topology, Section Al, for a discussion of nets.)

4.2 THE DANIELL INTEGRAL

171

If H is as above, a(H) is defined as the smallest a-field of subsets of 12 making every function in H Bore] measurable, that is, the minimal a-field containing all sets f -'(B), where f ranges over H and B over the Borel sets. If 9 is a class of subsets of K2, a(f) as usual denotes the minimal or-field over T.

The following will be assumed throughout: Hypothesis A: If the functions fn form a sequence in L decreasing to 0, then E(fn) decreases to 0. Equivalently, if the fn belong to L and increase to f e L, then E(fn) increases to E(f).

We are also going to carry through a parallel development under the following assumption: Hypothesis B: If the functions fn form a net in L decreasing to 0, then E(fn) decreases to 0 (with the equivalent statement just as in hypothesis A).

Hypothesis A is always assumed in the statements of theorems. Corresponding results under hypothesis B will be added in brackets. 4.2.2 Lemma. Let {fm} and { fn'} be sequences in L increasing to f and f',

respectively, with f < f' (f and f need not belong to L). Then Iim E(fm)
n

Hence E may be extended to L' by defining E(limn fn) = limn E(fn). [Under

hypothesis B (and with "sequence" replaced by "net" in the above statement), E may be extended to L" in the same fashion.] PROOF. As n --> 00,Jm A./n' I Jm Af' =Jme hence limn E(fn')

limn E(fm Afn') =

E(fm). Let m -), oo to finish the proof. [Under hypothesis B, the proof is the same, with sequences replaced by nets.] I We now study the extension of E. 4.2.3 Lemma. The extension of E to L' has the following properties:

(a) 0< E(f)
(b) Iff,geL',f
E(f i g)=E(fvg)+E(fng)=E(f)+E(g).

172 (e)

4 THE INTERPLAY BETWEEN MEASURE THEORY AND TOPOLOGY

increases

If If,,) is a sequence in L' increasing to f, then f e L' and

to E(f). [Under hypothesis B, the extension of E to L" has exactly the same properties; " sequence " is replaced by " net " in (e).] PROOF. Properties (a)-(c) follow from 4.2.2. To prove (d), let fn T f g,, ? g. Then

g a L+, with

E(fn v gn) + E(fn A g,,) = E[(f,, v gn) + (fn A g,,)]

by linearity = E(f,, + gn) (the sum of two numbers is the larger plus the smaller)

= E(fn) + E(gn) by linearity.

Let n --> co to obtain (d). To prove (e), let J,,m e L, f,,, If. as in -,, oo. If g,,, = f,m v .. vfmm, the gm form an increasing sequence in L+, and f.. :5: gm:! fm

for n<m;

(1)

hence E(fnm) < E(gm) < E(fm),

n < in.

(2)

In (l) let in - oo to obtain fn < limm gm < limmfm =f; then let n -+ oo to obtain gm If, hence E(gm) I E(f). Now let in oo in (2) to find that E(fn) S E(f) < limm E(fm); then let n --> oo to establish that limn E(f) [Under hypothesis B, (a)-(d) are proved as above, but (e) is more complicated. Let {fn} be a net in L" with f I f; for each n there is a net If..) in V with f,m T fn , so that f = sup,,, mfnm Let D consist of all finite sets of pairs (n, m) a M}, ME D. (n, m). Direct D by inclusion, and define gm = The g,K form a monotone net in e and gm If-, therefore f e L" and, by 4.2.2, E(gm) I E(f ). Now for each M, E(gM) < sup,,,,. E(f,, m), and sincef,,m < f, for

all in, it follows that E(gm) < sup,, E(.) = lim,, E(fn). But M is arbitrary, The reverse inclusion holds because f', t f, and the hence E(f) < lim,, proof is complete.] We now begin the construction of a measure derived from the linear functional E. Fron now on we assume that all constant functions belong to L and, as a normalization, E(l) = 1 (hence E(c) = c for all c).

173

4.2 THE DANIELL INTEGRAL

4.2.4 Lemma. Let 9 = {G c 0: IG cL'} and define µ(G) = E(IG), G E mar. Then 9 satisfies the four conditions of 1.3.2, namely: (a) 0, n e QF, u(o) = 0, µ(0) = 1, 0 < µ(A) < 1 for all A e 99. (b) If G G2 e W, then G, u G2, G, n G2 c 9r, and

µ(G, u G2) +µ(G, n G2) = µ(GI) + µ(G2). (c) If G,, G2 e 91 and G, c G2, then µ(G,) < µ(G2). (d) If G a g, n = 1, 2,... and G I G, then G E T and p(Ga) j p(G).

Thus by 1.3.3 and 1.3.5, u*(A) = inf{µ(G): G e 9, G e A} is a probability measure on the a-field W = {H c 0: µ*(H) + p*(H`) = 1), and p* = p on 9. [Under hypothesis B, we take 9 = {G a !Q: Ic e L"} and replace sequences by nets in 4.2.4(d). The classIi then has exactly the same properties.] PROOF. Part (a) follows since L contains the constant functions and E(c) = c. Part (b) follows from 4.2.3(d) with f = IG, , g = IG2 . Part (c) follows from 4.2.3(b), and part (d) from 4.2.3(e). [The proof under hypothesis B is identi-

cal.] I We now investigate Borel measurability relative to the a-field a(T).

4.2.5 Lemma. 1f f E L' and a e R, then {w: f (c)) > a} e g. Hence

f (A, aM) --> (R, 9(R)). [The same result holds for f e L" under hypothesis B.] Let f,, L+, f, If Then (fn - a)+ = (f - a) v 0 e L+ c L', and (f, - a)+ j (f - a)+ By 4.2.3(e) (or the definition of L'), (f - a)+ E L'; hence k(f - a)+ e L' for each positive integer k, by 4.2.3(c). But as k -+ oo, PROOF.

1A

k(f-a)+

I I(f>);

so by 4.2.3(e), I{ f>a) e L', and therefore {f > a) e 9. [Under hypothesis B, the proof is the same, with { fa} a net instead of a sequence.] I

4.2.6 Lemma. The a-fields a(L), a(L'), and a(ir) are identical. [Under hypothesis B we only have a(V) = a(f) and a(L) c a(L").] PROOF. By 4.2.5, a(gr) makes every function in L' measurable; hence a(E) c a(gr). If G E 9r, then IG e L'; hence G = (IG = 1) a a(L'); therefore a(sr) e a(L'). [Under hypothesis B, a(L") = a(W) by the same argument.]

174

4

THE INTERIsLAY BETWEEN MEASURE THEORY AND TOPOLOGY

If f e L, then f = f + -f-, where f +, f - e L' c :L'. Since f + and f- are a(L')-measurable, so is f. In other words, u(L') makes every function in L Bore] measurable, so that a(L) c a(L'). [Under hypothesis B, a(L) c a(L") by the same argument.] Now if f e L', then f is the limit of a sequencef a L, and since the f are a(L)-measurable, so is f. [This fails under hypothesis B because

the limit of a net of measurable functions need not be measurable; see Problem 1.] Thus a(L') e a(L). I Now by definition of the set function p* (see 4.2.4) we have, for all A c Sl, p*(A) = inf{E(IG): G e QV,

G

A)

=inf{E(f):f=IGE L', f>IA} > inf{E(f): fe L', f> IA}. In fact equality holds.

4.2.7 Lemma. For any A c 0, p*(A) = inf{E(f): f e L', f > IA). [The result is the same under hypothesis B, with L' replaced by L".) PROOF. Let f e L', f >- IA. If 0 < a < 1, then A e (f > a), which belongs to 9 by 4.2.5. Thus p*(A) < p{ f > a) = E(I f>a)). But since f >- 0 we have f Z aI(f>Q); hence E(I(f>Q)) <E(f)la. Let a-+ 1 to conclude that p*(A) S E(f). [The proof is the same under hypothesis B.]

We may now prove that p* is a measure on a(l):

4.2.8 Lemma. If .e = {H c fl: p*(H) + p*(H`) = 1), then 9r c f e, hence a(19) c 0. [The result is the same under hypothesis B.] PROOF. Let G e T; since 1c e L', there are functions f, e e with f, T 1G. Then

y*(G) = p(G) = E(IG) = lim E(f)

by 4.2.3(e).

n

By 4.2.7, p*(G`) = inf{E(f ): f e L', f

But if fn < IC', then 1 - fn >- Ian ;

since I - fn > 0, we have I - fn e L' c L'; hence p*(G`) < inf E(1 n

I - lim E(f) = 1 - E(IG). n

Thus p*(G) + p*(G`) < 1; since p*(G) + p*(G`) is always at least I by 1.3.3(b), we have G E .-Y. [Under hypothesis B, the proof is the same, with { fn} a net

instead of a sequence.] I

4.2

175

THE DANIELL INTEGRAL

We now prove the main theorem; for clarity we state the results under hypotheses A and B separately (in 4.2.9 and 4.2.10). Daniel! Representation Theorem. Let L be a vector space of real-valued functions on the set 92; assume that L contains the constant functions and is closed under the lattice operations. Let E be a Daniell integral on L, that is, 4.2.9

a positive linear functional on L such that E(fa) J. 0 for each sequence of functions f, e L with f 10; assume that E(1) = 1. Then there is a unique probability measure P on a(L) (=a(L') = a(T) by 4.2.6) such that each f e L is P-integrable and E(f) = In f dP. PROOF. Let P he the restriction of p* to a(L). By 4.2.6 and 4.2.8, P is a probability measure. If G e 19, then

E(I0) = p(G) = p*(G) = P(G) = f IG 0. n

Now if f e L', we define

k-1 l{(k-I)/2^,)

hn = k=1

n

2

Since I(a< f:5 b) = I(f>a) - I{ f>b} for a < b, and {f > a}, {f > b} e W by 4.2.5, it

follows that

In h dP. But the h form a sequence of nonnegative

simple functions increasing to f [see 1.5.5(a)], so by the monotone convergence

theorem, E(f) = In f dP.

Now let feL; f = f + -f-, where f +, f - c -L' c L'. Then E(f) _ E ( f ') - E(f -) _ ,$n f + d P - fn f d P = f n f dP. (Since f +, f - e L, the integrals are finite.)

This establishes the existence of the desired probability measure P. If P' is another such measure, then So fdP = 5 n f dP' for all f a L,and hence for all f e L', by the monotone convergence theorem. Set f = Ia, G e W1, to show that P = P on W. Now Ir is closed under finite intersection by 4.2.4(b); hence by 4.1.3, P = P' on a(lr), proving uniqueness. I 4.2.10

Theorem. Let L be a vector space of real-valued functions on the set

i2; assume that L contains the constant functions and is closed under the

lattice operations. Let E be a positive linear functional on L such that E(,,) 10 for each net of functions f e L with f 10; assume that E(1) = I. Then there is a unique probability measure P on o(L") (=a(W) by 4.2.6) such that: (a) Each f e L is P-integrable and E(f) = In f dP. (b) If is a net of sets in g and Ga TG, then G e

and P(G) T P(G).

176

4 THE INTERPLAY BETWEEN MEASURE THEORY AND TOPOLOGY

PROOF. Let P be the restriction of M* to a(L'). The proof that P satisfies (a) is done exactly as in 4.2.9, with L' replaced by L'; P satisfies (b) by 4.2.4(d).

The uniqueness part cannot be done exactly as in 4.2.9 because the monotone convergence theorem fails in general for nets (see Problem 2). Let P' be a probability measure satisfying (a) and (b). If f e L', there is a net of functions fa e L+ with fa If. We define 1

n2"

n = 1, 2, ... .

hna = 2n,I1 I{Ia>

If (k - 1)/2" < fa(w) < k/2n, k = 1, 2, ..., n2", then hna(w) =(k - 1)/2", and if fa(co) > n, then hna(w) = n. Thus the hna, n = 1, 2, ..., are in fact the standard

sequence of nonnegative simple functions increasing to fa [see 1.5.5(a)]. Similarly, if 1

nr22"

2nj=1 L I{j> j2-")

h"

the h" are nonnegative simple functions increasing to f. Now

f h" dP' a

n2"

1

y P'{f > j2-"}

2nj=I o2"

1

E lim P'{ fa > j2-"} ' 2n j=1 a

by (b)

n2"

1

=lim- Y P'{fa>j2r"} a

2" j=1

since the sum on j is finite

= lim f hna dP'; a

n

but

f f dP' = lim f ha dP' n

n

n

by the monotone convergence theorem

= lim lim f hna dP' n

a

n

= lim lim f hna dP' a

n

a

since hna is monotone in each variable,

so that "lim" may be replaced by "sup"

=lim f fadP' a

a

by the monotone convergence theorem

4.2

THE DANIELL INTEGRAL

177

Equation continues

= lim a

fa dP

Ja

by (a)

= fnf dP by the above argument with P' replaced by P.

Set f = Ic, G e 9r, to show that P = P' on 9; hence, as in 4.2.9, on a(w). The following approximation theorem will be helpful in the next section. Theorem. Assume the hypothesis of 4.2.9, and in addition assume that L is closed under limits of uniformly convergent sequences. Let 4.2.11

9'={G-=0:G={f>0) for some feL}. Then : (a)

9' = 9r.

(b) If A e 6(L), then P(A) = inf{P(G): G e,r', G :DA}. (c)

If G e W, then P(G) = sup{E(f): fe L, f < Ic}.

PROOF. (a) We have 5r' c Ir by 4.2.5. Conversely, suppose G e 1, and let f E L+ with f, T Ic (e L'). Set f = 2:1 1 2-nf . Since 0
{f>0)= U

1}=G.

n=1

Consequently, G e 9'. (b) This is immediate from (a) and the fact that P = u* on Q(L). (c)

If f e L+, f < Ic, then E(f) < E(IG) = P(G). Conversely, let G E 1,

with f e L, f T 1G. Then P(G) = E(!) = limn

sup

hence

P(G) < sup{E(f): fe e, f < IG}. Problems

Give an example to show that the limit of a net of measurable functions need not be measurable. 2. Give an example of a net of nonnegative Borel measurable functions fa increasing to a Bore] measurable function f, with lima f fa du -0 f f du. 3. Let L be the class of real-valued continuous functions on [0, 1], and let E(f) be the Riemann integral off. Show that E is a Daniel] integral on L, and show that o(L) _ -4[0, 1] and P is Lebesgue measure. 1.

178

4 THE INTERPLAY BETWEEN MEASURE THEORY AND TOPOLOGY

4.3 Measures on Topological Spaces

We are now in a position to obtain precise results on the interplay between measure theory and topology.

Definitions and Comments. Let fI be a normal topological space (0 is Hausdorff, and if A and B are disjoint closed subsets of Q, there are disjoint 4.3.1

open sets U and V with A c U and B c V). The basic property of normal spaces that we need is Urysohn's lemma: If A and B are disjoint closed subsets

of 0, there is a continuous function f: r2 -> [0, 11 such that f = 0 on A and f = 1 on B. Other standard results are that every compact Hausdorrf space is normal, and every metric space is normal. The class of Borel sets of 0, denoted by R(fl) or simply by.4, is defined as the smallest a-field of subsets of r2 containing the open (or equally well the closed) sets. The class of Baire sets of 0, denoted by V(fl) or simply by sad, is defined as the smallest a-field of subsets of S1 making all continuous realvalued functions (Borel) measurable, that is, sat is the minimal Q-field contain-

ing all sets f -'(B) where B ranges over R(R) and f ranges over the class C(f2) of continuous maps from f to R. Note that sd is the smallest a-field making all bounded continuous functions measurable. For let .F be a a-field that makes all bounded continuous functions measurable. If f e C(r2), then f + An is a bounded continuous function and f + An If + as n -+ oo. Thus

f + (and similarly f -) is .F-measurable, hence f =f' -f - is .F-measurable. Thus d c F, as desired. The class of bounded continuous real-valued functions on 0 will be denoted by Cb(Q).

If V is an open subset of R and f e C(Q), then f -'(V) is open in 11, hence

f -(V) e 2(c2). But the sets f -'(V) generate si(Q), since any a-field containing the sets f -'(V) for all open sets V must contain the sets f -'(B) for all Borel sets B. (Problem 6 of Section 1.2 may be used to give a formal proof, with ' taken as the class of open sets.) It follows that W(r) a R(r2). An F, set in rZ is a countable union of closed sets, and a G. set is a countable intersection of open sets. Theorem. Let r2 be a normal topological space. Then sl(r2) is the minimal a-field containing the open F, sets (or equally well, the minimal a-field containing the closed G6 sets). 4.3.2

PROOF. Let f e C(Q); then If > a) = Un , If >- a + (1/n)) is an open F,, set. As above, the sets {f> a}, a e R, f e C(O), generate V; hence .4 is included

4.3

179

MEASURES ON TOPOLOGICAL SPACES

in the minimal o-field ,Y over the F, sets. Conversely, let H = Un , F,,, F. closed, be an open F. set. By Urysohn's lemma, there are functions f" e C(f)) onF,,.Iff=Y', 2-"f", then feC(Q), with

00}=Un ,{f">0}= H.Thus He.4,sothat . c.4.

4.3.3 Corollary. If fl is a normal topological space, the open F, sets are precisely the sets (f > 0) where f e Cb(fl), f >_ 0.

PROOF. By the argument of 4.3.2. 1 4.3.4

Corollary. If Cl is a metric space, then .sd(Cl) = R(i2).

PROOF. If F is a closed subset of Cl, then F is a G. 1

F = n00{c o: dist(w, F) < "=i

;

n

hence every open subset of Cl is an F, . The result now follows from 4.3.2. 1 Corollary 4.3.4 has a direct proof that avoids use of 4.3.2; see Problem 1. We may obtain some additional information about the open F, sets of a normal space. 4.3.5 Lemma. Let A be an open F. set in the normal space S2, so that by 4.3.3,

A = (f > 0) where f C- Cb(fZ), f > 0. Then IA is the limit of an increasing sequence of continuous functions. PROOF. We have (f > 01 = Un , {f> 1/n), and by Urysohn's lemma there are functions f" e C(C) with 0< .fn :g 1, f" = 0 on (f = 0),f. = 1 on {f > I /n}. If 9" = max(f , ... ,./"), then 9" T Itt> o) .

The Daniell theory now gives us a basic approximation theorem. 4.3.6 Theorem. Let P be any probability measure on .W(f2), where Cl is a

normal topological space. If A e d, then (a) P(A) = inf{P(V): V n A, Van open F. set}, (b) P(A) = sup{P(C): C c A, C a closed Ga set}.

180

4 THE INTERPLAY BETWEEN MEASURE THEORY AND TOPOLOGY

PROOF. Let L = Cb(f) and define E(f) = In f dP, f e L. [Note that a(L) = ,V,

so each f e L is d-measurable; furthermore, since f is bounded, In f dP is finite. Thus E is well-defined.] Now E is a positive linear functional on L, and by the dominated convergence theorem, E is a Daniell integral. By 4.2.11(b), P(A) = inf{P(G): G e 9r',

G z A},

where

9'={Gc0:G={f>0} forsomefeL}. By 4.3.3, T' is the class of open F, sets, proving (a). Part (b) follows upon applying (a) to the complement of A. I Corollary. If f2 is a metric space, and P is a probability measure on R(f ), then for each A e R(f2), 4.3.7

(a) P(A) = inf{P(V): V A, V open}, (b) P(A) = sup{P(C): C c A, C closed}.

PROOF. In a metric space, every closed set is a Gs and every open set is an F, (see 4.3.4); the result follows from 4.3.6. 1

Under additional hypotheses on f2, we obtain approximations by compact subsets.

4.3.8 Theorem. Let ) be a complete separable metric space (sometimes

called a "Polish space"). If P is a probability measure on R(fl), then for each A e R(f2),

P(A) = sup{P(K): K compact subset of A}.

PROOF. By 4.3.7, the approximation property holds with "compact" replaced by "closed." We are going to show that if e > 0, there is a compact set K. such that P(Ke) > 1 - s. This implies the theorem, for if C is closed, then C n K, is

compact, and P(C) - P(C n K,) = P(C - K,) < P(c2 - K,) < e. Since fl is separable, there is a countable dense set {w1, w2, ...}. Let B(cv,,, r} (respectively, B(w,,, r)) be the open (respectively, closed) ball with

center at w and radius r. Then for every r > 0, Q = Un 1 B(w,,, r) so that Uk=, B(Wk, 1/n) T f2 as m.-oo (n fixed). Thus given e > 0 and a positive integer n, there is a positive integer m(n) such that P(Uk 1 B(wk, 1/n)) z 1 - e2` for all m Z m(n).

4.3 MEASURES ON TOPOLOGICAL SPACES

181

Let K, = nni Uk("1 B(uwk, 1/n). Then K, is closed, and !m() _

X00

L

P(K,`)

'PI

n=1

k=1

1

B [Cok'

c

n _J)

m

< n=1 Y

s2-" = s.

Therefore P(K,) >- I - s. It remains to show that K, is compact. Let {x1ix2, ...} be a sequence in K,. Then xP E no Uk "i R04, 1/n) for all p; hence x,, e Uk`=11 B(wk, 1) for all p. We conclude that for some k,, xP E B(wk,, 1) for infinitely many p, say, for p c- T,, an infinite subset of the positive integers. But xP E B(wk, #) for 1

all p, in particular for all p e T1; hence for some k2 , XP E B(cok,, 1) r B((vk2, #)

for infinitely many p e TI, say, for p e T2 e T1. Continue inductively to obtain integers k,, k2, ... and infinite sets TI T2 such that I

x1,Ej=1 f BLwk',J 1

for all peT,.

LL

Pick pi e Ti, i = 1, 2, ..., with PI < P2 < -. Then if j < i, we have xP,, x,,, E B(cokj , 1 /j), so d(xp, , XP) < 2/j - 0 as j - oo. Thus {xp,) is a Cauchy sequence, hence converges (to a point of K, since K, is closed). Therefore {xp} has a subsequence converging to a point of K,, so K, is compact. I

We now apply the Daniell theory to obtain theorems on representation of positive linear functionals in a topological context. Theorem. Let n be a compact Hausdorff space, and let E be a positive linear functional on C(Q), with E(l) = 1. There is a unique probability measure P on sd(Q) such that E(f) = In f dP for all f e C(CZ). 4.3.9

PROOF. Let L = C(Q). If fn e L, fn 10, then f -+0 uniformly (this is Dini's theorem). For given b > 0, we have fl! = Un I {f,, < S}; hence by compactness, N

0 = U {f, < S}

for some N

n=1

= { fN < S}

by montonicity of {fj.

Thus n >_ N implies 0 < f"(w) < fN(co) < b for all co, proving uniform convergence.

Thus if S > 0 is given, eventually 0 < f" < S, so 0 < E(f") < E(S) = S. Therefore E(fn) 10, hence E is a Daniell integral. The result follows from 4.2.9. 1

182

4

THE INTERPLAY BETWEEN MEASURE THEORY AND TOPOLOGY

A somewhat different result is obtained if we use the Daniell theory with hypothesis B.

Theorem. Let S2 be a compact Hausdorff space, and let E be a positive linear functional on C(Q), with E(l) = 1. There is a unique probability measure P on R(S2) such that 4.3.10

(a) E(f) = fa f dP for all f e C(O), and (b) for all A e R(S2),

P(A) = inf{P(V): V z) A,

V open)

or equivalently,

P(A) = sup{P(K) : K (-_ A, K compact). (" Compact" may be replaced by " closed " since n is compact Hausdorff.) PROOF. Let L = C(c2). If {f, n e D} is a net in L and f 10, then, as in 4.3.9, for any 6 > 0 we have fl = U j E F { f j < S} for some finite set F c D. If N e D and N z j for all j e F, then n = { fN < S) by monotonicity of the net. Thus n >_ N implies 0 < f < fN, < S, and it follows as in 4.3.9 that 10.

By 4.2.10 there is a probability measure P on a(L") = a(lr) such that E(f) = fn f dP for all f e L. But in fact ## is the class of open sets, so that a(lr) = R(Q). For if f e L", there is a net of nonnegative continuous functions

f T f; hence for each real a, {f > a) = U f f, > a) is an open set. Thus if G e 9, then IG e L", so that G = (Ia > 0) is open. Conversely if G is open and co e G, there is a continuous function f.: Q -+ [0, 11 such that fjw) = 1 and f,,, = 0 on G`. Thus Ic = supw.fu so that if for each finite set F c G we define gF = max{ fu, : w e F}, and direct the sets F by inclusion, we obtain a mono-

tone net of nonnegative continuous functions increasing to I.. Therefore IGeL", so that Ge4. Thus we have established the existence of a probability measure P on 9(0) satisfying part (a) of 4.3.10; part (b) follows since P = µ* on a(W), and V is the class of open sets. To prove uniqueness, let P' be another probability measure satisfying (a) and (b) of 4.3. 10. If we can show that P' satisfies 4.2.10(b), it will follow from be a net of open sets the uniqueness part of 4.2.10 that P' = P. Thus let with G T G; since G is the union of the G,,, G is open. By hypothesis, given

S > 0, there is a compact K e G such that P'(G) s P'(K) + S. Now G. U K` T G U K` = fl; hence by compactness and the monotonicity of Gm U K` = SZ for some m, so that K e G.. Consequently,

P'(G) T P'(G).

4.3 MEASURES ON TOPOLOGICAL SPACES

183

Property (b) of 4.3.10 is often referred to as the "regularity" of P. Since the word "regular" is used in so many different ways in the literature, let us state exactly what it will mean for us. Definitions. If p is a measure on .(S)), where S2 is a normal topological space, p is said to be regular if for each A E R(S2), 4.3.11

p(A) = inf{p(V): V

A,

V open)

and

p(A) = sup{p(C): C c A,

C closed).

Either one of these conditions implies the other if p is finite, and if in addition, .0 is a compact Hausdorff space, we obtain property (b) of 4.3.10. If p = p+ - jr is a finite signed measure on N(Q), S2 normal, we say that p is regular iff p+ and p- are regular (equivalently, if the total variation (p I is regular). The following result connects 4.3.9 and 4.3.10. 4.3.12

Theorem. If P is a probability measure on sa1(S2), fl compact Haus-

dorff, then P has a unique extension to a regular probability measure on eJ(1). PROOF. Let E(f) = In f dP, f E C(i2). Then E is a positive linear functional on L = C(S2), and thus (see the proof of 4.3.10) if f f.) is a net in L decreasing to 0, then j 0. By 4.3.10 there is a unique regular probability measure P' on

'(t2) such that In f dP = fn f dP' for all f e L. But each f in L is measurable: (S), d) - (R, V(R)), hence by 1.5.5, In f dP' is determined by the values of P'

on Baire sets. Thus the condition that f a f dP = In f dP' for all f e L is equivalent to P = P' on W(Q), by the uniqueness part of 4.3.9. 1 In 4.3.9 and 4.3.10, the assumption E(l) = 1 is just a normalization, and if it is dropped, the results are the same, except that "unique probability measure" is replaced by "unique finite measure." Similarly, 4.3.12 applies equally well to finite measures.

Now let fl be a compact Hausdorff space, and consider L = C(fl) as a vector space over the reals; L is a Banach space with the sup norm. If E is a positive linear functional on L, we can show that E is continuous, and this will allow us to generalize 4.3.9 and 4.3.10 by giving representation theorems for continuous linear functionals on C(O). To prove continuity of E, note that if f e L and If 11 _< 1, then -1 < f(co) < 1 for all co; hence -E(l) :5 E(f)

184

4 THE INTERPLAY BETWEEN MEASURE THEORY AND TOPOLOGY

E(l), that is, I E(f) I < E(1). Therefore IIEII < E(l); in fact IIEII = E(l), as may be seen by considering the function that is identically 1. The representation theorem we are about to prove will involve integration with respect to a finite signed measure y = p+ - u-; the integral is defined in the obvious way, namely,

f of dp = f .f dµ+

- f of d9

,

assuming the right side is well-defined. 4.3.13 Theorem. Let E be a continuous linear functional on C(SI), SI compact Hausdorff. (a)

There is a unique finite signed measure p on d(Q) such that E(f) _

In fdpfor all feCO). (b) There is a unique regular finite signed measure A on R(Q) such that E(f) = In f dA for all f c- C(SI).

Furthermore, any finite signed measure on .01(SI) has a unique extension to a regular finite signed measure on .(SI); in particular A is the unique extension of p. PROOF. The existence of the desired signed measures y and A will follow from

4.3.9 and 4.3.10 if we show that E is the difference of two positive linear functionals E+ and E-. If f Z 0, f e C(SI), define

E+(f) = sup(E(g): 0
g e C(S))}.

If f < M, then E+(f) 5 MIIEII < oo by continuity of E; hence E+ is finite. If 0:!g g ; < f;, i = 1, 2, with all functions in C(Q), then E(9i) + E(g2) = E(g, + 92) <- E+(.f, +f2).

Take the sup over g, and g2 to obtain E+(f,) + E+(f2) < E+(f, +f2). Now if O :g g<.f, +.f2 ,.fi,.f2 >: 0, define g, = 9 A f,, 92 = 0V(9 -f1)

Then 0 < g, <-.fi, 0 <- 92:5f2, and 9=9,+92. Thus E(g) = E(g,) + E(92) :!g E+(fl) + E+(f2); hence E+(f, +f2) < E+(f,) + E+(f2), and consequently

E+(f, +f2) _ E+(fi) + E+(f2)

(1)

Clearly E+(af) = aE+(f) if a 0. Thus E+ extends to a positive linear functional on C(SI). Specifically, if f =f + -f -, take E+(f) = E+(f+) E+(f-). If f can also be represented as g - h, where g and h are nonnegative

4.3

185

MEASURES ON TOPOLOGICAL SPACES

functions in C(c2), then E+(f +) - E''(f -) = E+(g) - E+(h) by (1), so the extension is well-defined. If f >- 0, f c C(Q), define

E-(f) = -inf{E(g): 0
E-(f) _ -inf{E(f) - E(f -g): 0
g e C((2)}

-E(f) + sup(E(h): 0 < h < f, h e C(Q)}

_ -E(f) +

E+(f).

Thus E=E+-E-. To prove that p is unique, assume that t is a finite signed measure on .W(S2) such that In f dr = 0 for all f e C(Q). Then So f dT+ = In f dT- for all

f e C(Q), hence T+ = t- (so t = 0) by 4.3.9. Uniqueness of A is proved similarly, using 4.3.10.

Now if p is any finite signed measure on .4(Q), E(f) = fn f dp defines a continuous linear functional on C(Q), so by what we have just proved there is a unique regular finite signed measure A on 94(Q) such that Sa f du = In f dl for all f e C(Q). But as in 4.3.12, this condition is equivalent to p = A on .W(Q), so p has a unique extension to a regular finite signed measure A on 94(Q). 1 Theorems 4.3.9, 4.3.10, and 4.3.13 are referred to as versions of the Riesz representation theorem. [This name is also given to the somewhat different result 3.3.4(a).] A result of this type for complex-valued functions and complex measures is given in Problem 6. We are going to show that II E 11 = I µ (i2)

A 1 (Q) in 4.3.13. To do this

we need a result on the approximation of Borel measurable functions by continuous functions. Theorem. Consider the measure space (S2, F, p), where Q is a normal topological space, 51-' = 94(Q), and p is a regular measure on F. If 0 < p < oo, 4.3.14

s > 0, and f e L"(Q, F, p), there is a continuous complex-valued function g e L°(S2, .°r, p) such that Ilf - gllp < e; furthermore, g can be chosen so that sup I g I <_ sup I f 1. Thus the continuous functions are dense in L". PROOF. This is done exactly as in 2.4.14, except that the application of Prob-

lem 12, Section 1.4, is now replaced by the hypothesis that p is regular. I

4

186

4.3.15

THE INTERPLAY BETWEEN MEASURE THEORY AND TOPOLOGY

Theorem. In the Riesz representation theorem 4.3.13, IIEII = I N I (Q)

_

I21(L)).

PROOF. If f e C(S2), E(f) = fa f d2 = Jn f d l+ - fa f d2 -; hence

IE(f)I < fnIll where II

II

d2+

+fnIJI

d2 = fn IfI dI21
is the sup norm. Thus IIEII < IAI(Q)

Now let A,, ... , A,, be disjoint measurable subsets of S2, and define f = L;=, MA,, where x; = 1 if 2(A) > 0, and x; = -1 if 2(A) < 0. By 4.3.14, if s > 0, there is a continuous function g such that Ilgll <- 1 and Jn If - g J dI 2I < e; hence E(g) = Ja g d2 > f a f dl - s. But fn

f d2 =

I2(A) I 1

hence

IIEII >- E(g) >- Y I2(A;)l - c.

But Y;_, 12(A;) J may be taken arbitrarily close to 12I (b2) (see Problem 4, Section 2.1); hence IIEII >- J 2I (S2) Since 12I (f) = I p I (fl), the result follows. I

If S2 is compact Hausdorff and M(0) is the collection of finite signed measures on d(L) (or equally well the regular finite signed measures on 2(f2)), Theorems 4.3.13 and 4.3.15 show that the map E -+,u (or E -+ A) is an isometric isomorphism of the conjugate space of C(S2) and M(il), where the norm of an element u e M(fl) is taken as I u I (0).

We close this section with some further results on approximation by continuous functions.

Theorem. Let u be a regular finite measure on normal. If f is a complex-valued Borel measurable function on S2 and 6 > 0, there is a continuous complex-valued function g on S2 such that 4.3.16

u{w:f(w) # g(o)} < 6. Furthermore, it is possible to choose g so that sup I g 1 < sup I f I .

PROOF. First assume f is real-valued and 0 5f < 1. If h"(w) = (k - 1)2-" when (k - 1)2-" < f(w) < k2-", k = 1, 2, ... , n2", h"(w) = n when f(w) >- n, the h" are nonnegative simple functions increasing to f. Let f" = h" - h"_,,

4.3

MEASURES ON TOPOLOGICAL SPACES

187

n = 1, 2, ... (with ho = 0), so that f = Y , f". Note that f" has only two possible values, 0 and 2-". If A" _ {f" # 01, let C. be a closed subset of A,, and V,, an open overset of A. such that µ(V" - C") < 62-". Since S2 is normal, there is a continuous g": S2 -+ [0, 1] such that g" = I on C. and g" = 0 off V". If g = Yn , 2-"g", then by the Weierstrass M-test, g is a continuous map of n into [0, 1]. We claim that if w 0 U' , (V" - C"), a set of measure less than 6, then f(w) = g(w). To see this, observe that for each n, w e C. or co 0 V.. If co e C,, c A,,, then 2-"g"(co) = 2-" = f"(co), and if co 0 V", then 2-"g"(co) = 0 = "(w) since co 0 A,,. This proves the existence of g when 0 < f < I ; the extension to a complex-

valued bounded f is immediate. If f is unbounded, write f = fI{ifI<"} + .f I { I f 1 >_ "} =11 + f2, where f, is bounded and µ{ f2

0} = µ{ I f I > n}, which

can be made less than 612 for sufficiently large n. If g is continuous aid µ{ f, # g} < 6/2, then µ{ f g} < 6, as desired. Finally, if If I < M < oo, and g approximates f as above, define gl(w) _ g((9) if I g(w) l < M, g1(w) = Mg(w)I I g(co) I if I g(w) I > M. Then g, is continuous, I g, I < M, and f(w) = g(w) implies I g(w) I < M; hence gl(w) _ g(w) = f (w). Therefore µ{ f 96 g1} < p{f # g} < 3, completing the proof.

4.3.17

Corollaries. Assume the hypothesis of 4.3.16.

(a) There is a sequence of continuous complex-valued functions f" on 0 converging to f a.e. [µ], with If. I < sup If I for all n. (b) Given e > 0, there is a closed set C c S2 and a continuous complexvalued function g on S2 such that µ(C) ? µ(S2) - e and f = g on C, hence the restriction off to C is continuous. If p has the additional property that µ(A) = sup{p(K): K e A, K compact) for each A e 64(Q), then C may be taken as compact. PROOF. (a) By 4.3.16, there is a continuous function f" such that If. I < M = sup If I and µ{ f" 96 f } < 2 -". If A,, = {f, f) and A = lim sup" A" , then

p(A) = 0 by the Bore[-Cantelli lemma. But if co 0 A, then f(w) =f(w) for sufficiently large n. (b) By 4.3.16, there is a continuous g such that µ{ f # g} < e/2. By regu-

larity of µ, there is a closed set C e f f = g} with µ(C) >- µ{ f = g} - e/2. The set C has the desired properties. The proof under the assumption of approximation by compact subsets is the same, with C compact rather than closed. Corollary 4.3.17(b) is called Lusin's theorem.

4

188

THE INTERPLAY BETWEEN MEASURE THEORY AND TOPOLOGY

Problems I.

2.

Let F be a closed subset of the metric space Q. Define f (w) = e-'(°" F) where d(w, F) = inf{d(w, y): y e F}. Show that the f are continuous and f 1 IF. Use this to give a direct proof (avoiding 4.3.2) that in a metric space, the Baire and Bore] sets coincide. Give an example of a measure space (0, #;, µ), where 12 is a metric space and .F = R(S2), such that for some A e .F, u(A) 96 sup{p(K): K c A, K compact}.

3.

In 4.3.14, assume in addition that S2 is locally compact, and that µ(A) _ sup{p(K): K compact subset of A} for all Borel sets A. Show that the

continuous functions with compact support (that is, the continuous functions that vanish outside a compact subset) are dense in L°(S2, W, g), 0 < p < co. Also, as in 4.3.14, if f e LP is approximated by the continuous

function g with compact support, g may be chosen so that supIgI 5 supifI. 4.

Let fl be a normal topological space, and let H be the smallest class of real-valued functions on a that contains the continuous functions and is closed under pointwise limits of monotone sequences. Show that H is the class of Baire measurable functions, that is, H consists of all f: (0, sd) - (R, R(R)) (use 4.1.4). (b) If H is as in part (a) and a(H) is the smallest a-field 5 of subsets of n (a)

making all functions in H measurable (relative to 5 and R(R)), show that a(H) = s?(f); hence a(H) is the same as a(C(a)). 5.

Let S2 be a normal topological space, and let Ko be the class of all continuous real-valued functions on Q. Having defined K. for all ordinals jf less than the ordinal a, define

K,, =U{K,,:/3
Baire measurable functions. 6.

Let E be a continuous linear functional on the space C(Q) of all continuous complex-valued functions on the compact Hausdorff space Q. Show that there is a unique complex measure p on d(S2) such that E(f) = !Q f du for all f e C(a), and a unique regular complex measure A on R(Q) such that E(f) = fn f dA. for all f e C(fl). Furthermore, IIEII =

4.4

MEASURES ON UNCOUNTABLY INFINITE PRODUCT SPACES

189

p I (Q) = 12I (S2), so that C(S2)* is isometrically isomorphic to the space of complex measures on .4(S2), or equally well to the space of regular complex

measures on R (Q). (If p = p, + iµ2 is a complex measure, in f dp is defined as J0 f dµ, + '10 f 4142 provided f is integrable with respect to p1 1

and p2. Also, p is said to be regular if p, and p2 are regular; since I P1 1, I P2 I < I p 1 <- 191 I + I P2 1, this is equivalent to regularity of I P See Section 2.2, Problem 6, and Section 2.4, Problems 10 and 11 for the basic properties of complex measures.) 7. Let M(S), 9) be the collection of finite signed measures on a a-field .9 of subsets of 0. Show that if we take JIM 11 = l y I (fl), it e M(S), JF), M(f), f) becomes a Banach space. (A similar result holds for the collection of complex measures.)

4.4 Measures on Uncountably Infinite Product Spaces

In Section 2.7 we considered probability measures on countably infinite product spaces. The result may be extended to uncountable products if certain topological assumptions are made about the individual factor spaces.

The product of uncountably many a-fields is formed in essentially the same way as in the countable case.

Definitions and Comments. For t in the arbitrary index set T, let (Q, Jr,) be a measurable space. Let Fit E T Q, be the set of all functions 4.4.1

w = (w(t), t e T) on T such that w(t) e !Q, for each t e T. If t1, ..., t" e T and B" e fl_ , fl,, , we define the set B"(t,, ... , t") as {w a f , E TO,: (w(tl), .... w(t")) e B"}. We call B"(t,, ... , t") the cylinder with base B" at (t,, ... , t"); the cylinder is said to be measurable if B" e fi=, If B" = B, x x B,,, the cylinder is called a rectangle, a measurable rectangle if B; a i = 1, ... , n. Note that if all Q, = Q, then Fit E T 0, = S T, the set of all functions from T

too.

For example, let T = [0, 1], Q, = R for all t e T, B2 = ((u, v): u > 3, 1
B2(4, j)_(xaRT:x(f)>3, 1 <x(j)<2} [see Fig. 4.1, where x, a B2(f, *) and x2 0 B2(-, i)]. Exactly as in 2.7.1, the measurable cylinders form a field, as do the finite

disjoint unions of measurable rectangles. The minimal a-field over the measurable cylinders is denoted by lit c T F,, and called the product of the o-fields .F,. If Q, = S and 9, = So for all t, fl, c T 9, is denoted by ,9T Again as in 2.7.1, fl, E,. 9, is also the minimal a-field over the measurable rectangles.

4 THE INTERPLAY BETWEEN MEASURE THEORY AND TOPOLOGY

190

3

2

x,

3

0

2

4

Figure 4.1.

We now consider the problem of constructing probability measures on be a finite , The approach will be as follows: Let v = {t1, ..., < t . (If T is not a subset of R, some fixed subset of T, where t, < t2 < total ordering is put on T.) Assume that for each such v we are given a probability measure P on J];_, . ',,; is to represent P{w e f, e T ill: (w(t1), ... , w(tn)) e B}. We shall require that the P be "consistent"; to see what kind of consistency is needed, consider an example. Suppose T is the set of positive integers and 0, = R, F, = R(R) for all t. Suppose we know P1 2345(B5) = P{w: (w,, w2, w3, w4i ws) E B5) for all B5 a R(R5). Then P{w: (w2, cu3) e B2) = P{uw (w1, (02, (03, w4, (05) e R x B2 x R2} = P12345(R x B2 x R2), B2 a R(R2). Thus once probabilities of sets involving the first five coordinates are specified, probabilities of sets inH, E r

.

volving (C02, w3) [as well as (Co,, (03 , w4), and so on], are determined. Thus the original specification of P23 must agree with the measure induced from

P12345 We are going to show that under appropriate topological assumptions, a consistent family of probability measures P determines a unique probability measure on fl, E T ."T, .

Now to formalize: If v = {t,, ..., to}, t, < .. < t , the space (fir

52,,,

If u = (ti,, ..., tik) is a nonempty subset of v and y = (y(t1), ... , y(tn)) a fl,,, the k-tuple (y(til),... , y(tik)) is denoted by yn . H1=1 .F,) is denoted by

Similarly if w = (w(t), t e T) belongs to fit E T S2 the notation to, will be used for (w(t1), ..., w(tn)). the projection of P on F. is the If PE is a probability measure on probability measure [n,,(P,,)](B) =

n

defined by e a,,: yn e B),

B e .F..

4.4

191

MEASURES ON UNCOUNTABLY INFINITE PRODUCT SPACES

Similarly, if Q is a probability measure on fl E T F , the projection of Q on SS is defined by r

-7

l

[7ru(Q)](B) = Q(w c- 11 Sgt: w e B} = Q(B(v)), tET I

B E F,,.

1111

We need one preliminary result.

Theorem. For each n = 1, 2, ..., suppose that .y" is the class of Borel sets of a separable metric space fl.. Let n = H. fl., with the product topology, 4.4.2

and let F = a(Q). [Note that S2 is metrizable, so that the Baire and Borel sets of S2 coincide; we may take

-

d(x, y) _

n

d"(x", y.) 2" 1 + d"(x", y")

where d" is the metric of fl,,. Also, if each fl. is complete, so is S2.]

Then F is the product a-field f" F. PROOF. The sets {w E S2: w, e A,, ... , (0" E A"}, n = 1, 2, ... , where the A;

range over the countable base for S2i (recall that separability and second countability are equivalent in metric spaces), form a countable base for 0. Since the sets are measurable rectangles, it follows that every open subset of 0

belongs to f" F"; hence .F c [" F.. On the other hand, for a fixed positive integer i let' = (Be R(S2;): {co e n: w; E B} E F}. Then ct° is a u-field containing the open sets of 0;, hence ' = R(f). In other words, every measurable rectangle with one-dimensional base belongs to F. Since an arbitrary measurable rectangle is a finite intersection of such sets, it follows that

We are now ready for the main result. 4.4.3

Kolmogorov Extension Theorem. For each t in the arbitrary index set T,

let 0, be a complete, separable metric space, and F, the Borel sets of Sgt . Assume that for each finite nonempty subset v of T, we are given a probability measure P,, on Assume the P are consistent, that is, P. for each nonempty u t= v. Then there is a unique probability measure P on .F = fit

cT

F, such that

7ct,(P) = P,, for all v.

PROOF. We define the hoped-for measure on measurable cylinders by P(B"(v)) =

B" a F,, .

192

4

THE INTERPLAY BETWEEN MEASURE THEORY AND TOPOLOGY

We must show that this definition makes sense since a given measurable cylinder can be represented in several ways. For example, if all S2, = R and B2 = (- oo, 3) x (4, 5), then B2(tl, 12) = {w: (0(t,) < 3,

4 < w(12) < 5) _ {w: w(t,) < 3, 4 < w(t2) < 5, = B3(t,, t2 i 1, )

w(13) a R}

where B3 = (- oo, 3) x (4, 5) x R.

It is sufficient to consider dual representation of the same measurable cylinder in the form B"(v) = B"(u) where k < n and u e v. But then

by the consistency hypothesis P"(B") = by definition of projection. = PJy e 52,,: y,, e Bk)

But the assumption B"(v) = B'(u) implies that if y e Q,,, then y e B" if y a Bk, hence P"(B") = P is well-defined on measurable cylinders; the class .moo of measurable cylinders forms a field, and a(.Fo) = ,F. Now if A,, ..., A. are disjoint sets in moo, we may write (by introducing

extra factors as in the above example) A; = B;"(v), i = 1, ..., m, where v = {t,, ..., r"} is fixed and the B;", i = 1, ..., m, are disjoint sets in SF, Thus

P(`U A;) = PI `U1B;"(v)) m

UB, n

Pv +

by definition of P

=1

P is a measure m

_

P(A;)

again by definition of P.

Therefore P is finitely additive on fo . To show that P is countably additive on .tea , we must verify that P is continuous from above at 0 and invoke 1.2.8(b). The Caratheodory extension theorem (I.3.10) then extends P to F. Let Ak, k = 1, 2, ... be a sequence of measurable cylinders decreasing to 0. If P(Ak) does not approach 0, we have, for some e > 0, P(A,,) >_ e > 0 for all k. Suppose Ak = B"k(vk); by tacking on extra factors, we may assume that the numbers nk and the sets vk increase with k. By 4.4.2, each S2"k is a complete, separable metric space and `Fvk =

*010.

It follows from 4.3.8 that we can find a compact set C"k c B"k such that Pvk(B"k - C"k) < Define Ak' = C"k(vk) e Ak. Then P(Ak - Ak') e/2k+1.

4.4 MEASURES ON UNCOUNTABLY INFINITE PRODUCT SPACES

193

P,,,, (B "k - C-) < e/2k+'. In this way we approximate the given cylinders by cylinders with compact bases. Now take nA2n...n,gk=Ak.

Dk=A1'nA2'n...nAk'cA,

Then P(Ak - Dk) = Pi Ak

U Aj) < > P(Ak (-i A;`) 1=1

1=1

k

k

Y P(A, - A;) < 1=1

e/21+1 < e/2. 1=1

Since Dk c: Ak' c Ak, P(Ak - Dk) = P(A,,) - P(Dk), consequently P(Dk) > P(Ak) - e/2. In particular, Dk is not empty.

Now pick xkeDk,k=1,2,....SayA1'=t") =C"'(v1)(note

all Dk c A 1'). Consider the sequence

(L xt,, ... , xt",), 1

(X2, (x,,, ... , X2,), xr",),

3 (4,, ... , x ),

that is, x/'/,, x2,,

Since the x', belong to C"', a compact subset of 1 0,,, we have a convergent subsequence x,,;" approaching some x,,, e C"'. If A2' = C"2(v2) (so Dk c A2' for k z 2), consider the sequence X11 1, e,", ... e C"2 (eventually), and extract a convergent subsequence xv2" -+ e C"2. Note that (x'2"), , = x12"; as n -+ oo, the left side approaches and since {r2n} is a subsequence of {r1 }, the right side approaches x,,,. Hence (xU2)V, = x,,,.

Continue in this fashion; at step i we have a subsequence x';" -- x,,, e C"',

and

for j < i. x,, for all j = 1, 2, ... (such a

(x,,)f =

Pick any co e fl: E T 0, such that j < i). Then co,,, e C"' for each j; hence

choice is possible since (x,,,)vj =

we nA,'c nA,=Qf,

j=t a contradiction. Thus P extends to a measure on F, and by construction, P for all v. j=1

Finally, if P and Q are two probability measures on .` such that for all finite v c T, then for any B" e P(B"(v)) = [ir,,(P)](B") =

Q(B"(v)).

Thus P and Q agree on measurable cylinders, and hence on Sr by the uniqueness part of the Caratheodory extension theorem.

4

194

THE INTERPLAY BETWEEN MEASURE THEORY AND TOPOLOGY

Problems

In Problems 1-7, (S) .F,) is a measurable space for each t e T, and 1.

Let p, be the projection map of () onto SZ that is, p,(w) = w(t). If (S, 9o) is a measurable space and P. S -+ S2, show that f is measurable iff p, o f is measurable for all t.

2.

If all n, = R, all F, = R(R), and T is a (nonempty) subset of R, how many sets are there in F? Show that if B e y , then membership in B is determined by a countable

3.

number of coordinates, that is, there is a countable set Tae T and a set B0 E FT, = j 1, E T. 'F, such that co e B if CT,, E B0 , where OT0 = (w(t), t E To). 4.

If T is an open interval of reals and fl, = R (or R), F, = .1(R) (or R(R)) for all t, use Problem 3 to show that the following sets do not belong to ,F: (a) {co: co is continuous at to}, where to is a fixed element of T. (b) {w: SUP. t;b w(t) < c}, where cc R and [a, b] c T

5.

Assume each Q, is a compact metric space, with .F, the Baire (= Borel)

subsets of 0, . Then by the Tychonoff theorem, f2 is compact in the topology of pointwise convergence. Show that F is the class of Baire sets of fl, in other words, .<J, E T ,it) = [: E T d411), as follows: (a)

If A = {co e n: w(t) E F}, F a closed subset of Q,, show that A = {w a ): f(co) = 0) for some f c- C(S)). Conclude that F a .ra1(S2).

(b)

6.

Use the Stone-Weierstrass theorem to show that the functions

f e C(Q) depending on only one coordinate [that is, f(w) = g,(co(t)) for some t, where g, e C(S2,)] generate an algebra that is dense in C(S2). Conclude that 4(S2) c ,F. Assume T is an open interval of reals, and (f2 F,) = (R, R(R)) for all t; thus n = RT, .f = A(R)T. Let A = {co e 92: co is continuous at t0), where to is a fixed point of T. (a) Show that A is an F,a (a countable intersection of F, sets); hence A e R(f2). (b)

7.

Show that A 0 d(fl), so that we have an example of a compact

Hausdorff space in which the Baire and Borel sets do not coincide. (Alternative proof of the Kolmogorov extension theoremt) Assume the hypothesis of 4.4.3, with the stronger condition that each fl, is a compact metric space. Put the topology of pointwise convergence on Q. (a) Let A be the set of functions in C(S2) that depend on only a finite number of coordinates; that is, there is a finite set v c T and a cont Nelson. E., Ann. of Math. 69, 630 (1959).

4.4 MEASURES ON UNCOUNTABLY INFINITE PRODUCT SPACES

195

tinuous g: 12 -+ R such that f(w) =

w e 12. Use the StoneWeierstrass theorem to show that A is dense in C(12). (b) If f e A, define E(f) = fag dP,,. Show that E is well-defined and extends uniquely to a positive linear functional on C(Q). Since E(l) = 1, the Riesz representation theorem 4.3.9 yields a unique probability

measure P on An) (= f I, E T .F, by Problem 5) such that E(f) _ (c)

fn f dP for all f e C(Q). Let v be a fixed finite subset of T, and let H be the collection of all functions g: (fl, F.,) --+ (R, E(R)) such that if f(w) = co e 12,

then J, f dP = in g

8.

Show that IA e H for each open set

A e 12,,, and then show that H contains all bounded Borel measurable functions on (Q., ,F.). Conclude that P,,. (Uniqueness of P is proved as in 4.4.3.) The metric space Q is said to be Borel equivalent to a subset of the metric space Q' i1T there is a one-to-one map f: Q -+ Q' such that E = f (Q) e R(12)

and f and f -' are Bore] measurable. [Measurability of f -' means that f -' : (E, R(E)) -+ (0, R(Q)).] (a) Let Q he a complete separable metric space with metric d. (It may be assumed without loss of generality that d(x, y) < 1 for all x, y, since the metric d' = d1(1 + d) induces the same topology as d and is also complete.] Denote by [0, 1]°° the space of all sequences of real numbers with components in [0, 1], with the topology of pointwise convergence. (This space is metrizable; explicitly, we may take the metric as °° I I xn do(X, y) = nE 2n 1 + Ix,,

yn I

If D = {w,, (02 , ...) is a countable dense subset of 12, define ... ). Show that f is continuous and one-to-one, with a continuous inverse.

f: 12 --+ [0, 11' by f (w) = {d(w, wn), n = 1, 2,

(b) Show thatf(1)) is a Borel subset of [0, 1]°°; thus n is Borel equivalent

to a subset of [0, 1]'. (c)

Let S = {0, 1}, and let 6" consist of all subsets of S. Show that there is a map g: ([0, 1), 4[0, 1)) --+ (S°°, .9"°) such that g is one-to-one,

g[0, 1) e Y', and both g and g`' are measurable. (d) Show that ([0, 1], R[0, 1]) and (S°°, S'°) are equivalent, that is,

there is a map h: [0, 1] -+ S°° such that h is one-to-one onto, and h and h-' are measurable. (e)

If (fl, Pn) is equivalent to (Se, 9n), with associated map hn, n = 1, 2, ..., show that (ln 12n, H. ,F.) is equivalent to (fln Sn, r l. .9'n). Thus by (d), ([0, 1]°°, (-4[0, 11)°°) is equivalent to (S'°, Y').

4 THE INTERPLAY BETWEEN MEASURE THEORY AND TOPOLOGY

196

Now by 4.4.2, (9[0, 1 ])°° is the minimal a-field over the open sets of [0, 1]Z, that is, (-4[0, 1])°° is the class of Borel sets _4([0, 1]°°). It follows from these results that if 0 is a complete, separable metric space, S1 is Borel equivalent to a (Bore]) subset of [0, 1].

4.5 Weak Convergence of Measures

By the Riesz representation theorem, a continuous linear functional on C(S2), where S2 is compact Hausdorff, may be identified with a regular finite signed measure on R(f2). Thus if {p,,} is a sequence of such measures, weak*

convergence of the sequence to the measure p means that In f dµ - In f dp for all f e C(O). In this section we investigate this type of convergence in a somewhat different context; 91 will be a metric space, not necessarily compact, and all measures will be nonnegative. The results form the starting point for

the study of the central limit theorem of probability.

Theorem. Let p, Pi, 102, ... be finite measures on the Borel sets of a metric space Q. The following conditions are equivalent:

4.5.1

(a) $ n f dun -> Jn f dp for all bounded continuous f: 51- R. (b) lim inf, In f dun >_ fa f dp for all bounded lower semicontinuous

f.n->R.

(b') urn sups . oo in f dun < Jn f du for all bounded upper semicontinuous f: S2 - R. (c) 5 n f dun -f n f du for all bounded f: (s2, (fz)) - (R, R(R)) such that f is continuous a.e. [p]. (d) lim inf, pn(A) >_ p(A) for every open set A (-_ S2, and un(Q) -' u(n) (d') lim sup, u (A) 5 u(A) for every closed set A c f , and

inn0 -.41).

(e) u,,(A) - u(A) for every A e.4(Q) such that p(aA) = 0 (aA denotes the boundary of A).

PROOF. (a) implies (b):

If g < f and g is bounded continuous,

liminf f fdp,,_liminf f n-oo

12

n

by (a).

But since f is lower semicontinuous (LSC), it is the limit of a sequence of con-

tinuous functions, and if If < M, all functions in the sequence can also be

taken less than or equal to M in absolute value. (See the appendix on general topology, Section A6, for the basic properties of semicontinuous

4.5

WEAK CONVERGENCE OF MEASURES

197

functions.) Thus if we take the sup over g in the above equation, we obtain (b). (b) is equivalent to (b'): Note that f is LSC if -f is upper semicontinuous (USC). (b) implies (c): Let f be the lower envelope off (the sup of all LSC

functions g such that g < f) and f the upper envelope (the inf of all USC functions g such that g >: f). Since f(x) = lim infy.x f(y) and f (x) = lim sup.,-.,f (y), continuity off at x implies f (x) =f(x) =J(x). Furthermore, f is LSC and f is USC. Thus if f is bounded and continuous a.e. [µ],

f f dµ = f f dµ 5 Jim inf f f dµn n

n-oo

s1

n

Slim inf f f dµn n

by (b)

since f 5 f

S lim sup f f dµn a n- 00

Slim sup f f dµn n-c

n

since f < f

S fn fdi

by (b')

= f f dp,

proving (c).

(c) implies (d): Clearly (c) implies (a), which in turn implies (b). If A is open, then IA is LSC, so by (b), lim inf,,-. pn(A) >- µ(A). Now In - 1, so pn(Q) -+ fi(Q) by (c)-

(d) is equivalent to (d'): Take complements. (d) implies (e): Let A° be the interior of A, A the closure of A. Then iim sup pn(A) S lim sup pn(A) S µ(A) n- oo

by (d')

n- 00

= µ(A)

by hypothesis.

Also, using (d), lim inf pn(A) >_ lim inf pn(A°) > p(A°) = p(A). n- ao

n- cc

Let f be a bounded continuous function on Q. If If < M, let A = {c a R: p(f -1{c)) # 0}; A is countable since the f -'{c} are disjoint and p is finite. Construct a partition of [- M, MI, say - M = to < t, < < tj = M, with t; 0 A, i = 0, 1, ..., j (M may be increased if necessary). If B, = {x: t; 5 f(x) < ti.,.1}, i = 0, 1, ..., j - 1, it follows from (e) that J-1 i-^1 /_ ti pn(Bi) - Y- ti y(Bi) i=0 (e) implies (a) :

i=0

4 THE INTERPLAY BETWEEN MEASURE THEORY AND TOPOLOGY

198

[Since f -'(ti, 1.+i) is open, of -' [ti, ti, I) c.f -'{ti, ti+i}, and {if -'{ti, ti+i} = 0 since ti, ti+, 0 A.] Now

fafdp -fnfdp

S

ti pn(Bi) J J dju. - Li=0 S2 2

+

I'Y-

+

[tiIt(Bi) - Jn f d4

t=0

ti p,,(B.) - y_ ti p(B.)I i=0

The first term on the right may be written as j_1

E

1

i=0

J Bt

(f(x) - ti)

-

and this is bounded by maxi(ti+i which can be made arbitrarily small by choice of the partition since p (S1) -+ p(fl) < oo by (e). The third term on the right is bounded by max.(ti+i - ti)p(S2), which can also be made arbitrarily small. The second term approaches 0 as n -+ co, proving (a). I 4.5.2

Comments.

Another condition equivalent to those of 4.5.1 is that

In f dp - fn f du for all bounded uniformly continuous f: it -* R (see Problem 1).

The proof of 4.5.1 works equally well if the sequence is replaced by a net. The convergence described in 4.5.1 is sometimes called weak or vague convergence of measures. We shall write

p

p are defined on e(R), there are corresponding

distribution functions F and F on R. We may relate convergence of measures to convergence of distribution functions.

Definition. A continuity point of a distribution function F on R is a point x e R such that F is continuous at x, or ± oo (thus by convention, oo and - oo are continuity points). 4.5.3

Theorem. Let p, Al, P2' ... be finite measures on £(R), with corresponding distribution functions F, F,, F2, .... The following are equivalent:

4.5.4

(a) p w' p.

(b) F (a, b] - F(a, b] at all continuity points a, b of F, where F(a, b] = F(b) - F(a), F(co) = lim, . F(x), F(-oo) = limx.._,,, F(x).

4.5

199

WEAK CONVERGENCE OF MEASURES

If all distribution functions are 0 at - oo, condition (b) is equivalent to the statement that F,,(x) -+ F(x) at all points x E R at which F is continuous, and F (oo) --, F(oo).

(a) implies (b): If a and b are continuity points of F in R, then (a, b] is a Borel set whose boundary has p-measure 0. By 4.5.1(e), µn(a, b] -+ µ(a, b], that is, F,,(a, b] -+ F(a, b]. If a = - oo, the argument is the same, and if b = oo, then (a, oo) is a Borel set whose boundary has µ-measure 0, and the proof proceeds as before. (b) implies (a): Let A be an open subset of R; write A as the disjoint PROOF.

union of open intervals I1, 12, .... Then lim inf µn(A) = lim inf Z .U.(") n- '0

k

Y lim inf Pn(Ik)

by Fatou's lemma.

n-co

k

Let E > 0 be given. For each k, let Ik' be a right-semiclosed subinterval of Ik such that the endpoints of Ik' are continuity points of F, and u(Ik') >- µ(Ik) - c2-k; the Ik' can be chosen since F has only countably many discontinuities. Then by (b). lim inf µn(Ik) >- lim inf µn(Ik) = P(IO') n-oo

n- 00

Thus

liminfµn(A) n- oo

- Y_ p(Ik) - Y_µ(Ik)-e=µ(A)-e. k

k

Since a is arbitrary, we have µn '-+,u by 4.5.1(d). I

Condition (b) of 4.5.4 is sometimes called weak convergence of the sequence {Fn} to F, written Fn - F. Problems 1.

(a)

If F is a closed subset of the metric space Q, show that IF is the limit

of a decreasing sequence of uniformly continuous functions fn, with O!gf.
uniformly continuous P. S2 - R. Show that in 4.5.1, µn-" µ if µn(A) - µ(A) for all open sets A such that µ(OA) = 0.

3.

Let f be a locally compact metric space. If A c Q, A is said to be bounded

if A c K for some compact set K.

200

4

THE INTERPLAY BETWEEN MEASURE THEORY AND TOPOLOGY

If p, pi, P2' ... are measures on R(fl) that are finite on bounded Borel sets, we say that p,, p if in f dp -+ f a f dp for all continuous functions f: 0 -+ R with compact support. (The support off is defined by supp f = {x: f(x) 0 0); note that f has compact support if f vanishes outside a compact set.) Show that the following conditions are equivalent: (a)

pn w* p.

(b) p (A) -> p(A) for all bounded Borel sets A with µ(8A) = 0. (c) 5- f dµ, -> f n f dp for all bounded f:.0 -. R such that f has compact support and is continuous a.e. [p]. 4.6

References

The approach to the Daniell integral and the Riesz representation theorem given in Sections 4.2 and 4.3 is based on Neveu (1965). If the hypothesis that L contains the constant functions is dropped in the development of the Daniell integral, the situation becomes more complicated. However, a representation theorem somewhat similar to 4.2.9 may be proved under the assumption that f e L implies I A f c- L. For a detailed proof of this result (known as Stone's theorem), see Royden (1968, Chapter 13). There are other versions of the Riesz representation theorem appropriate in a locally compact Hausdorff space Q. If E is a positive linear functional

on the set CjS2) of continuous real-valued functions on f2 with compact support, there is a unique regular measure p on the Borel sets such that E(f) = fn f dp for each fin QS2). Here, regularity means that p is finite on compact sets, y(A) = i-if{p(V): V A, V open) for all Borel sets A, and p(A) = sup(p(K): K e A, K compact) for all Borel sets A which are either open or of finite measure. Also, if E is a continuous linear functional on the Banach space C0(Q) of continuous real-valued functions on 12 that vanish at oo (f vanishes at oo if, given e > 0, there is a compact set K such that If I < s off K), there is a unique regular finite signed measure on the Bore] sets such that E(f) = in f dp for all f e C0(Q); furthermore, 1(E11 = I P1 ((2).

As in the compact case, the result may be extended to complex-valued functions and complex measures, as in Problem 6, Section 4.3. Proofs are given in Rudin (1966).

In probability theory, there is usually no reason to replace compact by locally compact spaces in order to take advantage of a more general version of the Riesz representation theorem. For if we have a random experiment whose outcomes are represented by points in a locally compact space, we simply compactify the space.

For an account of the theory of weak convergence of measures, with applications to probability, see Billingsley (1968) and Parthasarathy (1967).

Appendix on General Topology

Al

Introduction

The reader is assumed to be familiar with elementary set theory, including basic properties of ordinal and cardinal numbers [see, for example, Halmos (1960)]. Also, an undergraduate course in point-set topology is assumed; Simmons (1963) is a suitable text for such a course. In this appendix, we shall

concentrate on aspects of general topology that are useful in functional analysis and probability. A good reference for collateral reading is Dugundji (1966).

Before proceeding, we mention one result, which, although usually covered in a first course in topology, deserves to be stated explicitly because of its fundamental role in the construction of topological vector spaces (see 3.5.1). Throughout the appendix, a neighborhood of a point x is an open set containing x, an overneighborhood of x is an overset of a neighborhood of x.

Al.l Theorem. Let S2 be a set, and suppose that for each x e n, we are given a nonempty collection V(x) of subsets of f) satisfying the following: (a) x e each U e 'K(x). (b) If U1, U2 E'Y''(x), then U1 n U2 E 7''(x). (c) If U e y''(x) and U c V, then V e '4''(x). 201

202

APPENDIX ON GENERAL TOPOLOGY

(d) If U e 'V (x), there is a set V e 'Y'-(x) such that V c U and U E 'K(y) for each y e V.

Then there is a unique topology on fl such that for each x, Y'(x) is the system of overneighborhoods of x. PROOF. If such a topology exists, a set U will be open if U is an overneighborhood of each of its points, that is, U e 1"(x) for each x e U. Thus it suffices to show that .f = { U (-_ Q: U e 71(x) for each x e U} is a topology, and for each

x, the overneighborhood system . V (x) coincides with 'V(x). If U, V e .l, then by (b), U n V E 1'(x) for each x e U n V, so U n V e J. If U, e .f for each i e I, then (c) implies that U{U;: i e I} E ; (c) yields S2 e 9' as well. Since 0 e 9- trivially, .fT is a topology. If U E .K(x), there is a set V e with x e V c U. But then U e 'V (x) by (c). Conversely, if U e *(x), let W = {x' e U: U e Y''(x)}. If x' a W, then by (d), there is a set V E *(X') with V c U and U e 11(y) for each y e V. But then V c W, so by (c), W e Y' (x'); consequently, W e by definition of I, and furthermore x e W by (a). Thus if U e Y''(x), there is a set W e .T with x e W c U; hence U e .K(x). For the remainder of the appendix, V(x) will always stand for the collection of neighborhoods of x.

A2 Convergence

If Q is a metric space, the topology of f) can be described entirely in terms of convergence of sequences. For example, a subset A of S2 is closed n = 1, 2, ...} is a sequence of points in A and x - x, we have iff whenever x e A. Also, x e.4, the closure of A, if there is a sequence of points in A converging to x. This result does not generalize to arbitrary topological spaces. A2.1

Example. Let a be the first uncountable ordinal, and let S2 be the set of

all ordinals less than or equal to a (recall that for ordinals, a < b means a e b). Put the order topology on 0; this topology has as a base the sets S2 n (a, b) = {x e 92: a < x < b} where a and b are arbitrary ordinals.

We show that a belongs to the closure of f2 - {a}, but no sequence in S2 - {a} converges to a. If U is a neighborhood of a, then for some a, b, we have a r: S2 n (a, b) a U. Now a < a < b; hence a is countable, and therefore so is a + 1 . Thus a + 1 E U n (S2 - {a}), proving that U E 0 - {a}. But if x,, n 0 - {a}, n = 1, 2, ..., then each x is countable; hence so is c = sup x,,.

A2 CONVERGENCE

203

Since c < a, V = (c, a] is a neighborhood of a and x is never in V, so that the sequence cannot possibly converge to or. In order to describe an arbitrary topology in terms of convergence, we must consider objects more general than sequences. A sequence is a function on the positive integers; the desired generalization, called a "net" or " generalized sequence," is a function on a directed set.

Definitions. A directed set is a set D on which there is defined a preordering (a reflexive and transitive relation), denoted by <-, with the property that whenever a, b c- D, there is a c e D with a < c and A2.2

b < c. A net in a topological space f is a function from a directed set D into Q. A net will be denoted by {x, n e D} or simply by The net is said to converge to the point x if for every neighborhood U of x, the net is eventually in U, that is, there is an no e D such that x e U for all n e D such that

n>-n0. The basic relations between convergence and topology in metric spaces can now be generalized to arbitrary topological spaces. A2.3

Theorem. Let A be a subset of the topological space fi.

(a) A point xe S2 belongs to A if there is a net in A such that x - x. (b) A is closed if for every net in A such that x,, -+ x, we have x c- A. (c) A point x e r) is a cluster point of A (that is, every neighborhood of x contains a point of A other than x) if there is a net in A - {x} converging to X.

If x e A and x - x, let U E IW(x). Then x is eventually in U, in particular U n A 0, and thus x e A. Conversely, if x e A, then for each U e *(x), choose xv e U n A. Then {xv} becomes a net in A (with U:!5; V if PROOF. (a)

U=> V) and xv -x. (b) Suppose that A is closed. If x e A, x - x, then by (a), x e A; thus by hypothesis, x e A. Conversely, if A is not closed, let x e A - A. By (a) we can find x a A, x,, x. Since x 0 A, the result follows. (c) If x e A - {x}, x -> x, then if U E W(x), x is eventually in U, hence U n (A - {x}) 96 0, and thus x is a cluster point of A. Conversely, if x is a cluster point of A, then for each U e Qe'(x), choose xv c- U n (A - {x}); the xv form a net in A - {x} converging to x. Comments. The reason that sequences are not adequate in A2.3 is that in choosing a point xv in each neighborhood U of x, we are in general forced to make uncountably many choices. This would not be necessary if S2 were A2.4

APPENDIX ON GENERAL TOPOLOGY

204

first countable. (A first countable space is one with a countable base of neighborhoods at each point; that, is, if x e S2, there are countably many neighborhoods V1, VZ , ... of x such that for each neighborhood U of x, we have V. c U for some n; by replacing V. by fl : i V;, we may assume without loss of generality that Vn+1 (-- V. for all n.) In A2.3, the proof will go through if we

simply choose x e V for n = 1, 2, .... Thus in a first countable space, sequences are adequate to describe the topology; in other words, A2.3 holds

with "net" replaced by "sequence." A criterion for openness in terms of nets may also be given: The set V in 92 such that x --> x e V, we have x e V evenis open if for every net tually. [The " only if " part follows from the definition of convergence; for the

"if" part, apply A2.3(b) to V`.] A2.5 Definitions. Let {x,,, n e D} be a net, and suppose that we are given a directed set E and a map k -* nk of E into D. Then {x,,,,, k e E } is called a subnet of {x,,, n e D}, provided that " as k becomes large, so does nk" ; that is, given no a D, there is a ko e E such that k >_ ko implies nk >_ no. If E = D = the positive integers, we obtain the usual notion of a subsequence. If {xn , n e D} is a net in the topological space .0, the point x e Q is called

an accumulation point of the net if for each neighborhood U of x, x is frequently in U; in other words, given n e D, there is an m e D with m z n

and xa U. A2.6 Theorem. Let {xn , n e D} be a net in the topological space Q. If x e S2, x is an accumulation point of (xn) if there is a subnet {xnk , k e E) converging to X.

PROOF. If xnk - x, U c- °I1(x) and n c- D, then for some ko a E, we have xnk E U for k >_ ko. But by definition of subnet, there is a k1 E E such that k _ k1 implies k n. Thus jfk _ koand k _ k1,we have nk>:nand x an accumulation point of {xn, n e D). Let E be the x

collection of pairs (n, U), where n e D, U is a neighborhood of x, and x E U. Direct E by setting (n, U) < (m, V) if n < m and U V. If k = (m, V) e E let nk = m. Given n and U, then if k = (m, V) >_ (n, U), we have nk = m >_ n, so that {xnk , k c- E} is a subnet of {xn , n c- D). Now if U isa neigh-

borhood of x, then x e U for some n e D. If k = (m, V) z (n, U), then xnk = x, e V c U; therefore

x. I

For some purposes, it is more convenient to specify convergence in a topological space by means of filters rather than nets. If {xn, n e D) is a net

A2 CONVERGENCE

205

in fl, and a e D, let Ta = {n a D: n >_ a}, and let x(T,) be the set of all x,,, n > a. The x(T0), a e D, are called the tails of the net. The collection sd of tails is an example of a filterbase, which we now define. Definitions and Comments. Let sd be a nonempty family of subsets of a set r). Then sd is called a filterbase in fl if

A2.7

(a) each U e sd is nonempty; (b) if U, V e sd, there is a W e sd with W e U n V.

If F is a nonempty family of subsets of 0 such that (c) each U E .°F is nonempty, (d) if U, V e F, then U n V e P, and (e) if U c.F and U c V, then V e Pr,

then .F is called a filter in n. If sd is a filterbase, then F = {U c S2: U e V for some V e d) is a filter, called the filter generated by sd. If sd is the collection of neighborhoods of a given point x in a topological space, sd is a filterbase, and the filter generated by sd is the system of overneighborhoods of x. A filterbase sd in a topological space fl is said to converge to the point x (notation sd x) if for each U E 611(x) there is a set A e sd such that A C U. A filter .F in ) is said to converge to x if each U e 611(x) belongs to F. Thus a filterbase sd converges to x if the filter generated by sd converges to X. If {xa , ii a D} is a net, then x -- x if for each U c-611(x) we have x(Ta) c U

for some a e D, that is, x -+ x if the associated filterbase converges to x. Convergence in a topological space may be described using filterbases instead of nets. The analog of Theorem A2.3 is the following: A2.8 Theorem. Let B be a subset of the topological space Q.

(a) A point x E S2 belongs to B if there is a filterbase din B such that -+ X.

(b)

B is closed if for every filterbase sd in B such that .4 ---, x, we have

xaB. (c) A point x E n is a cluster point of B if there is a filterbase in B - (x) converging to x. PROOF. (a)

If sd -+ x and U E 011(x), then A c U for some A E sd, in partic-

ular, U n B 0 0; thus x e B. Conversely, if x e B, then U n B 96 Q for each U e 611(x). Let sd be the collection of sets U n B, U e 611(x). Then sd is a filter-

base in Band sd -- x.

206

APPENDIX ON GENERAL TOPOLOGY

(b) If B is closed, W is a filterbase in B, and sl --+ x, then x e B by (a), hence x e B by hypothesis. Conversely, if B is not closed and x e B - B, by

(a) there is a filterbase sad in B with sad -+ x. Since x 0 B, the result follows. (c)

If there is such a filterbase a and U e °h(x), then U A for some

A e .W; in particular, U n (B - {x}) 96 0, so x is a cluster point of B. Conver-

sely, if x is a cluster point of B, let a consist of all sets U n (B - {x}), U e 1&(x). Then d is a filterbase in B - {x} and d -+ x. I

If fl is first countable, the filterbases in A2.8 may be formed using a countable system of neighborhoods of x, so that in a first countable space, the topology may be described by filterbases containing countably many sets. A2.9

Definitions. The filterbase 9 is said to be subordinate to the filterbase sad

if for each A e d there is a B e.4 with B c A; this means that the filter generated by d is included in the filter generated by B. If k e E} is a subnet of {x, n e D}, the filterbase determined by the subnet is subordinate to the filterbase determined by the original net. For if no e D, there is a ko e E such that k - ko implies nk >_ no. Therefore

k>-ko} c{x,,:neD, nzno}. If sat' is a filterbase in the topological space 0, the point x e r) is called an accumulation point of sat if U n A # 0 for all U e &Ii(x) and all A e W, in other words, x e A for all A e sad. We may now prove the analog of Theorem A2.6. A2.10 Theorem. Let sat be a filterbase in the topological space 0. If x e 0,

x is an accumulation point of d if there is a filterbase -4 subordinate to d with x; in other words, some overfilter of d converges to x.

a

PROOF. If -4 is subordinate to d and 1--+ x, let U e 61l(x), A e sat. Then U => B and A B, for some B, B, a a; hence U n A B n B,, which is nonempty since .4 is a filterbase. Therefore x e A. Conversely, if x is an accumulation point of sad, let .4 consist of all sets U n A, Ue °le(x), A e sad. Then

sat c 9 (take U = 91), hence 9 is subordinate to a; since a -+ x, the result follows. I A2.11

Definition. An ultrafilter is a maximal filter, that is, a filter included in

no properly larger filter. (By Zorn's lemma, every filter is included in an ultrafilter.)

A2 CONVERGENCE

A2.12

207

Theorem. Let .F be a filter in the set Sl.

(a) .F is an ultrafilter if for each A c Sl we have A E .F or A` a F. (b) If .F is an ultrafilter and p: Sl - Sl', the filter T generated by the

filterbase p(F) = {p(F): F e .F} is an ultrafilter in il'. (c) If S is a topological space and .F is an ultrafilter in 0, F converges to each of its accumulation points. If ,F is an ultrafilter and A 0 F, necessarily A n B = 0 for some B e F. For if not, let .4 consist of all sets A n B, B e .F; then sad is a filterbase PROOF. (a)

generating a filter larger than .F. But A n B = 0 implies B c A°; hence A` e .F. Conversely, if the condition is satisfied, let F be included in the filter W. If A e W and A 0 F, then A` a .F c T, a contradiction since

AnA`=0. (b)

Let A c SY; by (a), eitherp-1(A) e IF orp-1(A`) e .F. Ifp^1(A) a .F,

then A

pp-1(A) e p(.F); hence A e T. Similarly, if p-1(A°) a .F, then A` e T. By (a), 9 is an ultrafilter. (c) Let x be an accumulation point of .F . If U e R1(x) and U # .F, then U° e F by (a). But U n U° = 0, contradicting the fact that x is an accumulation point of .F. We have associated with each net {x,,, n e D} the filterbase {x(T4): a e D} of tails of the net, and have seen that convergence of the net is equivalent to convergence of the filterbase. We now prove a converse result.

Theorem. If sad is a filterbase in the set fl, there is a net in a such that the collection of tails of the net coincides with sad. A2.13

PROOF. Let D be all ordered pairs (a, A) where a e A and A e sad; define (a, A) S (b, B) iff B e A. If (a, A) and (b, B) belong to D, choose C e and with C e A n B; for any c e C we have (c, C) >: (a, A) and (c, C) Z (b, B), hence D is directed. If we set X(., A) = a we obtain a net in Sl with x(T(a, A)) = A. I

We conclude this section with a characterization of continuity. A2.14 Theorem. Let f: it -+ Q', where 0 and fl' are topological spaces. The

following are equivalent: (a) The function f is continuous on Sl; that is, f -1(V) is open in Sl whenever V is open in Sl'.

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APPENDIX ON GENERAL TOPOLOGY

(b)

For every net

f(x).

in Q converging to the point x e S2, the net {

(c) For every filterbase d in fZ converging to the point x c- ), the filterbase f (sad) converges to f (x). PROOF. Let

be a net and 4 a filterbase such that the tails of the net

coincide with the elements of the filterbase. If, say, x(TQ) = A E s4, then n e D, n >_ a}. Thus the tails of the net { f coincide with f (A) _ { f the elements off (.a?). It follows that (b) and (c) are equivalent. If f is continuous and x -+ x, let V be a neighborhood of fl x). Then

f -'(V) is a neighborhood of x; hence x,, is eventually in f -'(V), so that f (x.) is eventually in V. Thus (a) implies (b). Conversely, if (b) holds and C is

closed in if, let

be a net in f -'(C) converging to x. Then f(x)

(b), and since C is closed we have f (x) e C by A2.3(b). Thus x e f -'(C), hence

f -'(C) is closed, proving continuity off. I A3 Product and Quotient Topologies

In the Euclidean plane R2, a base for the topology may be formed from sets U x V, where U and V are open subsets of R; in fact U and V can be taken to be open intervals, so that U x V is an open rectangle. If {(xn , y.),

n=1,2,...)

is a sequence in R2, then (x., (x, y) if x -> x and y -+ y, that is, convergence in R2 is "pointwise" or "coordinatewise" convergence. In general, given an arbitrary collection of topological spaces S2i, i E 1, let

Q be the cartesian product fi'EI f ,, which is the collection of all families (x;, i e I) ; that is, all functions on I such that x, a 0, for each i. We shall place a topology on S2 such that convergence in the topology coincides with pointwise convergence.

A3.l Definition. The product topology (also called the topology of point wise convergence) on S2 = fl; E r f2, has as a base all sets of the form

{xE):x1keUIk, k=1,...,n) where the Uik are open in S

and n is an arbitrary positive integer. (Since the

intersection of two sets of this type is a set of this type, the sets do in fact form a base.)

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A3 PRODUCT AND QUOTIENT TOPOLOGIES

If pi is the projection of f2 onto f2i, the product topology is the weakest topology making each pi continuous; in other words, the product topology is included in any topology that makes each pi continuous. The product topology has the following properties: A3.2

Theorem. Let 0 =F11 E 1 fl,, with the product topology.

(a) If {x("), n c D} is a net in 0 and x e f2, then x(") -+ x iff x;") -+ xi for each i. (b) A map f from a topological space L20 into n is continuous if pi of is continuous for each i. (c) If fi: f2o --+ 1i , i e I, and we define f: 1o -> 0 by f (x) = (fi(x), i e I ), then f is continuous iff each fi is continuous. (d) The projections pi are open maps of f2 onto L2,.

If x(") --+ x, then x;") = pi(xl)) --+ pi(x) = xi by continuity of the pi. Conversely, assume x;"> - xi for each i. Let PROOF. (a)

V= {ye0:yikE Uik, k = 1,...,r}, be a basic neighborhood of x. Since xik E Uik , there is an nk e D with x(,k) E Uik for n >_ nk. Therefore, if n e D and n >_ nk for all k = 1, ..., r, we have x e V, so that xt"i -+ X. (b) The "only if" part follows by continuity of the pi. Conversely, assume each pi of continuous. If x(") -+ x, then pi(f (x("))) -+ pi(f (x)) by hypo-

thesis; hence f(P)) -f(x) by (a). (c) We have fi = pi o f, so (b) applies. (d) The result follows from the observation that pi{x a S2: xik E Uik ,

k = 1, ... ,

n}

{1i Uik

if i # any ik, if i = ik for some k.

Note that if 0 is the collection of all functions from a topological space S to a topological space T, then 0 = fii E 1 Qi, where I = S and 1i = T for all i. If {f,,) is a net of functions from S to T, and f: S --+ T, then f" -+f in the product topology iff f"(s) --+f(s) for all s E S. We now consider quotient spaces.

Definition. Let 0o be a topological space, and p a map of 0, onto a set f2. The identification topology on 0 is the strongest topology making p continuous, that is, the open subsets of S2 are the sets U such that p-1(U) is A3.3

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open in no. When n has the identification topology, it is called an identification space, and p is called an identification map.

A quotient topology may be regarded as a particular identification topology. Let R be an equivalence relation on the topological space no, S2o/R the set of equivalence classes, and p: 0..0 - S2o/R the canonical projection: p(x) = [x], the equivalence class containing x. The quotient space of S2o by R is S2o/R with the identification topology determined by p. In fact, any identification space may be regarded as a quotient space. To see this, we need two preliminary results.

A3.4 Lemma. If n has the identification topology determined by p: 0o - S2, then g: S2 -> Q, is continuous iff g o p: 12o -- Q, is continuous. PROOF. The "only if" part follows from the continuity of p. If g e p is continuous and V is open in n1, then (g o p)-1(V) = p-1(g-1(V)) is open in no. By definition of the identification topology, g-1(V) is open in 0. 1

A3.5 Lemma. Let p: CIO -+ 0 be an identification, and let h: no -1, S21 be

continuous. Assume that h -p-' is single-valued, in other words, p(x) _ p(y) implies h(x) = h(y). Then h c p'1 is continuous.

PROOF. Since h = (h o p-1) o p, the result follows from A3.4. (Note that P_' is defined on all of 0 since p is onto.) Theorem. Let f: no - fi be an identification. Define an equivalence relation R on no by calling x and y equivalent iff f(x) = fly). Let p be the A3.6

canonical projection of no onto S2o/R. Then S2o/R, with the quotient topology,

is homeomorphic to n. PROOF. We have f(x) = f(y) if p(x) = p(y), so by A3.5, fop-1 and p of -1 are both continuous. Since these functions are inverses of each other, they define a homeomorphism of n and S2o/R. I The following result gives conditions under which a given topology arises from an identification.

Theorem. Let p be a map of 00 onto 0. If p is continuous and either open or closed, it is an identification, that is, the identification topology on r) determined by p coincides with the original topology on U. A3.7

A4 SEPARATION PROPERTIES

211

PROOF. Since p is continuous and the identification topology is the largest one

making p continuous, the original topology is included in the identification topology. If p is an open map and U is an open subset of 0 in the identifica-

tion topology, p-'(U) is open in fl,, hence p(p-'(U)) = U is open in the original topology. If p is a closed map, the same argument applies, with "open" replaced by "closed." A4 Separation Properties and Other Ways of Classifying Topological Spaces

Topological spaces may be classified as to how well disjoint sets may be separated, as follows. (The results of this section are generally discussed in a first course in topology, and will not be proved.)

Definitions. A topological space S2 is said to be a To space if given any two distinct points x and y, at least one point has a neighborhood not containing the other; f2 is a T, space if each point has a neighborhood not containing the other; .0 is a T2 (or Hausdorff) space if x and y have disjoint neighborhoods. This is equivalent to uniqueness of limits of nets (or filterbases). Also, C is said to be a T3 (or regular) space if i2 is T2 and for each closed set C and point x 0 C, there are disjoint open sets U and V with x e U and C c V; it is said to be a T4 (or normal) space if i2 is T2 and for each pair of disjoint closed sets A and B there are disjoint open sets U and V with A4.1

AcUand BcV.

It follows from the definitions that Ti implies T;_1i i = 4, 3, 2, 1. Also, the T1 property is equivalent to the statement that {x} is closed for each x. The

space f) is regular if it is Hausdorff and for each open set U and point x e U, there is a V c- 0&(x) with V c U. The space S is normal if it is Haus-

dorff and for each closed set A and open set U

with AcVc VcU.

A, there is an open set V

A metric space is T4, for if A and B are disjoint closed sets, we may take U = {x: d(x, A) - d(x, B) < 0}, V = {x: d(x, A) - d(x, B) > 0}, where d(x, A) = inf{d(x, y) : y e A}.

Urysohn's Lemma. Let S) be a Hausdorff space. Then 0 is normal if for each pair of disjoint closed sets A and B, there is a continuous function f: 0 -+ [0, 1] with f = 0 on A and f = 1 on B. A4.2

Tietze Extension Theorem. Let t) be a Hausdorff space. Then S2 is normal iff for every closed set A c C1 and every continuous real-valued A4.3

APPENDIX ON GENERAL TOPOLOGY

212

function f defined on A, f has an extension to a continuous real-valued function F on n. Furthermore, if if I < c (respectively, If 1 < c) on A, then 1FI can be taken less than c (respectively, less than or equal to c) on Q. Theorem. Let A be a closed subset of the normal space n. There is a continuous f: t -> [0, 1] such that A =f -1{O} if A is a Gd , that is, a countable intersection of open sets. A4.4

A4.5

Definitions and Comments. A topological space n is second countable

if there is a countable base for the topology, first countable if there is a countable base at each point (see A2.4). Second countability implies first countability but not conversely. Any metric space is first countable.

If 0 is second countable, it is necessarily separable, that is, there is a countable dense subset of Q. Furthermore, if n is second countable, it is Lindelof, that is, for every family of open sets V1, i e I, such that Ut V1 = S2, there is a countable subfamily whose union is Q (for short, every open covering of 0 has a countable subcovering). In a metric space, the separable, second countable, and Lindelof properties are equivalent as follows. Second countability always implies separability and Lindelof. If n is separable with a countable dense set {x1, x2 , ...), then the balls B(x;, r) = {y a f : d(y, xi) < r}, i = 1, 2,..., r rational, form acountable base. If n is Lindelof, the cover by balls B(x, 1/n), x e S2, has a countable

subcover {B(x,,;, 1/n), i = 1, 2, ...}, and the sets B(x,,;, 1/n), i, n = 1, 2, ..., form a countable base. This result implies that any space that is separable but not second countable (or not Lindelof), or Lindelof but not second countable (or not separable)

cannot be metrizable, that is, there is no metric whose topology coincides with the original one. A4.6

Definitions and Comments. A topological space .0 is said to be com-

pletely regular if Q is Hausdorff and for each x e n and closed set C 0 with x 0 C, there is a continuous f: fZ

[0, 1] such that f (x) = 1 and f = 0 on

C.

If A is a subset of the normal space n, then A, with the relative topology, is

completely regular (this follows quickly from Urysohn's lemma). Also, complete regularity implies regularity. Thus complete regularity is in between regularity and normality; for this reason, completely regular spaces are sometimes called T34 spaces. Now if CIO is a Hausdorff space and F is the family of continuous maps f: no -' [0, 1 ], let n = fl (If : f e , r) where each If = [0, 11. Let e:.00 -+ f be the evaluation map, that is, e(x) = (f (x), f e ,F).

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213

With the product topology on fl, e is continuous, and e is one-to-one if F distinguishes points, in other words, given x, y e ao, x :A y, there is an f e F such that f (x) 56f (y). Finally, if .F distinguishes points from closed sets, that is, if 1Q, is completely regular, then e is an open map of !no onto e(flo) c Q. [If is a net and e(x) x ++ x, there is a neighborhood U of x such that x is not eventually in U; that is, given m, there is an n >_ m with

x 0 U. Choose f e °F with f (x) = 1 and f = 0 on flo - U. Then for each m, we have f (x,,) = 0 for some n z m, so that f (x.) ++f (x). But then a contradiction.] e(x) It follows that if f2o is completely regular, it is homeomorphic to a subset of a normal space. [Since 0 is a product of Hausdorff spaces, it is Hausdorff;

the Tychonoff theorem, to be proved later, shows that f2 is compact, and hence normal (see A5.3(d) and A5.4).] Since e(Q0) is determined completely by F, we may say that the continuous functions are adequate to describe the topology of flo.

A5 Compactness

The notion of compactness appears in virtually all areas of mathematics. The original compactness result was the Heine-Borel theorem: If [0, 1] c U; Vi, where the Vi are open subsets of R, then in fact [0, 1) is covered by finitely many V, , that is, [0, 1 ] c Uk= 1 V.,, for some V;. , ... , V;.. In general, we have the following definition: Definition. The topological space f2 is compact if every open covering of fl has a finite subcovering. A5.1

There are several ways of expressing this idea.

Theorem. If fl is a topological space, the following are equivalent: (a) f2 is compact. (b) Each family of closed sets C, c f2 with the finite intersection property (all finite intersections of the Ci are nonempty) has nonempty intersection. Equivalently, for every family of closed subsets of 0 with empty intersection, there is a finite subfamily with empty intersection. A5.2

(c)

Every net in f2 has an accumulation point in 0; in other words

(by A2.6), every net in f2 has a subnet converging to a point of fl. (d) Every filterbase in fl has an accumulation point in f2, that is (by A2.10), every filterbase in fl has a convergent filterbase subordinate to it, or equally well, every filter in f2 has an overfilter converging to a point of 0. (e) Every ultrafilter in 0 converges to a point of Q.

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PROOF. Parts (a) and (b) are equivalent by the duality between open and closed sets. If is a net and sad is a filterbase whose elements coincide with the tails of the net, then the accumulation points of the net and of the filterbase coincide, so that (c) and (d) are equivalent. Now (d) implies (e) by A2.12(c), and (e) implies (d) since every filter is included in an ultrafilter. To prove that (b) implies (d), observe that if sd is a filterbase, the sets A e s:1, hence the sets A, A c sad, have the finite intersection property, so by (b), there

is a point x E n{A: A c-.4}. Finally, we prove that (d) implies (b). If the closed sets C; have the finite intersection property, the finite intersections of the C; form a filterbase, which by (d) has an accumulation point x. But then x

belongs to all C,. I It is important to note that if A c Q, the statement that every covering of A by sets open in f has a finite subcovering is equivalent to the statement that every covering of A by sets open in A (in the relative topology inherited from S2) has a finite subcovering. Thus when we talk about a compact subset of a topological space, there is no ambiguity. The following results follow quickly from the definition of compactness. Theorem. (a) If S2 is compact and f is continuous on 11, then f (Q) is compact. A5.3

(b) A closed subset C of a compact space ) is compact. (c) If A and B are disjoint compact subsets of the Hausdorff space Q, there are disjoint open sets U and V such that A c U and B e V. In particular (take A = {x}), a compact subset of a Hausdorff space is closed. (d) A compact Hausdorff space is normal. (e) If A is a compact subset of the regular space 0, and A is a subset of the open set U, there is an open set V with A c V e V e U. PROOF. (a) This is immediate from the definition of compactness. (b) If C is covered by sets U open in Q, the sets U together with a -- C

cover 0. By compactness there is a finite subcover. (c) If x 0 B and y e B, there are disjoint neighborhoods U,(x) and V(y) of x and y. The V(y) cover B; hence there is a finite subcover V(y), i = 1, ... , and V' = U'=1 V(y;) are disjoint open sets with n. Then U' =n7 I x e U' and B c V'. If we repeat the process for each x e A, we obtain disjoint open sets U(x) and V(x) as above. The U(x) cover A; hence there is a finite subcover U(X), i = 1, ... , m. Take U = U"' I U(x,), v = n,,"= 1 V(x,). (d) If A and B are disjoint closed sets, they are compact by (b); the result then follows from (c).

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(e)

If x e A, regularity yields an open set V(x) with x e V(x) and

V(x) c U. The V(x) cover A; so for some x...... xn, we have n

n

n

A c U V(xi) c U V(xi) = U V(xi) C U. i=1

i=1

i=1

The following is possibly the most important compactness result. If S2i is compact for each i e 1, then ill _ fl , iii is compact in the product topology. A5.4 Tychonoff Theorem.

PROOF. Let 97 be an ultrafilter in 0. If pi is the projection of i2 onto Ui, then by A2.12(b), pi(.F) is a filterbase that generates an ultrafilter in fli. By hypothesis, pi(F) converges to some xi a Di, and it follows that ." -+ x = (xi, i e 1). [To see this, observe that if sal is a filterbase in i2 and {x"} is a net whose tails are the elements of a, then the tails of {pi(x")} are the elements of

pi(d). Since x" -+ x if pi(x") -+ pi(x) for all i, by A3.2(a), it follows that sd - x iff pi(d) -' pi(x) for all i.] The Tychonoff theorem now follows from A5.2(e).

The following result is often used to infer that the inverse of a particular one-to-one continuous map is continuous. A5.5 Theorem. Let f: i2 - i2, where i2 is compact, 01 is Hausdorff, and f is continuous. Then f is a closed map; consequently iff is one-to-one onto, it is a homeomorphism.

PROOF. By A5.3(a), (b) and (c). I A5.6

Corollary. Let p: i2o - 0 be an identification, and let h: ffo - ill be

continuous; assume hop-' is single valued, and hence continuous by A3.5. Assume also that f)o is compact (hence so is i2 because p is onto) and f2, is Hausdorff. If h op-1 is one-to-one onto, it is a homeomorphism. PROOF. Apply A5.5 with f = h o

`. I

Corollary A5.6 is frequently applied in constructing quotient spaces. For example, if one pair of opposite edges of a rectangle are identified, we obtain a cylinder. Formally, let 12 = {(x, y): 0 < x < 1, 0:5 y < 1). Define an

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equivalence relation R on I2 by specifying that (0, y) be equivalent to (1, y), 0 < y < 1, with the other equivalence classes consisting of single points. Let h(x, y) = (ei2Rx, y), (x, y) ,E I2 ; h maps I2 onto a cylinder C. If p is the

canonical projection of I2 onto I2/R, then A5.6 implies that h o p-1 is a homeomorphism of I2/R and C. In some situations, for example in metric spaces, there are alternative ways of expressing the idea of compactness. Definition. A topological space 12 is said to be countably compact if every countable open covering of 0 has a finite subcover. A5.7

A5.8 Theorem. For any topological space S2, the following properties (a)(d) are equivalent, and each implies (e). If S2 is T1, then all five properties are

equivalent.

(a) Q is countably compact. Each sequence of closed subsets of 92 with the finite intersection property has nonempty intersection. (c) Every sequence in fl has an accumulation point. (d) Every countable filterbase in f) has an accumulation point. (e) Every infinite subset of fl has a cluster point. (b)

PROOF. The equivalence of (a), (b), and (d) is proved exactly as in A5.2, and (d) implies (c) because the tails of a sequence form a countable filterbase. To prove that (c) implies (d), let sad = {A1, A2, ...} be a countable filterbase,

and choose x e n"=1 A; , n = 1, 2. .... If x is an accumulation point of and U e °IE(x), then for each n there is an m >- n such that

a U, hence

U n nm 1 A i 0 0. It follows that U n A # 0 for all n, and consequently x is an accumulation point of W. To prove that (c) implies (e), pick a sequence

of distinct points from the infinite set A and observe that if x is an accumulation point of the sequence, then x is a cluster point of A. Finally, we show that (e) implies (a) if S2 is T1. Say U1, U2, ... form a countable open covering of S2 with no finite subcover. Choose x1 0 Ul ; having chosen distinct x1, ... , xk with x; 0 U/= I Ui , j = 1, ..., k, then xl,... , xk all belong to

some finite union Ui=1 Ui, n

k + 1; choose Xk+10 Ui=I U1 (hence xk+I 11 U1 U. and x1, ..., xk+I are distinct). In this way we form an infinite set A = {x1, X2, ...} with no cluster point. For if x is such a point, x belongs to U for some n. Since n is T1, there is a set U e V(x) such that U c U and xi 0 U, i = 1, 2, ..., n - 1 (unless xi = x). Since Xk 0 U. for k >- n, U contains no point of A distinct from x. I

A5 COMPACTNESS

A5.9

217

Definitions and Comments. The topological space Q is said to be

sequentially compact iff every sequence in ) has a convergent subsequence. By A5.8(c), sequential compactness implies countable compactness. In a first countable space, countable and sequential compactness are equivalent. For if x is an accumulation point of the sequence and V1, V2, ... (with

V,,.., = V. for all n) form a countable base at x, then for each k we may find nk >_ k such that

e Vk. Thus we have a subsequence converging to x.

A5.10 Theorem. In a second countable space or a metric space, compactness,

countable compactness, and sequential compactness are equivalent. PROOF. A second countable space is Lindelof (see A4.5), so compactness and

countable compactness are equivalent. It is first countable, so countable compactness and sequential compactness are equivalent; this result holds in a metric space also, because a metric space is first countable. Now a sequentially compact metric space fl is totally bounded, that is, for each e > 0, 0 can be covered by finitely many balls of radius e. (If not, inductively pick x1, x2, ... with xi+1 0 U;=, B(xi, s); then can have no convergent subsequence.) Thus for each positive integer n, ) can be covered l 1n), i = 1, 2, ..., k,,. If { U; , j e J} is an arbitrary by finitely many balls open covering of SZ, for each ball I/n) we choose, if possible, a set URi of the covering such that B(x,,i, 1/n). If x e fl, then x belongs to a ball B(x, a) included in some U;; hence x e B(x,,i, 1/n) a B(x, e) c U. for some n and i; therefore x e Thus the U,i form a countable subcover, and 1), which is countably compact, must in fact be compact. I Note that a compact metric space is Lindelof, hence (see A4.5) is second countable and separable.

Definition. A Hausdorff space is said to be locally compact if each x E !Q has a relatively compact neighborhood, that is, a neighborhood whose A5.11

closure is compact. (Its follows that a compact Hausdorff space is locally compact.) A5.12

Theorem. The following are equivalent, for a Hausdorff space i2:

(a) D is locally compact. (b)

For each x e S2 and U e all(x), there is a relatively compact open set V

with x e V c V c U. (Thus a locally compact space is regular; furthermore, the relatively compact open sets form a base for the topology.)

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APPENDIX ON GENERAL TOPOLOGY

(c) If K is compact, U is open, and K c U, there is a relatively compact open set V with K c V c V c U.

PROOF. It is immediate that (c) implies (a), and (b) implies (c) is proved by applying (b) to each point of K and using compactness. To prove that (a) implies (b), let x belong to the open set U. By (a), there is a neighborhood V1 of x such that K = V1 is compact. Now K is compact Hausdorff, and hence

regular, and x e U n V1, which is open in f2, and hence open in K. Thus (see A4. 1) there is a set W open in C1 such that x e W n K and the closure of W n K in K, namely, W n K, is a subset of U n V1.

Now xe Wn V1 and Wn V1 c WnKcU, so V=Wn V1 is the desired relatively compact neighborhood.

The following properties of locally compact spaces are often useful: Theorem. (a) Let S2 be a locally compact Hausdorff space. If K c U c 52, with K compact and U open, there is a continuous f:.0 -+ [0, 1] A5.13

such that f = 0 on K and f = 1 on 0 - U. In particular, a locally compact Hausdorff space is completely regular. (b) Let S2 be locally compact Hausdorff, or, more generally, completely regular. If A and B are disjoint subsets of i2 with A compact and B

closed, there is a continuous f: 11 -+ [0, 1] such that f = 0 on A and f = 1 on B. (c) Let S2 be locally compact Hausdorff, and let A c U c 0, with A compact and U open. Then there are sets B and V with A c V c B e U,

where V is open and a-compact (a countable union of compact sets) and B is compact and is also a G6 (a countable intersection of open sets). Consequently (take A = {x}) the a-compact open sets form a base for the topology. PROOF. (a)

Let K c V c V c U, with V open, V compact [see A5.12(c)).

P is normal, so there is a continuous g: V-+ [0, 11 with g = 0 on K, g = 1 on

T7- V. Define! = g on V. f = 1 on Q- V. (On V- V, g = 1 so! is welldefined.) Now f is continuous on 0 (look at preimages of closed sets), so f is the desired function. (b)

By complete regularity, for each x e A, there is a continuous

f,,: 0 -+ [0, 1] with f,(x) = 0, fx = 1 on B. By compactness, n

A C U {x: fx,(x) < H i=1

for some x1,...,xn.Let

;then g=1onBand0
so f = max(2g - 1, 0) is the desired function.

219

A5 COMPACTNESS

(c) By A.5.12(c), we may assume without loss of generality that U is included in a compact set K. By (b) there is a continuous f: Q -+ [0, 11 with f = 0 on A, f = I off U. Let V = {x: f(x) < ];), B = {x: f(x) < 1}. Then V is open, B is closed, a.id A c V c B c U c K; hence B is compact. Now B is a

Ga because

B=

n{x:f(x)<1+and 2 n

V=

n=1

U{x:f(x)
n=1

so V is a-compact. (In fact V is a countable union of compact Ga's.) I Corollary. Let C be completely regular, A a compact Ga subset of Q. There is a continuous f: cZ - [0, 11 such that f -1{0} = A. A5.14

PROOF. If A = n', Un , the U" open, by A5.13(b) there are continuous n : S2 - [0, 1 ] wi th fn = 0 on A and fn = 1 off Un . Let f =Yn 1 f, 2-n. I

f

A5.15

Theorem. If 0 is Hausdorff, the following are equivalent:

(a) 0 is locally compact Lindelof. (b) KI can be expressed as UOO 1 Un where U. is a compact subset of Un+1 for each n. (c) rl is locally compact and a-compact.

PROOF. (a) implies (b): The relatively compact open sets form a base by A5.12(b), in particular they cover Q. Extract a countable subcover { V1, V2 , ...}

using Lindelof, and let U1 = V1, and for n > 1, U. = Vn u W,, where W. is a relatively compact open set and Wn =) U"-' Vj. (h) implies (c): We have g = Un U", proving a-compactness. If x e 0, then x e U. for some n, with U. compact, proving local compactness. (c) implies (a): If 0 = Un 1 Kn, K. compact, and {U;} is an open covering of Q, extract a finite subcovering of each Kn, and put the sets together to form a countable subcovering of 0. 1 1

The Urysohn metrization theorem, which we shall not prove, states that for a second countable space Q, metrizability and regularity are equivalent. This result yields the following corollary to A5.15. A5.16 Theorem. If n is locally compact Hausdorff, then S2 is second count-

able if f) is metrizable and a-compact.

220

APPENDIX ON GENERAL TOPOLOGY

Proof. If n is second countable, it is metrizable by A5.12(b) and the Urysohn metrization theorem. Also, f) is Lindelof (see A4.5), hence is a-compact by A5.15. Conversely, if f) is metrizable and a-compact, then f) is Lindelof by A5.15, hence second countable (see A4.5). Finally, we consider the important one point compactjcation. A5.17 Theorem. Let 0 be locally compact Hausdorff, and let fl* = 0 U {oo},

where oo stands for any element not belonging to II. Put the following topology on fl*: U is open in n* if U is open in .0 or U is the complement (in Q*) of a compact subset of .0. Then: (a) If U c 0, U is open in it if U is open in S2*; thus S2* induces the original topology on 0. (b) fl* is compact Hausdorff.

PROOF. Part (a) follows from the definition of the topology. To prove (b), let { U3 be an open covering of fl*. Then oo belongs to some U, and the remaining Uj cover the compact set a* - U;, so the cover reduces to a finite subcover. Since f is Hausdorff, distinct points of it have disjoint neighborhoods. If

x e 0, let V be an open subset of it with x e V and V c Q, where V, the closure of V in d2, is compact [see A 5.12(b)]. Then V and Q* - V are disjoint neighborhoods of x and co. A6 Semicontinuous Functions

If f,, f2, ... are continuous maps from the topological space .0 to the extended reals R, and f (x) increases to a limit f (x) for each x, f need not be continuous; however, f is lower semicontinuous. Functions of this type play an important role in many aspects of analysis and probability. Definition. Let .0 be a topological space. The function f: ) -+A is said to be lower semicontinuous (LSC) on Q if {x a S2: f(x) > a} is open in f for each a e R, upper semicontinuous (USC) on 1Z iff {x a fl f (x) < a} is open in n for each a e R. Thus f is LSC if -f is USC. Note that f is continuous if it is both LSC and USC. We have the following criterion for semicontinuity. A6.1

A6.2

Theorem. The function f is LSC on S2 if, for each net

to a point x e n, we have lim inf f

f (x), where lim inf f

converging means

221

A6 SEMICONTINUOUS FUNCTIONS

f(x) whenever x.-x. (In

sup,, infk, f (xk). Hence f is USC iff lim sup f

a first countable space, "net" may be replaced by "sequence.") (b, oo], an open PROOF. Let f be LSC. If x - x and b < f (x), then x e f `(b, b eventually. subset of S2, hence eventually x,, e.f -'(b, oo], that is, f (x). Conversely if x --+ x implies Thus Jim inf f

_f(x), we show that V = {x: f(x) > a} is open. Let x,, --> x, where f(x) > a. Then lim inf,,f(x,,) > a, hence f (x,,) > a eventually, that is, x e V eventually. Thus (see A2.4) V is open. I We now prove a few properties of semicontinuous functions. A6.3

Theorem. Let f be LSC on the compact space fl. Then f attains its

infimum. (Hence iff is USC on the compact space S2, f attains its supremum.)

PROOF. If b = inff, there is a sequence of points x,, a 0 with f (x,,) -+ b. By compactness, we have a subnet x,,, converging to some x e 92. Since f is LSC, lim infk f (x.,) >_ f (x). But f (x,,,,) -+ b, so that f (x) < b; consequently f (x) _ b. I

A6.4 Theorem. If f, is LSC on 12 for each i e I, then sup; f, is LSC; if I is finite, then min; fj is LSC. (Hence iff; is USC for each i, then inf; f, is USC, and if I is finite, then max, f, is USC.)

PROOF. Let f = sup; f,; then {x: f(x) > a) = Ui E j {x: f,(x) > a); hence {x: f (x) > a) is open. If g = min(fl, f2 , ... , then {x: g(x) > a) =n {x:f,(x) > a)

I=1

is open. I A6.5

Theorem. Let f: S2 -+ R, SZ any topological space, f arbitrary. Define

f (x) = lim inf.f(y), y-.x

x e n;

that is, f (x) = sup inf f (y), V yeV

where V ranges over all neighborhoods of x. [If S2 is a metric space, then f(x) = suP.= I,2, ... infd(x,,),11 f(y)]

222

APPENDIX ON GENERAL TOPOLOGY

Then f is LSC on 0 and f 5 f; furthermore if g is LSC on Sl and g < f, then g < f. Thus f, called the lower envelope off, is the sup of all LSC functions that are less than or equal to f (there is always at least one such function, namely

the function constant at - oo). Similarly, if J(x) = lim sup,.x f (y) = infy sup,. y f(y), then f, the upper envelope off, is USC andf >- f; in fact f is the inf of all USC functions that are

greater than or equal to f. PROOF. It suffices to consider f. Let

be a net in Sl with x -, x and

lim inf f (x,,) < b < f (x). If V is a neighborhood of x, we can choose n b. Since V is also a neighborhood of x, we have

such that x,, e V and f

b > f (x,.) >- inff(y), so

f (x) = sup inf f (y) < b < f (x), V yeV

a contradiction. By A6.2, f is LSC, and f < f by definition off. Finally if g is LSC, g < f, then f (x) = lim infy,x f(y) >_ lim infy, g(y) > g(x) since g is

LSC. [If sups infy. v g(y) < b < g(x), then for each V pick xy e V with g(xv) < b. If Vl < V2 means that V2 c V1, the xy form a net converging to x, while lim infy g(xy) < b < g(x), contradicting A6.2.]

It can be shown that if Sl is completely regular, every LSC function on it

is the sup of a family of continuous functions. If Q is a metric space, the family can be assumed countable, as we now prove.

Theorem. Let Sl be a metric space, f a LSC function on C. There is a sequence of continuous functions f.: Sl --* R such that f T f. (Thus if f is USC, there is a sequence of continuous functions f If) If if I < M < co, the M for all n. f may be chosen so that I A6.6

PROOF [following Hausdorff (1962)]. First assume f t 0 and finite-valued.

If d is the metric on n, define g(x) = inf{f(z) + td(x, z): z e Sl}, where t > 0 is fixed; then 0 < g < f since g(x) f (x) + td(x, x) = f (x). If x, y e Sl, then f(z) + td(x, z) < f(z) + td(y, z) + td(x, y). Take the inf over z to obtain g(x) < g(y) + td(x, y). By symmetry, I9(x) - 9(y)I < td(x, y),

hence g is continuous on Q.

223

A7 THE STONE-WEIERSTRASS THEOREM

inf{f(z) + nd(x, z): z E S2}. Now set t = n; in other words let Then 0 < f,, T h < f. But given e > 0, for each n we can choose z e S2 such that f .(x) + s > f (z.) + nd(x,

nd(x, z.).

0. Since f is LSC, But f ,,(x) + E < f (x) + e, and it follows that d(x, lim inf,, . f >f(x) - e eventually. But now f (x) ; thus f

f (x) -

for large enough n.

2E

It follows that 0
hof. Letf.

Tf A7 The Stone-Weierstrass Theorem

In this section, we consider the family C(12) of continuous real-valued functions on a compact Hausdorff space Q. The set C(S1) becomes a Banach space under the sup norm IIf II = sup{I f(x)I : x E C2}. If A c C((I), A is called a subalgebra of C(S2) if for all f,, f2 e A, a, b e R, we have afl + bf2 e A and fl f2 e A. Thus A is a vector space under addition and scalar multiplication, and a ring under ordinary multiplication. The main theorem of this section gives conditions under which A is dense in C(S2). We assume throughout that S2 has at least two points; the Stone-Weierstrass theorem will be seen to be trivial if S2 has only a single point.

A7.1 Lemma. The function given by f(x) = IxI, -1 < x:!5; 1, can be uniformly approximated by polynomials having no constant term.

PROOF. If 0 < y < a <- 1, then 0 < y < y + J(a2 - y2) <- a. (For y2 - 2 y + 2a - a2 decreases when 0 < y < 1, and its value when y = a is 0.) If we I(a2 the above argument shows that define yo = 0, i = y,, + 0 < y,, 5 yrt+, < a for all n. Let n -+ oo to obtain y. - y, where y = y + J(a2 - y2); hence y = a.

-

Thus, changing the notation, define PO(X)

= 0,Pn+,(x) =

(x2

-

P.2(x)),

- 1 -< x < I.

APPENDIX ON GENERAL TOPOLOGY

224

By induction, pn is a polynomial having no constant term, and 0 <- pn(x) <(x2)'12 = Ixi. Since the domain [-1, 1] is compact andpn converges monotonically to a continuous limit p, the convergence is uniform by Dini's theorem (see 4.3.9 for the details).

A7.2 Lemma. Let A be a subset of C(S2) that is closed under the lattice operations; that is, if f, g c- A, then max(f, g) e A and min(f, g) E A. If f E C(Q) and f can be approximated at each pair of points by functions in A [if x 96 y, there is a sequence of functions fn e A with f,(x) -+f(x) and

f.(y) -fly)], then f e A. PROOF. Given S > 0, x, y E 0, there is a function e A with If -fxyl < b at the points x and y. Let U(x, y) = {z c -!Q; fsy(z)
{z e 0: ,,Y(z) > f (z) - S). Since x E U(x, y), the U(x, y) cover f) for each fixed y; by compactness, D = U1=1 U(xj, y) for some x,, ..., x,. Let fy = min{f,y: 1 < i:5 n} e A by hypothesis, and let n

V(y) = n V(x,, y) i=1

If xc fl, then xc U(x;,y)for some i,sof,(x) f (x) - S for all i, hence fy(x) > f (x) - b. But the V(y) cover S2 [note y e V(y)] so by compactness, S2 = Um I V(yi) for

some yl, ..., y.. Define fa = max{ fy, : 1 <- i:!5; m} e A by hypothesis. If x e 12, then fa(x) f (x) - S. Thus 11f -fall <- S and, since 6 is arbitrary, it follows that f e A. 1 A7.3 Lemma. Let A be a subalgebra of C(S)), and assume that A is a closed subset of C(SI). Then f e A implies If I e A; also, f g E A implies max(f, g) e A and min(f, g) e A. PROOF. If g e A and p is a polynomial with no constant term, then p c g e A.

If f e A and Ilf II = M, set g =f/M. Then Ig(x)I < 1 for all x, so by A7.1, Ilp ° 9 - 191 can be made arbitrarily small by appropriate choice of p; II

consequently IgI, hence If 1, belongs to A.

Nowf + g = max(f, g)+min(f, g), If - gI = max(f, g) - min(f, g). Addition and then subtraction of these equations completes the proof. I A7.4 Lemma. Let A be a subalgebra of C(S2) that separates points; that is, if x y, there is an f e A with f (x) f (y). Assume that for each x c- S2, there is an f e A with f (x) 0 0.

225

A7 THE STONE-WEIERSTRASS THEOREM

If x, y e 92, x # y, and r and s are arbitrary real numbers, there is an f e A withf(x) = r,f(y) =s. PROOF. If this is not the case, then for all f1i f2 e A, the equations

af1(x) + bf2(x) = r

and

af1(y) + bf2(y) = s

have no solution for a and b. Thus the determinant fl(x)f2(y) - f1(y)f2(x) is 0. In particular, if fl =f, f2 =f2' we obtain f(x)f2(y) = f (y)f 2(x), or f (x) f (y)(f (y) - f (x)) = 0 for all f e A. Thus if g e A and g(x) # g(y), we have g(x)g(y) = 0. If, say, g(x) = 0, then (takingf1 = g in the above determinant) g(x)f2(y) = g(y)f2(x) for all f2 a A. Since g(x) = 0 we must have g(y) # 0; hence f2(x) = 0 for all f2 a A, contradicting the hypothesis. The argument is symmetrical if g(y) = 0. 1

Stone- Weierstrass Theorem. Let A be a subalgebra of C(S2) that separates points. A7.5

(a) If for each x e 0, there is an f e A with f (x) # 0 (in particular, if the constant functions belong to A), then A = C(1); thus A is (uniformly) dense in C(S2).

(b) If all f e A vanish at a particular x e !Q (there can be at most one such point since A separates points), then A = {g a C(Q): g(x) = 01.

Since A is an algebra, so is A; by A7.3, A is closed under the lattice operations. Let g e C(S2); if x, y e S2, x # y, then by A7.4 there is an PROOF. (a)

f e A e A such that f(x) = r = g(x) and f(y) = s = g(y). Thus g can be approximated at each pair of points by functions in A. By A7.2, g c -A; in other words, A = C(S2). (b) If f (x) = 0 for all f e C(Q), let B be the algebra generated by A and the constant functions; thus B = { f + c: f e A, c a constant function}. Then B satisfies the hypothesis of (a), so B = C(S2). Now if S > 0, g e C(Q), and g(x) =

0, then for some f e A and some c we have Ilg - (1 + c)II < 6/2. But g(x) = f(x) = 0; hence Icy < 6/2, and therefore 11g -III < S. Thus g e A, so that {g a C(S2); g(x) = 0} c A. But the reverse inclusion holds by hypothesis, completing the proof. I A7.6

Corollary. The Stone-Weierstrass theorem holds equally well with

C(Q) taken as the collection of all continuous complex-valued functions on 0, if we add the hypothesis that whenever f e A, the complex conjugate f also belongs to A.

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APPENDIX ON GENERAL TOPOLOGY

Let B c A be the set of functions in A that assume only real values. If x # y, there is an f =f, + if2 e A with f(x) # fly). Thus either f,(x) #f,(y) orf2(x) #f2(y) Nowf, = (f + f)/2 andf2 = (f - f)/2i belong to A by hypothesis; hence fl, f2 E B since f, and f2 are real-valued. Thus B separates points. Furthermore, if x e 0, then f(x) # 0 for some f e A; but PROOF. (a)

If I2

= ff a B, so B satisfies the hypothesis of A7.5(a). If f = fl + if2 e C(Q), then f, and f2 can be uniformly approximated by functions in B; hence f can be uniformly approximated by functions in A. (b) This follows from (a) exactly as in A7.5.

If 0 is the unit disk in the complex plane and A is the collection of all analytic functions on S2, then A is a closed subalgebra of C(52) that separates points and contains the constant functions; but A # C(Q). Thus the hypothesis f e A implies f E A cannot be dropped in A7.6.

A8 Topologies on Function Spaces In this section we examine spaces consisting of functions from an arbitrary set 52 to another set 521. Some structure will be imposed on 52, ; it can be a metric space or, more generally, a gauge space, which we now define.

Definitions and Comments. A gauge space is a space 52, whose topology is determined by a family -9 = -9(01) of pseudometrics; thus a A8.1

subbase for the topology is formed by the sets Bd(x, b) = (y a 521: d(x, y) < S),

x e 52,, S > 0, d e .9. [A pseudometric has all the properties of a metric except that d(x, y) can be 0 for x :A y.] Convergence of x to x in 521 means

all de-9. d(x,,, Since Bd,(x, S) n BdZ(x, S) = Bd(x, S),

where d is the pseudometric max(d,, d2), it follows that the sets Bd. (x, S), d' = max(di...... di ), n = 1,

2, ... , the dik a _Q, form a base for the topology. Denote by -9 + =9 ' (rl,) the collection of all pseudometrics d+. The gauge space 01 is Hausdorff if whenever x # y, there is a d e 0 with d(x, y) # 0; in this case, -9 is said to be separating. The Hausdorff gauge spaces coincide with the completely regular spaces. To see this, let x belong to the open set U in the gauge space 52,. If de 9+ and Bd(x, S) e U, letf(y) = min(l, d(x, y)/S). Then f is continuous, 0:5f:5 1, f(x) = 0, and f = 1 off U. Conversely, if 0, is completely regular, it can be embedded in a product of closed unit intervals (see A4.6). Now a product of

gauge spaces is a gauge space [take d(x, y) = di(xi, yi), where the d; are

227

A8 TOPOLOGIES ON FUNCTION SPACES

pseudometrics for the ith coordinate space], and a subspace of a gauge space is a gauge space, and the result follows. If -9 consists of countably many pseudometrics dl, d2, ... , let 00

d=

"=1

min(d", 2-");

the single pseudometric d incudes the topology of 521; hence S21 is pseudometrizable (metrizable if Q, is Hausdori). We now discuss function space topologies.

Definitions and Comments. Let F(12, 521) be the collection of all functions from the set 0 to the gauge space C. If d is a family of subsets of A8.2

52, the topology the sets

,,, of uniform convergence on members of d has as subbase (g e F(12, 121): sup d(g(x), f (x)) < 8}, xeA

Aed,

b>0,

de-9.

Convergence relative to .d means uniform convergence on each A e sad. For example, if W = {12}, we obtain the topology of uniform convergence on 52, also called the uniform topology; if fl is a topological space and sad consists of all compact subsets, we obtain the topology of uniform convergence on compact sets. If sd is the collection of all singletons {x}, x e S2, we have the topology of pointwise convergence. Adaptations of standard proofs in metric spaces show the following: A8.3 Theorem. If C(1), (21) is the collection of all continuous maps from the topological space Ci to the gauge space 01, then C(12,121) is closed in F(12,121)

relative to the topology of uniform convergence on 52. In other words, a uniform limit of continuous functions is continuous. PROOF. If f" e C(52,121), f" -+f uniformly on 12, x0 a 92, and d e ..9, then

d(f(x),f(x0)) <_ d(f(x),f"(x)) + d(,,(x),f"(x0)) + d(f (x0),f(x0)) The result follows just as in the metric space proof. I A8.4 Theorem. If f is a continuous mapping from the compact gauge space

12 to the gauge space Q1i then f is uniformly continuous; that is, if d e 3(521)

and s > 0, there is a d+ e -9 +(1)) and a 8 > 0 such that if x1i x2 a 12 and d+(x1, x2) < S, then d(f(x1), f(x2)) < e.

228

APPENDIX ON GENERAL TOPOLOGY

PROOF. If f is not uniformly continuous, there is an e > 0 and e e 2(Sfl) such that for all S > 0, d+ a 2+(S2), there are points x, y e S2 with d+(x, y) < (5 but e(f (x), fly)) >_ 8. If we choose such (x, y) for each S > 0 and d e 2(12), we obtain a net with d(x,,, 0 but e for all n. (Take (61 i d) >_ (62, d') iff S1 5 a2 .) By compactness we find a subnet with yn,.) -> (xo, yo) for some xo, yo a fl; thus f(x,,,) -->f(xo), f(Y,,k) -'f(Yo)

by continuity. But 0, so d(xo, yo) = 0 and consequently, e(f(xo), f(yo)) = 0 by continuity off, a contradiction. I We now prove a basic compactness theorem in function spaces.

Arzela-Aseoli Theorem. Let 0 be a compact topological space, S21 a Hausdorff gauge space, and G c C(12, f21), with the uniform topology. Then G is compact iff the following three conditions are satisfied: A8.5

(a)

G is closed,

(b) {g(x): g e G} is a relatively compact subset of S21 for each x e 12, and (c) G is equicontinuous at each point of 92; that is, if e > 0, dc- 2(S)1),

x0 a 92, there is a neighborhood U of x0 such that if x e U, then d(g(x), g(xo)) < e

for all geG. PROOF. We first note two facts about equicontinuity. (1) If M e F(0, S21), where Q is a topological space and l1 is a gauge space, and M is equicontinuous at x0, the closure of M in the topology of pointwise convergence is also equicontinuous at x0. (2) If M is equicontinuous at all x E S2, then on M, the topology of pointwise convergence coincides with the topology of uniform convergence on compact subsets.

To prove (1), let f e M, f - f pointwise; if d e Q(S21), we have

d(f(x),f(x0)) <

d(1f(x),f,(xo)) + d(1f(xo),.f(xo))

If S > 0, the third term on the right will eventually be less than 6/3 by the pointwise convergence, and the second term will be less than 6/3 for x in some

neighborhood U of x0, by equicontinuity. If x e U, the first term is eventually less than 6/3 by pointwise convergence, and the result follows. To prove (2), let f, e M, f f pointwise, and let K be a compact subset of n; fix 6 > 0 and d e 2' (S21). If x c- K, equicontinuity yields a neighborhood

A8 TOPOLOGIES ON FUNCTION SPACES

229

U(x) such that y c U(x) implies d(f,(y), f (x)) < 6/3 for all n. By compactness,

K c U'= U(x;) for some x1, ..., x,. Then 1

d(f(x),f.(x)) < d(f(x),f(x;)) + d(f(x1),f,(xi)) + d(f.(x1),f (x)). If x e K, then x e U(x;) for some i; thus the third term on the right is less than 6/3 for all n, so that the first term is less than or equal to 6/3. The second term

is eventually less than 6/3 by pointwise convergence, and it follows that f -+f uniformly on K. Now assume (a)-(c) hold. Since G fXEn{g(x): g e G}, which is pointwise compact by (b) and the Tychonoff theorem A5.4, the pointwise closure is a net in G, there is a subnet Go of G is pointwise compact. Thus if converging pointwise to some g e Go. By (c) and (1), Go is equicontinuous at each point of f2; hence by (2), the subnet converges uniformly to g. But g is continuous by A8.3; hence g e G by (a). Conversely, assume G compact. Since S21 is Hausdorff, so is C(f2, f ) [as well as F(f, S21)], hence G is closed, proving (a). The map g -+ g(x) of G

into f is continuous, and (b) follows from A5.3(a). Finally, if G is not equicontinuous at x, there is an E > 0 and a d e k(ill) such that for each neighborhood U of x there is an xu e U and gu e G with d(gv(x), g11(x11)) >_ E. If U >_ V means U V, the gu form a net in G, so there is a subnet converging uniformly to a limit g e G. But xu --> x; hence g11(xu) - g(x), a contradiction. [The last step follows from the fact that the map (x, g) -, g(x) of fl x G into

f21 is continuous. To see this, let x -> x, and g -+g uniformly on fl; if d e .9(S21), then d(gn(xn), g(x)) <_ d(gn(xn), g(xn)) + d(g(x ), g(x)).

The first term approaches 0 by uniform convergence, and the second term approaches 0 by continuity of g.] If G satisfies (b) and (c) of A8.5, but not necessarily (a), the closure of G in the uniform topology satisfies all three hypotheses, and hence is compact. In the special case when S21 is the set of complex numbers, C(C2, .01) = C(f) is

a Banach space, and in particular, compactness and sequential compactness are equivalent. We thus obtain the most familiar form of the Arzela-Ascoli theorem: If f1, f2 , ... is a sequence of continuous complex-valued functions on the compact space fl, and if the f are pointwise bounded and equicontinuous, there is a uniformly convergent subsequence. In fact the f must be uniformly bounded by the continuity of the map (x, g) -- g(x) referred to above.

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A9 Complete Metric Spaces and Category Theorems A9.1 Definitions and Comments. A metric space (or the associated metric d) is said to be complete iff each Cauchy sequence (that is, each sequence

such that d(x", xm) --> 0 as n, m - oo) converges to a point in the space. Any compact metric space is complete since a Cauchy sequence with a convergent subsequence converges. It is important to recognize that completeness is not a topological property. In other words, two metrics, only one of which is complete, may be equivalent, that is, induce the same topology. As an example, the chordal metric d' on the complex plane [d'(z1, z2) is the Euclidean distance between the stereographic

projections of z, and z2 on the Riemann sphere] is not complete, but the equivalent Euclidean metric d is complete. Note also that the sequence z" = n

is d'-Cauchy but not d-Cauchy, so the notion of a Cauchy sequence is not topological. The subset A of the topological space 0 is said to be nowhere dense if the

interior of A is empty, in other words, if the complement of A is dense. A set B c 0 is said to be of category 1 in (1 if B can be expressed as a countable union of nowhere dense subsets of a; otherwise B is of category 2 in Q. The following result is the best known category theorem. Baire Category Theorem. Let (S), d) be a complete metric space. If A.

A9.2

is closed in (I for each n = 1, 2, ..., and U"'=1 A. = S2, then the interior A"° is nonempty for some n. Therefore i2 is of category 2 in itself.

PROOF. Assume A"° = 0 for all n. Then Al # f), and since S2 - A, is open,

there is a ball B(x1, S1) c 0 - A, with 0 < a1 < 1. Now B(x1i 2a1) ¢ A2 since A2° = 0; therefore B(x1,161) - A2 is a nonempty open set, hence includes a ball B(x2, l52) with 0 < b2 < 1. Inductively, we find B(xn, an), 0
B(x", 8n)rAn=0]. Now ifn<m, m-1

d(xn

a

x,,,):5 F d(xi i=n

m-1

a xi,. 1) < Y, 2 -'-+ 0 i=n

as

n, m --' oo,

so by completeness, xn converges to some x e 0. Since xk a B(xn, 48") for k > n, we have x e B(xn, Sn) for all n, hence x cannot belong to any An, a con'tradiction.

In A9.2, if for each n = 1, 2, ..., U. is an open dense subset of S2, then

U.0 0, for if the intersection is empty, then 0 is the union of the n, nowhere dense sets f) - Un. We prove a stronger result: 1

=

A9 COMPLETE METRIC SPACES AND CATEGORY THEOREMS

A9.3

231

Theorem. If U. is an open dense subset of the complete metric space 12

(n = 1, 2, ...), then n 1 U is dense. PROOF. Consider any ball B. Then B is a closed subset of the complete metric

space n, hence B is complete. Now U. n B is dense in B since B is open; hence U n B (and therefore U n B) is dense in B. By the remark after A9.2,

(n 1 Un) n Bo 0. Thus n' 1 U. meets every closed ball, hence every open ball, hence every nonempty open set. I Other important spaces satisfy the conclusion of A9.3: A9.4 Theorem. If U is an open dense subset of the locally compact Hausdorff space fl (n = 1, 2, ...), then nn 1 U. is dense.

PROOF. Let U be a nonempty open subset of f1; we must show that

U n nn U,,# 0. Since U1 is dense, U n U1 # 0, and by local compactness (see A5.12), there is a nonempty, relatively compact, open set V1 with V1 c U n U1. 1

Now V1 n U2 0 so we may choose a nonempty, relatively compact, open set V2 with V2 c V1 n U2. Inductively we select nonempty, relatively compact, open sets V. with V. c V decrease with it, they have the finite intersection property, and

since all V are subsets of the compact set V1. n-, V, # 0. If x belongs to this intersection, then x e U n n 1 U.. Definition. The topological space f2 is said to be a Baire space if the intersection of countably many open dense subsets of n is dense in f2. By A9.3 and A9.4, complete metric spaces and locally compact Hausdorff A9.5

spaces are Baire spaces.

SZ is a Baire space if for each A of category 1 in Q, A° = 0; that is, fl - A is dense. (b) Let f2 be a Baire space. If fZ = Un 1 A. where the A. are closed,

A9.6 Theorem. (a)

then

0 0 for some n. Thus f2 is of category 2 in itself.

PROOF. (a)

Assume fZ is a Baire space, and let A = UH 1 A,,, where the An

are nowhere dense. Then K) - A = n 1 (0 - A.) D n 1(Q - A.), which is dense by hypothesis. Conversely, if every set of category 1 in fZ has empty interior, let U. be

dense in f2, n = 1, 2..... Then El - U,, is (closed and) nowhere dense, so

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Un 1 (S2 - U") is of category 1 in S2. By hypothesis, the complement of this

set, namely (1 , U., is dense. (b) If A"° = 0 for all n, then fl is of category 1 in itself, contradicting

(a). I There are spaces that cannot be expressed as a countable union of nowwhere dense sets, but are not Baire spaces. (As an example, consider the free union of a Baire space and a non-Baire space.) Definition. The topological space Q is said to be topologically complete iff there is a complete metric don 0 that induces the given topology. The following sequence of results is aimed at characterizing topologically complete spaces. A9.7

A9.8 Theorem. Let (0, d) be a complete metric space. If U is an open subset

of S2, U is topologically complete. PROOF. If x, y e U, let

d'(x, y) = d(x, y) + If(x) -f(y)I, where f (x) =

[d(x, i) - U)]-'. Then d' is a metric on U, and d-convergence is equivalent to d'-convergence by continuity off. Thus d' induces the original topology. If {x"} is d'-Cauchy, it is d-Cauchy; hence it converges (relative to d) to some x e !Q. But x e U; for if not, then f(x") -+f(x) = oo, contradicting {x"} d'-Cauchy. Since d and d' are equivalent on U, x,, converges to x relative to

d'. I A9.9

Theorem. Let (0, d) be a complete metric space, and U., n = 1, 2, ... ,

open subsets of n. If A = l , U", that is, if A is a G5, then A, with the relative topology, is topologically complete. PROOF. Let d,, be a complete metric for U,, with d,, equivalent to d [d,, exists

by A9.8; we may assume that d,, < 1, for if not replace d,, by d"/(1 + d").] Define g: A B = fl 1 U; by g(x) = (x, x, ...), and define a metric d° on B by d°(x, y)= Y;°_ 2-`d,(x,, y;). Then d° induces the product topology on B, and a sequence {P)} in B is d°-Cauchy if {x;")} is d; Cauchy for each i. It follows that d° is complete. (Thus we have shown that a countable product of topologically complete spaces is topologically complete.) Now g is a homeomorphism of A and g(A) c B, and in fact g(A) is closed

in B. For if (x", x", ...) - y e B, then for each i we have d,(x", y,) - 0 as n - oo, so that x" -+ y, . But then y, is the same for all i, say y, = yo. Since

A9 COMPLETE METRIC SPACES AND CATEGORY THEOREMS

233

yi e U; for each i we have yo e n? , U; =A, hence y = (yo, yo, ...) eg(A), proving g(A) closed. Since B is complete, so is g(A), and since g is a homeomorphism, A is topologically complete. I

It follows from A9.9 that the irrational numbers are topologically complete. In fact, a complete metric for the set of irrationals in (0, 1) is given by

d(.a, a2 ..., b, b2, ...) = 1/n, where n is the first integer for which a 0 b (in the decimal expansion). A9.10 Theorem. Let A be a topologically complete subspace of the Hausdorfl space 0. If A is dense in Q, then A is a Ga .

PROOF. Let U. = {x a S2: for some neighborhood V of x, the diameter of V n A is less than 1/n}. (The diameter is calculated using a particular complete metric d for A.) If x e U,,, then x c- V c U,,, so U,, is open in Q. We are going to show that A =no , U,,. Let x e A, and let Vo = {y e A: d(x, y) < 1/4n). Then Vo is open in A; hence Vo = V n A, where V is open in S2. Now x e Yo c V, and if y, z e Vo, we have d(y, z) < d(y, x) + d(x, z) < 1/2n, so that Vo has diameter less than or equal to 1/2n < 1/n. It follows that xe nn , U,,. Conversely, assume x e nn , Un. Since A is dense, there is a net xi e A with xi -+ x. For each n, let V be a neighborhood of x such that V n A has

diameter less than 1/n. Eventually xi e V, and it follows that for some index i,, we have d(xi, x;) < 1 In for i, j >: in. (In other words, the net is Cauchy.) It may be assumed that i,,< in,, for all n; then the xi form a Cauchy sequence which converges to a limit y e A by completeness. Since S2 is Hausdorfi, we have y = x; hence x e A.

We now prove the main theorem on topological completeness. A9.11

Theorem. Let n be a metrizable topological space. Then S2 is topolog-

ically complete if it is an absolute Ga; in other words, 0 is a G,, in every metric space in which it is topologically embedded.

PROOF. If S2 is topologically complete and is embedded in the metric space Q,, then 0 is dense in 31; so by A9.10, S2 is a Ga in 31. Thus 12 can be expressed

as nR , W,,, where W = U,, n I2, U,, open in 52,. But 0 is a closed subset of the metric space 52,, so S2 is a G,, in Q, (rl = nn , {x a S2, : d(x, S2) < 1/n)). It follows that S2 is a Ga in 52,.

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APPENDIX ON GENERAL TOPOLOGY

Conversely, let fl be an absolute G. Embed (I in its completion (we assume as known the standard process of completing a metric space by forming equivalence classes of Cauchy sequences). By A9.9, f) is topologically

complete. I We now establish topological completeness for a wide class of spaces. A9.12

Theorem. A locally compact metric space is topologically complete.

PROOF. If Q is such a space and KI is its completion, then 4 is open in S2. For if x e ), there is, by local compactness, an open (in S2) set V with x e V n S2 and

V n (I compact. In fact V c S2, proving D open. For if y e V and y 0.0, then for each Ue °W(y), U n V n f A0 since 0 is dense; consequently, U n V n fl # 0. The sets U n V n f, U e all(y), form a filterbase -4 in V n fl converging to y, and since S1 is Hausdorff, y is the only possible accumulation point of .411. But y 0 V n 0, contradicting compactness of V n S2 [see A5.2(d)]. By A9.8, f2 is topologically complete.

A10 Uniform Spaces We now give an alternative way of describing gauge spaces.

Definitions and Comments. Let V be a relation on the set fl, that is, a subset of fl x (I. Then V is called a connector if the diagonal A10.1

D = {(x, x): x e 0} is a subset of V. A nonempty collection .

of connectors is called a uniformity

if

(a)

for all VL, V2 e .., there is a We Y with W c VL n V2 (in other

words W is a filterbase), and (b)

for each V e, there is a W e .

with WW -' c V.

[The relation WW-' is the composition of the two relations W and W -'; that is, (x, z) e WW -' if for some y e Q we have (x, y) e W -' and (y, z) a W; (x, y) e W ' means (y, x) e W.]

If in addition, .*' is a filter, that is, if V e .e and V c W, then W e .y, then .*' is called a uniform structure. In particular, if _* is a uniformity, the filter generated by ..£° is a uniform structure.

A10 UNIFORM SPACES

235

As an example, let -9 be a family of pseudometrics own, and let W consist of all sets of the form V = {(x, y): d+(x, y) <S}, b > 0, d+ e _Q+; then . is a uniformity on fl. [To see that condition (b) is satisfied, let

W = {(x, y): d+(x, y) <S/2};

then WW - 1 c V by the triangle inequality.] By analogy with the pseudometric case, if V belongs to the uniformity W, we say that x and y are V-close if (x, y) e V. Two uniformities W, and .7°2 are said to be equivalent if they generate

the same uniform structure; in other words, given V2 a -W2, there is a V, e *, with V, c V2 and given W, a .W,, there is a W2 E

2

with W2 c W1.

Two uniform structures that are equivalent must be equal; hence there is exactly one uniform structure in each equivalence class of uniformities.

If Y is a uniform structure and V, W e W, WW -1 c V, then if D is the diagonal, we have

W-1 = DW-1 cWW-1 hence W c V -1, so that V VnV

V,

a .. But then e

(and V n V-1

V).

Now U = V n V is symmetric, that is, U = U -1. Thus if . is a uniform structure, the symmetric sets in .; form a uniformity that generates 0, so that every uniformity is equivalent to a uniformity containing only symmetric sets. We are going to show later that every uniform structure is generated by a uniformity corresponding to a family of pseudometrics. As a preliminary, we establish the following result:

Metrization Lemma. Let U" , n = 0, 1, 2, ..., be a sequence of subsets of 0 x S2 such that each U,, is a connector, U0 = 0 x f), and U,3+1 A10.2

(= U"+1U"+lU"+1) e U,, for all n. Then there is a function d from S2 x S2 to the nonnegative reals, such that d satisfies the triangle inequality and U. c {(x, y): d(x, y) < 2-") c U"_1 for all n = 1, 2. .... If each U. is symmetric, there is a pseudometric satisfying this condition.

PROOF. Define f(x, y) = 0 if (x, y) e U. for all n; f(x, y) = # if (x, y) a U0 U1 ; f (x, y) = I if (x, y) e U, - U2, and in general, f (x, y) = 2-n if

(x, y) e U"_, - U". Since U"+, c U.+, c U,,, f is well-defined on C2 x fl; furthermore, (x, y) e U,,

if f (x, y) <2 -".

(1)

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APPENDIX ON GENERAL TOPOLOGY

If x, y e 0, let d(x, y) = inf Y"=0 f(xi, xi+1), where the inf is taken over all finite sequences x0, x1, ..., xn+1 with xo = x, x"+1 = Y Since a chain from x to y followed by a chain from y to z yields a chain

from x to z, d satisfies the triangle inequality, and since d(x, y) <- f(x, y), U. c {(x, y): d(x, y) < 2 -") by (1). If the U. are symmetric, then so is d; hence d is a pseudometric. Now we claim that n

f(xo, xn+1) <- 2 Y f(xi, xi+1)

for any xo, ... , xn+1

i=O

(2)

This is clear for n = 0, so assume the inequality for all integers up to

n - 1. Let a = Y"=0f(xi, xi+1) If a = 0, then (xi, xi+1) e n o Un, and since Ui3 c Ui_1 we have (xo, xn+1) e U. for all i, and there is nothing further to prove. Thus assume a > 0. Let k be the largest integer such that Yk=o f(xi, x;+1) < a12 [iff(xo, x1) >

a/2, take k = 0]. Then Y+=o f(xi, xi+1) > a/2, so y"=k+1f(xi, xi+1) <_ a/2.

By induction hypothesis, f(xo, xk) < 2 Ek=o f(xi, xi+1) < 2a/2 = a and Y-i=k+ls(xi, xi+l) 2a/2 = a Also, (xk+l , xn+1) C 2 n

f(xk, xk+1) ! Y f(xi, xi+l) = a. i=0

Let m be the smallest nonnegative integer such that 2-' :!9 a. If m _< 2, then (2) holds automatically since f <_ 3; thus assume m > 2. Then 2-' :5; a < 2`(in-1), so f(xo, xk) < a < 2-(rn-1), and therefore by (1), (xo, xk) [and similarly, (Xk, xk+l), (xk+1, xn+1)] belongs to Uni_1. Thus (xo, xn+1) e which implies that f(xo, xn+1) < 2-("'-2); that is, Un3,_1 C

f(x0, x"+1)

:2-(m-1) < 2a,

proving (2).

Finally, if d(x, y) < 2-", then for any chain from x to y, (2) yields (with x0 = x, xn+1 = y)f(x, y) <_ 2d(x, y) < 2-("-1), so (x, y) e Un-1. 1 A10.3 Metrization Theorem. If ,° is a uniformity, the following conditions

are equivalent: (a) . is pseudometrizable; that is, there is an equivalent uniformity generated by a single pseudometric. (See the example in A10.1 for the construction of the uniformity corresponding to a family of pseudometrics.) (b) . has a countable base; that is, there are countably many sets V in the uniform structure generated by .; ' such that if U e A, then V" c U for some n.

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A10 UNIFORM SPACES

PROOF. If .' is pseudometrizable with pseudometric d, the sets {(x, y): d(x, y) < r},

r rational, form a countable base. Conversely, assume Al has a countable base Vo, V1, V2, .... We may assume that the V" are symmetric (if not, consider

V. n V.'), V. c Vn+1 for all n (if not, consider n:=o V;), and V. = 0 x I. Take U0 = V0, U1 a symmetric set in the uniform structure .°F generated such by . such that U13 c V1 (c Uo); let U2 be a symmetric set in that U2' r-- V2 n U1. In general, let U. be a symmetric set in ..y such that U"3 c V" n U"_1. (To see that the U" exist, let W. be a symmetric set in F W", with W"2 C V" n U"_ I. If U. is a symmetric set in F such that U"2 W"2 C V" n U"_1. Also, since U. (- UU3 c V", then U"3 = U"3D c U,,4 c the U,, form a base for .*°.) By A10.2, there is a pseudometric d such that U. c {(x, y): d(x, y) < 2-"} e U"_1i n = 1, 2, .... It follows that . is equivalent to the uniformity of d. I

We now construct a topology arising from a uniformity. Definitions and Comments. Let . be a uniformity. The topology induced by or is defined by specifying the system Yl'(x) of overneighborhoods

A10.4

of each point x. If V e -°, let V(x) = {y: (x, y) e V}. We take $ (x) as the class of all oversets of the sets V(x), V e . °. It is immediate that the first three axioms for an overneighborhood system are satisfied (see A1.1). We verify the last axiom [A1.l(d)]. If V e .*', let U be a symmetric set in the uniform structure generated by Y e such that U2 C V.

If W e .3t° and W c U, we claim that W(y) c V(x) for each y e W(x). For if z e W(y), then (y, z) e W; but also (x, y) e W, and hence (x, z) E W' c U 2 c V, so z e V(x), as asserted. Thus V(x) e V(y) for ally e W(x) C V(x), as required in A I. I (d).

It follows that there is a unique topology having (for each x) the sets V(x), V e .y°, as a base for the overneighborhoods of x. Furthermore, any uniformity equivalent to W will induce the same topology. A topological space whose topology is determined in this way is called a uniform space. In particular, let .9 be a family of pseudometrics on 0, and let .-t° consist then of all sets V = {(x, y): d+(x, y) < S}, S > 0, d+ E

V(x)={y:d+(x,y)<S}. Thus the topology induced by Y e coincides with the gauge space topology determined by -9 (see A8.1). In other words, every gauge space is a uniform space. Conversely, every uniform space is a gauge space, as we now prove.

238

APPENDIX ON GENERAL TOPOLOGY

Theorem. Let 3F be a uniform structure, and let A' be the uniformity generated by the pseudometrics d such that the sets Vda = {(x, y): d(x, y) < 8} belong to .F for all b > 0. Then F and .*' are equivalent, so that the topology A10.5

induced by :" is identical to the gauge space topology determined by the pseudometrics d.

PROOF. It follows from the definition of .*' that . c .F. Now if U is a symmetric set in F, set U0 = S2 x f2, U, = U, U2 a symmetric set in F such that U23 c U1, and in general, let U. be a symmetric set in .°F such that C By A10.2, there is a pseudometric don d2 with U. C Vd.2-^ C Un-1

for all n. Since .F is a uniform structure, V2 - e .F for all n, hence V db a F' for all S > 0. But U2 C_ Vd, 114 c U1 = U, proving the equivalence of .F and

.r. I

Since uniform spaces and gauge spaces coincide, the concept of uniform continuity, defined previously (see A8.4) for mappings of one gauge space to

another, may now be translated into the language of uniform spaces. If f. 12 -- 521, where f2 and n, are uniform spaces with uniformities ?l and f-, f is uniformly continuous iff given V e V, there is a U e'W such that (x, y) e U implies (f(x), f(y)) e V. We now consider separation properties in uniform spaces. Theorem. Let f) be a uniform space, with uniformity .*'. The following are equivalent:

A10.6

(a) (1{ V: V e 0} is the diagonal D. (b) C1 is T2. (c) (d)

f)isT1. S2 is To.

PROOF. If (a) holds and x 96 y, then (x, y) 0 V for some V e Y. If W is a symmetric set in the uniform structure generated by .', and W2 c V, then W(x) and W(y) are disjoint overneighborhoods of x and y, proving (b). If (d) holds and x y, there is a set V e .W such that y 0 V(x) [or a set W e ay with x 0 W(y)]. But then (x, y) t V for some V e *, proving (a). I Finally, we discuss topological groups and topological vector spaces as uniform spaces. A topological group is a group on which there is defined a topology which

makes the group operations continuous (x,, -* x, y - y implies x y -+ xy;

A10 UNIFORM SPACES

239

x -+ x implies xK ' -+ x '). Familiar examples are the integers, with ordinary addition and the discrete topology; the unit circle {z: Izi = 1) in the complex plane, with multiplication of complex numbers and the Euclidean topology; all nonsingular n x n matrices of complex numbers, with matrix multiplication and the Euclidean topology (on R"2). If a = Ili S , where the Di are topological groups, then n is a topological group with the product topology if multiplication is defined by xy = (x,y;, i e I).

Theorem. Let n be a topological group, and let .*' consist of all sets VN = {(x, y) E 11 x 0: yx-' e N}, where N ranges over all overneighborhoods of the identity element e in Q. Then -ye is a uniformity, and the topology induced by .f° coincides with the original topology. In particular, A10.7

if 12 is Hausdorif, it is completely regular. PROOF. The diagonal is a subset of each VN, and Vs n VM = VN n M, so only the last condition [A10.1(b)] for a uniformity need be checked. Letf(x, y) _ xy-I, a continuous map of 92 x 0 into g (by definition of a topological group). If N is an overneighborhood of e, there is an overneighborhood M

of e such that f(M x M) c N. We claim that VM VM' c VN. For if (x, y) e VM' and (y, z) a VM, then

(y, x) e VM; hence xy-' e M and zy-' a M. But zx-' = (zy-')(yx-') = (zy-')(xy-')-' ef(M x M) c N. Consequently, (x, z) e VN, so ,Y is a uniformity. Let U be an overneighborhood of the point x in the original topology. If

N = Ux-' ={yx-':ye U), then VN(x)={y:(x,y)eVN}={y:yx-'eN)=

Nx = U; therefore U is an overneighborhood of x in the topology induced by

Conversely, let U be an overneighborhood of x in the uniform space topology; then VN(x) c U for some overneighborhood N of e. But VN(x) = Nx, and the map y - yx, carrying N onto Nx, is a homeomorphism of S2 with itself. Thus Nx, hence U, is an overneighborhood of x in the original topology. I A10.8

Theorem. Let f) be a topological group, with uniformity .;t° as

defined in A 10.7. The following are equivalent: is pseudometrizable (see A10.3). S2 is pseudometrizable; that is, there is a pseudometric that induces the given topology. (c) f2 has a countable base of neighborhoods at the identity e (hence at every x, since the map N - Nx sets up a one-to-one correspondence between neighborhoods of e and neighborhoods of x). (a) (b)

.

240

APPENDIX ON GENERAL TOPOLOGY

PROOF. We obtain (a) implies (b) by the definition of the topology induced by ,e (see A10.4). Since every pseudometric space is first countable, it follows that (b) implies (c). Finally, if (c) holds and the sets N1, N2, ... form a countable base at e, then by definition of Ye, the sets VN,, VNz .... form a countable

base for 0; hence, by A10.3, 0 is pseudometrizable. I A topological vector space is a vector space with a topology that makes

addition and scalar multiplication continuous (see 3.5). In particular, a topological vector space is an abelian topological group under addition. The uniformity -ye consists of sets of the form VN = {(x, y): y - x e N), where N is an overneighborhood of 0. Theorem A10.8 shows that a topological vector space is pseudometrizable if there is a countable base at 0. The following fact is needed in the proof of the open mapping theorem (see 3.5.9): A10.9

Theorem. If L is a pseudometrizable topological vector space, there is

an invariant pseudometric [d(x, y) = d(x + z, y + z) for all z] that induces the topology of L. PROOF. The pseudometric may be constructed by the method given in A10.2

and A10.3, and furthermore, the symmetric sets needed in A10.3 may be taken as sets in the uniformity Jr itself rather than in the uniform structure generated by Ye. For if N is an overneighborhood of 0, so is -N =

{-x: x e N) since the map x -p -x is a homeomorphism. Thus M = N n (- N) is an overneighborhood of 0. But then VM is symmetric and VMcVN. Now by definition of VM , we have (x, y) e V. iff (x + z, y + z) e VM for

all z, and it follows from the proof of A10.2 that the pseudometric d is invariant.

Bibliography

Apostol, T.M., " Mathematical Analysis." Addison-Wesley, Reading, Massachusetts, 1957.

Bachman, G., and Narici, L., " Functional Analysis." Academic Press, New York, 1966. Billingsley, P., "Convergence of Probability Measures." Wiley, New York, 1968. Dubins, L., and Savage, L., "How to Gamble If You Must." McGraw-Hill, New York, 1965.

Dugundji, J., "Topology." Allyn and Bacon, Boston, 1966. Dunford, N., and Schwartz, J. T., " Linear Operators." Wiley (Interscience), New York, Part 1, 1958; Part 2, 1963; Part 3, 1970. Halmos, P. R., " Measure Theory." Van Nostrand, Princeton, New Jersey, 1950. Halmos, P. R., " Introduction to Hilbert Space." Chelsea, New York, 1951. Halmos, P. R., " Naive Set Theory." Van Nostrand, Princeton, New Jersey, 1960. Hausdorff, F., " Set Theory." Chelsea, New York, 1962. Kelley, J. L., and Namioka, I., " Linear Topological Spaces." Van Nostrand, Princeton, New Jersey, 1963. Liusternik, L., and Sobolev, V., "Elements of Functional Analysis." Ungar, New York, 1961.

LoBve, M., "Probability Theory." Van Nostrand, Princeton, New Jersey, 1955; 2nd ed., 1960; 3rd ed., 1963.

Neveu, J., " Mathematical Foundations of the Calculus of Probability." Holden-Day, San Francisco, 1965. Parthasarathy, K., "Probability Measures on Metric Spaces." Academic Press, New York, 1967.

Royden, H. L. "Real Analysis." Macmillan, New York, 1963; 2nd ed., 1968. Rudin, W., " Real and Complex Analysis." McGraw-Hill, New York, 1966. Schaefer, H., " Topological Vector Spaces." Macmillan, New York, 1966.

Simmons, G., "Introduction to Topology and Modern Analysis." McGraw-Hill, New York, 1963.

Taylor, A. E., "Introduction to Functional Analysis." Wiley, New York, 1958. Titchmarsh, E. C., "The Theory of Functions." Oxford Univ. Press, London and New York, 1939. Yosida, K., "Functional Analysis." Springer-Verlag, Berlin and New York, 1968.

241

Solutions to Problems

Chapter 1

Section 1.1 2. 3.

We have iim sup. An = (-1, 1 ], lim inf,, A = {0}. Using lim sup,, An = {w: co e A. for infinitely many n}, lim infra An {co: co e An for all but finitely many n), we obtain

_

liminfAn ={(x,y): x2 +y2 < 1}, lim sup A. ={(x, y): x2 + y2 < 1) -{(0, 1), (0, -1)}. n

4.

If x = lim sup., x., then lim supra An is either (- co, x) or (- oo, x]. For if y e An for infinitely many n, then xn >y for infinitely many n; hence x >- y. Thus lim sup. An c (- oo, x]. But if y < x, then xn > y for infinitely many n, so y e lira sup. An. Thus (- co, x) c lim sup. A., and the result follows. The same result is valid for lim inf; the above analysis applies, with "eventually" replacing "for infinitely many n." 243

244

SOLUTIONS TO PROBLEMS

Section 1.2

4.

5. 8.

(a)

If - oo < a < b < c < oo, then µ(a, c] = µ(a, b] + µ(b, c], and

p(a, co) = p(a, b] + µ(b, oo); finite additivity follows quickly. If A" = (- oo, n], then A" I R, but µ(A") = n+). µ(R) = 0. Thus µ is not continuous from below, hence not countably additive. (b) Finiteness of p follows from the definition; since p(- oo, n] -> co, i is unbounded. We have p(U A,) >_ u(U°=1 A,) = Y; I p(A;) for all n; let n -- oo to obtain the desired result. The minimal a-field F (which is also the minimal field) consists of the 1

collection ig of all (finite) unions of sets of the form B1 n B2 n - n B., where B. is either A, or A,`. For any a-field containing A1, ..., A. must contain all sets in 9r ; hence c .F. But Ir is a a-field; hence F c 9r. n B", and each such Since there are 2" disjoint sets of the form B1 n

set may or may not be included in a typical set in .F, .F has at most 22" members. The upper bound is attained if all sets B1 n - - n B. are 9.

nonempty. (a) As in Problem 8, any field over ' must contain all sets in (; hence

F. But W is a field; hence F e W. For if A; = nj=1 B,j, then (Ui=1 A:)` = ni=1 U;=1 B`'J, which belongs to W because of the distributive law A n (B v C) = (A n B) v (A n C). (b) Note that the complement of a finite intersection ni=1 B;j belongs to -9; for example, if B1i B2 a W, then (B1 n B2")"

=B1`uB2 = (B1` n B2) U (B1` - B2c) u (B2 n B1) v (B2 r B1`) a !R. Now :a is closed under finite intersection by the distributive law, and it follows from this and the above remark that -9 is closed

under complementation and is therefore a field. Just as in the proof that F = 9r, we find that F = -9. 11.

(c) (a)

This is immediate from (a) and (b). Let A,, e ..', n - 1, 2, .... Then A,, belongs to some W,,., and we Let a = sup" of. < , so '',,, c 'a2 c may assume a1 < a2 < F'1. Then all Wa" a W., hence all A. e (ga. Thus U" An e Wa+1 e be, so U,, A e .9. If A e 91, then A belongs to some WQ; hence A` e 1ea+1 C Y.

(b) We have card 'a S c for all a. For this is true for a = 0, by hypothesis. If it is true for all 1 < a, then Up,,, Wa has cardinality at

245

CHAPTER I

most (card a)c = c. Now if -9 has cardinality c, then _9' has cardinality at most cx° = (2"0)x0 = 2'° = c. Thus card Wa < c. It follows that Up1 W. has cardinality at most c. Section 1.3

3.

(a)

Since A(O) = 0 we have n e ..#, and A' is clearly closed under complementation. If E, Fe . ' and A c 0, then

A[An(EvF)]=A[An(EuF)nE] + ALA n (E U F) n E']

since

E E .'

=A(AnE)+A(AnFt E`). Thus

A[An(EuF)]+A[An(EuF)`]

=A(AnE)+A(AnE`nF)+A(AnE`nF) = A(A n E) + A(A n E`) since Fe.ff = A(A)

since

E e M.

This proves that ,' is a field. Also, if E and F are disjoint we have

A[An(EvF)]A[An(EuF)nEI+A[An(EuF)nE`] =A(Ar-E)+A(Ar-FnE`) =A(AnE)+A(AnF) since EnF=Q. Now if the E are disjoint sets in ..1 and F. = Ui=1 E, T E, then 2(A) = A(A m F,,) + A(A r FCC) since F,, belongs to the field A'

A(AnF;)+2(Ar E`) since E` c Fn` and A is monotone n

i=1

A(AnE;)+2(Ar E`)

by what we have proved above. Since n is arbitrary, 00

A(A)

I A(A n En) + A(A n E") n=1

A(A n E) + A(A n E')

by countable subadditivity of A.

Thus E e . #, proving that . f' is a a-field.

246

SOLUTIONS TO PROBLEMS

Now A(A n E) + I(A n E`) > .1(A) by subadditivity, hence 00

A(A) = Z A(A n En) + ).(A n E`). n=1

Replace A by A n E to obtain I(A n E)

).(A n En), as de-

sired. (b) All properties are immediate except for countable subadditivity. If

A = U- 1 A., we must show that u*(A) < Y 1 µ*(A,), and we may assume that p*(A,,) < oo for all n. Given e > 0, we may µ(E"k) < µ*(An) + choose sets Enk e Fo with A. c Uk Enk and e2-". Then A e Un, k Enk and En, k µ(Enk) < [.,n µ*(An) + e. Thus µ*(A) < Yn µ*(An) + e, a arbitrary.

Now if A e ., then µ*(A) < µ(A) by definition of µ*, and if A c Un E,,, E. e .moo, then µ(A) < >Jn µ(E,) by 1.2.5 and 1.3.1. Take the infimum over all such coverings of A to obtain µ(A) < µ*(A); hence µ* = µ on .Fo. (c)

If Fe Fo , A c fl, we must show that µ*(A) >_ µ*(A n F) + + µ*(A n F`); we may assume µ*(A) < oo. Given E > 0, there are u*(A) + e. Now sets En e Fo with A e U. E. and E"= 1

(U (En n F))

µ*(A n F) <

by monotonicity

µ(EnnF) n

is countably subadditive and µ* = µ on

since p*

'0 Similarly,

µ*(A n F°) < y µ(E,, n F°). n

Thus

µ*(A n F) + p*(A n F°) < Y_µ(E,) < µ*(A) + s, and the result follows. (d) If A = B u N, where B e a(Fo), N c M e a(.Fo), µ*(M) = 0, then

B e 4 {note F0 e 4 and .,# is a or-field, so a(.°F0) c 4]. Also, any set C with p*(C) = 0 belongs to 4 by definition of p*-measurability; hence A e

Therefore the completion of o(F0) is included

in 4. Now assume y u-finite on Fo , and let A c-./#. If (I is the disjoint union of sets A. e JFo with µ(A,) < oo, then by definition of p*, there

is a set B. a a(.Fo) such that A n A. e B. and µ*(B, - (A n A.))= 0.

247

CHAPTER I

[Note that if A 0 .off we obtain only µ*(B") = µ*(A n A"); however if A e M (so that A n A" also belongs to M), we have

µ*(B"- (AnA"))=A*(B")-p*(AnA")=0.] If B = U" B", then B e a(FO), A c B, and u*(B - A) = 0. This argument applied to A` yields a set C e a(.F0) with C c A and µ*(A - C) = 0. Therefore A = C u (A - C) with C e

A-CcB-Cea(.Fo), and u*(B - C) = u*(B - A) + u*(A - C) = 0. Thus A belongs to the completion of a(. 0) relative to p*. Section 1.4 1.

Using the formulas of 1.4.5, the following results are obtained: (a) (d)

4.

(c) 3, (b) 8.5, (e) p(}, co) 7.25,

5,

+ p(- co, -D = 7.25.

Let Ck = {x a R": -k < x,:5 k, i = 1, ..., n}. Then µ is finite on Ck; hence the Borel subsets B of Ck such that µ(a + B) =µ(B) form a monotone class including the field of finite disjoint unions of right-semiclosed intervals in Ck ; hence all Borel subsets of Ck belong to the class (see 1.2.2). If B e J(R"), then B n Ck I B; hence a + (B n Ck) T a + B, and it follows that µ(a + B) = µ(B).

Now ifBe!(R"), then B=AvC,Ae (R"), CaDe9J(R"),with µ(D) = 0. Thus a + B = (a + A) u (a + C), and, by Problem 3, 5.

a + A e J(R"), a + C c a + D e R(R"). By what we have proved above, µ(A + D) = u(D) = 0; hence a + B e.4(R") and µ(a + B) = µ(B). Let A be the unit cube {x a R": 0 < x; < 1, i = 1, ..., n}, and let c= µ(A). For any positive integer, r, we may divide each edge of A into r equal parts, so that A is decomposed into r" subcubes A1, ..., A,, each with volume r-". By translation-invariance, µ(A,) is the same for all i, so if ) is Lebesgue measure, we have p(A) = r-"µ(A) = r-"c = r-"c.?(A) = c11(A),

i = 1, ..., r".

Now any subinterval I of the unit cube can be expressed as the limit of an increasing sequence of sets Bk, where each BA, is a finite disjoint union of subcubes of the above type. Thus y = cA on subintervals of the unit cube, and hence on all Borel subsets of the unit cube by the Caratheodory extension theorem. Since R" is a countable disjoint union of cubes, it follows that p = c A on Q(R"). 6. (a) If r + x1 = s + X2, x1, x2 a A, then x1 is equivalent to x2 , so that x1 = x2 since A was constructed by taking one member from each distinct Bx. Thus r = s, a contradiction.

SOLUTIONS TO PROBLEMS

248

If x e R, then x e B.,; if y is the member of B,, that belongs to A,

then x - y is a rational number r, hence x e r + A. (b) If 0< r< 1, then r + A c [0, 2]; thus

Y(µ(r + A): 0 < r < 1, r rational) = µ(U{r + A: 0 < r < I, r rational})

by (a)


LetF(x,y)=1 ifx+y>0;F(x,y)=0ifx+y<0.Ifa1 =0,b1 =1, a2 = -1, b2 = 0, then Lb,a, Obi., F(x, Y) = F(bl, b2) - F(al, b2) - F(bl, a2) + F(al, a2)

=1-1-1+0=-1;

hence F is not a distribution function. Other examples: F(x, y) _ max(x, y), F(x, y) = [x + y], the largest integer less than or equal to

x+y. 12.

(a)

First assume that u is finite. Let' be the class of subsets of R" having the desired property; we show that '' is a monotone class.

For let B. a If, B. T B. Let K" be a compact subset of B. with µ(B") < p(KK) + s, e > 0 preassigned. By replacing K" by Ui=1 K,, we may assume the K. form an increasing sequence. Then µ(B) _ u(B") < µ(K") + e, so that

µ(B) = sup{µ(K): K a compact subset of B}, and B e W. If B. a W, B"

.

B, let K" be a compact subset of B" such

that y(B") < p(K,.) + e2 -", and set K =n, 1 K.. Then

µ(B)-p(K)=p(B-K)5 (OB n=1

- Kr)) <- 1N(Bn-Ka):5 e; n=1

thus B e f. Therefore' is a monotone class containing all finite disjoint unions of right-semiclosed intervals (a right-semiclosed interval is the limit of an increasing sequence of compact intervals). Hence W contains all Borel sets.

If p is a-finite, each B e !(R") is the limit of an increasing sequence of sets B, of finite measure. Each B, can be approximated from within by compact sets [apply the previous argument to the

measure given by ui(A) = p(A n B;), A a R(R")], and the above argument that ' is closed under limits of increasing sequences shows that B E W.

249

CHAPTER 1

(b) We have p(B) < inf{p(V): V

V open}

B,

B, W = K`, K compact}.

< inf{p(W): W

If p is finite, this equals p(B) by (a) applied to B`, and the result follows.

Now assume p is an arbitrary Lebesgue-Stieltjes measure, and write R" = Uk 1 Bk, where the Bk are disjoint bounded sets; then B. c Ck for some bounded open set C,. The measure pk(A) _ p(A n Ck), A e R(R"), is finite; hence if B is a Borel subset of Bk and e > 0, there is an open set Wk B such that pk(Wk) < pk(B) + e2-k. Now Wk n Ck is an open set V. and B n Ck = B since B e Bk a Ck; hence p(Vk) < p(B) + s2`k. For any A e M(R"), let V. be an open set with Vk z) A n Bk and p(Vk) < p(A n Bk) + e2-k.

Then V = Uk Vk is open, V A, and p(V) < ik 1 p(Vk) < 1

p(A) + e. (c) Construct a measure p on M(R) as follows. Let p be concentrated

on S = {1/n: n = 1, 2,...) and take p{1/n} = 1/n for all n. Since

R=U

1

{l/n} u S` and p(S`) = 0, p is a-finite. Since 1

µ10,I] = Y00 - = 00,

n=1 n

p is not a Lebesgue-Stieltjes measure. Now p{0} = 0, but if V is an open set containing 0, we have

p(V) > p(-s, e) >_

1

k=r k

for some s

for some r

=00.

Thus (b) fails. (Another example: Let p(A) be the number of rational points in A.) Section 1.5

2.

If B e R(R),

{w:h(co)eB} ={coeA:h(co)eB} u{(vaA`:h(c))eB} = [A of -1(B)] u [A` n g- 1(B)] 5.

which belongs to F since f and g are Borel measurable. (a) {x: f is discontinuous at x} = Un 1 D, where D" = {x a Rk: for all S > 0, there exist x1i x2 e Rk such that Ix1 --- xJ < S and 1x2 - xI < S, but If(XI) -f(X2)1 >- 1/n). We show that the D. are closed. Let {xa}

SOLUTIONS TO PROBLEMS

be a sequence of points in D. with xQ - x. If 6 > 0 and N = {y: Iy - xI < b), then x,, e N for large a, and since X. a D", there are points x,1 and xa2 e N such that I f(xa1) -f(xa2)I >- 1/n. Thus Ix", - x1 < b, Ix.2 - x1 < 8, but f(xal) - f(xa2)I > 1/n, so that

xe D". The result is true for a function from an arbitrary topological space S to a metric space (T, d). Take D,, = {x e S: for every neighborhood N of x, there exist x1, x2 e N such that d (f (xl), f (X2)) > I /n}. (the above proof goes through with "sequence" replaced by " net.") In fact T may be a uniform space (see the appendix on general topology, Section A10). If -9 is the associated family of pseudo-

metrics, we take D. = {x e S: for every neighborhood of N of x, there exist x1, x2 e N and d e 2 such that d(f(x1), f(x2)) >- 1/n) and proceed as above. The result is false if no assumptions are made about the topology of the range space. For example, let S2 = (1, 2, 3}, with open sets 0, fl, and {1}. Define P. Q -. f2 by f(1) =f(3) = 1, f(2) = 2. Then the set of discontinuities is {3}, which is not an F,,. (b) This follows from part (a) because the irrationals I cannot be expressed as a countable union of closed sets C,,. For if this were possible, then each C,, would have empty interior since every nonempty open set contains rational points. But then I is of category 1 in R, and since Q = R - I is of category 1 in R, it follows that R is of category 1 in itself, contradicting the Baire category theorem. 6. By Problem 11, Section 1.4, there are c Bore] subsets of R"; hence there are only c simple functions on R". Since a Borel measurable function is

the limit of a sequence of simple functions, there are c"0 = c Borel measurable functions from R" to R. By 1.5.8, there are only c Borel 7.

measurable functions from R" to Rk. (a) Since the P,, are measures, J:k P"(Ak) = P"(S2) = 1, and it follows quickly that the a"k satisfy the hypotheses of Steinhaus' lemma. If {x"} is the sequence given by the lemma, let S = {k: xk = 1) and let B be the union of the sets Ak, k e S. Then t" =

L. [P,,(Ak) - P(Ak)J = 1 - CC keS 1

l

l-a

[P.(B)

-

l keS

P(Ak)J

and it follows that t" converges, a contradiction. Thus P is a prob-

ability measure. If Bk e .F, Bk j 0, then given e > 0, we have P(Ak) < e for large k, say for k >- ko. Thus P"(Ak0) < e for large n, say for n >- no. Since the Ak decrease, we have supnzrtp P"(Ak) < e

fork >- ko, and since Ak 10, there is a k1 such that for n = 1, 2, ... , no - 1, P,,(Ak) < e fork > k1. Thus sup" P"(Ak) < e, k >- max(ko, k1).

251

CHAPTER I

1 for all n. Add a point

(b) Without loss of generality, assume

1-

(call it oo) to the space and set

1 - P(fl) =

P{oo}. The P are now probability measures, and the result follows from part (a). Section 1.6

2.

is integrable In Y-1 Ifnl dp = Y' 1 fn Ifnl dp < oo; hence Y , and therefore finite a.e. Thus Y , f converges a.e. to a finite-valued

function g.

Yk 1 Ifkl, an integrable function. By Let g _ Yk=, fk. Then the dominated convergence theorem, In g dp ffl g dp, that is, 00

> f. fn

n=1

3.

ffln=1 fn du.

f2

Let xo a (c, d), and let xn -+ xo , x 96 xo . Then b

b

1

(Xn

-XO)

[f f(x., y) dy - faf(xO,Y)dy] -

= f b rf(xn , Y) - f (xo , Y)] xn - XO

a L

dY.

By the mean value theorem,

f(xn, Y) -f(x0, Y) =f1(An, Y) xn - xo

8.

between x and xo . By hypothesis, Ifi(2n, Y)I 5 h(y), for some A = where h is integrable, and the result now follows from the dominated convergence theorem (since y) - f(xo, y)]/[x - x0] -f1(xo, y), f,(x, ) is Borel measurable for each x). Let it be Lebesgue measure. If f is an indicator IB, B e M(R), the result to be proved states that p(B) = p(a + B), which holds by translationinvariance of it (Problem 4, Section 1.4). The passage to nonnegative

simple functions, nonnegative measurable functions, and arbitrary measurable functions is done as in 1.6.12. Section 1.7

2.

(a)

If f is Riemann-Stieltjes integrable, a =f = /3 a.e. [p] as in 1.7.1(a). Thus the set of discontinuities off is a subset of a set of p-measure 0,

together with the endpoints of the subintervals of the Pk. Take a different sequence of partitions having the original endpoints as

252

SOLUTIONS TO PROBLEMS

interior points to conclude that f is continuous a.e. [µ]. Conversely, if f is continuous a.e. [p], then a =f = /3 a.e. [p]. [The result that f

continuous at x implies a(x) =f(x) = /3(x) is true even if x is an endpoint.] As in 1.7.1(a), f is Riemann-Stieltjes integrable. 3.

(b) This is done exactly as in 1.7.1(b). (a) By definition of the improper Riemann integral, f must be Riemann

integrable (hence continuous a.e.) on each bounded interval, and the result follows. For the counterexample to the converse, take f (x) = 1,

n< x + oo as a -+ - oo, b -+ oo.) (b) Define

fi(x) =

(.1(x)

if -n<x
Then f T f; hence f is measurable relative to the completed a-field; also, In f,, dp I In f dp by the monotone convergence theorem. But In f dµ = r_,,,,,(f) by 1.7.1(b), and r_,,.,,(f) r(f) by hypothesis; the result follows. For the counterexample, take (- 1)n

f(X) n+1' 0,

n<x
Wehaver(f)=1-+++- --,but fRJfldµ=l+j+}+...=oo,

so that f is not Lebesgue integrable on R.

Chapter 2 Section 2.1

2.

Let D = {c o: f(w) < 0}; then 1.(A n D) < 0, 2(A n D`) z 0 for all A e F. By 2.1.3(d),

2+(A) = A(A n D°) = f f + dµ A

253

CHAPTER 2

since f + =f on D` and f + = 0 on D. Similarly,

2(A)= - A(A n D) = fA f - dµ -f on D, and f - = 0 on D`. The result follows. If El, ..., E. are disjoint sets in F, with all E. c A,

since f 4.

n

n

n

12(E1) I =

+(Ei) - 1(E1)1 <-

I

i=t

[A+(Ei) + 2 (E1)]

= IAI(UE;) <- IAI(A). Thus the sup of the terms Yi=1 1A(E)I is at most I2I(A). But 12I(A) = 2+(A) + A -(A)

= A(A n D`) - A(A n D) = IA(A n D`)I + I1(A n D)I since A(A n D`) >- 0 and

A(A n D).-5 0

= I1(E1)I + IA(E2)I

and the result follows. Section 2.2

1/n},n=1, 1,2, ...,sothatA=Un'=1A,,. Now

2.

00 > f I9I dµ z A

n

hence p(An) < oc. For the example, let p be Lebesgue measure on P1(R), and let g(x) be any strictly positive integrable function, such as g(x) _ e- I"l. In this case, A = R, so that p(A) = oo. 4.

If f is an indicator IA, the result is true by hypothesis. If f is a nonnegative simple function Y;=1 x j IA, , the A; disjoint sets in F, then Jnn

fd2=

;_1

x;1(A;)=J=i >x;J gdp=

= fig dy

Aj

x; f

1 ,gdp

i=1

by the additivity theorem.

If f is a nonnegative Borel measurable function, let fl, f2, ... be nonnegative simple functions increasing to f. By what we have just proved

254

SOLUTIONS TO PROBLEMS

J a fn d2 = fn fn g dµ; hence J a f d l= In fg dp by the monotone convergence theorem. Finally, if f is an arbitrary Borel measurable function,

write f = f - f -. By what we have just proved,

fnf-dl= f f-gdµ

f f+d2= f f+gdp, n

6.

n

n

and the result follows from the additivity theorem. (a) In the definition of IAI, we may assume without loss of generality that the E1 partition A. If A is the disjoint union of sets A1, A2, ... a .r, then n

n

j=1

11(E) I=

n

ao

Co

Ai)I < E Y_ 12(Ej n A,) I j=1 1i=1i A(Ej r j=1 i=1 00

n

00

E 12(Ej n Ai)I s

= i=1 j=1

i=1

IAI(A).

121(A). Now to show the reverse inequality, we may assume I2I(A) < oo; hence I21(A) < IAI(A) < co. For each i, there is a partition {E11, ..., Ein,} of A. such that Thus IAI(A)

Y I A(Ei) I > I A I (A) -

j=1

a

2'

,

E> 0 preassigned.

Then for any n, n

n

ni

I A I (A) ? E E 12(Eij) I? E 12 I (A) - a. i=1 j=1

1=1

Since n and a are arbitrary, the result follows. (b) If E1, ..., En are disjoint measurable subsets of A, n

n

n

1(21 + 22)(E1) I < F.. 121(E1) I + Y- 122(E,) I i=1 i=1 i=1

< 121 I (A) + 122 I (A),

proving IA1 +' 21 -< IA1I + 1121; IaAI = lal 121 is immediate from the (c)

definition of total variation. If µ(A) = 0 and 1211(Ai°) = 0, i = 1, 2, then µ(A1 u A2) = 0 and by

(b), lei + 221(Alc .\ A20):!9 1A1I(A1`) + I22I(A2`) = 0(d) This has been established when A is real (see 2.2.5), so assume A com-

plex, say, 2 = Al + i22. If µ(A) = 0, then A1(A) =22(A) = 0; hence A <<,u implies Al 4p and d2 4.u. By 2.2.5(b), I A I I 4 p, 1221 410; hence by (b), IAI 4 p. The converse is clear since I2(A)I <- 121(A). (e)

The proof is the same as in 2.2.5(c).

(f)

See 2.2.5(d).

255

CHAPTER 2

(g)

The "if" part is done as in 2.2.5(e); for the "only if" part, let p(A,,) -> 0. Since JAI < p by (d), I1.I(A,) --+ 0 by 2.2.5(e); hence A(A,,) --* 0.

Section 2.3 2.

We have

f bf'(x) dx = blim

f, h-'o I

f (x + h) -f (x) dx h

where we may assume h > 0 `f(x+h)-f(x)l


a

L

dx

fbf(x)dx + - fa+f(x)dx]

=limionfh h

h

i

[definef(x) = f(b), x > b; f(x) = f(a), x < a] < lim inf 1 [hf (b + h) - hf(a)] h-o h

since f is increasing

= f(b) - f(a). Alternatively, let It be the Lebesgue--Stieltjes measure corresponding to f (adjusted so as to be right continuous), and let m be Lebesgue measure. Write µ = i +,u2, where p1 < m and P2 1 m. By 2.3.8 and 2.3.9, fb

b

f'(x)dx= ff'dm a

a

= fDµdm b a

= 6.

(a)

f b dm1 dm = N1(a, b]

< µ(a, b] =f(b) -f(a)

Since A is linear and m is translation-invariant, so is ..; hence by Problem 5, Section 1.4, A = c(A)m for some constant c(A). Now if A, and A2 are linear transformations on R', then m(A,A2 E) = c(A1)m(A2 E) = c(A1)c(A2)m(E) and

m(A,A2 E) = c(A1A2)m(E);

256

SOLUTIONS TO PROBLEMS

hence

c(A1A2) = c(A1)c(A2)

Since det(A,A2) = det A, det A2, it suffices to assume that A corresponds to an elementary row operation. Now if Q is the unit cube {x: 0 < xi 5 1 ,i=1, ..., k}, then m(Q) = 1; hence c(A) = m(A(Q)). If e1, ..., ek is the standard basis for Rk, then A falls into one of the following three categories: (1)

Ae1 = ej, Ae; = e;, Aek = ek, k

i or j. Then c(A) = 1 =

Idet AI. (2) (3)

Ae! = kej, Ae; = e;, j i. Then c(A) = Iki _ Idet AI. Ae1 = e; + e;, Aek = ek, k 0 i. Then det A = 1 and c(A) is I also, by the following argument. We may assume i = 1, j = 2. Then ff

k

A(Q)={AEa;e;:0
f(

i=1

i=l,...,k}

k

={y= Yb;et:0
1= 1

111

If B, _ {y a A(Q): b2 < 1), B2 = {y a A(Q): b2 > 1}, and B2 - e2 = {y - e2:y a B2}, then B, n (B2 - e2) = Q ; for if y e B,, then b1 < b2 and if y + e2 a B2 c A(Q), then b2 + 1
= m(B, U (B2 - e2)) = m(Q) = 1. (b) Fix x e V and define S(y) = T(x + y) - T(x), y e {z - x: z e V).

Then if C is a cube containing 0, we have

T(x+C)={T(x+y):yeC}=T(x)+{S(y):yeC} = T(x) + S(C).

Therefore, m(T(x + C)) = m(S(C)), so that differentiability of Tat x is equivalent to differentiability of S at 0, and the derivatives, if they exist, are equal. Also, the Jacobian matrix of S at 0 coincides with the Jacobian matrix of T at x, and the result follows. (c)

Let A = A(0), and define S(x) = A-'(T(x)), x e V; the Jacobian matrix of S at 0 is the identity matrix, and S(0) = A-'(0) = 0. If we show that the measure given by m(S(E)), E e 2(V), is differentiable

257

CHAPTER 2

at 0 with derivative 1, then m(S(C)) m(C)

-

I

a

I < Idet AI

for sufficiently small cubes C containing 0. Thus by (a), m(T(C)) _ m(AS(C)) = Idet Alm(S(C)); hence m(T(C))

m(C) -

I det Al

I

I m(S(C)) m(C)

I

1

l det A I< e,

so that t is differentiable at 0 with derivative ldet Al. (d) (i) If x e C, then lxi < Jk f3 < S; hence I T(x) - xl < afl _ (QZ - $). Therefore T(x) e C2 . (ii) If x e 8C, then I T(x) - xl < afi as above, and a/3 = :(p - fi1). Therefore T(x) C,. (iii) We have IT(x) - xl < a/3, and

Of

1

< 1-2a-1-jR'-2P" and the result follows. (iv)

If y e C, - T(C) but y e T(C), then y e T(8C), contradicting (ii).

Now C, is the disjoint union of the sets C, n T(C) and C, - T(C); the first set is open since T is an open map, and the second set is open by (iv). By (iii), C, n T(C) # 0, so by connectedness of C1, we have

C, = C, n T(C), that is, C, c T(C). Also, T(C) c C2 by (i). Therefore m(CI) < m(T(C)) < m(C2), that is, (I - 2a)'13' < m(T(C)) < (I + 2a)kjJk.

Thus I - s < (1 - 2a)k < m(T(C))/m(C) < (I + 2a)k < I + s as (e)

desired. Assume m(E) = 0 and A(E) >0. By Problem 12, Section 1.4, and the

fact that 00

I(Cr) < nil E= U En(C,,sup r < I I% m(Cr) n, i = I

we can find a compact set K and positive integers n and j such that in(K) = 0, ).(K) > 0, and A(C) < nm(C) for all open cubes C containing a point of K and having diameter less than 1/j. If r, > 0, choose an open set D c K such that m(D) < e.

258

SOLUTIONS TO PROBLEMS

Now partition Rk into disjoint (partially closed) cubes B of dia-

meter less than I/j and small enough so that if B n K 96 0, then B e D. If the cubes that meet K are B1, ..., B, we may find open cubes C, z) Bi, I < i < t, such that m(Ci) < 2m(B,) and diam C-, < I /j. Then

2(K) <

,l(Bi) < i=1

i=

1(C;) < n

m(Ci) < 2n r=1

m(B;) =1

< 2nm(D) < 2ns.

Since a is arbitrary, 1(K) = 0, a contradiction. (f)

If f is an indicator IB, let E = T -'(B), so that B = T(E). Then fW f(y) dy = m(T(E))

and

fV f(T(x)) IJ(x)I dx = fE IJ(x)I dx,

and the result follows from part (e). The usual passage to simple functions, nonnegative measurable functions and arbitrary measurable functions completes the proof. Section 2.4 1.

(a)

If f = {al, ... , a, , 0, 0,...), then fn f dp = Yk=1 ak by definition of the integral of a simple function. If f = {an, n = 1, 2, ...}, with all

a >- 0, fn fdp =

1 a,, by the result for simple functions and the

monotone convergence theorem. If the a are real numbers, then fnfdp=fnf+dp-fnf-dp=Y' la.+-Y l an- if this is not of the form + oo - oc. Finally, if the a are complex,

fn-

dp=

=1

Rea.+iIma,, n=1

provided Re f and Im f are integrable; since IRe aJ, IIm anI
oo.

If f(a) = 0 except for a e F, F finite, than in f dy = &f(x) by definition of the integral of a simple function. If f >- 0, then In fdp >- I, f dp for all finite F; hence in fdp >- Y. f (a). If f (a) > 0

for uncountably many a, then Y. f (a) = oo ; hence In fdp = oo also. If f (a) > 0 for only countably many a, then in fdp = Y. f (a) by the monotone convergence theorem. The remainder of the proof is as in part (a). 8.

Apply Holder's inequality with g = 1, f replaced by If jr, p = s/r, 1/q = I - r/s, to obtain

CHAPTER 2

259

IfI"Pdµ)11p(f

I"dµ<(f 9.

I

dµ)1/9

as desired. Therefore If 11, < If II We have fn If l" dp <_ Ilf 11 xµ(S2), so lim supp-,, If IIp <- IIIII,,. Now lets > 0, A = {co: I f(w)l >- I!f li 0 - s) (assuming if 11 w < oo). Then fi1 Ifl"dµ>-

If l"dµ fA

- (II! II.-E)"µ(A)

Since p(A) > 0 by definition of IIIII, we have lim infp_. Iif IIp >- IIf II M If Ilf II = co, let A = {w: I f(w)I >- M} and show that

lim inf IIf IIp ? M; p-oo

since M can be arbitrarily large, the result follows.

If p(S2) = oo, it is still true that lim infp.,, Illllp ? Ilfil"',; for if p(A) = co in the above argument, then IIf IIp = oo for all p < oo. How-

ever, if p is Lebesgue measure on p(R) and f(x) = 1 for n < x < n + (1 /n), n = 1, 2, ... , and f (x) = 0 elsewhere, then 11f 11p = oo for

p
(a) We have I JE (f - z) dpi < fE l f - zl dp. Butw a Eimpliesf(w)e D; hence If(w) - zl < r; and thus J E If- zl 4:5 rp(E). If p(E) > 0, then 1

= p(E)

dp p(E) fE f µ 1

f '(f - z) dµ

r;

[l /µ(E)] JE f dµ e D c S`, a contradiction. Therefore p(E) = 0, that is, µ{w: f(w) a D} = 0. Since {w: f(w) 0 S} is a countable union of sets f '(D), the result follows. (b) Let p = iA I ; if El, ..., En are disjoint measurable subsets of A hence

j=1

I2(E;)I =

j=1

If E1

h dp <

j=1

fEf I h I dp

n

< r E µ(E j) < rp(A,). j=1

Thus µ(A,) < rµ(A,), and since 0 < r < 1, we have µ(A,) = 0. If A = (co: Ih(w)l < 1} = U {A,: 0 < r < 1, r rational}, then µ(A) = 0, so that Ihi >- I a.e. Now if p(E) > 0, then [ 1 /µ(E) ] J E h dµ = A(E)/µ(E) e S, where

S = (z e C: lzl < 1). By (a), h(w) e S for almost every co, so IhI < 1 a.e. [121].

260

SOLUTIONS TO PROBLEMS

(c)

If E e ,°F, In IE h dl2I = JE h dJAI = A(E) by (b); also, I. IEg dµ =

J, g dp = 2(E) by definition of A. It follows immediately that In fh diAI = f n fg du when f is a complex-valued simple function. If

f is a bounded, complex-valued Borel measurable function, by 1.5.5(b), there are simple functions ff -+f with I f,, <_ Ill. By the dominated convergence theorem, In fh diAl = In fg dµ. If f = hIE, we obtain 121(E) = f E hg dµ. (d) In (c), 121(E) >- 0 for all E; hence hg > 0 a.e. [p] by 1.6.11. But if

g((o) = Ig(w)le'O() and h(w) = e`*(W), then e'(9-v) = 1 a.e. on {g :A 0}, so that hg = Ig1 a.e., as desired. 12.

If I(a, b) can be approximated in L°° by continuous functions, let 0 < E < I and let f be a continuous function such that l'(a, b) -f IIco < e;

hence I'(a, b) -f I <- e a.e. For every 6 > 0, there are points x E (a, a + 5)

and y e (a - 5, a) such that I 1 - f (x)I < s and If() I < e. Consequently,

lim supra, f(x) >- 1 - e and lim infra- f(x) < e, contradicting continuity off. Section 2.5 4.

Let Bjkb = (I fj -fkl >- 5), Ba = nn

UT

J,

1

Bjka Then

oo

U Bjka } Ba

j,k=n 5.

and the proof proceeds just as in 2.5.4. Let (f,,,) be a subsequence converging a.e., necessarily to f by Problem 1.

By 1.6.9, f is p-integrable. Now if In f du +- In f dp, then for some e > 0, we have I In f, dp - 1, f dp >- s for n in some subsequence {mk). But we may then extract a subsequence {f,,) of { f a.e., so that In f,, du - In f du by 1.6.9, a contradiction. Section 2.6 4.

By Fubini's theorem, 1

N(C) = ff Ic dµ = n

(

[jig dµ2] dpl = Jn,NZ(C(wi)) dpI((o 1). in,

Similarly, p(C) = JRi pl(C((02)) dp2((02). The result follows since f >- 0,

fnf=0implies f=0a.e.

261

CHAPTER 2

7.

(a)

Let

Ank={xefl': (

Bnk = y e 522:

n


k k-1 < y< n) n

(n = 1, 2, ... , k = 1, 2, ... , n; where k = n, include the right endpoint as well). Then CO

n

G= l I U (Ank x Bnk) E .F. I

n=1 k=1

If f is only defined on a subset of 521, replace S21 by the domain off in the definition of Ank . (b) Assume B c C1. Each x e S21 is countable and (x, y) e B implies y < x; hence there are only countably many points Yx1, Yx2 , ... C-.02 such that (x, yxn) e B. (If there are only finitely many points yx1,

yx., take yxk = y,,, for k Z n.) Thus B = U 1 G., where G is the graph of the function fn defined by fn(x) = yxn . By part (a), B e.. [Note that yxn a K12, which may be identified with [0, 11; thus (a) applies.] If B e C2, each y E 522 is countable, and (x, y) e B implies x < y, so there are only countably many points xyn c-01 such that (xyn, y) a B. The result follows as above. (c) If F e 0, then F = (F n C1) u (F n C2) and the result follows from part (b). 9. Assuming the continuum hypothesis, we may replace [0, 1 ] by the first uncountable ordinal Y1. Thus we may take 521 = C12 = Y1, F1 = 2 = the image of the Borel sets under the correspondence of [0, 1 ] with Y1, and µl = µ2 = Lebesgue measure. Let f =1c, where C = {(x, y) e S21 x Q2:

y < x) and the ordering " 5 " is taken in fl, not [0, 1]. For each x, (y: f(x, y) = 1) is countable, and for each y, {x: f(x, y) = 0} is countable; it follows that f is measurable in each coordinate separately. Now fo f (x, y) dy = 0 for all x; hence So [ S' f(x, y) dy] dx = 0. But S o [1 - f (x, y)] dx = 0 for ally; hence 10 ' [J , f(x, y) dx] dy = 1. It follows that f is not jointly measurable, for if so, the iterated integrals would be equal by Fubini's theorem. Section 2.7 1.

Let T be the smallest a-field containing the measurable rectangles. Then I cfly Fj since a measurable rectangle is a measurable cylinder. But the class of sets A =F11= I 52; such that {w a 52: (w1, ..., a1n) e A } e I is a 1

262

SOLUTIONS TO PROBLEMS

a-field that contains the measurable rectangles of IIj=1 Qj, and hence contains all sets in f; = 1 Fj. Thus all measurable cylinders belong to T,

so Flit I y;

W.

5. k

r

{xeR°°:f(x)=n}={x:

n

i=1

111

ll

xi<1, k=1,2,...,n-1, Y_ x,>_1} i=1

1111

if n=1,2,..., n

{xeR°°:f(x)=oo)= {x:

i=i

x;<1, n=1,2,...). 111111

In each case we have a finite or countable intersection of measurable cylinders.

Chapter 3 Section 3.2 6.

(a)

Let z e K, a compact subset of U. If r < do, a standard application of the Cauchy integral formula yields 2n

1

f(z) = 2n

f

f (z

o

+ re") dB.

Thus if0
d

1

a

2n

f(z) 2 = f orf(z) dr = 2n for dr f. f(z + rei°) d6, or 2n

d

+ reie)r dr dO f(z) = nd2 fB=o f=of(z , =n d2

f f f(x+iy)dxdy

where D is the disk with center at z and radius r. An application of the Cauchy-Schwarz inequality to the functions 1 and f shows that If(z)I _<

as desired.

nd2

(nd2)'1211f 11,

CHAPTER 3

263

(b) If f , f2 , ..., is a Cauchy sequence in H(U), part (a) shows that fn

converges uniformly on compact subsets to a function f analytic on U. But H(U) c L2(f , .F, p) where 12 = U, .°r is the class of Borel

sets, and y is Lebesgue measure; hence fn converges in L2 to a function g e H(U). Since a subsequence of converges tog a.e., we have f = g a.e. Therefore f e H(U) and fn -+f in L2, that is, in the

H(U) norm. 7.

(a)

If 0
1

fo If(Yero)12 dO =

2n o

am rme-im° d©

1

27r

foo

n

1

Y an am rn+m

21r n, m

f

2n

J0

ei(n-m)° dO

since the Taylor series converges uniformly on compact subsets of D CO

E n=0

Ian

12 Y

2n,

which increases to Y' O Ian12 as r increases to 1. (b)

f f If(x + iy)I 2dxdy = fordr f 2.If(rei°)12 dO D

< 21rN2(f) fl r dr = irN2(f). 0 2n

(c)

f f Ifn(x + iy 1 2 dx dy D

= fo fo

r2nr dr d8 =

{2n + 2) -' 0,

but f2nlfn(YeiB)I2

dO =

2n

I

f2nr2n dO = r2n;

2n

hence N(fn) = 1 for all n. (d) If { fn} is a Cauchy sequence in H2, part (b) shows that { fn} is

Cauchy in H(D). By Problem 6, fn converges uniformly on compact subsets and in H(D) to a function f analytic on D. Now if 0 0, the Cauchy property in H2 gives 1

2n

I f,

If(Ye°) -fmrei°12 d©S e

264

SOLUTIONS TO PROBLEMS

for r < ro and n, m exceeding some integer N(e). Let m - oo; since fm -- f uniformly for Izi 5 ro,

f 2n

2n

r < ro, n >- N(s).

fn(re'B) - f(re'B) 12 dO < s,

o l

Since ro may be chosen arbitrarily close to I, N 2(f" - f) < e for n >- N(F), proving completeness. (e)

Since e" corresponds to (0, ... , 0, 1, 0, ...), with the I in position n, in the isometric isomorphism between H2 and a subspace of 12, the e" are orthonormal. Now if f c- H2, with Taylor coefficients a, n = 0, 1, ... , then = an , again by the isometric isomorphism. Thus

N2(f) _ Y lanl2 = Y II2 n=0

n=0

and the result follows from 3.2.13(f).

(f)

<en, em> = if

iy)em(x + iy) dx dy = f f en(rete)em(reie)r dr dO

D

D o2

f

[(2n + 2)(2m +

"f

1

r n+me'("-m)er dr dO o

0,

Fz

n :A m,

n=m.

1,

Thus the en are orthonormal. Now if f e H(D) with 00

f(z)=Eanz",

"=0

then 2n ro

f ff o

(re'a)em(reie)r dr dO

o

OD

_ E a. n-o

f

2"

0

ro

ame f r"ey"e (2m + 2)1/2 rme-r dr 2n 0

dO

since the Taylor series converges uniformly on compact subsets of D 2m + 2 1/2

-am(

2R

)

r0 2m+2

2

2m+22n=am(2m+2)

1/2 2m+2

ro

265

CHAPTER 3

Now feis integrable on D (by the Cauchy-Schwarz inequality),

so we may let ro -I and invoke the dominated convergence theorem to obtain

(T2" 1/2 m +2)

{{'

But the same argument with

replaced by f shows that rro

IIf IIH(D) = lim ro-'1 n..=0

f

J0 J0

rneinerme-imer dr dO

a 1227C

n4o 2n + 2

The result now follows from 3.2.13(f). 9.

(a)

Let g be a continuous complex-valued function on [0, 2n] with g(0) = g(2n). Then g(t) = h(e'r), where h(z) is continuous on 1). By the Stone-Weierstrass theorem, h can be {z e C: z For uniformly approximated by functions of the form _kthe algebra generated by z", n = 0, ± 1, ±2, ..., separates points, contains the constant functions, and contains the complex conjugate of each of its members since 2 = 1/z for IzI = 1. Thus g(t) can be uniformly approximated (hence approximated in L2) by trigonometric polynomials Y-k cke'k'. Since any con-

tinuous function on [0, 27z] can be approximated in L2 by a continuous function with g(O) = g(2n), and the continuous functions are dense in L2, the trigonometric polynomials are dense in L2

(b)

By 3.2.6, S 20n If(,)

cke'k`I2 dt is minimized when ck = ak = k"= (1/2n) f o" f (t)e-'k` dt. Furthermore, some sequence of trigonometric

polynomials converges to f in L2 since the trigonometric polynomials are dense. The result follows. This follows from part (a) and 3.2.13(c), or, equally well, from part (b) and 3.2.13(d). 10. (a) Let be an infinite orthonormal subset of H. Take M = (c)

{x1, x2

i

...}, where x = (I + l

n = 1, 2, .... To show that

M is closed, we compute, for n 0 m,

Ilxn-x.112=

12

=(1+n) +(i+»t) 1

2

>2.

SOLUTIONS TO PROBLEMS

266

Thus if y" e M, y" -+ y, then y" =y eventually, so y e M. Since (b)

IIx"112 = I +(I In), M has no element of minimum norm. Let M be a nonempty closed subset of the finite-dimensional space

H. If x e H and N =M n {y: IIx - yll < n}, then N 0 for some n. Since y IIx - yIi, y e N, is continuous and N is compact, N} = IIx -Yoll for some yo e N c M. But the inf over N is the same as the inf over M; for if y e M, y 0 N, then 11x - yo 11 <- n < II x -YII. Note that yo need not be unique; for example, let H = R, M = (- 1, 1), x = 0. inf{IIx - YII : Y e

Section 3.3 3.

Since fa IK(s, t)I di is continuous in s, it assumes a maximum at some point u e [a, b]. If K(u, t) = r(t)e10" , r(t) > 0, let z(t) = e- "(0. Let x,, x2 , ... be a sequence in C [a, b] such that fa IX"(t) - z(t)l dt -+0; we may assume that Ix"(t)I < 1 for all n and t (see 2.4.14). Since K is bounded,

f bK(s, t)x"(t) dt

lim

f 6K(s, t)z(t) dt

n-.

"

III

I

= lim I(Ax")(s)I < IIAII

a

co

Set s = u to obtain

fbIK(u,t)Idt
7.

as desired. (a) If x e L, then n

11x111=

(b)

x;e;

Ix,l Ile.hli < (niax IIe,l11)

Ix,l

But >i=1 IX,I < 11nkkX112, so we may take k = max; 11e,111. Let S = {x: lix112 = 1}. Since (L, II 112) is isometrically isomorphic to C", S is compact in the norm 11 112. Now the map x - 11x111 is a

continuous real-valued function on (L, 11 Ill), and by part (a), the topology induced by II II1 is weaker than the topology induced by 112. Thus the map is continuous on (L, II 112); hence it attains a minimum on S, necessarily positive since x e S implies x 96 0. 11

(c)

If x e L, x 96 0, let y = x/ 11x112; then IIYIl1 > M11YI12 by (b); hence 11 x1I1 >: rn11xIl2 . By (a) and Problem 6(b), 11 11, and 11 112 induce the

same topology. (d)

By the above results, the map T: D = 1 x; e; - (x1, ... , x") is a one-to-one onto, linear, bicontinuous map of L and C" [note that I1Y_:=1 x,112 is the Euclidean norm of (x1i ..., x") in C"]. If y; e L,

267

CHAPTER 3

y j --> y e M, then y ; - y k - 0 as j, k -* oo; hence T(y; - yk) = Ty; - Tyk -> 0. Thus {T),,} is a Cauchy sequence in C". If Ty; zE C", then y' - ->T-1zeL. 9. For (a) implies (b), see Problem 7; if (c) holds, then {x: Ilxl1 < c} is compact for small enough e > 0; hence every closed ball is compact (note that the map x -+ kx is a homeomorphism). But any closed bounded set is a subset of a closed ball, and hence is compact. To prove that (f) implies (a), choose x1 e L such that IIx1 I{

Suppose we have chosen x1i...,xkeL such that

IIxi11 = I

and

I1xi - x;ll >- I for i, j = 1, ..., k, i j. If L is not finite-dimensional, then S{xl, ..., xk} is a proper subspace of L, necessarily closed, by Problem 7(d). By Problem 8, we can find xk+1 eL with Ilxk+111 = 1 and Ilxi - xk+l II >- 1, i = 1, ..., k. The sequence x1, x2 , ... satisfies IIxn11 = 1

for all n, but Ilxn - x,n11 ? -1 for n 0 m; hence the unit sphere cannot possibly be covered by a finite number of balls of radius less than 4. 11.

(a)

Define ).(A) = f (IA), A e F. If A1i A2, ... are disjoint sets in whose union is A, then ).(A) = Y? f(IA,) since f is continuous 1

and"=1 IA, L' >

IA . [Note that n

ao

I 1 > IA, - IAI ° dµ = E µ(Ai) Sl i=1

i=n+1

by finiteness of p.] Thus

0

is a complex measure on F. If u(A) = 0,

then IA = 0 a.e. [µ], so we may write IA L°+ 0 and use the

continuity of f to obtain .1(A) = 0. By the Radon-Nikodym theorem, we have ).(A) = JA y du for some p-integrable y. Thus Ax) = J. xy dy when x is an indicator; hence when x is a simple function. Since f is continuous, y is p-integrable, and the finitevalued simple functions are dense in L°, the result holds when x is a bounded Bore] measurable function. Now let y1i Y2, ... be nonnegative, finite-valued, simple functions increasing to lyl. Then IIY"Ilq = f Yn9 dk

n

fn

IYI dp = f

n

since Il.f

1111 Y9 - I

li° = I;f 1111 Y.11gl°

yn-'e-iargyy dp

is bounded since (q - 1)p = q.

Thus IlYnllq < If II ; hence by the monotone convergence theorem, IIYIIq <- ii! 11; in particular, y e Lq. But now Holder's inequality and

the fact that finite-valued simple functions are dense in L° yield

Ax) = fn xy dµ for all x e V. Holder's inequality also gives

268

SOLUTIONS TO PROBLEMS

Ilf11 < IIYIIq; hence IIf II = IIYIIq. If y, is another such function, then

g(x) = In x(y - y,) dy = 0 for all x e LP. By the above argument, IIY - Y, IIq=0, soy=y, a.e. [µ]. (b) (i) If, say, yA - yB > 0 on the set F c A n B, let x = IF; then xIA = xIB; hence Sn x(yA - yB) dp = 0, that is,

f (YA-YB)dp=0. (ii)

But then µ(F) = 0. YA a.e. on A. we have

Since

11YA. I q -<

fA,,, - A

II

dµ.

approaches k4 as n - oo, so does IIYA v Amllq and it follows that dy - 0 as n -, oo. By symmetry, we may interchange m and n to obtain Since

f

fAm - A

f2

f

A,,,

as n, m -' oo. Thus YAn converges in LQ to a limit y, and since is another sequence [If 11 for all n, IIYIIq < Ilf II. If of sets with k, the above argument with A, replaced by B. shows that IIYA YB -+ y also. (iii)

Let A e,, µ(A) < co. In (ii) we may take all A,, A, so that YA = YA a.e. on A; hence y = yA a.e. on A. Thus if

x = IA, then f (X) = f (xIA) = Sn xyA dy = in xy dµ. It follows that f (x) = Sn xy dµ if x is simple. [If µ(B) = oo, then x must be 0 on B since x e LP.] Since y e V, the continuity off and Holder's inequality extend this result to all x e LP. (c) The argument of (a) yields a µ-integrable y such that f (x) =In xy dy for all bounded Borel measurable x. Let B = {(o : ly(w)I >- k); then

kp(B) < f IYI du = f IBe- iargyydp B

n

=f(IBe-`a`gy):5 IIYII IIIBII, = IIfll,(B).

Thus if k > Ilf II, we have p(B) = 0, proving that y e L°° and IIYII < If 11. As in (a), we obtain f(x) =Snxydp for all xeL', (d)

III II = IIYII ., and y is essentially unique. Part (i) of (b) holds, with the same proof. Now if 12 is the union of oo, define y on .0 by taking y = yAn disjoint sets A,,, with II for all n, we have y e L°° and IIYII. -< on A,,. Since

269

CHAPTER 3

IIf II . If x e L1, then Y,."=1

f(x)_ n=1 Since IIf II

<_

ll

XI" + x; hence

ll

f xydp= f xydp.

f

n n=1 A. by Holder's inequality (with p = 1, q = oo), the n=1

S2

result follows. Section 3.4 3.

Let {yn} be a Cauchy sequence in M, and let xo be any element of L with norm 1. By 3.4.5 (c), there is an f e L* with IIf II = I and f (xo) = 11 xo ll = I ; we define A,, a [L, M ] by A. x = f (x)yn . Then ll (A. - A,n)xIl

=

If(x)l Ily,, - ymll <- Ilyn - y,nll llxIl, so that IIAn - A,nll s Ilyn - ynll --> 0.

By hypothesis, the An converge uniformly to some A e [L, M ]; therefore I(yn - Axoll = IIA,,xo - AxoII <- IIAn - All -+ 0. 7.

Let L be the set of all scalar-valued functions on 0, with sup norm, and let M be the subspace of L consisting of simple functions x = 2i Xi IA3 ,

where the Ai are disjoint sets in .. Define g on M by g(x) = Y1 x1 yo(Ai).

Now Ig(x)I <- maxi (xil Ei yo(A)< maxi Ixilµo(Q) = po(Q)IIxll; hence Ilgll <- po(c) < eo. By the Hahn-Banach theorem,g has an extension to a

continuous linear functional f on L, with If II = Ilgll Define µ(A) = f(IA), A c 0. Since f is linear, p is finitely additive, and since f is an extension of g, p is an extension of yo. Now if µ(A) < 0, then f(IAc) = µ(A°) = p(O) - p(A) > µ(S]!).

But IIIAJ = 1, so that If II > p(Sl), a contradiction. 8.

Since An x -+ Ax for each x, sup,, IIAnxII < co. By the uniform boundedness principle, sup,, II An II = M < oo ; hence IlAxll = limllAnxll = lim inf IIAnxII "_00

Ilxll lim inf IIA,,II 5 Mllxll n-.ao

12.

Let L be the set of all complex-valued functions x on [0, 1 ] with a continuous derivative x', M the set of all continuous complex-valued functions on [0, 1 ], with the sup norm on L and M. If Ax =x', x e L, then A is a linear map of L onto M, and A is closed. For if x -' x and x,,' y, then since convergence relative to the sup norm is uniform convergence,

270

SOLUTIONS TO PROBLEMS

!o x"'(s) ds -+ Jo y(s) ds = z(t). Thus xn(t) - xn(0) -- z(t); hence x(t)

- x(O) = z(t). Therefore x' = z' = y. But A is unbounded, for if xn(t) _ sin nt, then Ilx"II = 1, IlAxnll 13.

oo.

By the open mapping theorem, A is open; hence A{x a L: IIxII < 1} is a neighborhood of 0 in M; say {y e M: IIYII < b} c A{x e L: Ilxll < 1}. If y e M, y 0 0, then Sy/2IIYII has norm less than 6, hence equals Az for some z e L, Ilzll < 1. If x = (2IIYII/8) z, then Ax = y and Ilxll < 2IIyII/8. Thus we may take k = 2/6.

Section 3.5 3.

Let xl, ..., x,, be a basis for L, and define A: C" -* L by A(al, ..., an) = + anxn. It is immediate that A is one-to-one onto, linear, and alxl + continuous; we must also show that A-' is continuous. If L is one-

dimensional, then A-'(ax) = a, so the null space of A-' is {0}, which is closed by the Hausdorff hypothesis. By Problem 2, A-' is continuous. If dim L = n + 1 and the result holds up to dimension n, let f be a nontrivial linear functional on L. Then [see 3.3.4(a)] N =f -'(0) is a maximal proper subspace of L; hence by the induction hypothesis, it is topologically isomorphic to C". It follows that N is closed in L; the argument is the same as in Problem 7(d), Section 3.3. By Problem 2, every linear functional on L is continuous, so if pi is

is the projection of C" on the ith coordinate space, p; -A-' is con5.

tinuous for each i, hence A -' is continuous. If L were metrizable, it would have a countable base at 0, which can be assumed to be of the form

Un={feL:If(x)I <Sn for all xeAn}, where Sn > 0 and An is a finite subset of [0, 1 ], n = 1, 2, .... Let y be a point in [0, 1 ] not belonging to Un I An, and let U = {f e L: I f(y)I < 1). Then U is a neighborhood of 0 in L but no U. c U, for there are (continuous) functions f with f = 0 on A. butf(y) = 1. 6. (a) implies (b): The sets {x: d(x, 0) < r}, r rational, form a base at 0. (b) implies (c): The base at 0 may be assumed to consist of circled

convex sets, and the proof of 3.5.7 shows that the corresponding Minkowski functionals generate the topology of L. (c) implies (a) : Up, p2, ... are seminorms generating the topology of L, define

d(x, y) = E

1

pn(x - Y)

"=12"1+pn(x-y)'

CHAPTER 3

271

It is easily checked that d is a pseudometric, and d(x;, x) - 0 if pn(x; - x) -+ 0 for each n, in other words, iff x1 --+ x in the topology of L. 7.

Let K" = {z: Izl < 1 - 1/n), n = 1, 2, .... If K is any compact subset of D, K c K. for large n; hence sup{I f(z)I : z e K) < sup{If(z)j: z e K"} =

p"(f). Thus the sets U. = {feL: p"(f) < 1/n}, n = 1, 2, ..., form a countable base at 0, so by Problem 6, L is metrizable. Suppose a single norm II II induces the topology of L. By continuity of pn, for each n there is a 8" > 0 such that 11f 11 < 6. implies p"(f) < I ; hence If II < 1 implies p"(f) < l/5,,. Therefore W = {feL: p"(f) < 1/6. for all n} is an overneighborhood of 0. For each n let z,, be a point in Knt1 but not in K", and let f" e U" such 1/S"+1 i a choice off" is possible because, for example, if that I

In
8.

then (z/zo)' approaches 0 uniformly on K. as k - oo, but approaches co at z,,. Now pn+1(.f") ? If"(z")I > 1/b"+1 , and since the U. decrease with n and fn e U", we have, for any k, fn e Uk for all n >- k; consequently fn -> 0. But then f" e W for large n; hence pn+1(f") < a contradiction. (a) We may write x = XICO, r) + XI[r, y + z, and 1

d(x, 0) = f0 Ix(t)I" dt = d(y, 0) + d(z, 0)

since [0, r) n [r, 1 ] = 0. Now d(y, 0) _ fo I xI" dt, which is continuous in r, approaches 0 as r -+ 0, and approaches d(x, 0) as r -> 1. By the intermediate value theorem, d(y, 0) = d(z, 0) _ .

d(x, 0) for some choice of r.

(b) By part (a) we can find yl with d(y1 i 0) d(x, 0) and I f (yl) I ? #. Let x1 = 2y1; then I f(x1)I >- 1 and d(x1, 0) = 2" d(y1i 0) = 2p-1 d(x, 0). Having chosen x1 i ..., x,, with I f (x1) I >_ 1 and

13.

15.

d(x1, 0) =2""- " d(x, 0), i = 1, ... , n, apply part (a) to x" to obtain =2"-1 d(x,,, with If(x,,+1)I z 1 and d(x"+1, 0) xn+1 0) = 2(n+1)("-1' d(x, 0). Since p < 1, d(x", 0) -> 0 as n -+ oo, as desired. (a) Let i be the identity map from (L, l2) to (L, 1). Since 1 2, i is continuous, and by the open mapping theorem, 9-2 ell. (b) If fT is the topology induced by II 11j, .1 = 1, 2, then 9-1 a 107"2 by hypothesis, and the result follows from part (a). Define T: L -+ C" by T(x) = (fl(x), ... , f,,(x)) and define h: T(L) --. C" by h(Tx) = g(x). Then h is well-defined; for if Tx1 = Tx2, then f1(x1) = f;(x2) for all i, so g(x1) = g(x2) by hypothesis. Since h is linear on

272

SOLUTIONS TO PROBLEMS

T(L), it may be extended to a linear functional on C", necessarily of the form h(y1, ..., y") = c1y1 + + c" y,,. Thus g(x) = h(Tx) = c1 fl(x)

all xeL. 16.

(a)

Assume fi(x) = 0 for all i. If k is any real number, f;(kx) = 0 for all i, hence J(kxlI < 1. Since k is arbitrary, x must be 0. (h) Since the yJ are linearly independent and T -1 is one-to-one, the x; are linearly independent. If x e L, then Tx = Yi=1 c; y, for some c1, ... , ci,; hence x = Y;= 1 ci x; .

Chapter 4 Section 4.2

2.

Let Sl be the first uncountable ordinal, with F the class of all subsets of fl, and µ(A) = 0 if A is countable, µ(A) = oo if A is uncountable. Define, for each o (c 0, f,#o) = I if co < a; fa((o) = 0 if co > a. Then fa T f where f = 1, and J, f, dp = 0 for all a since fa is the indicator of a countable set. But 1, f dp = oo, so the monotone convergence theorem fails.

Section 4.3 4.

(a)

First note that H is a vector space. For if g is continuous, f f e H: f + g e H} contains the continuous functions and is closed under pointwise limits of monotone sequences, and hence coincides with H. Thus if f e H and g is continuous, then f + g e H. A repetition of this argument (notice the bootstrapping technique) shows that if g e H, then {f e H: f + g e H} = H; hence H is closed under addition. Now if a is real, then {f e H: of e H) = H by the same argument; hence H is closed under scalar multiplication. Let 9 be the open F. sets. Then So is closed under finite intersection, and IA e H for all A e Y. (By 4.3.5, IA is the limit of an increasing sequence of continuous functions.) By 4.1.4, IA e H for all A e a($") [=d(fl) by 4.3.21. The usual passage to nonnegative simple functions, nonnegative measurable functions, and arbitrary

measurable functions shows that all Baire measurable functions belong to H. But the class of Baire measurable functions contains the continuous functions and is closed under pointwise limits of monotone sequences; hence H is the class of Baire measurable functions. (b) All functions in H are Baire measurable by part (a), hence a(H) a sad. But if A e sad, then IA is Baire measurable, so IA e H by part (a).

273

CHAPTER 4

But then IA is a(H)-measurable by definition of a(H), hence A E a(H). 5.

Let H be the class of Baire measurable functions. Then Ko c H, and if KK c H for all /3 < or, then Ka c H since H is closed under monotone

limits. Thus K c H. But K contains the continuous functions and is closed under pointwise limits of monotone sequences. For iffl,fz , ... EK f,eKa.,n=1,2,...,where a"
4), H c K. 6.

Existence of it and A follows from 4.3.13 applied to the real and imaginary parts of E. Now if J0 f dµ = 0 for all f e C(S2), then Sag dµ1 = Jag dµ2 =0

for all real-valued continuous functions on [2; hence Pi = µ2 = 0 by 4.3.13. This proves uniqueness of µ (and similarly, uniqueness of A). Now if f e C(S2), then In f dA = In fh dill for some complex-valued

Borel measurable h of absolute value 1; hence IE(f)l <-(n IfIdiAl; therefore IIE II -< IAI(f)) On the other hand, let E1, ..., E. be disjoint Borel

subsets of 0, with A(E) = rJe'01, j = 1, ..., n. Define f = I;=1 e-i° IE, and, for a given S > 0, let g be a continuous complex-valued function on 0 such that lgl < 1 and fn If - gI diAl < S (see 4.3.14). Now

f f d2_ r

J=1

rJ =

J=1

I A(E) 1,

hence

UfdA I-S= J=1

I E(g) I = fn gds, >_I f

A(EJ)I -S.

Since S is arbitrary, it follows that IIE II ? I2I(0). But just as in 4.3.13, A is the unique extension of p, so JAI(n) = Ipl(fl), completing the proof. Section 4.4

2.

Since there are c Borel subsets of R", there are only c measurable rec-

tangles in 9. Thus (Problem 11, Section 1.2) card 9 < c. But card 3.

.F >_ c (consider {co: co, = a), a e R); hence card 9 = c. Let W be the class of sets in F for which the conclusion holds. If A E ', then co e A` iff OT0 0 B0, that is, if wTa E Bo`; thus A` E W. If A. E c6', n = 1, 2, ... , with co e A. if wT E BTU, let To = Un 1 T", a countable

subset of T. Then co c- U' , A" if (or(, c BT., where BT. = (y E 0T0 : Yr. E BT" for some n). Now {y E r'To : YT" E BTU} = pn 1(BT"), where p" is

the projection of S2To onto Or.. Thus p": (0T0, -FT)) --*

(ClT,,, -T"), so

274

SOLUTIONS TO PROBLEMS

that BT. E .F r" , and consequently UM L, A. E W. Therefore' is a a-field containing the measurable cylinders, so ' _ .F. 4. (a) If the set A in question belongs to .F let To and Bo be as in Problem 3; assume without loss of generality that to E To. Since To is count-

able, there is a sequence t"

to with t" 0 To. Define x(t) = 0,

t E To; x(t) = 1, t To. Now if w(t) = 0 for all t, then co e A, hence wTo E Bo. But WTO = Xr,,, so that x e A. Since x is discontinuous at to, this is a contradiction. (b) Again assume the set A belongs to .F and let To and Bo be as in

Problem 3. Since To is countable, we can find s e T with s 0 To. Define x(t) = c - 1, t s; x(s) = c + 1, and let w(t) = c - I for all t. Then w e A and w(t) = x(t) for t E To ; hence x e A, a contradiction. 5.

(a)

Let F= n , V" ,

V,, open. Let g" be a continuous function from fl, to [0, 1 ] with g" = 0 on F, g" = I on V" Let g = Y' , 2-"g,,. Then g is continuous and g-'{0} = F. Define f: 0 -> R by f(w) = g(w(t)). Then f e C(92) and {w: f(w) = 0) = A; therefore A e But if % is the class of subsets B of f2, such that {w a 0: w(t) a B) E _d(S2), then ' is a a-field containing the closed sets; hence ' = .F,. It

follows that all measurable rectangles belong to d(fl); hence F c sa1(S2). (b) If x, y e fl, x # y, then for some t, we have x(t) 96 y(t). Let g be a

continuous function from ), to R with g(x(t)) # g(y(t)). Define f (co) = g(w(t)), co e n. Then f e C(S2), f depends on only one coordinate, and f(x) o fly). Since constant functions trivially depend on only one coordinate, the Stone-Weierstrass theorem implies that the algebra A generated by the functions in C(S) depending on one

coordinate is dense in C(Q). [The algebra A may be described explicitly as the collection of finite sums of finite products of functions depending on one coordinate.]

Now if f(w) = g (w(t)), g e C(.Q,), then f = g o p, where p, is the projection of Q onto Q. Thus every function in A is measurable

relative to F and R(R). But each function in C(f)) is a uniform (hence pointwise) limit of a sequence of functions in A, so that .F makes every function in C(S2) measurable. Therefore d(Q) c F. 8.

(a)

Since d is continuous, so is f. If co # w', let (co,,*} be a sequence in D with w"* - w. Eventually d(w, w"*) 96 d(w', w"*); hence f is one-to-

one. If f (w") -f (w), but d(w", w) > S > 0 for all n, let co* be an element of D such that d(w, co*) < 6/2. Since f (Con) -+ f (w), we have d(w", co*) -> d(w, co*) as n -+ co, and therefore d(w", (o) < d(w", w*)

+ d(w*, co) < S for large n, a contradiction. Thus f has a continuous inverse.

275

CHAPTER 4

(b) By part (a), f is a homeomorphism of !Q and f (S2) c [0, 1]'. Since i2

is complete, it is a G,, in any space in which it is topologically embed-

ded (see the appendix on general topology, Theorem A9.11). Thus f()) is a Ga in [0, 1 ]', in particular a Borel set. (c)

Let r e [0, 1) with binary expansion 0. a,a2

(to avoid ambiguity,

do not use expansions that terminate in all ones). Define g(r) = (a,, a2, ...); g is then one-to-one. If k/2" is a dyadic rational number with binary expansion O.a1 .. a.00 , then

['k'kk+1

g[

2"

{yeS"0:yi =a,,...,y"=a"}.

Thus g maps finite disjoint unions of dyadic rational intervals onto measurable cylinders, and since g[0, 1) = S' - {y e S°°: yn is eventually 1) = S `° - a countable set, we have g[0, 1) a Y ' and g and g-1 are measurable. (d) Let A = {y e S o0: yn is eventually 1), B = {y e S °°: y c- g[0, 1) and g-1(y) is rational}, where g is the map of part (c). Let q be a oneto-one correspondence of the rationals in [0, 1] and A u B. Define h:

[0, 1 ] -. S' by h(x) = g(x) q(x)

(e)

if x is irrational, if x is rational.

Then h has the desired properties. Define h: fln O2 -+11n Sn by h(w1, ()2 , ...) _ (M(all), h2(w2), ...). The mapping h yields the desired equivalence.

Section 4.5 3.

Assume µ"Z g, and let A be a bounded Borel set whose boundary has p-measure 0. Let V be a bounded open set with V A, and let Gj = {x e V: d(x, A) < 1/j}. If fj is a continuous map from it to [0, 1 ] such that fj = 1 on A and fj = 0 off Gj [see A5.13(b)], then

lµ"(A)-p(A)I -
+I f (fj-IA)dpi.

(1)

Since fj - IA is 0 on A and also on Gj`, the third term on the right side of (1) is bounded by p(Gj - A):!-< p(Gj - A°). Similarly, the first term is bounded by p (G j - A°). The second term approaches 0 as n --* co since the support offj is a subset of the compact set G.

276

SOLUTIONS TO PROBLEMS

Now let gjk be a continuous function from S2 to [0, 1], with supp g jk c V, such that as k -+ oo, gjk

{tip-A°}

To verify that the gjk exist, note that G j - A° = nn 1 U., where U. = {x e V: d(x, Gi - A°) < 1/n}. Let g.. be a continuous function from C1 to

[0, 1] such that g 'n = I on Cj

-

A°, g j. = 0 off U., and set gjk =

min(gj1, ... , gjk). Now

µn(Gj-A°)=fnlGj-AOdun<- fngjkdpn-' fngjkdµ

as

n -oo.

Thus

lim sup µn(G j - A°) < f gjk dp. n- 00 n

Since supp gjk c V and µ(V) < oo, we may let k -+ oo and invoke the monotone convergence theorem to obtain lim sup un(Gj - A°) S p(Gj - A°). n-ao

It follows from (1) and the accompanying remarks that lim sup I µn(A) - µ(A)I < 2p(G j - A°). n-I ao

As j -+ oo we have G j - A° 1 A - A° = 8A, and since µ(8A) = 0 we conclude that µn(A) -> µ(A), proving that (a) implies (b). Assume p .(A) -+ µ(A) for all bounded Borel sets with µ(8A) = 0, and

let f be a bounded function from 0 to R, continuous a.e. [µ] with suppf c K, K compact. Let V and W be bounded open sets such that K c V c V c W [see A5.12(c)]. Now

v= n {xeW:d(x, V)<8}= n Wa; 6>0

6>0

the Wa are open and 8W1 c {x a W: d(x, VV) = S}.

Thus the Ws have disjoint boundaries and µ(W) < oo, so µ(8W1) = 0 for some 8. Therefore we may assume without loss of generality that we have K e V, with V a bounded open set and µ(8V) = 0.

Now if A c V, the interior of A is the same relative to V as to the entire space Q since V is open. The closure of A relative to V is given by Av = A n V; hence the boundary of A relative to V is 81, A = (8A) n V.

277

CHAPTER 4

If A is a Borel subset of V and µ(8y A) = 0, then µ[(8A) n V] = 0; also

(8A)nV°cAnV°cV-V=V-V°=BV; hence µ[(8A) n V` ] = 0, so that u(8A) = 0. Thus by hypothesis, and µ' denote the restrictions of µ and p to V, we P(A). By 4.5.1, if have un' w, µ'. Since f restricted to V is still bounded and continuous a.e. [µ], we have f y f du.'- Jy f du', that is, fn f du. -+ Jn f dµ. This proves that (b) implies (c); (c) implies (a) is immediate.

Subject Index

A Absolute continuity of functions, 70 of measures, 59, 63 Absolute G,, 233 Absolute homogeneity, 114 Absorption of one set by another, 167 Accumulation point of filterbase, 206 of net, 204 Additivity theorem for integrals, 45 Adjoint of linear operator, 149 Algebra of sets, 4 Almost everywhere, 46 Annihilator of subset of normed linear space, 149 Approximation of Baire or Borel sets by closed, compact, or open sets, 34, 179-183 by continuous functions, 88, 185-188 by simple functions, 38, 88, 90 Arzela-Ascoli theorem, 228

B

Baire category theorem, 230 Baire sets, 178 Baire space, 231 Banach space, 114 Banach-Alaoglu theorem, 162 Bessel's inequality, 119 Bilinear form, 150 Borel-Cantelli lemma, 66 Borel equivalence, 195 Borel measurable functions, 35, 36 complex-valued, 80 properties of, 38-40 Borel sets, 7, 27, 178 Bornivore, 167 Bornological space, 167 Bounded linear operators, 128 weakly, 150 Bounded set, in topological vector space. 167

Bounded variation, 71

279

280

SUBJECT INDEX C

c, space of convergent sequences of complex numbers, 115, 136 Cantor function, 77, 78 Cantor sets, 33, 34 Caratheodory extension theorem, 19 Cardinality arguments, 13, 34, 42 Cauchy in measure, 93 Cauchy-Schwarz inequality, 82, 116 for sums, 87 Chain rule, 69 Change of variable formula for multiple integrals, 78 Chebyshev's inequality, 84 Circled set, 151 Closed graph theorem, 148, 166 Closed linear operator, 147 unbounded, 150 Cluster point, 203 Compact topological space, 213 countably, 216 locally, 217 relatively, 217 sequentially, 217 a, 218 Compactification, one-point, 220 Complete metric space, 230

Complete orthonormal set, 122 Completeness of L° spaces, 85, 90 Completion of measure space, 18 Complex measure, 69 Composition of measurable functions, 39 Conjugate isometry, 131 Conjugate linear map, 131 Conjugate space, 142 second,142 Consistent probability measures, 190, 191 Continuity of countably additive set functions, 10, 11 Continuity point of distribution function, 198

Continuous functions dense in L°, 88, 185, 188

Continuous linear functionals, 130, 135 extension of, 140, 156 representations of, 130-133, 136, 137, 184-186,188 space of, 131, 141

Convergence of filterbases, 205 of nets, 203 in normed linear spaces strong, 144 weak, 144 of sequences of linear operators, 134 strong, 134, 144, 149 uniform, 134 of sequences of measurable functions, 92ff.

almost everywhere, 47 almost uniform, 93 in L", 88 in L`°, 89 in measure, 92 in probability, 92 Convergence theorems for integrals, 44, 47, 49 Convex sets, 119

in topological vector spaces, 154ff. Countably additive set function, 6, 43, 62 expressed as difference of measures, 11, 44, 61

Counting measure, 7 Cylinder, 108, 189 D

Daniell integral, 170ff., 175 Daniell representation theorem, 175 Decreasing sequence of sets, I

De Morgan laws, I Density, 66 Derivative of function of bounded variation, 76 of signed measure, 74 Radon-Nikodym, 66 Difference operator, 27 Differentiation under integral sign, 52 of measures, 74ff. Dini's theorem, 181 Directed set, 203 Discontinuous linear functional, 135 Discrete distribution function, 76 Distribution function, 23, 29 decomposition of, 76 Dominated convergence theorem, 49 extension of, 96 Dynkin system, 168, 169

SUBJECT INDEX

281

E Egoroff's theorem, 94 Equicontinuity, 228 Extension of finitely additive set functions, 149

Extension theorems for measures, 13ff., 18, 19, 22, 183, 184

F F, set, 42, 178 Fatou's lemma, 48 Field of sets, 4 Filter, 205 Filterbase, 205 subordinate, 206 Finitely additive set function, 6 not countably additive, 11, 12 a-finite, 9 First countable space; 204, 212 Fubini's theorem, 101, 104 classical, 103, 106 Functional analysis, 113ff. basic theorems of, 138ff. G Go set, 178

Gauge space, 226, 237 Good sets principle, 5 Gram-Schmidt process, 125 Gramian, 125

H Hahn-Banach theorem, 139, 140, 149 Hausdorff space, 211 Heine-Borel theorem, 213 Hermite polynomials, 125 Hilbert spaces, 114, 116ff. classification of, 123, 124 separable, 124, 133 Holder inequality, 82 for sums, 87 I

Identification topology, 209

Increasing sequence of sets, I Indicator, 35 Inner product, 114 space, 114 Integrable function, 37, 81 Integral, 36ff. as countably additive set function, 43 indefinite, 59, 73 Integration of series, 46, 52 Internal point, 154 Isometric isomorphism, 123, 133, 137, 142, 163, 186, 189

J Jordan-Hahn decomposition theorem, 60

K

Kolmogorov extension theorem, 191, 194

L L' spaces, 80ff. completeness of, 85 continuous linear functionals on, 131133,137,165 /', l'(S2), 87 L°°, 89 1°°, , (c1), 90

Lattice operations, 170 Lebesgue decomposition theorem, 68, 76 Lebesgue integrable function, 51 Lebesgue integral abstract, 36ff.

comparison with Riemann integral, 53 Lebesgue measurable function, 39 Lebesgue measurable sets, 26, 31, 33, 54 Lebesgue measure, 26, 31, 100, 106 Lebesgue set, 78 Lebesgue-Stieltjes measure, 23, 27 Legendre polynomials, 125 Lim inf (lower limit), 2 Lim sup (upper limit), 2 Limit under integral sign, 52 of sequence of sets, 3, 12, 52

SUBJECT INDEX

282

Lindelof space, 212 Linear functionals, 130 continuous, 130, see also Continuous linear functionals positive, 170 Linear manifold, 119 generated by set, 121 Linear operator(s), 127ff.

bounded,128 closed, 147 continuous, 128 with discontinuous inverse, 147 idempotent, 130 range and null space of, 149 spaces of, 133 Lipschitz condition, 78 Locally compact space, 217 Locally convex topological vector space, 153

characterization of, 156 Lusin's theorem, 187

M Measurable cylinder, 108, 189 Measurable function, 35 Bore], 35 jointly, 101, 107 Lebesgue, 39 Measurable rectangle, 97, 108, 189 Measurable sets and spaces, 35 Measure(s), 6 absolutely continuous, 59, 63 complete, 18 complex, 69 extension of, 13ff., 183, 184 on field, 6 finite, 9 on infinite product spaces, 108ff., 189ff. Lebesgue, 26, 31, 100, 106 l.,ebesgue-Stieltjes, 23, 27 outer, 16, 22 probability, 6 product, 97, 100, 104, 106, 109, 111 regular, 183, 189 a-finite, 9 signed, 62 singular, 59, 66 spaces of, 186, 189

on topological spaces, 178ff. uniformly a-finite, 97 Measure-preserving transformation, 50 Minkowski functional, 154 Minkowski inequality, 83 for sums, 87 Monotone class theorem, 19 Monotone convergence theorem, 44 extended, 47 Monotone set function, 16

N

Negative part of countably additive set function, 62 of function, 37 Neighborhood, 201 Net, 203 Norm(s), 84, 114 on finite-dimensional space, 134 inducing same topology, 134-136 of linear operator, 128, 133 Normal topological space, 178, 211 Normed linear space, 114 linear operators on, 127ff.

0 Open mapping theorem, 147, 159 Orthogonal complement, 121 Orthogonal direct sum, 121 Orthogonal elements, 118 Orthogonal set, 118 Orthonormal basis, 122 Orthonormal set, 118 complete, 122 Outer measure, 16, 22 Overneighborhood, 201

P Parallelogram law, 118 Parseval relation, 122 Polarization identity, 124 Polish space, 180 Positive homogeneity, 138 Positive linear functional, 170

SUBJECT INDEX

283

Positive part of countably additive set function, 62 of function, 37 Pre-Hilbert space, 114 Probability measure, 6 Product measure theorem, 97, 104 classical, 100, 106, 111 infinite-dimensional, 109, 111 Product a-field, 97, 108, 189 Product topology, 208 Projection, 119, 120, 130, 148 Projection theorem, 121 Pseudometric, 84, 226 Pseudonorm,84 Pythagorean relation, 118 Q

Quotient space, 135 Quotient topology, 210

R Radial, 154 Radial kernel, 154 Radon-Nikodym derivative, 66 Radon-Nikodym theorem, 63 Rectangle, 96, 97, 108, 189 Reflexivity, 142, 163, 164 Regular measure, 183, 189 Regular topological space, 211 completely, 212 Riemann integral, 53-57 Riemann-Stieltjes integral, 56 Riesz lemma, 136 Riesz representation theorem, 130, 133, 181, 182, 184-186, 188 S

Second countable space, 212 Section of set, 98 Semicontinuous functions, 220 Semimetric, 84 Seminorm(s), 84, 113 family of, generating locally convex topology, 153 Separable Hilbert spaces, 124, 133

Separable topological spaces, 212 Separation properties for topological spaces, 211 Separation theorems, 159-161 strong, 160 Set function, 3 countably additive, 6 finite, 9 finitely additive, 6 Shift operator, 129 one-sided (unilateral), 129 two-sided (bilateral), 129 a-field (a-algebra), 4 countably generated, 111, 148 minimal, 5, 12 Simple functions, 36 dense in Lo, 88, 90 Singular distribution function, 77, 78 Singular measures, 66 Solvability theorem, 150 Space spanned by set, 121 Steinhaus' lemma, 42 Stone's theorem, 200 Stone-Weierstrass theorem, 225 Strong convergence in normed linear space, 144 of operators, 134,144, 149 Strong topology, 144, 161 Subadditivity, 114, 138 countable, 16 Sublinear functional, 138 Subnet, 204 Subspace, 119

closed, 119

T T, spaces, 211, 212 Tails of net, 205 Tietze extension theorem, 211 Topological isomorphism, 136 Topological spaces, 201ff, measures on, 178ff. Topological vector space, 114, 150ff. locally convex, 153 Topologically complete space, 232 Topology of pointwise convergence, 208, 227 of uniform convergence, 153, 227 Total variation, 62, 69

SUBJECT INDEX

284

Translation-invariance of Lebesgue measure, 33 Tychonoff theorem, 215 U

Ultrafilter, 206 Uniform boundedness principle, 143 Uniform convergence of operators, 134 Uniform space, 237 Uniform structure, 234 Uniformity, 234 Urysohn metrization theorem, 219 Urysohn's lemma, 211

V

Variation bounded,71 of function, 71 lower, 62 total, 62, 69 upper, 62 Vitali-Hahn-Saks theorem, 43

W Weak and weak* compactness, 162-164 Weak convergence, 144, 161 of distribution functions, 199 of measures, 196-199 Weak* convergence and weak* topology, 161

Vague (=weak) convergence of measures, 198

Weak topology, 144, 161