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A in y with (0) E u. But since " f.lq) determines a Hausdorff topology.
To show that the canonical projection q is both an open and closed mapping, first let U be an open set in X and suppose x* E q(U). Then q(x) = q(y) for some y E U. Therefore, x E y*. Let V E f.l such that S(y,v) cU. Then x E S(y,V) which implies x* E S(y*, q(V» c q(U). Hence q(U) is open in X/R 1.1 so q is an open mapping. If F is closed in X and if x* does not belong to q(F) then x E F which implies there exists aWE f.l such that S(x,W)nStar(F,W) == 0. Let V E f.l with V* < W. Suppose S(x*, q(V»nq(F)"#- 0. Then there exists some y* E X/R>, such that x* ,y* E q(V) for some V E V and such that y* E q(F). y* E q(F) implies that there exists some z E F with q(z) = q(y) which in tum implies y E z*, so y E S(z, V). Moreover, x* ,y* E q(V) implies there are r, s
5.3 Subspaces, Sums, Products and Quotients
117
E V with q(r) = q(x) and q(s) = q(y) so Y E S(s,V) and x E S(r,V). Consequently, there exist V I, V2, and V 3 E V such that x,r E V I, s,y E V2, and y,Z E V 3 . But then there are WI and W 2 E W with V I UV 2 C WI and V2 UV 3 cW 2 • NOWX,SE WI ands,zE W 2 impliesW l nW 2 "#0,W I cS(X,W) and W 2 C Star (F, W). Hence S(x,W)nStar(F,W) "# which is a contradiction. Therefore, S(x*, q(V»nq(F) = 0. Thus q(F) must be closed so q is also a closed mapping. -
°
Our last example of an inductive limit uniformity is the uniform sum of a collection {(X a, /-1a)} of uniform spaces. We define a new uniform space LX a by first defining the points of LX a to be the ordered pairs (x, a) where x E X a' The mappings i a:X a ---7 LX a defined by i a(X) = (x, a) are called the canonical injections. The uniformity a of LX a is defined to be the inductive limit uniformity determined by the collection {i a} of canonical injections. The uniform space (LX a,a) is called the uniform sum of the uniform spaces X a' The proof of the following proposition is left as an exercise (Exercise 6).
PROPOSITION 5.5 The topology associated with the uniform sum of a collection of uniform spaces is the topology of the disjoint topological sum of these spaces considered as topological spaces.
The canonical injections essentially transfer the uniformities /-la from the spaces X a onto the disjoint "pieces" i a(X a) of LX a and since a is the inductive limit uniformity determined by the i a's, a is the finest uniformity on LX a that makes all the i a 's uniformly continuous.
EXERCISES 1. Show that if (X, /-1) is a uniform space and Y c X, that the projective limit uniformity /-1, on Y determined by the function i:Y ---7 X such that i(y) = Y for each y E Y is identical with the subspace uniformity on Y introduced in Section 4.4. 2. Show that a net in a product space converges to a point p if and only if its projection in each coordinate subspace converges to the projection of p. 3. Show that a net in a product uniform space is Cauchy if and only if the projection of the net in each coordinate subspace is Cauchy. 4. Show that the product complete.
nx a
of complete uniform spaces (X a' /-1a) is
5. Let (X, /-1) be a uniform space and let R be an equivalence relation on X. Show that the topology associated with the quotient uniform space with respect
118
5. Fundamental Constructions
to R is the topology of the quotient (topological) space X/R introduced in the Introduction. 6. Prove Proposition 5.5. 7. Show that if (X, /.1) is a non-normal uniform space with the finest uniformity u, that there exists an equivalence relation R on X such that X/R is not Hausdorff. UNIFORM QUOTIENTS A uniform relation in a uniform space (X, /.1) is an equivalence relation R on X such that for each pair of non-equivalent points x,y E X, there exists a sequence {Un} C /.1 satisfying: (1) If xRx' and yRy', then x' and y' are in no common member of U I . (2) If Un +1 E U n +l , there exists a Un E Un such that if p E U n +1 and pRq, then Seq, Un+d C Un.
A uniform quotient is a quotient uniform space X/R where the equivalence relation R is a uniform relation. 8. Show that a uniform quotient of a Hausdorff uniform space is Hausdorff. 9. Show that the canonical projection q:X uniformly continuous.
--7
X/R of a uniform quotient is
UNIFORML Y LOCALLY COMPACT SPACES A uniform space is said to be uniformly locally compact if it has a uniform covering consisting of compact sets. For each countable ordinal a put X a = a + 1 and let '!a be the order topology on X a' Then X a is compact, so X a has a unique uniformity /.1a consisting of all open coverings. Let (L, /.1) be the sum of the collection {(X a, /.1a) I a < WI }. Let R be the equivalence relation on L defined by x - y if and only if x and y belong to the same X a' Let Y = W1 with the order topology. 10. [So Ginsburg and J. Isbell, 1959] Y has a unique uniformity with a basis consisting of all finite open coverings. R is a uniform relation on (L, /.1). L/R = Y. Let q:L --7 Y be the canonical projection. There exists a U E Il such that q(U) = {q(U) I U E U} is not a uniform covering in Y. L is complete, but q(L) = Y is not complete. L is uniformly locally compact.
5.4 Hyperspaces
119
5.4 Hyperspaces Let (X, /.1) be a uniform space and let X' denote the set of all non-void closed subsets of X. If H, K E X' with He Star (K, U) and K c Star (H, U) for some U E /.1, then Hand K are said to be U-close, denoted by IH - K I < U. Note that this relationship is reflexive, i.e., I H - K I < U implies I K - HI < u. If U E /.1 and F E X', put B(F,U) = {K E x'il F - KI < U) and let U' = {B(F,U) IFE X'). Then define /.1* = {U' IU E /.1). We want to show that /.1* is the basis for a uniformity /.1' on X' but first we establish some useful lemmas. LEMMA 5.] If u* < V.IH-FI < U.and IK-FI < U.then IH-KI
~. Then F Ii c F ~ which implies P E Star(F~,U). Therefore, K c Star(F~,U) which implies F~ is not contained in Star (K, U) or else F ~ E B(K,U) which is a contradiction. Similarly, Fa is not contained in Star (K, U) for each a ~~. Suppose y is not greater than or equal to~. Then there is a 8 such that y ~ 8 and ~ ~ 8 which implies F 8 is not contained in Star (K, U) and F 8 C F l' Hence F y is not contained in Star (K, U). Consequently, Fa is not contained in Star (K, U) for each a E A. Put U = Star (K, U) and for each a E A let H a = Fa - U. Then IHa} is a Cauchy net in X' that is directed by set inclusion. Let IGy}, yE B, be the collection of elements of X', directed by set inclusion, that contain a memher of IH a}. Then I G y} is a proper Cauchy net in X'. Let D = I (x, y) E X x B 1x E G y} and define ~ on D by (x, y) ~ (y, 8) if Y ~ 8. For each (x, y) E D put \jI(x, y) = x. Then \jI:D ~ X is a net which we will show is almost Cauchy. For this let WE /-1. Pick U E /-1 with U* < W. Since IGy} is stable, there exists a A with G y E B(GA,U) for each y ~ A. Pick x E Star(GA,U). Then some W E W contains Star(x,U). Let V = IW E wlw contains Star(x,U) for some x E Gd. Let R = I(y,~) E 01 ~ ~ A}. Then R is residual in D. Let (y,~) E R which implies ~ ~ A which in tum implies G~ c Star(GA,U) so y E Star(GA,U) which implies y E S(x,U) for some x EGA' Then there exists aWE V containing y. Therefore, \jI(y, ~) = yEW. Let C w = l(y,~)E RI\jI(y,~)E WforsomeWE V}. ThenulCwlWE V}=R. So I C w W E V} is a family of suhsets of D whose union is residual in D such that \jI(C w ) eWE V. It remains to show that C w is cofinal in D for each WE V. For this let WoE V and (y,~) E D. Then there exists an Xo E G A with S(xo,U)cW o. Pick(z,8)E Rwith(xo,A)~(z,8)and(y,~)~(z,8). ThenG8 eGA which implies Z EGA' But (z, 8) E R implies 8 ~ A which in tum implies G 8 E B(G A'U), so G A C Star (G 8,U). Therefore, x E S(s,U) for some S E G 8 which implies S E S(xo,U). Then (xo, A) ~ (s, 8), (y,~) ~ (s, 8), and \jI(s, 8) E S(xo,U) cWo. Hence Cwo is cofinal in D so \jI is almost Cauchy. 1
°
By hypothesis, \jI clusters to some P E X. Let V be a neighborhood of p. Then there exists a cofinal C c D with \jI(C) c V. Let a E A. Then Fa = G ~ for some p E B. Pick x E G~. Then there exists (y, 8) E C with (x, ~) ~ (y, 8) and \jI(y, 8) E V. Therefore, ~ ~ 8 which implies G 8 C G ~ and \jI(y, 8) = y E G 8 C G~ c Fa SO F arN *- 0. Hence p is a cluster point of IF a} so P E K. But \jI c X - U which is closed, so p E X - U which is a contradiction. Next we have the case K = 0. In this case, let D in the argument above be the set {(x, y) E X x A Ix E Fa}. Define:5; on D as before and define \jI:D ~ X by \jI(x, a) = x for each (x, a) E D. Then just as in the argument above, we can show that \jI is almost Cauchy and hence clusters to some p E X. But then as
124
5. Fundamental Constructions
the argument above shows, p E K, so K *' 0. Consequently, (X, /J.) must be supercomplete. Conversely, assume X is supercomplete. Let {xa)' a E A, be an almost Cauchy net in X. For each a put H a = {x~ IB ~ a) and Fa = CI(H a). Then {Fa) isanetinX'. Let VE /J.andpick UE /J.with U*< V. Since {xa) is almost Cauchy, there exists a collection {C y), Y E B, of cofinal subsets of A and a collection {U y) C U such that uyC y is residual in A and {x~ I BE C y} C U y for each y. Pick 0:0 E A such that B E uyC y for each B ~ 0:0. Then H ao C uyC y' For each y pick Vy E V such that Star (Uy,U) C V y' We want to show Fao CUyU y. LetYE Fao' IfYE Hao,clearlYYE UyV y . Supposeyisalimit point of H ao' Let U E U be a neighborhood of y. Then there exists an x ~ E Hao with x~ E U. Now x~ E Hao implies x~ E U y for some y so U C V y. Hence F ao C UyV y. Let a1 E H ao' Then F al C F ao which implies F al C Star (F ao' V). Pick y E B. Let BE C y such that B > a1' Then x~ E U y and x~ E F al which implies x~ E VynF al so uV y C Star(F al ,V). But F ao C UyV y so F ao C Star(F al ,V). Hence F al E 8(F ao,v) for each a1 E H ao so {Fa} is Cauchy. Since X' is complete, {Fa) converges to its set of cluster points K. Let p E K. Let W be a neighborhood of p in X. Then there exists a cofinal C C A with Fan W *' 0 for each a E C. If Y E F anW then y E H a or y is a limit point of H a' In either case, W contains some x~ with B ~ a. For each a E C pick K(a) ~ a such that x K(a) E Fan W. Then {x K(a) Ia E C} c W is cofinal in A. Hence. {x a} clusters top. COROLLARY 5.1 Cofinal completeness implies supercompleteness which in turn implies completeness. COROLLARY 5.2 (1. 1sbell. 1962) If X is paracompact. then it is supercomplete with respect to u. Notice that our proof that supercompleteness implies each almost Cauchy net clusters only relies on the fact that there exists a cluster point of {Fa). Consequently, the existence of a cluster point for each proper Cauchy net in X' is equivalent to supercompleteness. Furthermore, if p is a cluster point for the proper Cauchy net F = {Fa) and U E /J., then for each U E U containing p, UnF ~ *' 0 for cofinally many F ~'s in F. Let y be any index. Then there exists a 8 ~ y with UrlF B *' 0 and since F is directed by inclusion we have UnF y *' 0. Therefore, S(p, U)nF a*,0 for each Fa E F. Thus we can pick an Xu E S(P,U)nF a for each U E /J.. Then {xu} converges to p which implies p is a limit point of Fa for each a. Therefore, p E Fa for each a since each F a is closed. Hence nF a*' 0. Similarly, if nF a*,0 then any p E nF ex is a cluster point of F. We record these observations as
5.4 Hyperspaces
125
PROPOSITION 5.9 If (X. 11) is a uniform space then the following statements are equivalent: (I) (X. 11) is supercomplete, (2) Each proper Cauchy net in X' has a cluster point and (3) For each proper Cauchy net F in X', nF 0.
*'
EXERCISES 1. A function f:X -t Y from a uniform space (X, 11) into a uniform space (Y,v) determines a function j':X' -t Y' where X' and Y' are the hyperspaces of X and Y respectively. j' is defined by j'(A) = Cl y (f1Al) for each A E X'. j' is called the hyperfunction of f. Show that j' is uniformly continuous if and only if f is uniformly continuous. THE HYPERSPACE OF A METRIC SPACE 2. The Hausdorff distance h between two (closed) sets A and B in a metric space (X, d) is defined as the maximum of sup I dCa, B) I a E A) and sup {d(A,b) I b E B) where d(x, S) =in!{ d(x,y) lYE S) for any subset S. Show (a) h is a metric on HX (the hyperspace of X) that generates the topology of HX so that the hyperspace of a metric space is again a metric space. (b) If (X, d) is complete then (HX, h) is complete. THE HYPERSPACE OF A COMPACT SPACE 3. Show that the hyperspace of a compact space is compact. 4. Show that a discrete space of the power of the continuum, with the uniformity determined by all countable coverings, is complete but not supercomplete. 5. Show that real Hilbert space is supercomplete but not cofinally complete. 6. Define the limit inferior of a net {Fa) in X' to be the set inf{F a) = {x E X Ifor each neighborhood U of x, {Fa) eventually meets U}. Similarly, define the limit superior to be the set suplF a) = Ix E X I for each neighborhood U of x, {F a) frequently meets U}. If K is the set of cluster points of {F a} then (a) sup{Fa}=K=inf{F a }
(b) (X, 11) is supercomplete if and only if whenever IF a} is Cauchy then for each U E I..l there is a ~ with Fa C StarCK, U) for each ex ~ ~.
5. Fundamental Constructions
126
7. Use the results of Exercise 8 to show that if X is paracompact, then it is supercomplete with respect to u, without reference to cofinally Cauchy or almost Cauchy nets (i.e., do not appeal to Corollary 5.1).
5.5 Inverse Limits and Spectra We begin our discussion of inverse limits of topological and uniform spaces with a special case that will motivate the concept in general. Let {X n ) be a sequence of topological spaces where n ranges over the non-negative integers and suppose that for each positive integer n, there is a continuous function In:Xn --7 X n- 1 • The sequence of spaces and mappings {Xn,fn) is known as an inverse limit sequence. Inverse limit sequences are often represented by diagrams like the one below:
In
In-!
r::
If m < n, then the composition mapping = fm+1 © fm+2 © ... © fn is a continuous mapping from Xn to X m. Consider the sequence of points lx n ) such that for each n, Xn E Xn and Xn = In+l (x n+!). Then {xn ) can be identified with a point of the product space n;:'=oXn by means of the function g:l --7 uX n, where 1 denotes the non-negative integers, defined by g(n) = Xn- By means of this identification, the set Y of all such sequences can be considered to be a subset of n;:,=ox n. Then Y, equipped with the subspace topology, is called the inverse limit space of the sequence {X n, fn)' Y is denoted by lim.,...Xn or by X~. The functions fn are sometimes called bonding maps.
LEMMA 5.7 If {Xn,fn) is an inverse limit sequence with onto bonding maps and for some counrable set {an) of positive inlegers there is a set {x an } such that xa n E Xa n for each n and such that if m < n, then la~m (x an ) = x am ' then there eXlsrs a point in lim.,...X n whose coordinare in Xa n is xa n for each n. Proof: First assume {an) is infinite. For each positive integer m, there exists a least positive integer an such that an ~ m. If an = m put Xm = x an ' otherwise let Xm = f,:: (x an )· Then IXm) is a point of lim.,...X n such that xa n E Xa n for each fl. Next, assume Ian) is finite. Then there exists a greatest an, say aj. If m < aj put Xm = ;:;. (xa). For each m ~ aj, assume Xm has already been defined. Since fm+l is onto, pick Xm+! E Xm+! such that fm+l (x m+!) = Xm • Then by induction we can complete the sequence IXm) such that Ix m } E lim.,...Xn and for each n, xa n E XOn " Lemma 5.7 illustrates the fundamental property of inverse limit spaces, that for any coordinate Xb all elements of lim"",Xn having that coordinate have
5.5 Inverse Limits and Spectra
127
all other coordinates Xm with m < k determined by the inverse limit sequence, whereas there may be some room for choice of the coordinates Xm with m > k. If the bonding maps are not onto, limf-X n may be the empty set. An example of this occurs when the Xn are all countable discrete spaces, say Xn = {x;:'} and the bonding maps fn:Xn ~ X n -1 are of the form fn(x;:') = x;:,~il' Clearly {Xn> fn} is an inverse limit sequence, but if we begin with a point x~, it is only possible to pick the first m coordinates before we reach xi and there does not exist an X'J'+1 such that fm+l (x'J'+l) = Xl' Consequently, there do not exist any sequences {xnl such that Xn E Xn and fn(x n ) = Xn-l for each n > 1. Therefore, limf-X n = 0. Under certain conditions, we can assure that limf-X n l' 0. For instance. THEOREM 5.2 If each space Xn in the inverse limit sequence /XnJn} is a compact Hausdorff space then limf-X n i= 0. Proof: For each positive integer n let Yn C DXn be defined by {Yi} E Yn if for each j < n, Yj-l =Jj(y). Then limf-Xn = nYn' We will show that for each n, Yn is closed. Suppose p E DXn - Yn- Then for some j < n we have Jj+l (Pi+l) i= Pi' Since Xj is Hausdorff, there exists disjoint open sets Uj and Vj containing Pj and Jj+l (Pj+l) respectively. Put Vj+l = J;ll (Vi) and let Up be a basic open set in DXn containing p and having Uj and Vj+l as its /h and j+ Fi factors respectively. Then no point of Yn lies in Up since if q = {qn} E Up, then qj+l E V j +1 which implies qj E Vj so q does not belong to Up. Therefore, Yn is closed. Since {Yn } is a decreasing chain of closed sets in the compact space DXn, nYn i= 0 so limf-X n i= 0. •
There are many applications of inverse limit spaces of inverse limit sequences in topology. What we are interested in here is extending the concept to uniform spaces, and extending it in a more general setting. For this first notice that by changing the definition of an inverse limit sequence {Xn.fn} so that the Xn are now uniform spaces and the bonding maps are uniformly continuous, we get an inverse limit uniform space limf-Xn that is a uniform subspace of the uniform product space DXno This follows from the parallel between product topological spaces and product uniform spaces [see Section 5.3]. But what we have in mind is something more general than this. Let (D,~:X/)l ~ X/A defined by mapping an equivalence class in X/)l onto the equivalence class of X/A to which a member of the equivalence class in X/)l belongs. Then we have the following commutative diagram:
~
ill
X
~
X)l
.t ·A
II ~
X/)l
.t ~
t)l
~
iA X
-7'
XA
~
X/A
where i~ denotes the identity map on X considered as a mapping from X)l onto X A' It is left as an exercise [Exercise 3] to show that ~ is uniformly continuous with respect to the metric uniformities on X/)l and X/A respectively. Then {X/A' ~} is an inverse limit system of metrizable spaces. Let veX) = limrX/A and let 1tA denote the canonical projection of veX) onto X/A' Then the uniformity ofv(X) has a basis consisting of the coverings
veX) is called the weak completion of X with respect to v. A uniform space is said to be weakly complete if each co-directed Cauchy net clusters. We will show that (X,v) is uniformly homeomorphic to a dense uniform subspace of veX) and that veX) is weakly complete. The weak completion of a uniform space with respect to a uniformity was introduced by K. Morita in 1970 in a paper titled Topological completions and M-spaces published in Sci. Rep. Tokyo Kyoiku Daigaku 10, No. 271, pp. 271-288. To see that veX) is weakly complete, let 'I':D ~ veX) be an co-directed Cauchy net. Then for each A E A,1t A © 'I':D ~ X/A is an co-directed Cauchy net in the metric space X/A' The notion of weak completeness is different from the notion of completeness, because every metric space is weakly complete with respect to its metric uniformity. We prove it for X/A by showing that 'l'A = 1t1.. © 'I' clusters in X/A' For this let n be a positive integer and let Rn be residual in D such that 'l'1..(R n ) c S(xn , 2- n ) for some Xn E X/A' Then n;;'=I'I'A(R n ) f:. 0, so pick Y E n;;'=I'I'A(R n ). Suppose nIiEDCI('I'" (R 0» = 0 where R 0 = {y E D 18 ~ y}. Then there exists an a E D such that y does not
130
5. Fundamental Constructions
belong to CI('VA(R a». Let m be the least positive integer such that S(y,2-m)nStar(CI('V)...(R a», 2- m) = 0. Now for each n > m, y E 'V),JR n) which implies 'V".)... Now {yd defines a point in DAX/!!>).. and since 'V).. converges to y).. for each A, we see (by Exercise 2 of Section 5.3) that 'V must converge to {yd in D)..X/!!>)... But since 'V(D) c veX) = lim).. and since lim).. is closed in D)..X/!!>A by Lemma 5.8, we have that 'V clusters in veX). Consequently, veX) is weakly complete. To see that (X,v) is uniformly homeomorphic to a dense subspace of veX), notice that for any x E X, {)..(x)} for each x E X. It is easily shown [see Exercise 4] that
(Un)' Therefore, 1t~! (4))..(U~+!)) < ).. and S(1t)..(y), 4>)..(U~» C X/!!>).. so we have a
contradiction. Therefore, inf(X) converging to y. Since g and h are morphisms (uniformly continuous functions in the category of uniform spaces), g © q> converges to gCy) and h © q> converges to hCy). But g © f= h © fand q> cf(X) implies g © q> = h © q>. Since nets converge to unique limits in uniform spaces, we have gCy) = hCy). Therefore,fis an epimorphism. A morphism f:X ~ X is called a retraction if f2 =f © f =f. If a retraction is either an epimorphism or a monomorphism, it is an identity morphism. By a contravariant functor F:C ~ D of two concrete categories C and D, we mean a functor F that assigns to each object X of C an object F(X), but to each morphism f:x ~ Y of C, a morphism in the opposite direction F(j):F(Y) ~ F(X) of D to C that satisfies property (3) above, but instead of satisfying property (4), it satisfies (4') For each composed morphism g © fof C, F M(g © f) = F M(j) © F M(g). Observe that the composition of two contravariant functors is a covariant func-
5.7 Categories and Functors
149
tor. Also, functors of different variances can be composed with the resultant functor being a contravariant functor. A duality is a contravariant functor F:C ---* D such that there exists a contravariant functor F-1:D ---* C with both F- 1 © F and F © F- 1 being identity functors. If C and D are concrete categories, and F:C ---* D is a duality, then every concept, definition, or theorem about C gives rise to a dual concept, definition or theorem about D. To illustrate what is meant by dual concepts and definitions, we consider the concepts of sums and products of objects of an arbitrary concrete category C. As we shall see, these concepts are dual concepts. We have already considered the concepts of sums and products of uniform spaces. We now give categorical definitions for sums and products of objects in arbitrary concrete categories in such a way that the ordinary definitions of sums and products of uniform spaces are equivalent to the categorical definitions in the category of uniform spaces. Let {X a} be a collection of objects in the concrete category C and let {i a) be a collection of morphisms such that i a:X a ---* L for some L E C. Then L is called the sum of the objects {X,,} if the following conditions hold: (5) If f:L ---* Y and g:L ---* Yare distinct morphisms, then for some index a, the morphisms f © i a and g © i a are distinct. (6) For each family of morphisms f a:X a ---* Y, there exists a morphism f: L ---* Y such that f © i a = fa for each a. Similarly, an object rr E C is said to be a product of the objects {X a} if there is a collection {p a} of morphisms such that P a:rr ---* X a for each a, if the following conditions hold: (7) If f:Z ---* rr and g:Z ---* rr are distinct morphisms, then for some index a, the morphism P a © f and P a © g are distinct.
(8) For each family of morphisms f a:Z ---* X a, there exists a morphism f:Z ---* rr such that P a © f = fa for each a. To show that the categorical definitions of sum and product are dual, we must show that if L is the sum of {X a} in C, then rr = F(L) is the product of {F(X a)} in D and vice-versa. For this, assume L is the sum of IX a} in C and put rr = F(L). Suppose F(j) and F(g) are distinct morphisms such that F(j):F(Y) ---* rr and F(g):F(Y) ---* rr. Then f:L ---* Y and g:L ---* Y. Since F- 1 © F is the identity, F- 1 © F(j) = f and [F- 1 © F](g) = g. Therefore, f = g implies [F-l © Fl(j) = [F-I © F](g), so F(j) = F(g) which is a contradiction. Therefore, f and g are distinct. Since L is the sum of {X a}, there exists an a with f © i a ::j:. g © i a. Then by an argument similar to the above, F(f © i a) ::j:. F(g © i a) which implies F(i a) © F(j) ::j:. FU a) © F(g). For each a, put p a = F(i a) and let Z = F(Y). Then if F(j):Z ---* rr and F(g):Z ---* n are distinct morphisms, for some index a, the
150
5. Fundamental Constructions
morphisms Pu © F(j) and p u © F(g) are distinct Roughly speaking, condition (5) in C forces condition (7) in D. Next, suppose F(f u):Z --? F(X u) is a family of morphisms in D. Then --? Y is a family of morphisms in C, so by (6), there exists a morphismf:L --? Y with f © i u = fa for each a. But then F(j):Z --? D such that p u © F(j) = F(f u) for each a. Again, roughly speaking, condition (6) in C implies condition (8) in D. Hence D = F(L) is the product of {F(X u») in D.
f u: X u
The proof that if D is the product of (X u) in D, then L = r! (D) is the sum of IF- 1 (X u») is similar. This establishes the duality of the concepts of sum and product in arbitrary concrete categories. To illustrate what is meant by dual theorems we will show that the categorical theorem: sums are unique (in C), gives rise to the dual (categorical) theorem: products are unique (in D). To see this, let L be the sum of IXu) in C. Then D = F(L) is the product of I F(X u») in D as we have already seen. If D' is another product of IF(X u»), then L' = F- 1 (il,) is another sum of IX u) so there exists a uniform homeomorphism i:L' --? L (since sums are unique in C). Therefore, F(i):D --? D' and since i-I © i = IE' and i © i-I = IE, we have:
Similarly, F(i- l ) © F(i) = IF(E), so F(i) is a uniform homeomorphism of D = F(L) onto D' = F(L'). Therefore, products are unique in D. The theorem that products are unique in C implies sums are unique in D is proved similarly. This establishes the duality of these two theorems. The principle of duality says that any categorical concept, definition or theorem gives rise to a dual concept, definition or theorem. It is also of interest to give categorical definitions of the concepts of subspace and quotient space. For the concept of a subspace, we recall that for a uniform space X, a subspace can be considered to be a uniform space S such that there exists a one-to-one uniformly continuous function i:S --? X such that the uniformity of X relativized to i(S) is identical with the identification uniformity of i(S). By a uniform imbedding of S into X, we mean a uniform homeomorphism of S onto a subspace S' of X. Clearly, if S is a subspace of X, the mapping i:S --? X above is a uniform imbedding.
Then a one-to-one mapping i:S --? X is a uniform imbedding if and only if whenever i = g © f where g and f are uniformly continuous and f is one-to-one and onto, then f is a uniform homeomorphism. To see this first assume i:S --? X is a uniform imbedding and i = g © f where g and f are uniformly continuous and f is one-to-one and onto. Let v be the uniformity of Sand Il the uniformity of X restricted to i(S). For V E v, i(V) = {i(V) I V E V} E Il since i is a uniform imbedding. Let A be the uniformity of S' = f(S). Since g is uniformly
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continuous, g-I(i(V» E A. Butf= g-I © i because i is one-to-one and g = i © i which implies g is one-to-one. Hence f(V) E A, so f is a uniform homeomorphism.
r
Conversely, if the one-to-one mapping i = g © f where g and f are uniformly continuous and f is one-to-one and onto implies f is a uniform homeomorphism, then if V E v, f(V) E A. Assume i(V) does not belong to /.!. Then g(g-I U(V))) is not in /.! since g is one-to-one. But f(V) E A implies g -I U(V» E A since f = g-I © i. Therefore, the identification uniformity /.!* of g is strictly finer than /.!. Let A' be the uniformity induced on 5' by /.! with respect to g. Then A' is strictly coarser than A = gl (/.!*). Hence f is uniformly continUOUS with respect to v and A', but (5,v) and (5', A') are not homeomorphic which is a contradiction. Therefore. dV) E /.! so i is a uniform imbedding. Next. we use the characterization of uniform imbeddings to motivate the categorical definition of imbeddings. A monomorphism i:S --7 X in a concrete category C is said to be an imbedding if whenever i = g © f where g and f are morphisms, and f is one-to-one and onto. then f is an isomorphism. This means that if f:5 --7 5' then 5' has the same structure as 5. so that no structure can be inserted on 5 that is between the structure of 5 and that of i(5). In the case of uniform spaces, this means that the uniformity of 5 is the one induced by the function i. It is also clear that in any concrete category, two imbeddings with the same image have isomorphic domains. The dual categorical concept of an imbedding is that of a quotient morphism. An epimorphism q:X --7 Q in a concrete category D is said to be a quotient if whenever q = d © h where d and h are morphisms and d is one-toone and onto. then d is an isomorphism. To show that in the category of uniform spaces, a uniform quotient mapping satisfies this categorical definition, let (X, j..l) be a uniform space and q:X --7 Ya uniform quotient mapping. Then q is onto and Y has the identification unifonnity with respect to /.! generated by q. Let v be the uniformity of Y, then V E v if and only if l (V) E /.!.
r
Let q = d © h where d and h are uniformly continuous and d is one-to-one and onto. Let A be the uniformity of Q' = heX). If W E A, then h -I (W) E /.!. Since q is a uniform quotient, q(h- I (W» E v. But then d(h(h- I (W))) = deW) E v, so d is a uniform homeomorphism. Consequently, in the category of uniform spaces, unifonn quotient mappings satisfy the categorical definition of a quotient morphism. A subcategory S of a concrete category A is a category such that each object of S is an object of A and each morphism of 5 is a morphism of A. S is said to be a full subcategory of A if every morphism of A whose domain and range are both objects of S is a morphism of 5. A full subcategory R of a category A is said to be a reflection of A if for each object X E A, there is an object X* E R and a morphism r:X --7 X* (called the reflection morphism) such that each morphism f:X --7 Y where Y E R can be factored uniquely over X* by r
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(i.e., f = g © r for a unique g:X* -+ Y). The space X* is called the reflection of X in R or simply the R-reflection of X. There are many interesting reflections of the category of all uniform spaces. We have already encountered some of them unknowingly. For instance, the class of all complete uniform spaces C is a reflection of the category U of all uniform spaces. For each uniform space X E U, the completion X* E C is the C-reflection of X and the natural imbedding i:X -+ X* is the reflection morphism such that any morphism f:X -+ Y E C factors uniquely over X* by i. For any reflection R of a category A, an R-reflection of an object X E A is unique up to isomorphism. To see this, note that if X* and X+ are both R-reflections of X, then there exists reflection morphisms r*:X -+ X* and r+:X -+ X+ such that any morphism f:X -+ Y E R factor uniquely over both X* and X+ by r* and r+ respectively. Therefore, r* = g+ © r+ and r+ = g* © r* for some unique g* and g+. Now g*:X -+ X+ and g+:X+ -+ X*. Moreover, r+ = g* © (g+ © r+) = (g* © g+) © r+ so (g* © g+) = lx+. Similarly, g+ © g* = Ix'. Therefore, g + = g*-l so X* and X+ are isomorphic (in the class of uniform spaces this means uniformly homeomorphic). The fact that an R-reflection is unique allows us to make the simplifying assumption that if X E R, the only R-reflection of X is X itself and the only reflection morphism r:X -+ X is the identity morphism. This allows us to state the following fundamental theorem about reflective categories.
THEOREM 5.15 If R is a reflection of a concrete category A, X, YEA and X*, y* are their R-reflections respectively, then any morphism f:X -+ Y determines a unique morphism f* :X* -+ y* such that r © f = f* © s where r:X -+ X* and s:Y -+ y* are the reflections morphisms respectively. Proof: Since s © f:X -+ Y*, it can be factored uniquely over X* by r. Letf* © r be this factorization. Then s © f = f* © r and since f* is unique, it is the unique morphism that satisfies the theorem. Theorem 5.15 shows that the correspondences X -+ X* and f -+ f* determine a functor P called the reflection functor. The object function of P is Po:A -+ R defined by Po (X) = X* for each X E A and the morphism function PM is defined by PM(f) = f* for each f E A. To see that P is a functor, we need to show that Po and PM satisfy (3) and (4) of the definition of a functor. To demonstrate (3) we note that by Theorem 5.15, the identity Ix has a corresponding unique morphism l~:X* -+ X* such that l~ © r = r © Ix = r. Now Ix> © r = r so Ix> = 1~ since G is unique. Therefore, PM(1X) = Ipo(x), To demonstrate (4), if g © f is a morphism of A, then by Theorem 5.15, there exists unique morphisms g*, f* and (g © j)* such that (g © j)* © r =r © (g © j),f* © r = r © fand g* © r = r © g. But then r © (g © j) = g* © r © i=
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g* © 1* © r. Since (g © j)* is unique, (g © j)* = g* © 1*. Therefore, PM(g) © PM(/)'
The dual concept to the concept of a reflection is the concept of a corejlection defined as follows. A full subcategory K of a concrete category A is said to be a coreflection of A if for each object X E A, there exists an object X* E K and a coreflection morphism k:X* ~ X such that each morphism /:Z ~ X where Z E K can be factored uniquely over X* by k (i.e., / = k © g for a unique morphism g:Z ~ X*). The space X* is called the coreflection of X in K or simply the K-reflection of X. As with reflections, there are many interesting coreflections of the category of uniform spaces. Possibly the simplest and most useful is the fine corejlection that maps each uniform space (X, J..l) onto (X, u) where u is the finest uniformity for X. Here, the coreflection morphisms are the identity mappings. To see that the class F of fine uniform spaces is a coreflection of the category U of uniform spaces, it suffices to show that each morphism j:Z ~ X where Z is a fine space can be factored uniquely over (X, u). But since /:Z ~ X is continuous and Z is fine, j:Z ~ (X, u) is uniformly continuous, so / can be factored over (X, u) by the identity mapping. By the principle of duality, for any coreflection K of a category A, a K-coreflection X* of an object X is unique up to isomorphism. As with reflections, the uniqueness of K-coreflections allows us to make the assumption that if X E K, the only k-coreflection of X is X itself and the only coreflective morphism k:X ~ X is the identity morphism, so we have the dual theorem: THEOREM 5.16 1/ K is a corejlection 0/ a concrete category A, X, Y E A and X*, y* are their K-corejlections respectively, then any morphism/:X ~ Y determines a unique morphism 1* :X* ~ y* such that / © k = j © 1* where k:X* ~ X and j:Y* ~ Yare the corejlective morphisms respectively.
Theorem 5.16 shows that the correspondences X ~ X* and / ~ 1* determine a functor K called the coreflection functor. The object function of K is !Co:A ~ K defined by !Co (X) = X* and the morphism function KM is defined by KM(/) = 1* for each/E A.
EXERCISES 1. Show that the definitions of uniform sums and products are equivalent to the categorical definitions of sums and products in the concrete category of uniform spaces.
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DUAL THEOREMS 2. Show that the categorical theorem P: A retraction f that is a monomorphism is an identity implies the dual theorem Q: A retraction f that is an epimorphism is an identity. REFLECTIONS 3. Let U be the category of uniform spaces and for each (X, Jl) E U let p(Jl) denote the uniformity consisting of all coverings refined by finite members of Jl (see Exercise 6, Section 5.6). Then (X, p(Jl» is a precompact uniform space. Show that the class P of all precompact uniform spaces is a reflection of U with reflection morphisms being the identity functions. Let P denote the reflection functor (called the precompact reflection). Show that for each (X, Jl) E U that Po(X, Jl) = (X, p(Jl» and for each morphism f:(X, Jl) ~ (y, v) that PM(j) =f. 4. For each (X, Jl) E U let e(Jl) be the collection of all coverings of X refined by a countable member of Jl. Show that e(Jl) is a separable uniformity for X. [A uniform space is separable if it has a basis consisting of countable coverings.] Show that the class U w of all separable uniform spaces is a reflection of uniform with reflection morphisms being the identity functions. Let e denote the reflection functor (known as the separable reflection), and show that for each (X, Jl) E U, eo(X, Jl) = (X, e(Jl» and for each morphism j:(X, Jl) ~ (Y, v) that eM(j) = f·
5. Show that the separable uniformities are precisely the countably bounded uniformities. 6. Show that the reflections P and e commute, i.e., for each (X, Jl) p(e(X,Jl» = e(p(X,Jl». 7. Let 1t denote the completion reflection that maps each (X, Jl) Show that the reflections P and 1t commute.
E
E
U,
U into /lX.
8. It was long an unsolved problem (due to Ginsberg and Isbell) whether the reflections e and 1t commute. 1. Pelant (1974) showed that in general they do not. THE SAMUEL COMPACTIFICATION 9. Show that the composition 1t © P of the reflection 1t and P is again a reflection. Since for each X E U, p(X) is precompact, it is clear that 1t(p(X» is compact and p(X) is uniformly homeomorphic with a dense subspace of 1t(p(X». 1t(P(X» is called the Samuel compactification of X, or the uniform compactification of X.
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COREFLECTIONS 10. Let U be the category of unifonn spaces and for each X = (X, ~) E U let denote the locally fine coreftection of~. Put X* = (X, A(~» and let L be the category of locally fine unifonn spaces. Show that A:U ~ L defined by A(X) = X* for each X E U and A(j) = f is a coreflective functor so that the locally fine coreflection is really a coreftection categorically. A(~)
Chapter 6
PARACOMPACTIFICATIONS
6.1 Introduction In the late 1950s and during the 1960s, K. Morita and H. Tamano worked (independently) on a number of problems that involved the completions of uniform spaces. We state some of these problems here even though we have not yet defined some of the terms used in the statements of these problems. (1) (Tamano) Characterize the uniform spaces with paracompact completions" (2) (Morita) Characterize the Tychonoff spaces with paracompact topological completions. (3) (Morita) Characterize the Tychonoff spaces with LindelOf topological completions. (4) (Morita) Characterize the Tychonoff spaces with locally compact topological completions. (5) (Tamano) Is there a paracompactification of a Tychonoff space analogous to the Stone-Cech compactification or the Hewitt realcompactification? By the topological completion of a uniform space, we mean the completion of the space with respect to its finest uniformity. By a paracompactification of a topological space, we mean a paracompact Hausdorff space Y such that X is homeomorphic with a dense subspace of Y. If we identify X with this dense subspace (which is customary), then Y is a paracompact Hausdorff space containing X as a dense subspace. Certain paracompactifications are compact. We call these compactifications. The theory of compactifications is well developed. The theory of paracompactifications is not. Notable among compactifications is the Stone-Cech compactification. We will devote a significant amount of effort in this chapter to its development. We are not yet in a position to define what a rea[compact space is. Like paracompactness, realcompactness is a generalization of compactness. We will show in the next chapter thatfor all practical purposes, realcompactness is also a generalization of paracompactness. We will of course, formalize what we mean by "for all practical purposes," but for now, we only mention that this formalization is a set theoretic matter. It turns out that the statement All
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paracompact spaces are realcompact is known to be consistent with the axioms of ZFC, but it is not known if it is independent of ZFC. Furthermore. it is known that if there exists a Tychonoff space that is paracompact but not realcompact, that its cardinality is larger than any cardinal with which we are familiar.
Once we know what a realcompact space is, we will be able to define a rea/compactification analogously with the way we defined a paracompactification. Like the theory of compactifications, the theory of realcompactifications is well developed. Among the various realcompactifications a space may have, there is one, known as the Hewitt realcompactification that plays a role in the theory of realcompactifications analogous to the role played by the Stone-Cech compactification in the theory of compactifications. Since "for all practical purposes," paracompactifications lie between realcompactifications and compactifications, it is natural to ask if there exists a paracompactification that plays a role analogous to the role played by the Stone-Cech compactification and the Hewitt realcompactification in the theories of compactifications and realcompactifications respectively. This is precisely Problem 5 above that we will refer to as Tamano's Paracompactification Problem. In this chapter we will show that, in general, there does not exist such a paracompactification. We will also characterize those spaces for which such a paracompactification exists and examine some of their properties. In 1937, two papers appeared that characterized the Tychonoff (uniformizable) spaces as those having compactifications. The two papers were: Application of the Theory of Boolean Rings to General Topology that appeared in the Transactions of the American Mathematical Society (Volume 41, pp. 375-481) by M. H. Stone and On bicompact spaces by E. Cech that appeared in Annals of Mathematics (Volume 38, pp. 823-844). Their approaches were very different, but both exhibited a compactification that is the largest possible compactification of a Tychonoff space. Clearly every compactification of a Tychonoff space X is the completion of X with respect to some totally bounded uniformity (since we can relativize a compactification's unique uniformity to X to obtain the desired totally bounded uniformity on X). In 1948, P. Samuel published a study of compactifications obtained as completions of uniform spaces (see Ultrafilters and compactifications, Transactions of the American Mathematical Society, Volume 64, pp 100-132). As a result, compactifications obtained as the completion of a uniform space are sometimes called the Samuel compactification of the uniform space. In this chapter we will employ this approach. The Stone-Cech compactification will be obtained as the completion of a Tychonoff space with respect to its ~ uniformity (introduced in Chapter 3). If X is uniformizable, the
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completion (X',W) of (X,~) is usually denoted by central role in general topology.
~X.
The study of
~X
plays a
Since ~ is the finest totally bounded uniformity, the Stone-Cech compactification is realizable as the completion with respect to the finest uniformity for which a compactification exists. A similar situation occurs with respect to realcompactifications. We will find in the next chapter that realcompactifications are precisely the completions with respect to the countably bounded uniformities and the Hewitt realcompactification is the completion with respect to the finest countably bounded uniformity (namely the e uniformity). Consequently, if Tamano's Paracompactification Problem is to be answered affirmatively for a given space X, the solution needs to be the completion of X with respect to the finest uniformity for X that has a paracompact completion. K. Morita was interested in when the topological completion of a Tychonoff space is paracompact. He showed that if X is an M-space (to be defined later in this chapter), then X has this property. Morita called the topological completion of an M-space X the "paracompactification of X." Consequently, we will call a paracompactification that is a solution to Tamano's Paracompactification Problem the Tamano-Morita paracompactification. In studying Tamano's Paracompactification Problem, several questions come to mind. The first is: Which spaces admit a paracompactification? This question is easily answered (for T J spaces). Since compactifications are paracompactifications, the existence of the Stone-Cech compactification implies that all Tychonoff spaces have paracompactifications. Conversely, if a space has a paracompactification it must be Tychonoff since paracompact Hausdorff spaces are normal (hence uniformizable) and as we saw in Chapter 5, a subspace of a uniform space is also a uniform space. Consequently, the class of spaces is not expanded by considering those with paracompactifications as opposed to those with compactifications. The next question that arises is: Are all paracompactifications obtainable as a completion with respect to some uniformity? This question is also easy to answer. If X is a Tychonoff space and PX some paracompactification of X, then X is a dense subspace of PX. Since PX is paracompact, by Theorem 4.4, it is cofinally complete with respect to its finest uniformity u, and hence complete with respect to u. Now u induces the v (derived) uniformity on X (see Section 4.5). Since completions are unique (Theorem 4.10), (PX,u) is the completion of (X,v). Consequently, all paracompactifications are obtainable as completions with respect to some uniformity. A completion of a uniform space that is cofinally complete will be called a uniform paracompactification. We have already seen (Theorem 4.12) a necessary and sufficient condition for a uniform space to have a uniform paracompactification. Clearly, a Tychonoff space can have many paracompact_ ifications in the same manner that it can have multiple distinct Samuel
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compactifications. We begin our study of paracompactifications by first considering compactifications. We wish to obtain the Stone-Cech compactification of PX of a Tychonoff space X as the completion of X with respect to its P uniformity. Since Phas a basis A consisting of all finite normal coverings it is clearly precompact, so the completion of eX, P) is compact. But what makes the completion of eX, P) the Stone-Cech compactification? E. Cech characterized PX in the following way:
THEOREM 6.1 (E. Cech, 1937) Any Hausdorff compactification BX of X is the image of PX under a unique continuous mapping f that keeps X pointwise fixed and such that f(~X - X) = BX - X. M. H. Stone characterized ~X in another way:
THEOREM 6.2 (M. H. Stone. 1937) Iff is any continuous mapping of a Tychonoff space X into a compact Hausdorff space Y. then f has a unique continuous extension ff'J :~X -7 Y. We propose to show that the completion of (X, ~) satisfies Theorems 6.1 and 6.2 in the sense that if we replace ~X with the completion of (X, ~) in the statement of both these theorems, they both remain true. Then we will show that any compactification satisfying Theorems 6.1 and 6.2 is the completion of eX, P). Thus, among the various compactifications a Tychonoff space X may have, the completion of eX, P) is distinguished as the Stone-Cech compactification of X.
6.2 Compactifications In what follows, we will adopt the following notation: if (X, f..l) is a uniform space, then f..lX will denote its completion. For us then, ~X will denote the completion of (X, ~). We first prove Theorem 6.2 using this interpretation of ~X.
(Proof of Theorem 6.2) Let Wbe the unique uniformity of ~X (Theorem 2.8). Then (BX, W) is the completion of (X, ~). Let B be the unique uniformity on Y. By Lemma 3.8, each Tychonoff space admits a uniformity that has a basis consisting of all finite normal coverings. Let U E B be one of them. For each UE UputVu=Unj(X)andlet V= (VUIUE Ul. Then V is a finite normal covering of I(X) and 1 (V) = {f-l (Vu) IU E U} is a finite normal covering of X. By definition of B, r 1 (V) E B. But then f is uniformly continuous with respect to ~ and B. By Proposition 4.20, Y is complete, so by Theorem 4.9, I has a unique uniformly continuous extension/f'J:BX -7 Y.-
r
(Proof of Theorem 6.1) X can be identified with dense subspaces of both BX and BX. Define f:X c ~X -7 X c BX by I(x) = x for each x E X. Then, as
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in the proof of Theorem 6.2, B' (the unique uniformity on BX) has a basis consisting of all finite normal coverings. If U' E B' is one of these finite normal coverings, we have just seen that r 1 (Unf(X)) is a finite normal covering of X, so f is uniformly continuous. Therefore, we can apply Theorem 6.2 and obtain a unique continuous extension f~:PX ~ BX. Since PX is compact, f(PX) is closed in BX. But X c f(PX) c BX which implies f(PX) = BX since CI(X) = BX. It remains to show that f~(PX - X) c BX - X. For this let x E PX - X and let {x u} C X be the fundamental Cauchy net in X (see Theorem 4.9) that is used to define x' = j'(x). Suppose x' E X. Now {xu} converges to x and {f(x u )} converges to x'. But since f(x) = x for each x E X, {f(x u)} = {x u}. But then {xu} converges to x' E X and {xu} converges to x which is not in X which is a contradiction. Hence x' does not belong to X. Therefore,j~(pX - X) c BX - X. Since f~(X) = X, we have f~(PX - X) = BX - X.Finally, assume BX is a compactification of X that satisfies Theorem 6.1. Then there exists a unique uniformly continuous function f:BX ~ PX such that f(BX) = PX, f(x) = x for each x E X, and f(BX - X) = PX - X. Let U E P be a finite normal covering of X. Then there exists a finite normal covering U' of W such that U' nX = U. Since f is uniformly continuous,j-l (U') is a finite normal covering of BX that belongs to B'. But then r 1(U')nX E B. Since f(x) = x for each x E X and f(BX - X) = PX - X, r1(U')nX = r1(U'nX) = rl(U) = U. Hence B contains all finite normal coverings of X, so pcB. But by Lemma 3.8 and Theorem 2.8, the unique uniformity B' on BX has a basis consisting of all finite normal coverings (on BX). Consequently B has a basis consisting of (perhaps not all) finite normal coverings so B c p. But then B = p, so BX is pX. In order to demonstrate the utility and importance of the Stone-Cech compactification, this section and the next two will be devoted to an analysis of some of the most important topological properties of Tychonoff spaces in terms of subsets of pX. The approach we will follow is due to H. Tamano, and appeared in his 1962 paper titled On compactifications which appeared in the Journal of Mathematics of Kyoto University (Volume I, Number 2). Tamano's approach is usually thought of as characterizing topological properties of a Tychonoff space X in terms of the behavior of subsets of PX - X (which it does). But as we shall see, his approach is deeply involved with certain normal coverings and with extending uniformities of X into pX. In this section we build the tools needed for this development. In the next section we prove Tamano's Completeness Theorem, and in Section 6.4, we present Tamano's Theorem. In Chapter 2, we defined an open set A c X to be regularly open if intx(Clx(A)) = A. If in turn, X is a subspace of Y, we can generalize this concept by taking the closure and the interior in Y rather than in X in order to get what is called an extension of A over Y. In general, if U is open in X and D' is open in Y such that V = V'nX, then V' is said to be an extension ofD over Y. Put VE(y) = Y - Cly(X - V). Then vEry) is an extension of Dover Y since
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UE(Y)nX = (Y - Cly(X - U»nX = X - Cly(X - U)nX = X - (X - U) = U. We call UE(Y) the proper extension ofU over Y. It is clear from this definition that U c V implies UE(Y) e VEly). LEMMA 6.1 Let X be a dense subspace of Y and let U be open in X. Then UElY) is the largest extension of U over Y. Proof: Let U' be an extension of U over Y. If Y does not belong to UE(Y), then y E Cly(X - U). Therefore, every open set V'(y) c Y containing y meets X - U which implies V'(y)nX is not a subset of U = U' nX. Consequently, V'(y) is not contained in U' which implies y does not belong to U' since U' is open. Therefore, U' c UE(y). • LEMMA 6.2 Let X be a dense subspace of Y. Then for any A eX. we have Intx(Clx(A)) = Inty(Cly(A))nX. Proof: Let x E Intx(Clx(A». Then there exists an open set U(x) containing x such that U(x) c Clx(A) c Cly(A). But then there exists an open set U'(x) c Y such that U'(x)nX = U(x). Suppose U'(x) is not a subset of Cly(A). Then there exists ayE U(x) with y E Y - Cly(A) which is open. Since X is dense in Y, U'(x)n[Y - Cly(A)]nX oF 0. Let Z E U'(x)n[Y - Cly(A)]nX which implies Z E U'(x)nX = U(x) c Cly(A). But on the other hand, Z E Y - Cly(A) which implies Z does not belong to Cly(A) which is a contradiction. Therefore, U'(x) c Cly(A) which implies x E Inty(Cly(A»nX so Intx(Clx(A» c Inty(Cly(A»nX.
Conversely, if x E Inty(Cly(A»nX then x E X and there exists an open set U'(x) e Y containing x such that U'(x) c Cly(A) which implies U'(x)nX c Cly(A)nX. Put U(x) = U'(x)nX. Then U(x) is an open set in X containing x with U(x) e Cly(A)nX. Suppose y E Cly(A)nX. Then every open set V'(y) c Y containing y meets A. But then V'(y)nX meets A. Since A c X, every open set V(y) c X containing y meets A which implies y E Clx(A). Therefore, Cly(A)nX e Clx(A). Hence U(x) c Cly(A)nX c Clx(A). Therefore, x E Intx(Clx(A», so Inty(Cly(A»nX e Intx(Clx(A». • LEMMA 6.3 Let X be a dense subspace of Y. If u' is an extension of the open set U eX. then Cly(U') = Cly(U). Proof: Since U = U'nX we have U c U' which implies Cly(U) c Cly(U'). Suppose y E Cly(U'). Then every open set V'(y) c Y containing y contains a point of U'. Since V'(y)nU' oF 0 and since X is dense in Y, there exists an x E X such that x E V'(y)nU' which implies x E U. Therefore, every open set V'(y) c Y containing y meets U which implies y E Cly(U). Hence Cly(U') c Cly(U) so Cly(U,) Cly(U) .•
=
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6. Paracompactifications
LEMMA 6.4 Let X be a dense subspace of Y. Then VE(y) is regularly open in Y if and only if V is regularly open in X. Also, if V is regularly open in X, then VE(Y) = Inty(Cly(V)). Proof: If VE(Y) is regularly open, then V = VE(Y)nX = Inty(Cly(VE(Y))nX. Since VE(Y) is an extension of V, by Lemma 6.3 we have V = Inty(Cly(V»nX. Then by Lemma 6.2, V = Intx(Clx(V» so V is regularly open. Conversely, suppose V is regularly open. Then V = Intx(Clx(V» = Inty(Cly(U»nX. Hence V' = Inty(Cly(U» is a regularly open extension of V over Y. Therefore, V' c VE(Y) by Lemma 6.1. We can complete the proof by showing VE(Y) c V'. For this, we suppose on the contrary that VE(Y) is not a subset of V' which implies VE(y) is not a subset of Cly(U'). Since X is dense in Y we have [VE(Y)n(Y Cly(V,)]nX"# 0. Therefore, V = V£(Y)nX is not a subset of Cly(V') which is a contradiction. Hence VE(Y) c V' so VE(Y) = Inty(Cly(U».PROPOSITION 6.1 Let X be a dense subspace of Y and Y a dense subspace of Z. Let V and V be open sets in X and Y respectively. Then the following are valid: (1) VE(Z) nY = VE(Y) , (2) [VE(y)]E(Z) = VE(Z) and (3) VE(Z) c [V nX ]E(Z). If V is regularly open then VE(Z) = [V nX ]E(Z). Proof: To prove (1) we note that VE(Z)nY = [Z - Clz(X - V)]nY = Y [Clz(X-V)nY] = Y - Cly(X - V) = VE(Y). To prove (2) observe that [VE(y)]E(Z) = Z-Clz(Y - VE(Y) = Z - Clz[Y - (Y - Cly(X - V»] = Z - Clz(Cly(X - U» = Z-Clz(X-U) = VE(Z). To prove (3), note that V is an extension of vnX so by Lemma 6.1, V c (V nX)E(Y) and hence VE(Z) c [V nX ]E(Z) by (2) above. If V is regularly open, then vnX is also regularly open since vnX = Inty(Cly(V»nX = Inty(Cly(VnV»nX by Lemma 6.3. But then by Lemma 6.2, vnX = Intx(Clx(VnX». Therefore, V = (V nX)E(Y) by Lemma 6.4. Again by (2) above, VE(Z) = (V nX)E(Z) . _
If we take an open covering U of X and then take the proper extension of each member of U to BX, we get a new covering W(PX) of X in BX called the proper extension of U into BX and we call the set uUE(PX), denoted E u, the extent of U in BX. An open covering V of X is said to be stable if there exists a normal sequence {V n} of open coverings of X such that VI < V and E v = E Vn for each n. The first of Tamano's theorems that we prove shows that an open covering is stable if and only if its extent is paracompact. In general, if X is a dense subspace of Y, we can define the proper extension VE(Y) of V over Y and the extent uVE(Y) of V in Yanalogously.
Letf:X -+ R (reals). We denote by Z(j) the zero set of fdefined by Z(j) = X If(x) = O} and by O(j), the complement of Z(j) in X. We denote by Z(X) the set of all zero sets of X. In Section 1.1, O(j) was called the support of f. Also recall from Section 1.1 that a partition of unity on a space X is a family {x
E
6.2 Compactifications
163
= {A} of continuous functions ,,:X ~ [0,1] such that 11..,,(x) = 1 for each x E X, and such that all but a finite number of members of vanish inside some neighborhood of x. It is clear that {O( A)} is a locally finite covering of X. If {O(,,)} is star-finite, we say is a star-finite partition of unity on X. A partition of unity {,,} is said to be subordinate to a covering {Va} If each 0(,,) is contained in some Va. In the paper Vne generalization des espaces compacts by 1. Dieudonne, referenced in Section 1.2, he established the following:
THEOREM 6.3 (J. Dieudonne, 1944) For each locally finite open covering {Va} of a normal space, there exists a partition of unity {a} subordinate to {Va} such that O(a) C Vafor each a. Proof: Let X be a normal space and {Va} a locally finite open covering of X. By Lemma 1.3, {Va} is shrinkable to an open covering {Va}. By Theorem 0.3, for each a there exists a real valued continuous function f a:X ~ [0,11 with fa(CI(Va» = 1 andfa(X - Va) = O. Since {Va} is locally finite, so is IVa}. Hence for each p E X,fa(P):t 0 for at most finitely many a. Thus F:X ~ [OJ) defined by F(x) = La!a (x) for each x E X is well defined. To show F is continuous, let p E X. Then there exists an open W(P) containing p such that W(P) meets only finitely many Va, say V aj • • . Van' Therefore, F(x) = L7=da,(x) for each x E W(P). Since each fa, is continuous on W(P), so is F. But then F is continuous at each point of X. For each a, define a by a(x) = f a(x)/F(x). Clearly {a} is a partition of unity subordinate to {Va} such that O(a) C Va for each a. The first theorem of Tarnano that we prove not only determines stable coverings as those with paracompact extents, but also shows they are determined by star-finite partitions of unity. In K. Morita's paper Star-Finite Coverings and the Star-Finite Property referenced in Section 2.4, he showed that a locally compact space X is paracompact if and only if X has the star-finite property (every open covering has a star-finite refinement). We will need this result to prove the first of Tarnano's theorems, so we include Morita's development here.
PROPOSITION 6.2 A regular Linde16f space has the star-finite property. Proof: Let X be a regular Linde16f space and let U be an open covering of X. For each p E X, there exists an open set V(P) such that CI(V(P» c U for some U E U. Then V = {v(p)lp E X} is an open covering of X and as such has a countable subcovering say {V(pn)}' For each positive integer n let V n E U such that CI(V(pn» c Vn. Since X is normal by Lemmas 1.2 and 3.6, there exist real valued continuous functions fn on X such that fn(CI(V(pn))) = 0 and fn(X-Vn) = 1, so we can apply Theorem 2.16 which yields a star-finite
164
6. Paracompactifications
refinement of U. Consequently, a regular LindelOf space has the star-finite property. PROPOSITION 6.3 If X is a regular space and there exists an open covering U of X such that each member of U is LindelOf and U admits a starfinite refinement V, then X can be decomposed into a sum of disjoint open sets, each of which is Lin de IOf and X isfully normal and has the star-finite property. Proof: If x,y E X and there exists a positive integer n with y E Starn(x, V) then we put x - y. As was seen in Section 2.4, this relation can be used to decompose X into disjoint sets X a, a E A, such that x and y belong to the same X a if and only if x - y [see Theorem 2.15]. Then for each a E A, Starn(x, V) c X a for any x E X a' Since X a N ~ = 0 if a '# ~,X a is both open and closed.
To show a particular X a is LindelOf, we first show that for any open covering W of X and any pair (x,n) such that x E X a and n a positive integer, that there exists a countable subset of W covering Starn(x, V). We induct on n. First note that Star (x, V) c U for some U E U and U is LindelOf. Next assume Starn(x, V) has this property. Then there exists a countable collection {Vi} c V that covers Starn(x, V) which implies Star n+ 1 (x, V) C U;=I Star (Vi, V). But Star (Vi, V) C Vi for some Vi E U, so Starn +1 (x, V) C U;=I Vi' Since each Vi is LindelOf, Star n+ 1 (x, V) can be covered by countably many members of W. But then, for any open covering W of X, X a = U;;'=I Starn(x, V) can be covered by countably many members of W. Since X a is closed, X a is LindelOf. This proves the first part of the proposition. The second part follows from Lemma 3.6 and Proposition 6.2. A space X is said to be locally compact if each p compact neighborhood.
E
X is contained in a
LEMMA 6.5 A locally compact Hausdorff space is completely regular. Proof: Let X be a locally compact space and let p be a point of X contained in some open set W. Since X is locally compact, there exists an open set V containing p such that CI(V) is compact. Put U = VnW. Then CI(U) is also a compact neighborhood of p. Since CI(U) is a compact Hausdorff space, it is normal, so there exists a continuous function g:CI(U) ~ [0,1] such that g(P) = 1 and g(CI(U) - U) = O. Let h be the constant function on X - U defined by h(x) = o for each x E X - U. Define f on X by f(x) = g(x) if x E U or f(x) = h(x) otherwise. Clearly f:X ~ [0,1] is continuous, f(P) = 1 and f(X - W) = O. Consequently X is completely regular.LEMMA 6.6 If U is an open covering of a Hausdorff space X such that the closure of each member of U is compact, and if U admits a star-finite refinement V, then X isfully normal and has the star-finite property.
165
6.2 Compactifications
Proof: Clearly X is locally compact so by Lemma 6.5, X is completely regular. Moreover, any set U of U has the property that for any open covering W of X, U can be covered by countably many members of W. This was the only use made of the Lindelbf property in Proposition 6.3. Hence the same method of proof can be used to prove this lemma. THEOREM 6.4 (K. Morita, 1948) A necessary and sufficient condition for a locally compact Hausdorff space to be paracompact is that it possess the star-finite property. Proof: Let X be a locally compact Hausdorff space. Assume first that X possesses the star-finite property. Then each open covering has a star-finite refinement. Since a star-finite refinement is locally finite, we conclude that X is paracompact. Conversely assume X is paracompact. By Stone's Theorem (l.l), X is fully normal. Since X is locally compact, there exists an open covering U such that the closure of each member of U is compact. Since X is fully normal, U has a star refinement. Then by Lemma 6.6, X has the star-finite property. • E. Cech, in his paper On bicompact spaces referenced in Section 6.1, showed that a T 1 space X is normal if and only if Cll3x(E)nCll3x(F) = 0 for each pair of disjoint closed sets E, F c X. This useful result is needed to characterize the locally compact Hausdorff spaces as those spaces X such that BX - X is compact for any compactification BX of X. This result will also be needed in the proof of Taman 0 ' s characterization of stable coverings.
THEOREM 6.5 (E. Cech, 1937) A T] space X is normal if and only if CI 13x (E)nCI 13x (F) = 0 for each pair of disjoint closed sets E, F eX. Proof: Let X be a T 1 space and assume that for disjoint closed sets E, F c X, Cll3x(E)nCll3x(F) = 0. Since pX is normal, there exist disjoint open sets U,V c pX such that Cll3x(E) c U and Cll3x(F) c V. Then UnX and VnX are disjoint open sets in X containing E and F respectively, so X is normal. Conversely, assume X is normal and that E and F are disjoint closed sets in X. Then there exists a real valued continuous function f:X ~ [0,1] such that feE) = 0 and f(F) = 1. By Theorem 6.2, f can be extended to a continuous functionf 13 :pX ~ [0,1]. Since E is dense in Cll3x(E) andf(E) = O,fI3(Cll3x(E» = O. Similarly f 13 (CI 13x (F» = 1. Then U = {x E pX ifl3 (x) < 1/2} and V = {x E PX ifl3 (x) > 1/2} are disjoint open sets containing Cll3x(E) and Cll3x(F) respectively, so CI 13x (E)nCI 13x (F) = 0 .•
PROPOSITION 6.4 A Tychonojf space X is locally compact if and only
if BX - X is compact for any compactification BX of X. Proof: For each x
E
X, there exists a neighborhood U(x) of x such that
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Clx(U(x» is compact, if X is locally compact. Then Clx(U(x» is closed in BX so Clx(U(x)) = Cl ~x(U(x». Therefore. no point of BX - X is contained in Cl ~x(U(x)), so X is open in BX. Hence BX - X is closed and therefore compact. Conversely, if BX - X is compact, then it is closed so for each x E X, there exists a neighhorhood U(x) of y in ~X such that C!(U(x»n[BX - Xl = 0. But then Cl~x(U(x» = Clx(U(x». Hence Clx(U(x» is compact. We conclude that X is locally compact.We are finally in a position to prove the first major theorem from Tamano's 1962 paper. This theorem not only estahlishes the connection among stahle coverings, star-finite partitions of unity and coverings with paracompact extents, hut it is essential in the proof of Tamano's characterization of the LindelOf property in terms of compact subsets of BX - X (where BX is a compactification of X) that we will encounter later.
THEOREM 6.6 (H. Tamano. 1962) The following statements are equivalent for a Tychonojf space x: (1) V is a stable covering ofX. (2) the extent Ev of V in ~X is paracompact and (3) there exists a star-finite partition of unity {
by Proposition 6.1.(1). Therefore, V~nX' contains V~ so uV~ contains uV~ which implies E vn contains X' for each non-negative integer n. To show that Ev n c X' for each n, let p E Ev n . Then p E V~(x) for some x E X. Let y E
V~(x)nX. Then ~1
179
The points -00 and +00 are called points at infinity. In general, if (X, Jl) is a unifonn space and (X', Jl) is its completion, the set X' - X is called the set of points at infinity of X. Similarly, if P is a paracompactification of X, PX - X is referred to as the set of points at infinity of X. Points at infinity are always relative to some uniformity. Points at infinity with respect to precompact unifonnities (e.g., ~X - X) play an important role in general topology. If Jl is a unifonnity for X and H C JlX - X, then H is called a set at infinity. Compact sets at infinity will be important to us in what follows. Sets at infinity can be very large with respect to the original unifonn space. For example, the space N of positive integers with the usual topology is countable whereas ~N - N is not. We have already seen a characterization of local compactness in tenns of compact sets at infinity, namely, Proposition 6.4. Our next characterization of paracompactness in tenns of compact sets at infinity relies on a well known characterization of paracompactness by E. Michael that appeared in the Proceedings of the American Mathematical Society in 1953 (Volume 4, Number 3, pp. 831-838). LEMMA 6.14 (E. Michael, 1953) The following properties of a regular topological space are equivalent: (1) X is paracompact, (2) every open covering of X has a locally finite (not necessarily open) refinement and (3) every open covering of X has a locally finite closed refinement. Proof: (1) ~ (2) is obvious. To show (2) ~ (3) let U be an open covering of X. Since X is regular, there exists an open covering V of X such that the closures of elements of V is a refinement of U. By assumption, there exists a locally finite refinement W of V. Then the closures of elements of W is the desired locally finite closed refinement of U.
To show that (3) ~ (1), let U be an open covering ofX. Let V be a locally finite refinement of U and let W be a covering of X consisting of closed sets, each one of which intersects only finitely many members of V. Now let Z be a locally finite closed refinement of W. For each V E V, let V' = X - u{Z E Z IVnZ = 0}. Then V' is an open set containing V such that if Z E Z, then Z intersects V' if and only if Z intersects V. For each V E V pick a U v E U such that V C U v· Let U' = {V' nUv I V E V). Then U' is an open refinement of U and since each element of a locally finite covering Z intersects only finitely many elements of U', U, is locally finite. Notice the similarity between statement (2) of the following theorem and the statement of Theorem 6.9. By replacing points at infinity with compact sets at infinity in the hypothesis, the strength of the conclusion is raised from topological completeness to paracompactness.
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6. Paracompactifications
THEOREM 6.11 (H. Tamano. 1960) Let BX denote any compactification of X. Then the following statements are equivalent: (1) X is paracompact. (2) For each compact set C at infinity. there exists a partition of unity . For each )" E cI> put '1')" = )" InJ.!X.'Fhen 'I' = {'I'),,) is a partition of unity on nJ.!X such that Cl j3x(O('I',..» does not contain·~ p for each '1'", E '1'. Since this holds for each p E ~X - nJ.!X. we have that nJ.!X is topologically complete.
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6. Paracompactifications
Let v be the finest unifonnity on nl..lX and let v' be the unifonnity induced on X by v. Since v is the finest uniformity on n~..IX, v is finer than the unifonnity u' induced on n~X by u. But then v' is finer than u which implies that v' = u. Since n~X is topologically complete, n~X is the completion of X with respect to v'. Since completions are unique, n~X = uX. -
LEMMA 6.16 paracompact.
A perfect preimage of a paracompact space is
Proof: Let {Va} be an open covering of X. For each y E Y, let {Va, Ii = 1 ... n(y)} be a finite subcovering for the compact set 1 (y). Then F(y) = u7=C.jl Va, is closed in X so f(F(y» is closed in Y. Put V(y) = Y - f(F(y». Then V(y) is an open neighborhood of y and rl(V(y» c u7=Q)V a ,. Next let {W(y)} be a precise locally finite open refinement of {V(y) lYE Y}. For each i = 1 ... n(y), put W(y,i) = r 1 (W(y»nV a,' Then the collection {W(y,i) lYE Y and i = 1 ... n(y)} is an open covering of X that refines {Va}. Moreover, {W(y,i)} is locally finite since if p E X, there exists a neighborhood V of f(P) that meets at most finitely many of the W(y) and then the neighborhood 1 (V) of p meets at most finitely many of the W(y,i). -
r
r
John Mack sketched an outline of the proof of the following theorem for the author late one night at the 1991 Northeast Topology Conference after a lengthy discussion of paracompact subspaces of BX containing X. It plays a major role in the author's solution to Tamano's Paracompactification Problem.
THEOREM 6.16 (1. Mack, 1991) Let u be the finest uniformity for the space X. Then for each p E BX - uX there is a paracompact Y c BX - {p} that contains uX. Proof: By Theorem 6.13, X c uX c BX so by Theorem 6.14, BuX = BX. Since uX is topologically complete and BuX = BX, by Theorem 6.10, if p E BX - uX, there exists an F E C*(uX x BX) such that F(x, x) = 0 and F(x, p) = 1 for each x E uX. Putf= max{F,l} and for each pair x,Y E uX put 8(x,y)
= sup { If(x, q) - f(Y, q) II q E
BX}.
Then max{f(x, y),f(y, x)} :
Now put = j © i © ' where i:M -+ M* is the imbedding of M into its completion and j:M* -+ I3M* is the imbedding of M* into its Stone-tech
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6.S Paracompactifications
compactification. Then :uX -7 ~M*. Let ~ be the extension of over ~uX = ~X. Since ~X and ~M* are both compact, ~ is closed and the inverse image of each compact set in ~M* is compact in ~X. Put Y = (~tl (M*) and let ~ = ~ I Y. Then ~:Y -7 M* and since Y is the total inverse image of M*, ~ is also closed and the inverse image of each compact set in M* is compact in Y. Therefore, ~ is a perfect mapping. By Lemma 6.15, Y is paracompact. Clearly, uX c Y. By a proof similar to the proof of Lemma 6.12, we can show that d*:M* x [0,1] has a continuous extension d~:M* x ~M* -7 [OJ]. Set g = d~ © (~ x ~). Then g E C*(Y X ~X) and g(Y x ~X) c [0,1]. Let (x,y) E X X X. Then g(x,y) = d~(~(x),~(Y» = d~(x,y) = d*(x,y) = S(x,y) ~f(x,y). Therefore, M*
-7
fl X x X
~ g I X x X which implies
fl X x ~X
~ g IX x ~X.
Hence 1 = f(x, p) ~ g(x, p) for each x E X. Therefore, 1 ~ g(x, p) for each y E Y. Since g(x, x) = Sex, x) = 0 for each x E X, g(y,y) = 0 for each y E Y. Consequently, p does not belong to Y. Hence Y is a paracompact subset of ~X {p) containing uX. -
THEOREM 6.17. lfrr is the finest uniformity for a space X such that the completion rrX is paracompact, then rr = u (the universal uniformity for X). Proof: Put PX = Il{Y c ~xlx c Y and Y is paracompact). Clearly PX *- 0 and since ~X is paracompact, rr is finer than ~. Therefore, rrX c ~X which implies PX c rrX. Now, for each paracompact space Y such that X eYe ~X, it is easily shown that Y is the completion of X with respect to some uniformity 'A that is finer than ~. Then by Theorem 6.15, uX c PX. Therefore, PX = ux.
There are two uniformities on X that bound rr, namely, u (the finest) and sup{'AI'AX c ~X and 'AX is paracompact), which we denote by sup'A. Clearly sup'A c rr c u. Now rrX is paracompact and PX c rrX. We claim that the completion sup'AX of X with respect to sup'A is PX. Since X c PX c 'AX for each uniformity 'A such that 'AX is a paracompact subset of ~X, 'A' (the uniformity of the completion 'AX) induces a uniformity 'A+ on PX and the uniformity 'A on X. Consequently, sup'A+ = sup{'A+) is a uniformity on PX that induces the uniformity sup'A on X. Thus (X, sup'A) is a dense uniform subspace of (PX, sup'A+). To see that (PX, sup'A+) is complete, let \jI be a Cauchy net in PX with respect to sup'A+. Then \jI is Cauchy in PX with respect to each 'A+. But then \jI is Cauchy in 'AX which implies \jI converges to some P A. E 'AX c ~X for each 'A. Since ~X is Hausdorff, p A. = P ~ for each pair of uniformities 'A, ~ such that 'AX and ~X are paracompact subsets of ~X. Therefore, there exists apE ~X such that \jI converges to p and p E PX since p = p A. E AX for each A. Hence (PX, SUPA+) is complete, and since completions are unique, SUpAX = PX.
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Finally, we show that rrX = uX. Since PX c rrX, rr induces a unifonnity rr+ on PX that is finer than sup/...+ since rr is finer than sup/.... Then (X, rr) is a dense uniform subspace of (PX, rr+). Now (PX, rr+) is complete since (PX, sup/...+) is complete and rr+ c sup/...+, and since completions are unique, rrX = PX = uX. But then uX is paracompact, so by hypothesis, rr = u. Analogously to the definition of topologically complete, a space will be called topologically preparacompact if it is preparacompact with respect to its finest unifonnity.
THEOREM 6.18 (N. Howes, 1992) A necessary and sufficient condition for the existence of the Tamano-Morita paracompactification is topological preparacompactness. Proof: Suppose X is topologically preparacompact and let u be the universal unifonnity on X. Then uX is paracompact which implies that rr exists and equals u. Conversely, if rr exists, then by Theorem 6.17, rr = u which implies that uX is paracompact which in turn implies that u' is cofinally complete. Therefore, u is preparacompact which implies X is topologically preparacompact. We now show that some of the results in Chapter 4 can be generalized by replacing cofinally Cauchy nets with almost Cauchy nets. Part (I) of the following theorem is due to J. Isbell and appeared in his paper Supercomplete spaces referenced in Chapter 5.
THEOREM 6.19 (1. Isbell, 1962, N. Howes, 1992) Let X be a Tychonoff space and let u be the finest uniformity on X. Then (1) X is paracompact if and only if (X, u) is supercomplete, (2) X is LindelOf if and only if (X, e) is supercomplete and (3) X is compact if and only if (X, P) is supercomplete. Proof: (1) is essentially the equivalence of (1) and (2) in Theorem 5.8. To prove (2), we need only show that if (X, e) is supercomplete then X is Linde16f since by Theorem 4.4 and Corollary 5.1, a Lindelof space is supercomp1ete with respect to the e unifonnity. By Lemma 4.1 it will suffice to show that each (I)-directed (countably directed) net is almost Cauchy with respect to e. For this let ",:D -7 X be an (I)-directed net. Let U E e. Then there exists a countable nonnal covering {Vii that refines U. For each i put Cj = {d E D I",(d) E Vii and let E = u{ Cj ICj is not cofinal in D}. If E = 0 pick any d E D. Then {Cj } is a collection of cofinal subsets of D such that each d' ;::>: d is in some C j and for each i, ",(C j ) c V j c U for some U E U. If E "# 0, then for each i such that Cj is not cofinal in D, there exists a d j E D with R(dj)nC j = 0 where R(dj) = {d E Dldj ~ d}. Since", is (I)-directed, there exists a d' E D with d j ~ d' for each i such that Cj is not cofinal in D.
6.5 Paracompactifications
189
Hence R(d')ilC i = 0 for each i such that Ci is not cofinal in D. Therefore, R(d')ilE = 0. For each j put D j = CjilR(d'). Then {D j } is a collection of cofinal subsets of D such that each d+ ~ d' is in some D; and for each j, \jf(D j ) C Vj C U for some U E U. Since U was chosen arhitrarily, \jf is almost Cauchy. To prove (3), we need only show that if (X, ~) is supercomplcte then X is compact since by Theorem 4.4 and Corollary 5.1, a compact space is supercomplete with respect to~. For this, it will suffice to show that each net in X is almost Cauchy with respect to~. Let \jf:D ~ X be a net in X. Let U E ~. Then there exists a finite normal covering {V I ... V n} that refines U. For each i = 1 ... n put Ci = (d E D I \jf(d) E Vi}' Then an argument similar to the above can be used to show that there exists a d E D such that the collection {C i I Ci is cofinal in D) is a collection of cofinal subsets of D such that each d' ~ d is in some C i and for each i, \jf(C;) C Vi C U for some U E U. Therefore, each net in X is almost Cauchy with respect to ~. If instead of Theorem 4.4, we were to use Theorem 6.18 to motivate our definition of preparacompactness, then we would define preparacompactness to be the property that each almost Cauchy net has a Cauchy suhnet. To distinguish between these two notions we define a uniform space to he almost paracompact if each almost Cauchy net has a Cauchy subnet. Then we can prove: THEOREM 6.20 (N. Howes. 1992) Let (X. fJ.) be a Uniform space and let v be the uniformity on X derived from fJ.. Then (1) (X. fJ.) has a paracompact completion if and only if (X.V) is almost paracompact. (2) (X. fJ.) has a LindeLOf completion if and only if (X.V) is countably bounded and almost paracompact. Proof: To prove (1) let (X', fJ.') be the completion of (X, fJ.) and let u' he the finest uniformity for X'. Assume (X,v) is almost paracompact and that \jf:D ~
X' is an almost Cauchy net with respect to u'. Let E = D X u' and define < on E by (d,U') < (e,V') if d < e and V' ~)~} is an inverse limit system of unifonn spaces, and the inverse limit of this system, which we denote by W(~X) is the weak completion of ~X with respect to W. Since ~X is complete, by Theorem 5.4, W(~X) = ~X, so ~X is the inverse limit of {~X/\f'd<j>~)~}. If \fI" is a nonnal sequence of open members of Wthen there exists a \fI fl such that f.! E r with A < f.!. To see this let \fI" = {W~} and let Y E r. Let \fly = {V~}. Then V~ is finite and {W~nV~} is a normal sequence. Put U~ = V~ and for each n > I put U~ = W~nV~. Then let f.! be the element of r such that \fIfl = {U~}. Clearly, A < f.!. Therefore, {\fly lYE r} is cofinal in the collection of all nonnal sequences of open members of W. Consequently, by Exercise 5 of Section 5.5, ~X is the inverse limit space of the inverse limit system {~X/\f'y, (<j>~)~} where YE r. Now since lly:X/y ~ ~X/\f'y is the inclusion mapping, ll~ must be one-toone, so ~(X/y) can be identified with the subset ll~[~(X/y)] of ~X/\f'"y" But then ~X/\f'y is a compactification of X/y containing a copy of ~(X/y) so ~X/\f'r can be identified with ~(X/y). Moreover, for each pair A, f.! E r with A < f.!, <j>~:X/fl ~ X/" has a unifonnly continuous extension ~(<j>~):~(X/fl) ~ ~(X/,,). Since ~(X/fl) and ~(X/,,) can be identified with ~X/\f'fl and ~X/\f'" respectively, we can consider ~(<j>~) as a mapping from ~X/\f'fl into ~X/\f'". Then ~(<j>~) restricted to X/fl is the same as (<j>~)~ restricted to X/w Therefore, ~(<j>~) = (<j>~)~. Consequently, we have the following: THEOREM 6.22 (K. Morita, 1970) {~(X/,,), (<j>~)~} is an inverse limit system whose inverse limit can be identified with ~x. Proof: The proof of this theorem has essentially been proved in the preceding discussion. It remains to observe that since r is cofinal in A, the inverse system in the hypothesis of the theorem has the same inverse limit as the system where A ranges over r, namely, ~X (see Exercise 5, Section 5.5). Consequently, ~X has the spectrum W(X/,,), (<j>")~}. Recall from Section 5.5 that uX = u(X) has the spectrum {X/", <j>t}. Also recall the unifonn imbedding <j>:X ~ uX defined by <j>(x) =( <j>dx)} E uX. It's extension <j>~:~X ~ ~uX = ~X can be seen to be the mapping defined by <j>~(x) = (<j>~(x)} E
~X.
THEOREM 6.23 (K. Morita, 1970) For any uniformizable space X, uX c ~X and uX = n {(<j>~rl (X/,,) IA E A} = n {(f~rl (T) I f:X ~ T is continuous with T metrizable}. Proof: That uX c
pX follows from either Theorem 6.13 or Exercise 1.
To prove
6.6 The Spectrum of PX
195
the first equality, suppose x E uX. Then x E fhX/A such that for each pair A, Il E A with A < Il, ~(xh!) = x A. Hence x E (~rl(X/A) for each A E A. Therefore, uX c n{(~rl(x/A)IA E A}. Conversely, suppose x E n{(~rl(X/A) IA E A}. Then ~(x) E X/A for each A E A. Therefore, i3(x) = (~(x)} E ITAX/A' Suppose x does not belong to uX. Then there exists a pair A, Il E A with A < Il such that ~(x)J.) ;ex Awhich implies (~)i3(x)J.) ;e XA which in tum implies x does not belong to ~X. But this is a contradiction since for each A E A, x E (~rl (X/A) c ~X. Therefore, n{ (~rl (X/A) A E A) c uX. Consequently, uX = n{ (~rl (X/A) A E A). 1
1
To prove the second equality, let f be a continuous mapping of X into a metric space T. For each positive integer n, let Un = {Set, 2- n) t E T}. Then the sequence {Un} forms a basis for the metric uniformity of T. For each n put Vn = 1 (Un)' Then {V n} is a normal sequence of open coverings of X, so {V n} = A for some A E A. Then A: X ---7 X/A' Let x* E X/A and let Xl, X2 E X* so that A(xd = x* = A(X2). Then for each positive integer n, X2 E S(Xl,vn) which implies f(x 2) E S(f(x d,U n ). Therefore,f(x d =f(x 2)' Define g:X/A ---7 T by g(x*) = f(x 1 ) where Xl E x*. Then f = g © A and g is continuous. 1
r
To see that g is continuous let t E T and let U be an open neighborhood of t in T. Then there exists a positive integer n such that Set, 2- n ) c U in T which implies Set, Un) C U which in tum implies Star(f-l (t),V n) c F- 1 (U). Pick x E 1 (t). Then S(x,V n) is open in X A' Put V = S(x,v n). Then fey) c U, so (g © ~ © i A)(V) c U. Since i A(V) = V, fey) = (g © ~)(V). For any open W c X Awe have:
r
W
=
(XE XAIS(x,V n) c W for some positive integer n}.
From this it is easily seen that (~rl(A(W» = W so that ~ is also an open mapping. Hence Z = ~(V) is open in X/A and g(Z) c (g © ~)(V) =fey) c U. Moreover, A(X) = x* so x* E Z and g(x*) = f(x) = t. Therefore, g is continuous. Consequently, (~rl(X/A) c (~rl[(gJ3rl(T)] = (f J3 r 1 (T). Therefore, n{(~rl(x/A)IA E A} c n{(fJ3rl(nlf:X ---7 T with f continuous and T metrizable}. Since the opposite inclusion is obvious, we have the second equality. Let f:X ---7 Y be a continuous mapping and let g: Y ---7 T be continuous where T is a metrizable space. Then by Theorem 6.23, uX c © fh- 1 (T) = (fJ3 r J [(g J3rJ (n] which implies fJ3(uX) c (g J3rJ (T). Since this holds for all such g, again by Theorem 6.23,fJ3(uX) c uY. Therefore,fJ3 maps uX into uY. Consequently, we can define an extension r:uX ---7 uY of f by r = f J3 uX. Then if g:Y ---7 Z is a continuous function, (g © f t =r © gU and (lxt = luX'
«g
1
LEMMA 6.18 fiX eYe uX then uY = uX.
196
6. Paracompactifications
Proof: If g: Y -7 T is continuous where T is a metric space, then f = g IX: -7 T is continuous. Thenr:uX -7 uT, gU:uY ~ uTand (ix)u:uX -7 uY, where i x :X-7 Y is the inclusion (identity) mapping. Since (ixt = iuX , uX c uY. Since gUlux:ux -7 uT,r = gUlux. Butr =f~lux and gU = g~luY sof~lux = (g~ I uY) IuX = g~ I uX which impliesf~ = g~. Conversely, for any continuousf:X -7 T where T is a metric space,r:uX -7 uT. Since T is metric, T is weakly complete with respect to the metric uniformity which implies T is weakly complete with respect to u so uT = T by Theorem 5.4. Therefore, g = IY is a continuous mapping from Y into T such that g IX =f.
r
Then by Theorem 6.22, uY = ill (g~rl (T) Ig: Y -7 T where g is continuous and T is metrizable) c ill (f~rl (T) If:X -7 T where f is continuous and T is metrizable) = uX. Therefore, uY c uX. Similarly, uX c uY so uY = uX.-
THEOREM 6.24 (K. Morita. 1970) uX is characterized as a space Y with the following properties: (1) Y is topologically complete and contains X as a dense subspace. (2) Any continuous f:X ~ T where T is a metric space can be extended to a continuous mapping from Y into T. Proof: uX is the inverse limit of the inverse system of metric spaces I X/cD", ~} and hence by Theorem 5.5 is topologically complete. Thus uX satisfies (1). To show uX satisfies (2), letf:X -7 T where T is a metric space. Thenr:uX -7 uT. But since T is a metric space, T is weakly complete, so by Theorem 5.4, uT = T. Therefore, is a continuous mapping from uX into T.
r
Conversely, let Y be a space satisfying (1) and (2). Let g E C*(X). Then g(X) is a bounded subset of R and hence metrizable, so g:X -7 g(X) is a mapping of X into a metric space and by (2) g can be extended to Y. But then X is C*-imbedded in Y so by Theorem 6.14, BX = BY and X eYe BX. Again by (2), for any continuous mapping f from X into a metric space T, there is a continuous extension g:Y -7 T. Since BX = BY we have f~ = g~ where f~ and g ~ are the unique continuous extensions from BX and BY into BT given by Theorem 6.2, and g~ I Y = g. Now g~(Y) c T so f~(Y) c T which in turn implies Y c (f~rl (T). Therefore, by Theorem 6.23, Y c uX. Then by Lemma 6.17, uY = uX. But since Y is topologically complete, Y = uX.-
EXERCISES 1. If I Y, F) is an inverse limit system where Y = I Y a Ia E A} and F = {~:Y a -7 Y ~ Ia,B E A with B < a} and if for each a E A, X a C Y a , then the inverse limit of {X,G} is a subset of the inverse limit of {y, F} where X = {X a I a E A} and G
= {~IXal a,B
E
A with B < a}.
6.7 The Tamano-Morita Paracompactification
197
2. Show that uX is characterized as a topologically complete space Y which is the smallest with respect to properties (1) and (2) below: (1) Y contains X as a dense subspace. (2) Every real-valued bounded continuous function on X can be extended to a continuous function over Y. 3. Show that if X is the topological sum of IX A IA topological sum of I uX A I A E A}.
E
A}, then uX is the
6.7 The Tamano-Morita Paracompactification As an example of the Tamano-Morita paracompactification of a space X, we consider the case where X is an M-space. This example was given by K. Morita in his paper Topological completions and M-spaces referenced in Section 5.5, as an example of a space whose topological completion is paracompact. M-spaces were introduced in 1964 by Morita in a paper titled Products of Normal Spaces with Metric Spaces (Math. Annelen, Volume 154, pp. 365-382). Paracompact M-spaces, unlike paracompact spaces, are countably productive. M-spaces are defined as follows: a space X is said to be an M-space if there exists a normal sequence {Un} of open coverings of X satisfying the condition (M) below: (M)
If IKn} is a decreasing sequence of non-empty closed sets in X such that Kn C S(x, Un) for each positive integer n, and for some x E X, then nKn 0.
*
Let X be an M-space and let {$A I A E A'I be the collection of all normal sequences of open coverings of X satisfying condition (M). It is left as an exercise to show that A' is cofinal in A. Hence by Exercise 1 of Section 6.6, uX is the inverse limit of the inverse limit subsystem IXN.lI,., ~ IA E A'I of IX/CPA' ~ IA E AI. Recall that a continuous function f:X ~ Y is said to be a perfect mapping if f is closed and for each p E Y, 1 (P) is a compact subset of X. f is said to be quasi-perfect if f is closed and 1 (p) is countably compact for each pE Y.
r
r
LEMMA 6.19 For each A E A', A:X ~ X/CPA is a quasi-peifect mapping. Proof: Let A E A' and let CPA = {Un I. Then A:X ~ X/CPA is the mapping defined in Section 5.5 by 1.. = ~ © i A where i A:X ~ X A is the identity mapping on X viewed as a mapping from X onto X A and ~ is the quotient mapping from X A onto X/CPA' X A is the pseudo-metric space (X, d A) where d A is the pseudo-metric generated by ct>A' From the proof of Theorem 6.23, ~ is an open mapping. To show that 1.. is closed, let F be a closed subset of X and
198
let y* E CI«h.(F». Let x we have:
6. Paracompactifications E
~I(y*).
Since Star(5;(x, U n+l ), Un+d c sex, Un),
Hence V = ~(/n{XA (S(x. Un))) is open in Xj)1) = X/
A) - X/ct>A' Hence X/ct>)1 = «~h-l (X/ct>A) for each "A. E A'. Then (ar l (X/ct>)1) = (ar l «(~)~rl (X/ct>A» = (((~)~ © ari (X/ct>A) = (~rl (X/ct>A)' Thus (~rl (X/ct>A) = (ar l (X/ct>)1) for each "A., j..l E 1\.' with "A. < j..l. Combining this result with Theorem 6.23 it is easily shown that:
This fact allows us to establish the following theorem: THEOREM 6.25 (K. Morita, 1970) II X is a Tychonojf M-space then uX = (/~)-l (T)for any quasi-perfect mappingffrom X onto a metric space T.
6.7 The Tamano-Morita Paracompactification
199
Proof: The proof consists in showing that a quasi-perfect mapping f from X onto a metric space T coincides with ,,:X ~ X/cD" for some A E N. From the proof of Theorem 6.23, we already know that f = g © " for some A E A and for some continuous g:X/cD" ~ T defined by g(x*) = fez) where z E x*. Let! Un l and! Vn l be as defined in the proof of Theorem 6.23. Clearly g is onto since for l (t) which implies g(x*) = [(x) = t. To show g each t E T we can pick an x E is one-to-one, let x* and y* be distinct elements of X/cD". Then there exists a positive integer n such that S(x,Vn)nS(y,V n) = 0 for some x E x* and y E y*. Therefore, S(f(x) , Un)nS(f(y), Un) = 0. Since g(x*) = j(x) and g(y*) = fey), g(x*) g(y*) so g is one-to-one.
r
"*
Finally, to show g is open, let U be open in X/cD" and let x* E U. Then there exists a positi vc integer n such that S(x*, 2 -n) C U so g(S(x*, 2- n c g(U) and contains g(x*). Therefore, if g(S(x*, 2- n» is open in T then x* is an interior point of g(U). Since x* was chosen arbitrarily, g(U) is open. Let x E x*. Then
»
g(S(x*,2- n»
=
UE TIY*E S(x*,2- n),YE y*,andf(y)=tl =
Now X - SexY n) is closed so f(X - SexY n» = T - S(f(x), Un) is closed in T since f is a closed mapping. Therefore, S(f(x) , Un) is open in T so g is an open mapping. Hence g is a homeomorphism so we can identify T with X/cD" which implies f coincides with ". It remains to show that A E N. We do this by showing that {V n} satisfies (M). Let K = {Kn} be a decreasing collection of closed sets in X such that for
some p E X, Kn c S(P,V n) for each positive integer n. Put q =f(P). Thenf(K n) c Seq, Un-I) for each n since Kn c S(p,V n). Now the f(Kn) are closed subsets of T and {f(Kn)} is a decreasing sequence so q E f(Kn) for each positive integer n. Therefore, l (q)nK n 0 for each n. Hence {f-I (q)nKn l is a decreasing 1 (q) sequence of closed sets in X. Since is countably compact, I'In[j-l (q)nKnl 0 so nKn 0. Therefore, {V n l satisfies (M). •
r
"*
"*
r
"*
LEMMA 6.21 Let f:X ~ T where f is continuous and T is metric. If X is an M-space then :uX ~ T and the following assertions hold: ( J ) f is onto if and only if is onto. (2) f is closed if and only if is closed and (3) f is quasi-perfect if and only if is perfect.
r
r
r
r
Proof: From the proof of Theorem 6.23, we know there exists a continuous function g:X/cD" ~ T such that f = g © " for some A E A. The proof can be modified to show that A can be chosen in such a way that A E N. For this let {Un} and {V n} be as in the proof of Theorem 6.23. Then cD" = {V n }. Pick J..1. E N such that A < J..1. and let cDll = {W n ). Then 1l:X ~ X/"
w Then f = h © " which in tum implies u«" with f(X), we see that f agrees with 0 for each x E X. Consequently, g-1 defined by g-I(X) = l/g(x) is continuous. Therefore, g-1 E C. Since 1t is the product uniformity on P, 1t is the coarsest uniformity on P that makes all the canonical projections uniformly continuous. But for each f E C, Pt = f, so g-1 is uniformly continuous. Since for each U E 1t, there is a Z E Z with Z c U for some U E U, and since g -1 is uniformly continuous, then for each £ > 0, there is a Z E Z such that g-1 (Z) is contained in a set of diameter £. Now if x E Z(gn), then g(x) < 2- n+1. Hence there exists a sufficiently large n with g-I(X) > max{g-I(y)ly E Z} for any x E Z(gn)' This implies that Z(gn)nZ = 0. But Z(gn) = n7=1 Z(fi) which implies that Z(gn) E Z. Then Z does not satisfy the finite intersection property which is a contradiction. Therefore, Z is a CZ-maximal family. Consequently, by property (2), Z has a non-empty intersection. Let pEn {Z IZ E Z}. Let U E 1t' and choose V E 1t' such that V* < U. Suppose no Z E Z meets S(P,V). Since there exists a Z(V) E Z and a V E V with Z(V) c V, we see that p does not belong to Z(V) which is a contradiction. Therefore, Z(V)nS(p,V) "# 0 which implies there is an open neighborhood V' of p such that V'nV"# 0 so there is aU E U with V\..;V cU. But then Z(V) c S(p, U) and Z(V) contains H a for some index a. Hence, {x a} converges to p so (X, 1t') is complete. This completes the proof of (2) """"* (3). To show (3) """"* (1), let X be a closed subset of a product P = IlaR a where R a = R for each index a. Since R a is complete for each a, so is P. Let 1t denote the product uniformity on P and let 1t' denote 1t restricted to X. Since X is closed in P, (X, 1t) is complete. It will suffice to show that e is a basis for 1t'. For this let U be a basic member of 1t. Then
for some finite collection of uniform coverings Ua, of R a, where P a, denotes the canonical projection of Ponto Raj for each i = 1 ... n. Now each Uaj has a countable subcovering Vaj since Raj is LindelOf. Furthermore, each Va, is normal since Raj is paracompact. Therefore, P~~ (Val)n ... npu~(Va") is a
206
7. Reaicompactifications
countable normal covering of P that refines U. Hence U is a member of the e uniformity for P. But then e is a basis for the n' uniformity on X. Therefore, e =n', so X is e-complete. Now that we know realcompact spaces are precisely the e-complete spaces, we can use this fact to derive some interesting properties about realcompact spaces.
THEOREM 7.3 Every Lindel6f space is reaicompact. This is because Lindelbf spaces are cofinally complete with respect to the e uniformity by Theorem 4.4 and hence complete with respect to e hy Corollary 4.1.
THEOREM 7.4 A closed subspace of a reaicompact space is reaicompact. This is because if Y is a closed subspace of a realcompact space X, then Y is complete with respect to the e uniformity of X relativized to Y by Proposition 4.18. But the e uniformity of X relativized to Y is precisely the e uniformity of Y.
THEOREM 7.5 A product of reaicompact spaces is reaicompact. Proof: Let X = flaX a where each X a is realcompact and for each a, let e a be the e uniformity of X a' Then for each a, (X a' e a) is complete so X is complete with respect to the product uniformity by Exercise 4 of Section 5.3. Let /-l denote the product uniformity of the e a's. Then a basis element of /-l is of the form
for some finite collection of uniform coverings U a , of e a, where each U a , is countable and where p a, denotes the canonical projection of X onto X a, for each i = 1 ... n. Then U is also countable and therefore belongs to e (the e uniformity of X). Hence /-l c e. Since (X, /-l) is complete and e is finer than /-l, we have that (X, e) is complete so X is realcompact.We have already seen in Sections 2.2 and 5.5 how a normal sequence of open coverings can be used to construct a pseudo-metric. If we consider the family
0; and if dE D and e :::; d then e E
D. To show property (l) holds let d 'J..., deE D and let 'J... and e be the normal sequences from which they were constructed. Let 'J... = {U~} and e = {U~}. Then y = 'J...ne = {U~nU~) is a normal sequence of open uniform coverings that generates a pseudo-metric d y. Moreover, d y :::; d'J... and d y :::; de as can be seen by the method of construction of these pseudo-metrics [see proof of Theorem 2.5]. But then d y :::; d'J... vd e . Let ~ be the normal sequence of open coverings {V n} where Vn = {S(x, 2- n ) Ix EX} where Sex, 2- n) is the sphere about x of radius 2- n with respect to the pseudo-metric d 'J. . vd e. For each fl, U~nU~ < Vn since d y :::; d'J...vd e and since Sex, U~nU~) = Sy(x, 2- n ) and S(x,V n) = Sex, 2- n) where S vex, 2- n ) is the sphere about x of radius 2- n with respect to the pseudo-metric d y. Hence ~ E . But from the construction of d~, it can be seen that d~ = d 'J. . vd e since the definition of d ~ depends on the definition of the function 8(x,y) which is defined to be 2- n where fl is the largest index such that y E S(x,V n) or to be 0 if there is no largest index. But since SexY n) = sex, 2- n ), and sex, 2- n ) is the sphere about x of radius 2- n with respect to d'J... vd e , we have S vCx, 2- n ) = Sex, 2- n)for each positive integer n. [A complete proof that ~ generates d 'J. . vd e involves showing that if two pseudo-metrics d and d' have the same spheres of radius 2- n for each positive integer fl, then d = d' which we leave as an exercise.] To show that property (2) holds, let p be a pseudo-metric and suppose there exists a d'J... E D such that for each e> 0 there exists a 8> 0 with d')...(x,y) :::; 8 implies p(x,y) :::; e. As we saw from the above discussion, d is generated by the normal sequence of uniform coverings {Un I where Un = {S 'J...(x, 2- n ) IX EX} and p is generated by the normal sequence of open coverings {V n} where Vn = {S(x, 2- n ) Ix EX} where Sex, 2- n ) is the sphere about x of radius 2- n with respect to the pseudo-metric p. Then for each positive integer n, there exists a positive integer m such that Urn < Vn' Hence {V n } E which implies p ED. To show that property (3) holds when (X, ~) is a Hausdorff uniform space, notice that if x 7= y there exists a U E ~ with Sex, U)nS(y, U) = 0. Let 'J... = {U~} be a normal sequence of open coverings such that U} < U. Let d'J... be the pseudo-metric constructed from 'J..., Then d'J... E D and d'J...(x,y) -:/- O. From the above discussion it should corne as no surprise that uniformities can be studied as families of pseudo-metrics that satisfy the above properties (1)
208
7. Realcompactifications
- (3). In such a case, we call a family D of pseudo-metrics satisfying these properties a pseudo-metric uniformity. It is left as an exercise (Exercise 2) to show that if D is a pseudo-metric uniformity, then the family of coverings f.l = {U~ldE Dand£>O} where U~= {SAx,£)lxE X},isabasisforacovering uniformity on X. It is easily shown (Exercise 3) that the intersection of any collection of pseudo-metric uniformities is again a pseudo-metric uniformity. Since 0 belongs to every pseudo-metric uniformity, the intersection is never empty. Consequently, if S is a non-empty family of pseudo-metrics on X, there exists a smallest pseudo-metric uniformity D containing S. We say S is a subbase for D and that D is generated by S. If B is a subbase for D such that for each dE D and £ > 0, there exists a d' E B and a 8 > 0 with d'(x,y) :-::; 8 implies d(x,y) :-::; £ for each pair x,y E X, then B is called a base for D. It is left as an exercise (Exercise 4) to show that if S is a subbase for D, then the family B of all finite joins d I V ••• vdn such that d; E S for each i = 1 ... n is a base for D. The family C(X) of all real valued continuous functions on X and the family C*(X) of all real valued bounded continuous functions on X can be used to generate two pseudo-metric uniformities C and C* as follows: for each f E C(X) let df be defined as dlx,y)
=
If(x) - fCy) I
for each pair x,y E X. It is an easy exercise (Exercise 5) to show that df is a pseudo-metric on X. Let C = (df If E C(X)} and C* = (df If E C*(X)}. Let c and c* denote the covering uniformities associated with C and C* respectively. Then c and c* have bases of the form b = {U{ I df E C and £ > O}
b* = {U{ Idf E C* and £ > O} respectively, where U{ = (Six, £) I x EX} and where Six, £) is the sphere about x of radius £ with respect to the pseudo-metric df . We now show that c c e and c* c~. For this, let U E c*. Then U = U{ for some f E C*(X) and £ > O. Since f E C*(X) there is some inf a and sup b such that f(X) c [a,b]. f mayor may not assume the end points a and b. But since f is continuous, it assumes each point P such that a < P < b. Let 8 = £/2 and pick PIE X such that f(P I ) = a if f(x) = a for some x E X or else pick PIE X such that f(P I) = a + £. If a + £ > b pick any x E (a,b) and put P I = x. For each positive integer k pick Pk+1 E X such thatf(Pk+l) = f(Pk) + £ if f(Pd + £:-::; b. Otherwise put Pk+1 = b unless f does not assume the value b in which case put PhI =Pk. Let n be the least positive integer such thatf(Pn+l) =f(Pn).
7.2 Realcompact Spaces
209
Clearly {S/'pi, E) Ii = 1 ... n} is a finite subcovering of U{. It is also clear that ufs < U{ and U{; E c. Therefore, every U{ where f E C*(X) has a finite normal refinement and consequently belongs to~. This shows that c* c~. This argument can be modified to show that c c e (see Exercise 6). Consequently, if X is complete with respect to c, then X is real compact and if X is complete with respect to c* then X is compact. In what follows we will show that the completion with respect to c is the Hewitt realcompactification, and the completion with respect to c* is the Stone-Cech compactification, but for now we merely record the following: PROPOSITION 7.1 If a Tychonoff space is complete with respect to the c uniformity, then it is real compact. If it is complete with respect to c*, it is compact. How to translate the concept of uniform continuity into the terminology of pseudo-metric uniformities follows from the next proposition, whose proof we leave as an exercise (Exercise 7). PROPOSITION 7.2 Let (X, M) and (Y, N) be pseudo-metric uniform spaces and let (X, J..l) and (Y,v) be their associated covering uniform spaces. Then f:X -t Y is uniformly continuous with respect to J..l and v if and only if for each e E Nand E >0, there exists a d E M and a 0 > 0 such that if d(x,y) < 0 then e(j(x),f(y)) < E. Proposition 7.2 has the following useful implication:f:X -t R is uniformly continuous with respect to some pseudo-metric uniformity D if and only if df E D. Consequently,fis uniformly continuous with respect to c or c* if and only if df E Cor C* respectively. For C* this means that if f E C*(X) then df E C* so f is uniformly continuous with respect to c* and hence can be continuously extended to the completion of X with respect to c*. By Proposition 7.1, c*X is compact. But then by Theorem 7.1, c*X is the Stone-Cech compactification of X. Since PX has a unique uniformity, we have THEOREM 7.6 c* = ~.
EXERCISES 1. Let d and d' be two pseudo-metrics on X having the same spheres of radius 2- n for each positive integer n. Show that d = d'. PSEUDO-ME1RIC UNIFORMITIES 2. Let D be a pseudo-metrIc uniformity on X and let v = {U~ IdE D and E > 0 I where U~ = {Sd(X, E) Ix E Xl. Show that v is the basis for a covering
210
7. Realcompactifications
uniformity that generates the same topology on X as the pseudo-metric uniformity. 3. Show that the intersection of any collection of pseudo-metric uniformities is again a pseudo-metric uniformity. 4. Let S be a subbase for a pseudo-metric uniformity D and let B be the family of all finite joins d 1 V ... vd n such that d i E S for each i = 1 ... n. Show that B is a base for D. 5. Let f E C(X) and define df by dtCx,y) = If(x) - fey) I for each pair x,y Show thatfis a pseudo-metric on X.
E
X.
6. Show that the e uniformity is finer than the c uniformity.
7.3 Realcom pactifications By a realcompactification of a Tychonoff space X, we mean a realcompact space Y in which X can be imbedded as a dense subspace. From Theorem 7.2 it is easily seen that a realcompactification Y of X can be realized as the completion of X with respect to some countably bounded uniformity. To see this, note that if Y is realcompact, it is e-complete, and the e uniformity of Y relativized to X is countably bounded. The completion of X with respect to this uniformity is Y. Furthermore, e is the finest countably bounded uniformity on X. For this, first notice that the e uniformity is countably bounded since it has a basis consisting of countable normal coverings. Next, if J..l is a countably bounded uniformity of X and U is an open member of J..l, then there is a normal sequence {Un} of open members of J..l such that U 1 < U. Let V be a countable subcovering of U and for each positive integer n put Vn = Unn V. Then {V n} is a normal sequence of open coverings of X such that VI < V C U. Therefore, U E e so e is finer than J..l. Also, this shows that the countable normal coverings V such that V < U for some U E J..l form a basis for J..l, so J..l has a basis consisting of (perhaps not all) countable normal coverings. Therefore, the countably bounded uniformities are precisely the ones that have bases consisting of countable normal coverings (i.e., they are the separable uniformities). Shirota showed that if X is a Tychonoff space, that eX is precisely the Hewitt realcompactification uX. He did this by proving the following:
7.3 Reaicompactifications
211
THEOREM 7.7 (T. Shirota, 1951) Let X be a Tychonoff space. Then eX has the following properties: (1) X is dense in eX, (2) eX is realcompact, (3) eachfE C(X) can be continuously extended to eX. Also, any space satisfying these three properties is homeomorphic with eX. Clearly eX satisfies property (1) of Theorem 7.7. Shirota states in his proof that it is obvious that eX satisfies property (2) also. This would be the case if we knew, for instance, that e' (the e uniformity of eX) is the e uniformity of eX. Although it is not surprising that this should be so, the proof is not what we usually think of as being obvious. Consequently, we first prove this as a lemma before proving Theorem 7.7.
LEMMA 7.1 Let e' be the uniformity of eX. Then e' is the e uniformity of ex' Proof: Each U' E e' has a countable normal refinement. To see this, let V' = {V~} be a closed uniform refinement of U'. Then V = {v~nX} = {V J3} E e and therefore has a countable uniform refinement say {Vi}' For each Vi there is a VJ3, E V with Vi C VJ3 , so
since V' is a closed covering. Thus (CleX(Vi )} refines V' so W' = (lntex(CleX(Vi refines U'. Now W' E e'. To see this, note that {V;} = {V;nX} for some uniform covering {V;} of eX. Pick y E Then
»}
V;.
Therefore, V; c Clex(V;) for each i so (CleX(Vi )} is a uniform covering of eX. But then (Clex(V;)} has an open refinement in e' which implies W' E e'. Hence U' has a countable normal refinement. It remains to show that all countable normal coverings of eX belong to e'. For this let {W;} be a countable normal covering of eX. Then there exists a normal sequence {U~} of open coverings such that CleX(U'I) < {W;}. But then {U~nX} is a normal sequence of open coverings of X such that CleX(U'1 )nX refines {W;nX}. Hence, Clex(U'dnX E e. But for each VEe, CleX(V) = (CleX(V) I V E V} E e'. Therefore
Consequently, {W;}
E
e'. Therefore, e' is the e uniformity of eX. -
212
7. Reaicompactifications
Proof of Theorem 7.7: It remains to show that eX satisfies property (3) and that if Y is another Tychonoff space satisfying properties (1) - (3) then Y and eX are homeomorphic. That eX satisfies property (3) is easily seen from the remarks preceding Proposition 7.1 and those following Proposition 7.2. Iff E C(X) then dJ E C so that f is uniformly continuous with respect to c. Since c c e, f is uniformly continuous with respect to e and consequently can be uniquely extended to a uniformly continuous functionf':eX --7 R. To show that if Y is another Tychonoff space satisfying (1) - (3) then Y and eX are homeomorphic is a little more difficult. For this we will first show that any Tychonoff space Y satisfying properties (1) - (3) also satisfies the following property: (3') If Zn E Z(X) for each positive integer n, then nCly(Zn) = Cly(nZn). For this suppose nCly(Zn) i= Cly(nZn). Then there exists ayE nCly(Zn) Cly(nZn). For eaGh positive integer n, let fn E C(X) such that Zn = Z(fn) and such that Ifn I ::; 1. Since y does not belong to Cly(nZn), there exists a zero set Z c X such that y E Cly(Z) and Zn(nZn) = 0. LetfE C(X) such that Z = Z(f) and put g = L,gn where gn = 2-n( Ifn I + If I). Then g is strictly positive, i.e., g(x) > 0 for each x E X. Let g' be the extension of g over Y and for each positive integer n let g~ be the extension of gn over Y. Now g'l X = g = L,gn = L,g~ IX. By property (1), X is dense in Y so g' = L,g~. Also, for each positive integer n, g~ ::; lin I + 1f'1 where in is the extension of fn to Y and f' is the extension of f to Y. Consequently, g~(y) = 0 for each positive integer n, so g(y) = O. Hence there exists an h E C(X) such that h(x)g(x) = 1 for each x E X. Let h' be the extension of hover Y. Since (hg)' IX = hg = (h'l X)(g' IX) and since X is dense in Y, we have (hg)' = h'g' = 1. Hence g'(y) = g(y) i= 0 which is a contradiction. Therefore, nCly(Zn) = Cly(nZn). We now use property (3') to show that Y and eX are homeomorphic. For this we show that (Y, e) is the completion of (X, e). We need only show that every countable normal covering U of X can be extended to a countable normal covering U' of Y such that U = UP nX. This will show that (X, e) is a uniform subspace of the complete uniform space (Y, e), and since completions are unique, Y is homeomorphic with eX. Now there exists a normal sequence {Un} of countable open coverings of X such that U 1 < U. UE(Y) is an extension of U into Y and for each positive integer n, U~(Y) is an extension of Un into Y. By property (3'), for each positive integer n,
Consequently, U~(Y) is a covering of Y for each positive integer n, so
UE(y)
is a
7.3 Reaicompactifications
213
covering of Y. Furthermore, since Un+! n which is a contradiction. Therefore, EJ C XJ' 00
COROLLARY 7.8 uX = nEJ. THEOREM 7.10 (H. Tamano. 1962) The following statements about a Tychonojf space are equivalent: (1) X is realcompact. (2) For each p E ~X - X. there is a closed G Ii set C in ~X - X containing p. (3) For each p E ~X - X. there is a countable star-finite partition of unity
LEMMA 7.3 A non-empty class of cardinals C is closed whenever ME C, then: (1) N E C for each cardinal N < M. (2) the sum of any M members of C is in C, (3) 2M E C.
225
if and only if.
Proof: The necessity of conditions (1) - (3) is obvious. Conversely, assume C satisfies conditions (1) - (3). We need to show that C contains products, exponentials, suprema and successors. For this let M E C. Then M + 1 0, then Z' = {Z( g) E Z(Y) 10 '# Z(f)nZ c Z( g) for some Z E Z} is a CZ-maximal family on Y. Proof: It is easily shown that Z' is a non-empty subfamily of Z(Y) with the countable intersection property. Therefore, it suffices to show that Z' is maximal with respect to the finite intersection property in Z(Y). For this suppose Z' is not maximal, i.e., suppose W is a non-empty subfamily of Z(Y) with the finite intersection property that contains Z' as a proper subset. Let Z(g) E W - Z' for some g E C(Y) and put F 0 = Z(g)nZ(j). Then FoE W so F 0'# 0. NowputF 1 =Y.
227
7.5 Shirota's Theorem
For each rational r E [0, Illet Ur = (xlf(x) < r}, Zr = (xlf(x) ~ r}, U~ = (x I g(x) < r} and Z~ = (x Ig(x) ~ r}. Put G r = UrnU~ and Fr = ZrnZ~. Then G r is open in X and Fr is closed. For any pair of rational numbers r, s E [0, 1] with r < s we have Fr c Gs c Fs. For each x E X put lex) = sup{ r Ix does not belong to Fr}. Then l E C(X) and Z(g') = nFr = nr(ZrnZ~) = Z(j)nZ(g) = Fo. For each r, Fr E Z(X) and Z(j) c Fr. Therefore, each Fr E Z which implies nrFr E Z. Hence FOE Z. Then by the definition of Z', Z(g) E Z' which is a contradiction. Therefore, Z' is measurable after all. THEOREM 7.15 (T. Shirota, 1951) Let X be a Tychonoff space such that IX I is non-measurable. Then X is realcompact if and only if X is complete with respect to u. Proof: Assume X is complete with respect to u. It suffices to show that if Z is a CZ-maximal family of X and U E u, there is aBE Z and a U E U with B cU. The reason for this is that if this is the case, then a proof similar to (1) ~ (2) of Theorem 7.2 will imply that there is apE X with p E nZ. Thus X will be realcompact. For this, observe that U E U implies there is a normal sequence {Un} with U 1 1. Let {E 1 ••• EN} c R cover each point of E E R at least M times. Put U = {E 1 ••• EN~l } and let V be the set of points of E that are covered at least M times by U. Let p E V and let V be a collection of M members of U that contain p. Then p E nV E R. Moreover n V c V. Clearly, V is the union of the collection of non-void intersections of M Proof: We use induction on N. If N
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8. Measure and Integration
members of U. Since this collection is finite and each of these intersections is measurable, so are V and E - V. By the induction hypothesis we have j..l(V) :s; L~~l j..l(Enf) V)/M and j..l(E - V) :s; L~~l j..l(En - V)/(M - 1). Since E - V C EN we have j..l(E - V):S; j..l(EN)' Then (M - 1)j..l(E - V):S; L~~l j..l(En - V) which implies M j..l(E - V) :s; L~~l j..l(En - V) + j..l(EN)' Therefore, j..l(E) = j..l(V) + j..l(E - V) :s; L~;;l j..l(Enf) V)/M + L~~l j..l(En - V)/M + j..l(EN)/M
=
L~~f j..l(En)/M + j..l(EN)/M = L~=Ij..l(En)/M.PROPOSITION 8.5 Ifj..l is a measure on a ring Rand {E I ... EN} c R is a collection of subsets of E E R such that no point of E is contained in more than M of the En' then j..l(E) ~ (L~=Ij..l(En))/M.
=1 the proof is trivial so assume the proposition holds for each positive integer n < N. If M = 1 the proof is trivial so assume M > 1. Then by the induction hypothesis, j..l(E) ~ L~~f j..l(En)/(M - 1) so (M - 1)j..l(E) ~ L~;;l j..l(En). Therefore, Mj..l(E) ~ L~;;f j..l(En) + j..l(E) ~ L~=l j..l(En) which implies j..l(E) ~ L~=Ij..l(En)/M.-
Proof: Again, we induct on N. If N
PROPOSITION 8.6 If j..l is a a-additive measure on a a-ring Rand {An} is an ascending sequence of sets (i.e., An C An+l for each n) then j..l(uAn) = limn->~j..l(An)'
Proof: Put A 0 = 0. Then j..l(uAn) = j..l(U;;'=1 (An - An-d) = L;;'=l j..l(An - An-d limN->~L~=Ij..l(An - An-d = limN->~j..l(U~=1 (An - An-d) = limN->~j..l(AN)'-
=
PROPOSITION 8.7 If j..l is a a-additive measure on a a-ring Rand {An} is a decreasing sequence (i.e., An+l C An for each n) of members of R of which at least one hasfmite measure then j..l(nA n) = limn->~j..l(An)' Proof: If for some positive integer m, j..l(Arn) < 00 then j..l(An) :s; j..l(Arn) < 00 for each n > m and hence j..l(nA n) < 00. By Lemma 8.1 j..l is subtractive so j..l(Arn) j..l(nA n) = j..l(Arn - nAn) = j..l(u(Am - An». Now the sequence {Arn - An} of members of R is increasing, so by Proposition 8.6 we have: j..l(u(Arn - An» =
Combining these two equations yields:
from which the desired result can be obtained. -
8.3 Properties of Measures
237
A set function J..l:C -7 [0,00] where C is a collection of sets is said to be continuous from below at a set A if for every increasing sequence (An) C C for which uAn = A we have limn ..... ~J..l(An) = J..l(A). Similarly, J..l is continuous from above at A if for each descending sequence I An} C C for which nAn = A and IJ..l(Arn) I < 00 for some positive integer m, we have limn .....~J..l(An) = J..l(A). Propositions 8.7 and 8.8 show that a-additive measures on a-rings are both continuous from above and below. The following proposition shows that the converse also holds under appropriate conditions.
PROPOSITION 8.8 If J..l is a finite, non-negative, additive set function on a a-ring Rand J..l is either continuous from below at each A E R or continuous from above at 0, then J..l is a a-additive measure on R. Proof: Since J..l is additive, property M3 of the definition of a measure holds. By Exercise I of Section 8.2, properties MI and M2 hold so it only remains to show that J..l is a-additive. For this, first observe that the additivity of J..l and the fact that R is a ring implies (by induction) that J..l is finitely additive; i.e., if AI' . An E R with AinA j = 0 for i "# j then J..l(u?=lA;) = L?=lJ..l(A i). Let (An) be a disjoint sequence of members of R and put A = uAn- For each positive integer n put Fn = u?=lA i and Gn = A - Fn. If J..l is continuous from below, then since (Fn) is increasing and A = uFn, by Proposition 8.7 we have:
If J..l is continuous from above at 0, then since (G n) is decreasing and nG n = 0 we have for each positive integer, J..l(A) = L?=lJ..l(A;) + J..l(G n). Taking the limit of both sides of this equation gives:
by Proposition 8.8. Consequently, in either case, J..l is a-additive which is the desired result. -
EXERCISES 1. If J..l is a a-additive measure on a a-ring R and if (An) is a sequence of sets in R, and if we define
show that J..l(lim infn .....~An) $ lim infn .....~J..l(An) and if J..l(ui'=nAi) is finite for at least one value of n, then J..l(lim supn .....~An) ::::>: lim supn .....~J..l(An). 2. Let R denote the real line with the interval topology and let B denote the
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8. Measure and Integration
family of all bounded half open intervals of the fonn [a, b) in R. Let R be the collection of all finite disjoint unions of members of B. Define!-l on B by !-l([a,b» = b - a and observe that !-l(0) = !-l([a, a» = a - a =O. (a) Show that if {A 1 ••• An} are disjoint members of B and each Ai is contained in some given set A E B then L?=l!-l(A i ) :s; !-leA). (b) Show that if a closed interval [a 0, b 0] is contained in the union of a finite number of bounded open intervals, say (ai, bi) for i = 1 ... n then b o - ao < L?=l (b i - ai). (c) Show that if {An} is a sequence of sets in B and A E B with A c Ui=lA i then !-leA) :s; Li=l!-l(A i ). (d) Show that R is a a-ring in R.
8. 4 Outer Measures A non-empty collection of sets H in a set X is said to be hereditary if whenever A E H and B c A then B E H. Hereditary collections share the property with rings, a-rings, algebras and a-algebras that the intersection of any set of hereditary collections is again a hereditary collection and consequently, for any collection S of sets there is a smallest hereditary collection Hs containing S. In what follows, we will be interested in those hereditary collections that are also a-rings and a-algebras. Hereditary a-rings are the class of sets upon which we will define outer measures. Outer measures, sometimes referred to a Caratheodory measures (after C. Caratheodory who introduced them in a paper titled Ueber das lineare Mass von Punktmengen - eine Verallgemeinerung des Langenbegriffs (Nachr. Ges. Wiss. Gottingen) in 1914, are an important generalization of measures. Their importance arises from a standard technique employed when attempting to prove a set function is a measure. This technique is to first prove the function is an outer measure on some hereditary a-ring and then use a theorem from the theory of outer measures (to be demonstrated below) to conclude that the function is a measure on a suitably restricted subcollection that is a a-ring. If E is any collection of sets, then H(E) will denote the smallest hereditary L-ring containing E and H(E) will be referred to as the hereditary a-ring generated by E. H(E) is the collection of all sets that can be covered by countably many members of E. That H(E) is indeed a hereditary a-ring is left as an exercise (Exercise 1). Let R be a hereditary a-ring and let !-l*:R ~ [0,00]. If Ii* has the property
M4'. !-l*(UAi) :s; L!-l*(Ai) whenever {Ai} is a sequence in R then !-l* is said to be countably subadditive. If!-l* has the property that whenever A, B E R then 1-1*(AuB) :s; 1-1*(A) + 1-1*(B) then 1-1* is said to be
8.4 Outer Measures
239
subadditive. If!-l* has properties Ml and M2 that define a measure plus property M4' then !-l* is called an outer measure. Measures on rings can be extended to outer measures on the hereditary a-rings that these rings generate.
THEOREM 8.3 If!-l is a a-additive measure on a a-ring R and if for each A E H(R) we define !-l*(A) = inf{L!-l(An) I {An} is a sequence in R covering A} then !-l * is an outer measure on H( R) such that !-l*(A) = !-l(A) for each A E R. Proof: Let A E R and put A J = A and An = 0 for each positive integer n > I. Then {An} cRandAcuA n· By the definition of!-l*, !-l*(A):l!-l(A n) = !-leA) + O. Conversely, if {An} is a sequence in R covering A, then by Proposition 8.5 !-leA) : m 2 n which implies a nk > (m2n - 1)/2n = m - l/2 n which in turn implies a nk ~ m. Therefore, sn(x) ~ sm(x). Case 3: (l(x) does not belong to F m) Let k be the least positive integer such that x E E nk . Then x E Emj where j = k mod (2n-m). Hence
Thus sm(x) 5, sn(x). In all of these cases, sm(x) 5, sn(x). To show that for each n, Sn 5, f, first note that if f(x) = 00 for some x E X that sn(x) 5, f(x) for each n. Therefore, suppose x E X and f(x) < 00. Let k be the positive integer such that f(x) E [k, k + I). Then Sk(X) = k 5,f(x). For each positive integer n > k, there are 2n subintervals [(k
+ i - l)/2 n , (k + i)/2n]
where i = I .. . 2n that partition [k, k + I). Let i be the positive integer such that
8. Measure and Integration
248
f(x)
E
[(k + i -1)J2n, (k + i)/2n]. Then sn(x) = (k + i - 1)/2n ~f(x). Consequently, each positive integer n. This proves (1).
Sn ~ffor
To prove (2), first note that if f(x) = 00 then sn(x) = n for each positive integer n so {sn(x)} converges to f(x). Therefore, assume that f(x) < 00. As shown above, if k is the positive integer such that f(x) E [k, k + 1), then for each positive integer n > k, there are 2n subintervals [(k + i - 1)/2n, (k + i)/2n] that partition [k, k + 1) andf(x) belongs to one of them, say [(k + j - 1)J2n, (k + j)/2nj and sn(x) = (k + j - l)/2 n. Hence If(x) - sn(x) I < 2- n. Consequently, the sequence (f(x) - sn(x)} is Cauchy and therefore converges to O. Therefore, (sn(x)} converges to f(x). This proves (2).COROLLARY 8.7 Sums and products of measurable functions in£O [0.00] are measurable.
Proof: Letfand g be measurable functions from X into [0,00]. By Theorem 8.7, there exists sequences {fn} and {g n} of simple measurable functions such that 0 ~ fl ~ h ~ ... ~ f, 0 ~ g I ~ g2 ~ ... ~ g, (fn(x)} converges to f(x) and (gn(x)} converges to g(x) for each x E X. It is an easy exercise (Exercise 1) to show that sums and products of simple measurable functions are simple measurable functions. Consequently, {fn + gn} and {fngn} are sequences of simple measurable functions. We can also show (Exercise 2) that ([fn + gn](x)} converges to [f + g](x) and ([fngn](x)} converges to [fg](x) for each x E X. Hence f + g andfg are measurable.-
EXERCISES
I. Show that sums and products of simple measurable functions into [0, 00] are simple measurable functions. 2. Show that if {an} and {b n } are sequences in [0,00] such that 0 ~ a I ~ a 2 ~ •• . , 0 ~ b 1 ~ b 2 ~ , . . , {an} converges to a and {b n } converges to b, then I an + b n } converges to a + b and {anb n } converges to abo 3. Show that the set of points at which a sequence of measurable real valued functions converges is a measurable set. 4. Show that if f is a real valued function on X such that Ix measurable for each rational number r, then f is measurable.
E
X If(x) ~ r} is
5. Let IAn} be a sequence of measurable sets in a a-algebra M and let J..l be a measure on M. Define the limit inferior denoted lim infAn of the sequence
249
8.6 The Lebesgue Integral
(An 1 by lim infAn = U;;'=1 (nk=nAn) and define the limit superior denoted lim supAn by n;;'=1 (uk=nAn)' Show that (a) Il(iim injAn) ::::: lim injll(An) and (b) if Il(U;;'=1 An) < 00 then lim sUPIl(A n) ::::: Il(lim supAn).
8.6 The Lebesgue Integral In the theory of integration we often encounter the concept of infinity and the symbols 00 and -00. We have already defined a measure 11 to be a set function on a set X into [0, 00 L In order not to have to make special provisions for dealing with these concepts and symbols in some of the following theorems, we define addition (+) and multiplication (x) on [-00,00) as follows:
a + 00 = 00 + a = 00 for each a such that -00 < a, a - 00 = -00 + a = -00 for each a such that a < 00, a x 00 = 00 x a for each a such that < a, x 00 = 00 x = 0, a x 00 = 00 x a = -00 for each a such that a < 0.
°
°
°
With these definitions, it can be shown that the commutative. associative and distributive laws hold for [0, 00). Since -00 + 00 and (-00) x (00) are not defined, we cannot extend these laws to [-00,00], but fortunately, we will not need to. The cancellation laws also hold in [0,00) with the following modifications:
a + b = a + c implies b = c if a < 00, ab = ac implies b = c whenever < a < 00.
°
If s is a simple measurable function on X where (a 1 ••• an 1 c [0, 00) is the range of s and for each i = 1 ... n, Ei = (x E Xls(x) = ad, and if M is a a-algebra on X and 11 is a measure on M, then for E E M. we define the Lebesgue integral of s with respect to 11 as:
IEsdll = L?=1 aill(EnEi). If j:X ~ [0,00) is a measurable function, we define the Lebesgue integral of j with respect to 11 to be the supremum of all simple measurable functions s such that 0::::: s :::::j, i.e.,
I Efdll = sup {IEsdlll
°: :
s ::::: j and s is a simple measurable function I.
Clearly, fEfdll E [0,00] and the two definitions of Lebesgue integral given above for the case where j is a simple measurable function are equivalent. The Lebesgue integral behaves in the same manner as the Riemann integral, as the following theorem shows.
8. Measure and Integration
250
THEOREM 8.8 Let (X, M) be a measurable space and let E. F E M. Let j.! be a measure on M andfand g be measurablefunctionsfromX into [0,00]. Then (1) 1fO ~f~ g then Jtfdj.! ~ fEgdj.!. (2) If E c F then Jtfdj.! ~ JF/dj.!. (3)
If c E [0, (0) then JEcfdj.! = cftfdj.!.
(4)
Ifj(x) = 0 for each x
E
E then Jtfd!J. = O.
= 0 then Jtfd!J. = O. (6) Jtfd!J. = JxXtfdj.!. (5)
Ifj.!( E)
The proof of Theorem 8.8 is straightforward and is left as an exercise (Exercise I). The next proposition reveals an interesting property about certain integrals, namely, that they are also measures. PROPOSITION 8.18 Let M be a a-algebra on X, j.! a measure on M and s a simple measurable function on X. For each E E M put A(E) = JEsdj.!. Then A is a measure on M. Proof: Let the range of s be {a 1 . . . ak} and for each i = 1 ... k let Ai = {x E X Isex) = ai}' Suppose {En} is a sequence of disjoint members of M such that E =uEn • Then since j.! is a-additive we have:
Consequently, A is also a-additive so property M3 of a measure is satisfied. Clearly A(0) = 0 so MI is satisfied. By Theorem 8.8, M2 is satisfied. Therefore, A is a measure on M. PROPOSITION 8.19 If M is a a-algebra on X, and sand t are simple measurable functions on X, then: fx(s + t)dj.!
j.!
is a measure on M
= fxsdj.! = fxtdj.!.
Proof: Let the range of s be {ai ... am} and the range of t be {b 1 ••• bk }. For each i = I ... m and each j = I ... k put Ai = {x E X Isex) = ad and Bj = {x E X It(x) = bj }. Now for each pair i, j put Eij = AinBj • This yields fEi/s + t)dj.!
= (ai + b)j.!(Ei) = aij.!(Ei) + b/E;) = fEi/dj.! + fEij tdj.!.
Thus the conclusion of this proposition holds for each Eij in place of X. Then by Proposition 8.18, we have:
251
8.6 The Lebesgue Integral
fx(s + t)dJ.!
= Lrj~JE'J (s + t)dJ.! = Lrj~dfE'J sdj.l + fE,/dj.l] =
L7'J,~JE
" )
sdj.l + L7'J,~JE tdJ.! " )
= fxsdj.l + fxtdj.l. •
The great success of Lebesgue's definition of the integral is largely due to the ease of passing to the limit of certain sequences of measurable functions. One example of this is the following celebrated theorem.
THEOREM 8.9 (H. Lebesgue, 1904) If M is a a-algebra on X, Il a measure on M and {fn} a sequence of measurable functions on X such that 0 :,; Ji(x) :,; /j(x) :,; 00 for each pair i, j with i < j and for each x E X, then if {fn(x)} converges to f(x) for each x E X,f is measurable and
limn~JxfndJ.!
= fxfdj.l.
Proof: By Theorem 8.8(1), fxfndJ.! :,; fxfn+!dj.l for each positive integer n, so there exists an a E [0,00] such that lfxfndj.l} converges to a. By Proposition 8.17, f is measurable. Since for each x E X, {fn (x)} is a non-decreasing sequence, we have fn(x) :,; f(x) for each positive integer n. Therefore, by Theorem 8.8(1), fxfndj.l :,; fxfdJ.! for each positive integer n. Consequently, a :,; fx/dj.l because the convergence of dxfndj.l} to a implies a is the supremum of lfxfndj.l} since lfxfndj.l} is non-decreasing. Let s be a simple measurable function such that 0 :,; s :,; f and let c E (0,1). For each positive integer n put En = {x E X Ifn(x) ~ cs(x)}. Clearly each En is measurable and En C En+! for each n. Also, X = uEn, for if f(x) = 0 then x E E! whereas if f(x) > 0, cs(x) < f(x) since c < 1, and since {fn(x)} converges to f(x), there exists a positive integer k such that cs(x) < hex). Therefore, x E Ek • Then
for each positive integer n. By Proposition 8.18, A(E) = fEcsdj.l is a measure on M. Since {En} is an ascending sequence with uEn = X, by Proposition 8.7, {A(En)} converges to A(X). Then, since lfxfndJ.!} converges to a, lfEncSdJ.!} converges to fxcsdj.l, and for each n fEncsdj.l:'; fxfndj.l, we have dxsdj.l = fxcsdj.l :,; a. Since this is true for each c < 1, it is clear that fxsdj.l :,; a. Hence, for each simple measurable function s with 0 :,; s :,; f,
ixsdll :,; a :,; fxfdj.l. By definition, fx/dj.l is the supremum of all such simple measurable functions, we have a = Jxfdj.l. Therefore, Ifxfndj.l} converges to Jxfdj.l. •
252
8. Measure and Integration
In what follows, some of the sequences of numhers to which we will have occasion to refer will have rather complex representations such as the sequence of integrals {fxfnd/-ll in the proof of the preceding theorem. Rather than stating that this sequence converges to the numher fxfd/-l (we include DO as a numher) we will adopt a slightly simpler notation by writing
THEOREM 8.10 If Un} is a sequence of measurable functions from X
into [0. DO] andj(x)
= 'Ln!n(x)jor each x E
X, then fxfd/-l
= 'Lnfxfnd/-l.
Proof: By Theorem 8.7, there exist sequences {sn} and I (n} of simple measurable functions such that 0 : 1/2 and 1 otherwise, and putfn = g if n is even andfn = h if n is odd. Clearly lim inf fn = 0 so fx(lim inf fn)dJ..l = O. If J..l(A) > 0 for each A c X with nonempty interior, then both fxgdJ..l > 0 and fxhd/.! > 0 which implies lim inf fxfndJ..l > O. Hence strict inequality holds. If we define J..l(A) = f6X A dx where XA is the characteristic function of A and f6X A dx is the ordinary Riemann integral, then J..l is such a measure.
THEOREM 8.12 If M is a a-algebra on X and J..l is a measure on M, and iff:X ---7 [0,00] is measurable, then A defined by A(E) = fEfdJ..lfor each E E M is a measure on M and fxgdA = fxgfdJ..lfor each function g:X -7 [0,00]. Proof: Let {En} be a sequence of disjoint measurable sets in X with E = uEn. Then for each x E X, [XEJ](X) = ~n[XEnJ](X) since x can belong to at most one En. Therefore, XEf= ~nXEJ· Consequently, A(E) = ixXddJ..l and A(En) = fXXEJdJ..l.
254
8. Measure and Integration
By Theorem 8.l0 we have A(E) = ixXEfdll = LJXXEJdll = LnA(En).
Since 1l(0) = 0 implies A(0) = 0, we see that A is a measure on M. Let E and put g = XE. Then since XE is a simple measurable function,
E
M
Consequently, the equation in the conclusion of the theorem holds whenever g is the characteristic function of a measurable set. Next let g be the simple measurable function L~=l anXAn where An = {x E X Ig(x) = an}. Then fxgdA =
L~=l anA(An)
=
L~=l aJxMn dA
=
L~=l aJxXAJd!!
=
L~=l ixanXAJdll.
The last equality being attained by Theorem 8.8(3). By Theorem 8.10
L~=JxanXAJdll
=
fxL~=1 anXAJdll
= fxgfdll.
Hence the conclusion of the theorem also holds whenever g is a simple measurable function. If g is a measurable function, then by Theorem 8.7, there exists a sequence {sn} of simple measurable functions on X such that 0 ::; s I ::; S2 ::; ... ::; g and such that sn(x) ~ g(x) for each x E X. Then, from what we have already shown, for each positive integer n, JXSndA = JxsJdll. By Theorem 8.9, JXSndA ~ fxgdA so JxsJdll ~ fxgdA. But since 0::; s I ::; S2 ::; ... ::; g we have 0 ::; s If::; s2!::; ... ::; gf and since sn(x) ~ g(x) for each x E X, we have [s,J](x) = sn(x)f(x) ~ g(x)f(x) = [gj](x) for each x E X. Again by Theorem 8.9, fxsJdll ~ Jxgfd!!. Therefore, JxgdA = Jxgfdll.The equation in the statement of Theorem 8.12 is often written dA = fdll, or even as f = dA!dIl. The latter notation is suggestive of the role f plays but has no meaning as a ratio.
THEOREM 8.13 (H. Lebesgue) If {fn} is a sequence of complex, measurable functions on X with f(x) = limfn(x) for each x E X and if there exists a measurable function g on X with fx Ig Id!! < 00 such that Ifn(x) I ::; g(x) for each x E X, then fx If Idll < 00 and limn ~JX Ifn -fl dll
=0
and limn ~Jxfn dll
= fxfdll·
Proof: Clearly If I ::; g and since f is measurable by Corollary 8.5, Jx If Idll < 00. If - fn I ::; 2g, we can apply Theorem 8.11 to the functions 2g - If - In I to
Since get
8.6 The Lebesgue Integral
255
Now lim inln -4~ [2g - II - In I] = 2g and lim inln -4~Jx(2g - II - In 1)dJ..l = Jx 2gdf..l lim SUPn-4~Jx II - In I df..l. Hence lim SUPn-4~Jx II - In I dJ..l $; O. Therefore, limn-4~Jx II - In I df..l = O. By Exercise 2, IJx(f - In)df..ll $; Jx II - In I df..l which implies IJxldJ..l - JxlndJ..l1 $; Jx II - In I dJ..l for each positive integer n. But since limn -4~Jx II - In I df..l = 0 we have limn -4~ IJxldf..l - JxlndJ..l1 = 0 or in other words,
limn-4~Jxlndf..l = Jxldf..l.-
EXERCISES 1. Define L 1 (f..l) to be the collection of complex measurable functions on X for which Jx III df..l < 00. The members of L 1 (f..l) are called the Lebesgue integrable functions on X with respect to f..l. If IE L 1 (f..l) and I = u + iv where u and v are real measurable functions, define the integral of I (with respect to f..l) over E as
fddf..l = fEU+df..l- fEU-df..l+ ifEv+dll- dEv-dJ..l for each measurable E c X. Each of the four integrals on the right are finite, so the integral on the left is a complex number. Show that if I, gEL j (f..l) and a and b are complex numbers, then al + bg ELI (f..l) and
fx(al + bg)df..l = afxldf..l + bfxgdf..l. 2. Show that if IE L 1 (f..l) then IJxldf..ll
$;
Jx III df..l.
3. Let {fn} be a sequence of complex measurable functions on X such that In (x) -7/(x) for each x E X. Show that if there exists agE L 1 (f..l) such that lin (X) I $; g(x) for each x E X and each positive integer n, then: (a) IE L i (f..l). (b) limn--+~Jx lin - II dJ..l = O. (c) limn-4~Jxlndf..l = JxldJ..l. 4. Suppose In:X -7 [0,00] is measurable for each positive integer n and In+l ? In for each n. Also assume In (x) -7/(x) for each x E X and liE L 1 (f..l). Show that
ixl n dJ..l-7 ixldJ..l. 5. Show that the condition liE L 1 (f..l) is essential in Exercise 4. 6. Let I ELI (f..l). Show that for each < £ whenever f..l(E) < 8.
£
> 0 there exists a 8 > 0 such that JE III df..l
256
8. Measure and Integration
8.7 Negligible Sets If ~ is a measure on a a-algebra M and if E E M such that ~(E) = 0, then E is said to be a negligible set with respect to J..l. Negligible sets are negligible in the theory of integration. If P is a property that a point x may have (e.g., some function I may be continuous at x or differentiable at x) we say x has property P and write P(x). If A is a set and E is a negligible subset of A such that for each x E A - E we have P(x), we say that P holds almost everywhere on A. This is frequently abbreviated P holds a.e. on A. It is common to make statements like "I is continuous almost everywhere on A," in which case it is meant that the measure of the set of points of discontinuity of Ion A has zero measure. If I and g are measurable functions such that ~(E) = 0 where E = Ix I/(x) * g(x)} then X - E and E are disjoint sets whose union is X. Consequently,
fxld~ = fx-dd~+fd* = fX-Egd~+O = fx-Egd~+fEgd~ = fxg*. Hence, functions that are the same almost everywhere have equal integrals. This is what is meant by saying that negligible sets are negligible in the theory of integration. Since functions that are equal almost everywhere behave the same with respect to integration, we can generalize our definition of measurable function in the following way. Let E E M and let I:E ~ R. If ~(X - E) = 0 and 1 (U)nE E M for each open U in R then I is said to be measurable on E. Clearly, if we put/(x) = 0 for each x E X - E, we extend/to a function on X that is measurable with respect to the old definition. Intuitively, it should not matter what values we assign to I on X-E. We would like to be able to assign values to I on X - E in an arbitrary manner and still get a measurable function with respect to the old definition.
r
But here, a problem arises! It may be the case that certain subsets of X - E are not measurable. Fortunately, Corollary 8.4 states that every measure can be completed, i.e., every subset of a set of measure zero is itself a subset of measure zero. It will be to our advantage to just deal with complete measures. More measurable sets just mean more measurable functions. Then we can extend the function I:E ~ R above in any arbitrary manner to X and be assured that the extended function is measurable with respect to the old definition of measurability. This new definition of measurability has many consequences. For example, Theorem 8.10 can be modified to allow the sequence {In} to be a sequence of measurable functions on X that converges almost everywhere on X. With our new definition of measurability it is easily shown that the limit of (fn ) is still a measurable function I and Jxld~ = I.Jxlnd~, without having to restrict ourselves to the set on which the convergence actually takes place.
8.8 Linear Functionals and Integrals
257
EXERCISES 1. Show that if f:X ~ [0,00] is measurable, E almost everywhere on E,
2. Show that if f everywhere on X.
ELI (/1)
and
°
E
M and
Jdd/1 = for each E
E
fdd/1
=
M, then
°
then f =
°
f = 0 almost
3. Show that if f ELI (/1) and IJxfd!! I = Ix If Id!!, then there exists an a such that af= If I almost everywhere on X.
E
R
4. Show that if {fn} is a sequence of complex measurable functions defined almost everywhere on X such that Lnix Ifn i d!! < 00, then the function f(x) = LJn (x) converges almost everywhere on X,f ELI (Jl) and fxfdJl = LnfxfndJl. 5. Show that if Jl(X) < 00, f ELI (Jl), FcC is closed and the averages A£ = l/Jl(E)fddJllie in F for each E E M with Jl(E) > 0, then f(x) E F for almost all x EX. 6. Let {En} c M such that LnJl(En) < 00. Show that almost all x most finitely many of the En.
E
X lie in at
8.8 Linear Functionals and Integrals Recall from linear algebra that a vector space V over the scalar field F is a set V whose elements are called vectors and whose two operations are called addition and scalar multiplication. A linear transformation of V into another vector space W is a mapping A of V into W such that A(ax + ~y) = aA(x) + ~A(y) for all x,y E V and a,~ E F. In the special case W = F (the field of scalars), A is called a linear functional. Exercise 1 of Section 8.6 shows that L I (Jl) is a vector space whose scalar field is C. It is easily seen that the mapping I Jl:L I (Jl) ~ C defined by I Jl (f) = fxfdJl is a linear functional on L 1 (Jl). In the special case where V is the set of all continuous complex valued functions on the unit interval [0, 1], and F = R then the linear functional J:V ~ F defined by J(f) = f6f(x)dx (the ordinary Riemann integral) is clearly a positive linear functional. Since integrals are linear functionals, it is natural to ask: when are linear functionals integrals? In 1909, F. Riesz provided the following remarkable answer for the vector space C of all continuous complex valued functions defined on [0, 1]: for each positive linear functional A on C, there exists a finite positive Borel measure Jl on [0, 1] such that A(f) = f6fdJl. In fact, we now develop the celebrated Riesz Representation Theorem in a setting more general than the vector space C.
8. Measure and Integration
258
LEMMA 8.4 If X is a locally compact Hausdorff space, U is open in X and K c U is compact, then there exists an open V with compact closure such that K eVe CI(V) c U. Proof: Each point of K has an open neighborhood with compact closure, and K is covered by finitely many of them. Therefore, K lies in an open set W with compact closure. If U = X, put V = W. Otherwise, for each p E X - U let Wp be an open set containing K whose closure does not contain p and put Fp = WnWpn[X - U]. Then IFp} is a collection of compact sets such that nFp = 0.
Then IX - Fp} is an open covering of X. Pick Fq E {Fp}. Then some finite collection I (X - Fp I) ... (X - FpJ} covers Fq so FqnFp1 n ... nFpn = 0. Then V = UnWqnWpI n ... nWPn is open and contains K. Furthermore, CI(V) c CI(Wq)nCI(Wpl)n .. . nCl(WpJ since FqnFp1 n ... nFpn = 0.The collection of all complex valued functions f on a space X whose support (denoted O(/) has compact closure is denoted by CK(X). It is easily shown that CK(X) is a vector space under the operations of functional addition [(f + g)(x) = f(x) + g(x)] and scalar multiplication «aj)(X) = a[f(X)]). The notation K O. there exists a closed set F and an open set U such that FeE e U and f.!{U - F) < E. (2)!! IS a regular Borel measure on X. (3) IfE EM. there are sets G. H eX such that H is an F (J. G is a Gil. He E e G and f.!(G - H) = 0 Proof' The proof of this theorem is closely related to the proof of the Riesz Representation Theorem. For that reason it is also left as an exercise at the end of the section. THEOREM 8.16 If X is a locally compact Hausdorff space in which every open set is a-compact and f.! is a Borel measure on X with f.!(K) < 00 for each compact K. then f.! is regular.
8.8 Linear Functionals and Integrals
261
Proof: For each f E CK(X), put A(j) = fxfd/l. Clearly, A is a linear functional. Suppose f E C K(X) such thatf~ O. Then f is real valued and hence bounded by some real numher r on X. Moreover. f(X - K) = 0 for some compact K c X. Since X is open, there exists an ascending sequence {Kn I of compact sets such that X = uKw Since f is real valued, m(E) = fdd/l for each E E M is a measure on M by Theorem 8.12. By Proposition 8.6, m(X) = limn-.=m(K n). Hence Jxfd/l = limn -.=kfd/l. For each fl, kfd/l = fKnnKfd/l ~ fKid/l ~ r/l(K). Since /l(K) < 00. {JKJd/l} is an ascending sequence of non-negative real numbers bounded above by r/l(K). Hence fxfd/l is a non-negative real number. Therefore. A is a positive lineal functional. By Theorem 8.14, there exists a measure v such that for each f E CK(X), Jxfd/l = fxfdv.
Let U be an open set. Then U = U£=lFi where {F;} is a sequence of compact sets. Pick fl E CK(X) such that F 1 < fl < U. Let K 1 be the closure of the support of fl' Assume that for each m ~ fl thatfm E CdX) has been chosen such that [u 7'=1 1 F i ]U[u7'=ll KJ < fm < U. Pickfn+l with [Ui'=l F i]U[Ui'=l K J O. By Theorem 8.14, there exists a closed set F and an open set U with FeE c U and v(U - F) < E. Since U - F is open, this means /leU - F) < E. Hence inf{ /leU - F) IFeE c U where U is open and F is closed} = 0 which implies /leE) = inf{ /leE) lEe U and U is open} so /l is regular. -
EXERCISES THE RIESZ REPRESENTATION THEOREM
1. Let X be a locally compact Hausdorff space and let A be a positive linear functional on CdX). For each open U c X put /leU) = SUp{A(j) If < U}. Then for each E c X let /leE) = inf{ /leU) lEe U and U is open}. Let N be the class of all E c X such that /leE) < 00 and /leE) = sup {/l(K) IKe E and K is compact}. Let M be the class of all E c X such that EnK E N for each compact K. Show that: (a) /l is well defined (on open sets U eX). (b) /leA) ~ /l(B) if A c B. (c) /leE) = 0 implies E E Nand E E M. (e) f~ g implies A(j) ~ A(g). 2. Show that if {En} is a sequence of sets in X, then /l(uE n) ~ Ln/l(En). 3. Show that N contains all compact sets.
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8. Measure and Integration
4. Show that N contains every open set U with
~(U)
< 00.
5. Show that if E = uEn, where {En 1 is a sequence of disjoint members of N, then ~(E) = Ln~(En). 6. Show that if E = uE n, where {En 1 is a sequence of disjoint members of N, and if ~(E) < 00 then E E N. 7. Show that if E E N and E > 0, there exists a compact K and an open U such that K c E c U and ~(U - K) < E. 8. Show that if A, BEN, then A - B, AuB and AnB E N. 9. Show that M is a a-algebra on X that contains all Borel sets. 10. Show that N consists of those E E M such that ~(E) < 00. 11. Show that ~ is a measure on M. 12. Show that for each IE CK(X) that 'A(j) = Jxld~. [Hint: It suffices to prove this for I real, so it is enough to show 'A(j) ::; Jxld~ for each real I E CK(X). Let E be the support of some real I E CK(X), and let [a, b] contain the range of f. Let E > and choose a finite setYl ... Yn with Yi - Yi-l < E andyo < a < Yl < ... < Yn = b. For each i = 1 ... n put Ei = {x E X IYi-l < I(x) ::; Yi InE. Show there exists open sets VI' .. Vn with Ei c Vi for each i such that ~(Vi) < ~(EJ + Eln for each i and such that I(x) < Yi + E for each x E Vi. Then show there are functions hi < Vi with Lhi = 1 on E so I = LhJ. Use this to finish the proof.]
°
13. Complete the proof of Theorem 8.14. 14. Prove Theorem 8.15 LEBESGUE MEASURE Euclidean n-space R n is a real vector space with respect to coordinate-wise addition and scalar multiplication. If x, Y E R n are given by (x 1 •.• x n ) and (y 1 ... Yn) respectively, there is an inner product defined on R n by (x, Y) = L!=lXiYi and Ix I is defined as (x, X)1/2. The norm I I satisfies the triangle inequality; i.e., Ix - yl ::; Ix - zl + Iz - yl for any z ERn. This should be familiar to the reader from linear algebra. The function d(x, y) = Ix - Y I is a metric on Rn. IfE c R n and x E R n we define the translate of E by x as the set E + x = {y + x lYE E I. A set of vectors in R n of the form C = {x Iai < Xi < hi} or a set obtained by replacing any or all of the < with ::; in the n inequalities defining C will be called an n-cell. The volume or measure of an n-cell is defined to be v(C) = TI!=l (hi - a;). If a E R n and e > 0 we call the set B(a, e) = {xl ai ::; Xi < ai + e}
8.8 Linear Functionals and Integrals
263
the box at a with side £. We will also refer to such a box as an £-box. For each positive integer m let Am be the set of all vectors in R" whose coordinates are integral multiples of 2-" and let Bm be the collection of all 2-" boxes at vector& XE A". 15. Show the following: (1) For each m, each x E R" lies in one and only one member of Bm. (2) If U, E Bm and U 2 E Bk where m < k then either U leU 2 or
U,nU 2 =0.
(3) If U E Bm then v(U) = 2- mn (4) If m < k and U E Bm then A" has precisely 2(k~m)" vectors in U. (5) Each non-void open set in R" is a countable union of disjoint boxes, each belonging to some Bm. 16. Show that there exists a positive, complete measure m defined on a a-algebra Min R" having the following properties: (1) m(U) = v(U) for each £-box U. (2) M contains all the Borel sets in R".
(3) E E M if and only if there exists an F cr set F and a G Ii set G with FeE c G and meG - F) =o. (4) m is regular. (5) For each x E R" and E E M, m(E + x) = m(E). (6) The property (5) is called translation invariance. If)l is a positive, translation invariant Borel measure on R" with )l(K) < for each ompact set K, then there exists a real number r with )leE) = rm(E) for each Borel set E in R". 00
Chapter 9 HAAR MEASURE IN UNIFORM SPACES
9.1 Introduction In 1933, in a paper titled Die Massbegrifj der Theorie der Kontinuierlichen Gruppen published in the Annals of Mathematics (Volume 34, Number 2), A. Haar established the existence of a translation invariant measure in compact, separable, topological groups. Translation invariance of a measure f.l in a topological group G means that if E is a measurable set then f.l(E + x) = f.l(E) for each x E G. Here, E + x = {Y E G Iy = a + x for some a E G). E + x is called the x-translate of E. The transformation Tx defined on G by Tx(y) = Y + x is called the x-translation or simply a translation. Topological groups will be defined later in the chapter and these concepts will be developed formally. In 1934, in a paper titled Zum Haarschen Mass in topologischen Gruppen (Comp. Math., Volume 1), J. von Neumann showed the uniqueness of Haar's measure, and in 1940, A. Weil published L'integration dans les groupes topologiques et ses applications (Hermann Cie, Paris) where Haar's results were extended to locally compact topological groups. In 1949, I. Segal extended Haar's results to certain uniformly locally compact uniform spaces that we will call isogeneous uniform spaces (Journal of the Indian Mathematical Society, Volume 13). In 1958, Y. Mibu, evidently unaware of Segal's work, independently established similar results for this same class of spaces (Mathematical Society of Japan, Volume 10). The Haar measures of both Segal and Mibu were Baire measures rather than Borel measures. Recall from Chapter 8 that Baire measures are defined on a smaller class of sets than Borel measures, namely, on the smallest a-ring containing all the compact sets. In 1972, G. Itzkowitz extended Haar's results to the Borel sets of a class of locally compact uniform spaces (Pacific Journal of Mathematics, Volume 41) that he called equi-homogeneous uniform spaces. That equi-homogeneous uniform spaces are equivalent to isogeneous uniform spaces is the subject of Exercise 2 at the end of this section. Itzkowitz showed the existence of a Haar integral (translation invariant, linear functional on the set of real valued continuous functions with compact support) for locally compact equihomogeneous uniform spaces. His approach was to show that a locally compact equi-homogeneous uniform space (X, f.l) is homeomorphic to a quotient G/H of topological groups, where H is a stability subgroup of G, and then apply Weil's
9.1 Introduction
265
theory of invariant measures on these quotients. as recorded in Chapter 3 of L. Nachbin's book The Haar Integral. (Van Nostrand, New York, 1965) to obtain a unique Haar measure. Itzkowitz's approach involved showing the modular function on H is constant and then appealing to theorems in Weil's theory to show this implies the existence and uniqueness of a Haar measure. In Section 4 of his paper. he also presented an alternate proof of the existence part of his development of a Haar measure on the Borel subsets of a locally compact equi-homogeneous uniform space. His proof that locally compact equi-homogeneous uniform spaces are quotients of topological groups contains an error as pointed out by the author in a paper titled On Haar Measure in Uniform Spaces (Mathematica Japonica, 1995) with a counterexample to his proof of Lemma 2.1 which is used in an essential manner to prove his Theorem 2.2. However. his alternate existence proof of a Haar measure on the Borel subsets of a locally compact equi-homogeneous uniform space is valid. This leaves his extended theory of Haar measure (on the Borel subsets of a locally compact isogeneous uniform space) incomplete in the sense that the uniqueness part of the proof has not been established. In 1992, the author, unaware of Itzkowitz's work, showed the existence and uniqueness of a Haar measure on the Borel subsets of locally compact isogeneous uniform spaces and presented that development in a series of lectures at the 1992 Topology Workshop at the University of Salerno. The existence part of the author's development is essentially the same as Itzkowitz's alternate existence proof. But the author's uniqueness proof is a uniform space argument rather than an appeal to Weil's theory of invariant measures on quotients of topological groups. The author's development, as presented at the Salerno Workshop is given in this chapter. It turns out that Itzkowitz's theorem that locally compact isogeneous uniform spaces are quotients of topological groups is true. We will prove this in Section 3 using a modification of Itzkowitz's approach that allows us to avoid the use of his erroneous Lemma 2.1. We will nor show that the rest of ItZkowitz's approach can be corrected because the topology we get with our new proof is finer than Itzkowitz's topology on the group G and this necessitates additional work to straighten out his approach which is beyond the Scope of this chapter. What we will show is that the converse of this result is true, i.e .. that quotients of topological groups are isogeneous uniform spaces. This characterizes the locally compact isogeneous uniform spaces as locally compact quotients of topological groups and leads to necessary and sufficient conditions for locally compact uniform spaces to have a topological group structure that generates the uniformity or to have an abelian topological group structure that generates the uniformity. The Segal-Mibu approach uses a generalization of K. Kodaira's construction given in a paper titled Uber die Beziehung zwischen den Massen und den
266
9. Haar Measure in Uniform Spaces
Topologien in einer Gruppe, (Proc. Phys. Math. Soc. Japan, Volume 23, No.3,
1941, pp. 67-119) whereas the Itzkowitz-Howes approach uses a generalization of A. WeiI's technique published in his paper referenced earlier in this section. At the present moment it may appear that the Weil technique is more powerful in isogeneous uniform spaces in that it can be used to obtain a measure on a larger class of sets. However, it is probable that the two methods are equivalent. If this is the case, we would have a way of constructing the measure directly on the Borel sets using a simple combinatoric method. A uniform space (X, /.1) is said to be isogeneous if there exists a basis v for /.1 and a collection H of uniform homeomorphisms of X onto itself such that: H, and each pair of points x, y E X, Y E S(x, U) if and only if i(y) E S(i(x), U) for each U E v. (2) For each pair x, y E X, there exists an i xy E H that carries x onto y.
(1) For each i
E
The members of H are called isomorphisms with respect to v or simply isomorphisms. v is called an isomorphic basis for /.1. Clearly, if i is an isomorphism and U E v, then i(S(x, U» = S(i(x), U) for each x E X. Also, it is easily seen that compositions and inverses of isomorphisms are again isomorphisms. A topological space is said to be homogeneous if for each pair of points p, q E X there exists a homeomorphism of X onto itself that carries p onto q. Clearly isogeneous uniform spaces are homogeneous topological spaces. There are various types of isomorphisms. An isomorphism t:X ~ X is called a translation if t has no fixed points. If the isomorphism r:X ~ X has a proper subset F '" 0 of fixed points and F does not separate X - F, then r is called a rotation. If F separates X - F, r is called a reflection. In what follows, we will show that locally compact isogeneous uniform spaces have a unique integral that is not only translation invariant, but also invariant under rotations and reflections. All topological groups are isogeneous uniform spaces with respect to the classical group uniformities and the classical group translation Tx defined by Tx(y) =Y + x for each x '" 0 in a topological group satisfies the above definition of translation. Let C(X) denote the ring of real valued continuous functions on X and CK(X)' the members of C(X) whose support have compact closures. For any f E CK(X) and isomorphism i:X ~ X, denote f © i by /; E CK(X). By a Haar integral for X, we mean a positive linear functional! on CK(X) such that Iif;) = I(f) for each isomorphism i:X ~ X. A Haar measure for X is an almost regular, Borel measure m satisfying m(i(E» = m(E) for each Borel set E and each isomorphism i:X ~ X. The following lemma is left as an exercise.
E
LEMMA 9.1. If (X, /.1) is a locally compact uniform space, then eachf CK(X) is uniformly continuous.
9.2 Haar Integrals and Measures
267
Isogeneous unifonn spaces were introduced in a series of lectures by the author at the 1992 Topology Workshop, held at the University of Salerno, Italy. The remaining material in this chapter is from the Workshop lecture series.
EXERCISES 1. Prove Lemma 9.1. EQUI-HOMOGENEOUS UNIFORM SPACES
Let (X, U) be an entourage unifonn space. A function f:X ~ X is said to be nonexpansive with respect to a base B for U if for each U E B and (x, y) E U, the relation (f(x), f(y» E U also holds. By a B-nonexpansive homeomorphism f of a unifonn space (X, U) onto itself, we mean a homeomorphism f of X onto itself such that f is nonexpansive with respect to a base B for the unifonnity U. A unifonn space (X, U) will be called an equi-homogeneous space if there is a group G of homeomorphisms acting on X such that (i) G is transitive (i.e., given p, q E X, there is agE G such that g(P) = q, and (ii) there is a base B for U such that G is a group of B-nonexpansive homeomorphisms of the unifonn space. 2. Show that the equi-homogeneous unifonn spaces are precisely the isogeneous unifonn spaces. 3. A collection G of functions from a unifonn space (X, U) to a unifonn space (Y,V) is said to be equi-continuous if for each V E V, there is a U E U such that for each g E G, [g x g](U) c V. Show that if G is a group of homeomorphisms acting on a unifonn space (X,U), then the following are equivalent: (i) there is a base B for the unifonnity such that G is a group of B-nonexpansive homeomorphisms of (X, U), and (ii)
G is an equi-continuous group of unifonn homeomorphisms on the unifonn space (X, U). 4. Show that if (X, U) is a locally compact isogeneous unifonn space, then (X,U) is unifonnly locally compact.
9.2 Haar Integrals and Measures In this section, (X, J.I.) is assumed to be a locally compact isogeneous unifonn space. Let v be an isomorphic basis for J.I. and H a collection of isomorphisms with respect to v that satisfy condition (2) in the definition of an isogeneous unifonn space. Let g ~ 0 be in CK(X) such that g(x) ~ b for each x in some U-sphere S(y, U) where U E v. If f ~ 0 is in CK(X) with f(X - K) = 0 for some
268
9. Haar Measure in Uniform Spaces
compact K c X then there is a finite subset I x I . . . xn} of K such that {S(x I ' U) ... S(x n , U)} covers K. If Iflj = sup {f(x)Ix E S(Xj' U)} and aj ~ iflJib for each j = 1 ... n, then f(x) ~ L7=1 ajgxi(x) for each x E X where gXj = g © LXI} for some isomorphism ixjy:X -7 X that carries xi onto y. The finite collection {a I .. an} is said to dominate fwith respect to g. Put [fl g 1= inf{ Liaj I {aj} dominates fwith respect to g}.
THEOREM 9.1. The number Ifl g] is non-negative,finite and satisfies: [Ji Ig] = Ifl g]for each isomorphism i:X -7 X. (2) Ifl + 12 I g] ~ Ifl Ig] + 1f21 glfor each pair fl.h E CdX). (3) [afl g] = alfl glfor each a > O. (4)fl ~h implies Ifllg] ~ 1f2 lg]foreachfl ~ 12 E CK(X). (5) Ifl h] ~ [fl g][g I h]for each h E CdX) with h O. (6) [h Ij]-I ~ [fl g]/[h Ig] ~ [fl hlfor each h E CK(X) with h O. (1)
'*
'*
Proof: We prove only (1). (2) through (6) are left as an exercise. Now Ji(x) = [f©i](x) ~ L}=laj[gxj © n(x) for each {ail that dominatesfwith respect to g. Also, gxj © i = g © [ix j y © i] where [i xj y © i] is an isomorphism so {a;} , dominatesfi with respect to g. Hence [Ji Ig] ~ [fl g]. Now fi,-1 = f© i © i-I = f and from what has just been proved, we have [fi g] = [Ji,-J i g] ~ [Ji! gJ. Therefore, [fi Ig] = [fl g] for each isomorphism i.· Let CK(X) denote the non-negative members of CK(X) and choose some k E CK(X). For each g E CK(X), define 1~:CK(X) -7 [0,00) by Ig(f) = [flg]/[klg] for eachfE CK(X).
COROLLARY 9.1. For each g E C-;rXJ. (l)lg~O.
(2) [klj]-I ~/g(f)~ [flk]. (3) Ig(fi) Ig(f)for each isomorphism i:X -7 X. (4) Ig(rtf) = aIg(f)jor each a > O.
=
(5) Ig(fl + h) ~ 19(fl) + l~(h)·
Since X is locally compact, for each U E g(X - K) = 0 for some U-small compact set K.
LEMMA 9.2. For each 11,12
~
there exists agE C;(X) with
'* '*
CKrX) with II 0 12 and lor each e ~ Ig(11 + h) + elor each g E 0 and such that the support of g is U-small. E
> 0, there exists a U E v such that Ig(fl) + Ig(hJ ct;(X) with g
'*
Proof: Let e > O. Assume [fl + h](X - K) = 0 for some compact K and letfE CHX) such that/(K) = l. For any 8 > 0 put = II + fz + 8fand for i = 1,2 put hi = /;/ for 0, there exists a, p, y, b E [0,1] with C/(Aa) < Ar < A i3 , C/(C y) < Bs < C 8 and A(A i3 - C/(Aa» < e/2 and A(C 8 - CI(C y» < £/2. An argument similar to the one above yields A[(Ai3uC8) - (CI(Aa)uCI(C y))] < e, so AruBs is a continuity point of A(A aUC a) which implies sUPa 0, they must be identical for any point where anyone of them is
10.3 The Haar Functions
297
continuous (and hence identical with respect to their corresponding limits as opposed to their limits superior). For each (a, ~) E [0, 1] x [0, 1], put g'(a, ~) = lim supxh(A a, x)/h(B 13, x). Then the function defined by f(a, ~) = g'(a, ~) satisfies the hypothesis of Lemma 10.5, so its set of discontinuities lies on a countable family of decreasing curves which implies the set of discontinuities of g' lie on a countable family F of increasing curves. Then at all points (a, ~) of continuity, g'(a, ~) is the common limit of all four functions in (10.12) and (10.13). Pick ~o so that none of the curves of F contains a horizontal segment with ordinate ~o. Then (a, ~o) is a continuity point for g' for all a except a countable exceptional set that we denote by E 130' Then
is a function only of ~, say g(~), and is therefore independent of the interval of sets {Aa}, except for the original choice of ~o. Definefby f(a) = g(a, ~o). Then, since ~o is a continuity point of g, f(a)
=
g (~)
Now f depends only on the interval {A a }. Furthermore, for any index a, the set A a can be replaced by either its interior or closure without altering its position in the interval (due to the method of construction in Proposition 10.3), the convergence of the ratios, or the value of f(a). Consequently, any other continuity point of g could have been chosen (which would multiply both f and g by a common factor, leaving the ratio unchanged), and the proof would remain valid. Let ~o be the index in the hypothesis of Lemma 10.6 and let P' be the collection of sets A that belong to an interval {A a} C T such that if A = A r then y is a continuity point of the function f(a)/g(~o) of Lemma 10.6. Put A(A) = limxh(A, x)/h(B 130' x) = lim r H(A,r)/H(Bf3o' r). THEOREM 10.3 (L. Loomis. 1949) A is a Loomis content on T and hence can be extended to afinitely additive measure on Z. Moreover P' = ZnT.
Proof: We first show that P' is dense in T with respect to < and that if U, V E P' are uniformly separated then UuV E P'. If U, V E P', there exist intervals {Aa), {Bf3} c Twith Aa < Af3 and Ba < Bf3 whenever a < ~, and such that U = A e and V = B ~ for some 8, ~ that are continuity points of A(A a) and A(B 13) respectively. Let [0, 1] ----+ [0, 1] be a strictly increasing, onto function with
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10. Uniform Measures
(8) = ~ and for each a E [0, 1] put C a = B 0 and pick a,~, y, 8 E [0,1] with Aa < U < A f3 and By < V < B'6 such that A(A f3 - CI(A a» < £/2 and A(B'6 - CI(B y» < £/2. Now U = Ae and V = C e , so there exist r, s E [0,1] withA r < U < As, C r < V < Cs, A(As - CI(Ar» < £/2 and A(Cs - CI(C r » < £/2. Clearly AruC r < UuV < AsuCs and AsuCs - CI(AruC r ) c [As - CI(Ar)]u[C - CI(C r )] so A(AsuCs AruCr ) < A([As - CI(Ar)]u[Cs - CI(C r )]). Hence it only remains to show that A is sub-additive on p' to show that A(AsuCs - CI(AruC r » < £. For this note that for any A, B E p' that h(AuB, x) :S: h(A, x) + h(B, x) for each x E /-!, so A(AuB) :S: A(A) + A(B). Consequently, UuV is a continuity point of A(A aUC a) so UuV E P'. To show p' is dense in T, let A, BET with A < B. By Proposition 10.3, there is an interval {A a} C T with A 0 = A and A I = B and A a < A f3 whenever a < ~. By Lemma 10.6 we can choose a y such that A < A y < Band y is a continuity point of A(A a). Therefore A 'Y E pi so p' is dense in T. That A satisfies L1 on p' is clear from the definition of A. L2 and L3 follow from the fact already shown above that for A, B E pI, h(AuB, x) :S: h(A, x) + h(B, x) for each x E /-! and from the fact that h(AuB) = h(A, x) + h(B, x) if Star(A, x)nStar(B, x) = 0. L4 follows from the fact that if A E P' then H(A, x) = sup {H(A, y) Iy < x} (Proposition 10.6) and the fact that p' is dense in T. Therefore p' satisfies the conditions of Theorem 10.1 if we set P = T, so A can be uniquely extended to a Loomis content on T. Hence by Theorem 10.2, A is a finitely additive measure on the ring Z of zero-boundary sets with respect to A. Since A E p' implies A is a continuity point of A(A a) for some {A a} C T with A E {A a}, A is a zero-boundary set with respect to A by the remarks preceding Corollary 10.1. Therefore A E ZnT. Conversely, if A E Zn T then A is a continuity point of A with respect to some interval {A a} C T that contains A (also by the remarks preceding Corollary 10.1). Hence A E P'. Therefore, P'
=ZnT.This concludes the correct part of Loomis' development. In the final two sections, a development of an only marginally weaker version of Loomis' claims will be presented that is closer in spirit to the approach of Loomis' original construction of a uniform measure on a metric space.
EXERCISE 1. Show that a set of points in [0, 1] x [0, 1], no pair of which determines a positive slope, lie on a decreasing curve (allowing vertical and horizontal segments as curve arcs).
lOA Invariance and Uniqueness of Loomis Contents and Haar Measures
299
10.4 Invariance and Uniqueness of Loomis Contents and Haar Measures In this section we first prove the in variance and uniqueness of a Loomis content on its ring Z of zero-boundary sets. We will then use this result to prove the uniqueness of the Haar measure developed in the previous chapter. We have been using 11 not only as a basis for the uniformity v, but also as an index set for defining limits as in the definition of A in Theorem 10.3. We now want to expand our index set 11 to an index set 11* in such a way that we can have intervals of indicies from 11* analogous to the intervals of sets {Aa} such that CI(A a) < Int(A 13) whenever a < P in Proposition 10.3. For any x, yEll with x < Y, we use axiom A 1 to construct a collection {y a} C 11 where a ranges over the rationals in [0, 1], Yo = x, Y I = Y and Ya < Y 13 whenever a < p. Then for an irrational y E [0, 1] we let Yy be the uniform covering defined by Yy = {Yy(P) Ip E X} where Yy(p) = ua k. For each m > k,
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11. Spaces of Functions
Theorem 8.11 can be used to show that: JXliminfi-.=lfnj(x)-fm(x)lpdJ.!:S; liminfi-.Jxlfn,(x)-fm(x)lpdJ.!:S; fP.
But lim infi-.= Ifn,(x) - fm(x) Ip = If(x) - fm(x) Ip so Ix If - fm Ipd!! :s; £P. Hence If-fm Ip :s; £P so f - fm E U(!!) which implies f E U(!!) since f = fm + (f - fm)' Moreover, we have shown that for each £ > 0, there exists a positive integer k such that if m > k then If - fm Ip :s; £P so f is the limit of {fn} with respect to the LP-norm. The proof that the theorem holds when p left as an exercise. -
=
00
is considerably easier and is
COROLLARY 11.1 If {fn} is a Cauchy sequence in U(!!), where I :s; p :s; 00, and iff is the limit of {fn}, then {fn} has a subsequence that converges pointwise almost everywhere to f
The proof of the corollary is contained in the proof of the theorem. Let SM denote the class of complex valued, simple measurable functions on X whose support has finite measure. Then the members of U (!!) can be approximated by members of SM for 1 :s; P < 00. In fact: PROPOSITION 11.4 SM is dense in U (!!) for 1 :s; P < 00. Proof: Clearly SM C U(!!). First suppose f ~ 0 is in U(!!) and let {sn} be a sequence of simple measurable functions with 0 :s; s 1 :s; S 2 :s; ... :s; f and such that {sn(x)} converges to f(x) for each x E X. That such a sequence exists follows from Theorem 8.7. Since O:s; Sn :S;ffor each positive integer n, we have that Sn E LP(!!) for each n and hence Sn E SM for each n.
Now Ix If Ipd!! < 00 since fEU (!!). But f~ 0 implies If IP = fP = IJP I so Ix IJP Id!! < 00 which implies JP ELI (!!). Furthermore, If - Sn IP :s; JP for each n and {(f - snY} converges to 0 for each x E X. Then by Theorem 8.13, limn-.Jx I (f - snY - 0 Id!! = 0
which implies limn -.=Jx If - Sn IPd!! = 0 which in turn implies limn -.= If - Sn Ip = Consequently, SM is dense in LP (!!). -
o so {sn} converges to f in the LP -norm.
Now let us consider a more specialized class of measure spaces that includes the locally compact isogeneous uniform spaces equipped with Haar measure. Specifically, for the remainder of this section we assume X is locally compact and that!! is a positive, complete, almost regular Borel measure that is finite on compact sets (i.e., the type of measure obtained by applying the Reisz Representation Theorem to a positive linear functional on CK(X)). In such measure spaces we can approximate members of LP(!!), for 1 :s; p < 00 with
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11.1 LP-spaces
sequences of continuous functions in CK(X). In fact, for such spaces, we will show that CK(X) is dense in U (J..L) for 1 :s; P < 00, with respect to the U -nonn. To do this, we first need a result known in the literature as Lusin's Theorem. It is this result that requires the additional assumptions. Fortunately the Haar measures on locally compact unifonn spaces satisfy most of the restrictions of classical analysis necessary to build the theory of real analysis on locally compact metric spaces. This is why, once we have Haar measure on this class of unifonn spaces we can move the theory of differentiation of a measure to locally compact, isogeneous unifonn spaces.
THEOREM 11.3 If f is a complex measurable function on a locally compact space X and J..L is a positive, complete, almost regular, Borel measure that is finite on compact sets, and ifJ..L(E) < 00, andfis zero outside of E, thenfor each £ > 0 there exists agE CK(X) such that the measure of the set on whichf and g are not equal is less than £. Moreover, we can pick g such that sup 1 g(x) 1 :s; sup Ij(x) I. Proof: First assume E is compact and 0 :s; f(x) < 1 for each x E X. From the proof of Theorem 8.7, there exists a sequence {sn} of simple measurable functions with O:s; SI :s; S2 :s; ... f such that limsn(x) = f(x) for each x E X. Moreover, from the construction in the proof, the s" have the fonn
where E"j = {x 1 (i - 1)/2" :s; f(x) < i/2"} for each positive integer i = 1 ... n2" and F" = {x 1 n :s; f(x) :s; oo}. Since we assumedf < 1 we have F" = 0 for each n and E"j = 0 for i > 2" so s" simplifies to
for each n. Put Then
01
=
SI
and for each positive integer n let
0,,+1
=
S,,+1 -
s".
f(x) = 1:::=10,,(X) for each x E X. By inspection, it is easily seen that for each n > 1, s" - Sn~1 is a simple measurable function that equals 2~" on the set of points where f - S,,_1 ~ 2" and o otherwise. Consequently, 2"0" is the characteristic function of this set which we denote by E". Then E" c E for each n. Let U be an open set containing E such that CI(U) is compact. Then for each n, there exists a compact set K" and an open set Un with K" c En C Un C U and J..L(Un - K,,) < £/2". By Theorem 8.5, there exist functions gn E CK(X) with 0:5 g" :5 1 such that Kn < g" < U", Let
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1l. Spaces of Functions
Clearly this series converges unifonnly on X. An easy modification to Theorem 1.12 implies g is continuous. Since an = 2- nhn on Kn we have g = f except on u(Un - Kn) which is a set of measure less than £. Thus we have proved the existence part of the theorem for the case where E is compact and 0 s: f < 1. Consequently, it is easily shown that the existence part holds if E is compact andfis merely a bounded complex measurable function. Furthennore, we can remove our compactness assumption, for if Jl(E) < 00, there exists a compact K c E with Jl(E - K) < £ for any £ > O. To prove the general case, let f be a complex measurable function and for each positive integer n put Fn = (x Ilf(x) I > n). Then each Fn is measurable and N n = 0. Consequently, limnJl(Fn) = 0 by Proposition 8.7. For each positive integer n put En = X - Fn. Then f is a bounded function on En. For each n define fn by fn(x) = f(x) if x E En and fn (x) = 0 otherwise. Then f is a bounded measurable function on X and Jl«(xlf(x) =t- fn(x))) s: Jl(Fn). Now let £ > 0 and pick k with Jl(Fk ) < £/2. Then/k is a bounded function on X and Jl( (x [f(x) =t- fn(x»)) < £/2. Then there exists a g E CK(X) with Jl( (x Ig(x) =t- fn(x»)) < £/2 so Jl( (x Ig(x) =t- f(x))) < £. This proves the existence part of the proof. To conclude the proof we first observe that if supx If(x) I = 00 then supx I g(x) [ s: supx If(x) I so assume supx If(x) I < 00. Put b = supx If(x) I and let hex) = x if Ix I s: b and bx/ Ix I otherwise. Then h is a continuous function from C onto CI(S(O, b». If g' E CdX) such that f(x) = g'(x) except on a set of measure less than £ and g = h © l then f(x) = g(x) except on a set of measure less than £ and supx Ig(x) I s: supx If(x) I. This concludes the proof.THEOREM 11.4 If X is a locally compact space and Jl is a positive, complete, almost regular Borel measure on X then CK(X) is dense in LP(Jl)for I S:p < 00 0
Proof: Let SM be the class of complex valued, simple measurable functions on X whose support has finite measure. If s E SM and £ > 0, then by Theorem 11.3, there exists agE CK(X) with g(x) = sex) except on a set of measure less than £ and Igl S:suPxls(x)1 = Isl~. LetE= (xlg(x)=t-s(x»). Then Ig-slp
= [fElg-sIPdJll llP s: rfEl2lsl~lpdJllllP = 2Isl~£lIp.
Therefore, it is possible to find a sequence (gn) c CK(X) such that (gn) converges to s with respect to the LP -nonn metric. But then since SM is dense in U(Jl) by Proposition 11.3, so is CK(X).Theorems 11.2 and 11.4 together say that for 1 s: p < 00, U(Jl) is the completion of CK(X) with respect to the LP -nonn metric. The case where p =00 is different because the definition of L ~(Jl) is essentially different than the
11.1 LP-spaces
325
definition of LP(f..l) for p < DO. Now CK(X) is a metric space with respect to the L ~ -norm metric. To characterize the completion of C K(X) with respect to this metric, we need the following definition: A complex valued function f on X is said to vanish at infinity if for each £ > 0 there exists a compact set K such that If(x) I < £ for each x E X - K. The class of all continuous, complex valued functions that vanish at infinity is denoted by C ~(X). On CK(X) the L ~ -norm coincides with another norm called the supremum norm that is defined by If I
=supx If(x) I. THEOREM 11.5 If X is a locally compact space then C ~(X) is the completion of CdX) with respect to the L = -norm and the supremum norm.
Proof: It is left as an exercise to show that C =(X) is, indeed, a metric space with respect to the L = -norm. We need to show that CK(X) is dense in C =(X) and that C =(X) is complete. That CK(X) is dense in C =(X) follows from the fact that if f E C =(X) and £ > 0, then there exists a compact set K with if(x) I < £ outside K and an h E C K(X) with 0 ~ h ~ 1 and h(K) = 1. If we put g = fh then clearly g E CK(X) and If - g I = = supx If(x) - g(x) I < £. To show C ~(X) is complete let Ifn} be a Cauchy sequence in C =(X). For each £ > 0, there exists a positive integer N such that if m, n > N then Ifm - fn I < £ so supx Ifm(x) - fn(x) I < £. Then for each x E X, the sequence Ifn(x») is a Cauchy sequence in C and hence converges to some point f(x) E C. This pointwise limit function defined by f(x) = limnin(X) is well defined. Moreover, Ifn) converges uniformly to f since for each £ > 0 there exists a positive integer N such that if m, n > N then supx Ifm(x) - fn(x) I < £/2. Also, for each x E X, there exists a k > N with Ih(x) - f(x) I < £/2. But then if m > N, Ifm(x) - hex) I < £/2 so Ifm(x) - f(x) I < £ for each x E X. Now an easy modification of Theorem 8.12 shows thatfis continuous. It remains to show that f E C =(X). For this let £ > O. Then there exists an 1= < £/2 and there exists a compact set K such that Ifn(x) I < £/2 outside K. But then If(x) I < £ outside K so f E C ~(X).-
n with If - fn
EXERCISES 1. Prove Proposition 11.1. 2. Prove Proposition 11.3. 3. Show that d defined by d(J, g) =
If - g Ip
4. Prove Theorem 11.2 for the case p =
is a pseudo-metric on LP (f..l).
DO.
5. (Jensen's Inequality) If f..l is a positive measure on a a-algebra M in a space
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11. Spaces of Functions
X such that ~(X) = 1 and if g is a real valued function in L I (~) with a < g(x) < b for each x E X, and iffis convex on (a, b), show that:
6. If ai > 0 for each i = 1 ... n such that Lai = 1 and if Xi is a real number for each i = 1 . . . n, show that e(LaixJ :s;; Laiex, where e is the exponential function. 7. Show that if X is a locally compact space and ~ is a positive, complete, almost regular Borel measure on X and if the distance d(f, g) between two functions f, g E CK(X) is defined by d(f, g) = Jx If - g Id~ then (CK(X), d) is a metric space and the completion of (CK(X), d) is precisely the class of Lebesgue integrable functions on X. [Recall that the Lebesgue integrable functions were defined in Exercise 1 of Section 8.6.] 8. Show that if p < q that for some measures ~, U (~) c L q (~) whereas for other measures L q (~) c U (~), and that there are some measures such that neither U (~) or L q (~) contains the other. What are the conditions on ~ for which these situations occur?
11.2 The Space L2(~) and Hilbert Spaces The space L 2 (~) is known as the space of square integrable functions. It plays a major role in modem physics and in many other mathematical applications. In fact, it is the mathematical model that underlies the wave interpretation of quantum mechanics, when X is Euclidean space and ~ is Lebesgue measure on X. L 2(~) is a special case of a more general class of spaces called the Hilbert spaces. Other Hilbert spaces also play important roles in quantum mechanics, in fact, they are the mathematical models behind all quantum phenomenon. We will need some results about Hilbert spaces for our development of uniform differentiation in the next chapter. Hilbert spaces derive their name from David Hilbert who published a series of six papers between 1904 and 1910 titled Grundzuge einer allgemeinen Theorie der linearen Integralgleichungen I - VI in Nachr. Akad. Wiss. Gottingen Math.- Phys. Kl. that involved these spaces. They were republished in book form by Teubner Verlagsgesellschaft, Leipzig in 1912 and reprinted by Chelsea Publishing Co., New York in 1952. We now define these spaces. A complex vector space H is called an inner product space if for each pair of vectors u, v E H there is a complex number (u, v) called the inner product or sometimes the scalar product or dot product that satisfies the following axioms:
11.2 The Space L2()l) and Hilbert Spaces
327
(1) (u, v) = (v, u)* (the * representing complex conjugation). (2) (u + v, w) = (u, w) + (v, w) for u, v, w E H. (3) (cu, v) = c(u, v) for u, v E Hand c E C. (4) (u, u) ~ for each u E H. (5) (u, u) = if and only if u is the zero vector in H.
°
°
°
There are a number of observations we can make about these axioms. First, (3) implies (0, x) = for each x E H and (I) and (3) together imply (x, cy) = c*(x, y) for each pair x, y E H and c E C. Next we observe that (2) and (3) together imply that for each y E H, the mapping defined by A(X) = (x, y) for each x E His a linear functional on H. (1) and (2) can be combined to show that (x, y + z) = (x, y) + (x, z) for x, y, z E H. Finally, by (4) we can define a norm Ix I for each x E H by Ix I = (x, x)1!2 so that Ix 12 = (x, x). PROPOSITION 11.5 (Schwarz Inequality) For each x, y E H, I(x, y) I :s; Ix II y I where the norm on the left is the modulus of the complex number (x,y).
°
Proof: Ifx=Oory=Othen l(x,y)l:s; Ixllyl so assumext:O t:y. Let abean arbitrary complex number. Then (x + ay, x + ay) ~ and (x + ay, x + ay) = Ix 12 + I al 2 1y 12 + a(y, x) + a*(x, y) and a(y, x) + a*(x, y) = 2Re(a(y, x» so
Ix1 2 + laI2IyI2+2Re(a(y,x»~0.
°
Now each complex number a can be represented by a = re il for some real number r ~ and some complex number e il for some real number t. Recall that I eill = 1. Similarly, (y, x) = I(y, x) Ie is for some real number s. Hence Re(a(y,x» = Re(re it I(y, x) leis) = Re(rl (y, x) I ei(s+I) = rl (y, x) IRe(ei(s+I) and Re(e i (S+I) :s; 1. Consequently,
°
so Ixl2 + Irl21yl2 + 2rl(y, x)1 ~ for each real number r ~ 0. Put r = -I x I/ Iy I. Then substituting this value for r in the previous inequality yields Ix II y I :s; 1(x, y) 1.An immediate result of the Schwarz inequality is the so called triangle inequality: 1x + y 1 :s; 1x 1 + 1 y I. It follows from the observation that Ix + y 12 = (x + y,x + y) = (x, x) + (x,y) + (y,x) + (y,y):s; Ixl 2 + 2Re«x, y» + lyl2:s; Ixl2 +21(x,y)1 + lyl2:s; Ixl 2 +2lxllyl + lyI2=(lxl + lyl)2. Consequently, if we define the distance d(u, v) between two vectors u, v E H to be 1 u - vi, it is easily shown that d satisfies the axioms of a metric space. In particular, if u, v, w E H we see that d(u, v) :s; d(u, w) + dew, v) follows from 1 u - vi :s; 1 u - wi + 1 w - v I. If the metric space (H, d) is complete, we call H a Hilbert space. We Now observe that if J.l is a positive measure, then L 2(J.l) is a Hilbert space.
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11. Spaces of Functions
For this we define an inner product on L 2(Il) by (f, g) = fxfg*dll. Since g E L 2(Il) implies g* E L 2(Il) and the exponents p = 2 = q are conjugate exponents, Proposition 11.2 implies fg* ELI (Il) so (f, g) is well defined on H. Now we observe that if we define Ifl2 = (f, f) or equivalently, If I = (f, f)l!2 then we have
so the L2-norm I 12 on L 2 (Il) is equivalent to the inner product norm I I on L 2(Il). Since Il was assumed to be positive, by Theorem 11.2 we know L 2(Il) is complete with respect to the L 2-norm, so L 2(Il) is complete with respect to I I and therefore is a Hilbert space.
PROPOSITION 11.6 For a given y E H. the mappings defined by f(x) g(x) = (y. x), h(x) = Ix I for each x E X are uniformly continuous functions on H.
= (x,y),
Proof: To show f is uniformly continuous let E > 0 and put 0 = Ix I - x 21 < 0 we have by Proposition 11.5 that:
E/ I y I.
Then if
Therefore, f is uniformly continuous on H. A similar argument shows that g is also uniformly continuous on H. To show h is uniformly continuous, let E > 0 and put 0 = E. By the triangle inequality, if Ix I - x 21 < 0, then Ix I I :0:; Ix I - x 21 + IX21, so IXII - IX21:o:; IXI -x21. Similarly, IX21 - IXII = IX2 -xII = IXI-x21,so IlxI -lx211:o:; IXI -x21. Butthen
so that h is also uniformly continuous on H. Recall that a subset S of a vector space V is called a subspace of V if S is a vector space with respect to the addition and scalar multiplication operations defined on V. A necessary and sufficient condition for S c V to be a subspace is that u + v E S and cu E S for each u, v E S and c E C. If H is an inner product space, a closed subspace of H is a subspace that is closed with respect to the metric topology generated by the inner product norm. A set E in a vector space is said to be convex if for each u, vEE and 0 :0:; x:O:; 1, the vector (1 - x)u + xv is contained in E. One can visualize this property of convexity by imagining a "line segment" being traced out between u and v as x goes from 0 to 1, and that all the vectors on this "line segment" are contained in E. If (u, v) = 0 for some u, v E H we say u is orthogonal to v and denote this by u ..L v. Let u 1- be the collection of v E H which are orthogonal to u. If v, W E u1- then (u, v + w) = (u, v) + (u, w) = 0 and if c E C then (u, cv) = c(u, v) =0 so v
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11.2 The Space L 2(1l) and Hilbert Spaces
+ W E u1. and cv E u1.. Hence u1. is a subspace of H. Now u1. is the set of vectors x E H where the continuous function g(x) = (u, x) = 0, so u1. is a closed set in H. If S is any subspace of H, let S1. denote the collection of all v E H that are orthogonal to every u E S. Clearly S1. = n{u1.1 u E SI. Since each u1. is a closed subspace of H, so is S1.. PROPOSITION I1.7 Each nonempty, closed, convex set in a Hilbert space has a smallest element with respect to the inner product norm. Proof: Let E be a nonempty, closed, convex subset of the Hilbert space H. Put b = inf{ Iv II VEE I. Then there exists a sequence {vn IcE such that { Ivn I I
converges to b. For any pair x, y E H, Ix + yl2 + Ix - yl2 = (x + y, x + y) + (x-y,.x-y) = 21xl2 + 21y12. If we apply this identity to x/2 and y/2 we get (1/4)lx-yI2= IxI 2/2+ lyI2/2-I(x+y)/212. SinceEisconvex,(x+y)/2E E which implies I(x + y)/21 ~ b, so (11.2) for each pair x, y E H. If we replace x and y in this inequality by Vm and Vn we see that as m, n ~ 00, the right side of (11.2) tends to zero. Hence IVm - Vn I ~ o as m, n ~ 00 which implies {v n I is a Cauchy sequence in E c H. Since H is complete, {v n I converges to some v E H, and since E is closed, vEE. Also, since the norm function h(x) = Ix I is continuous on H by Proposition 11.6, we have I v I = limn IVn I = b. Consequently, there exists a vEE of smallest norm. It remains to show that v is unique. If u is another member of H such that lui = b = lvi, then the inequality (11.2) implies lu - vI2 ::; 0 so u = v. Therefore, E has a smallest element with respect to the inner product norm. Let S be a closed subspace of the Hilbert space H and let u E H. Then the set u + S = {u + v Iv E S I is closed and convex. To see that u + S is closed, let w be a limit point of u + S. Then there exists a sequence {u + Vn I in u + S that converges to w, so {u + Vn I is Cauchy. If E > 0, there exists a positive integer N such that if m, n > N then I u + Vm - U - Vn I < E which implies IVm - Vn I < E, so {v n I is Cauchy in S. Since S is a closed subspace of the complete space H, S is complete which implies {v n I converges to some v E S. Then if E > 0, there exists a positive integer N such that if n > N then Iv - Vn I < E which implies I(u+v) - (u + vn) I < E, so {u + Vn I converges to u + V E U + S. But since limits are unique in metric spaces, u + v = W so W E U + S. Hence u + S is closed. To see u + S is convex, let u + v and u + W E
U
+ S and let 0 ::; A::; 1. Then
(1 - A)(U + v) + A(U + w) = U + (1 - A)V + AW E U + S. Therefore, u + S is convex. Consequently, we can apply Proposition 11.7 to u + S and get a
smallest element Ps.L(u) in u + S with respect to the inner product norm. Next, putps(u) = u - Ps.L(u). Thenps andps.L are functions on H. Since Ps.L(u) E u +
330
11. Spaces of Functions
5, Ps(u) E 5. The function Ps is called the orthogonal projection of H onto 5. The function ps.l is called the orthogonal projection of H onto 5.1. For this later definition to make sense, we need to show thatps.l(u) E 5.1. For this let x = Ps.l(u). Then by the definition of the orthogonal projection onto 5.1 we have, for each y E 5 with Iy I = 1, Ixl2 ~ Ix-cyl2 = (x-cy,x-cy) = IxI 2 -c(y,x)-c*(x,y)+ Icl 2 for each scalar c. If we substitute c = (x, y) into this inequality, we get 0 ~ -I (x,y) 12 which implies (x, y) = 0 for each y E 5 with Iy I = 1. But then (x, y) = OforanYYE 5. Therefore,x=ps.l(u)E 5.1.
THEOREM 11.6 If 5 is a closed subspace of the Hilbert space H, then the orthogonal projections Ps and Ps.l of H onto 5 and 5.1 respectively have the following properties: (1) u = ps(u) + Ps.l(u)for each u E H. (2) The orthogonal projections are unique. (3) The orthogonal projections are linear. (4)lfu E 5 thenps(u) = u andps.l(u) = o. (5)lfu E 5.1 thenps(u) =0 andps.l(u) = u. (6) I u - Ps(u) I = inf{ Iu - v II v E 5j for each u E H. (7) Iu 12 = Ips(u) 12 + Ips.l(u) 12 for each u E H. Proof: (1) follows immediately from the definition of Ps(u). To show (2), first note that 5n5.1 = {OJ. This is because if x E 5n5.1 then (x, x) = 0 which implies x = O. Next let u = v + w where v E 5 and w E 5.1. Thenps(u) + Ps.l(u) = U = v + w which implies Ps(u) - v = w - Ps.l(u). Since the left side of this equation is in 5 while the right side is in 5.1, we conclude that both sides are the zero vector. Therefore, v = Ps(u) and w = Ps.l(u), so the orthogonal projections are unique. To show (3), let u, v E Hand c, dEC. Then by (l),ps(cu + dv) + Ps.l(cu + dv) = cu + dv = c[ps(u) + Ps.l(u)] + d[Ps(v) + Ps.l(v)] so Ps(cu + dv) - cps(u) dps(v) = cps 1. (u) + dps.l(v) - Ps.l(cu + dv). Again, the left side of this equation is in 5 while the right side is in 5.1 so we conclude both sides are the zero vector. Therefore, Ps(cu + dv) = cps(u) + dps(v) and Psl.(cu + dv) = Cps.l(u) + dps 1. (v) so the orthogonal projections are linear. To show (4) and (5), note that if u E 5, then (1) implies u - Ps(u) = Ps.l(u) so the left hand side of this equation is in 5 while the right hand side is in 5.1. Again, we conclude both sides are the zero vector, so Ps(u) = u and Ps.l(u) = O. This proves (4). A similar argument proves (5). To show (6), note that by definition of Psl.(u) we have Iu - Ps(u) I = Ips.l(u)1 =inf{lu+vllvE S} =inf{lu-vllvE S}. This proves (6). To show (7), observe that 1u 12 = (u, u) = (Ps(u) + Psl.(u), Ps(u) + Psl.(u» = Ips(u) 12 +
1l.2 The Space L2(~) and Hilbert Spaces
331
(Ps(u), Psl.(u)) + (Psl.(u), Ps(u)) + Ipsl.(u)12. Since (Ps(u), Psl.(u)) = 0 = (Psl.(u), Ps(u)) we have that 1u 12 = Ips(u) 12 + Ipsl.(u) 12 which proves (7).COROLLARY J J.2 If S :t- H then there exists a u E H such that u 1- S and u:t- O. Proof: Pick v E H - S. Put u = Psl.(v). Then v:t- Ps(v) so u:t- O. But u 1- S since
UES.l.In Proposition 11.6 we saw that the function f(x) = (x, y) for a fixed y E H is unifonnly continuous. Since (x, y) E C and since the definition of the inner product causes f to be linear, we see that f is a continuous linear functional. The Riesz Representation Theorem (Chapter 8) showed that positive linear functionals on CK(X) could be represented as positive measures on X. It is therefore natural to ask if continuous linear functionals on a Hilbert space can be represented as inner product functions with respect to a given vector. The answer is affinnative as the next theorem shows. THEOREM J J.7 If A is a continuous linear functional on a Hilbert space H, then there exists a unique v E H such that A(U) = (u, v)for each u E H. Proof: If A(U) = 0 for each u
E H put v = O. Otherwise put K = {u 1 A(U) = OJ. Since A is linear, K is a subspace of H and since A is continuous, K is closed. Since A is not identically zero, Theorem 11.6 shows that K.l :t- {O j. Choose w E K.l such that w :t- O. Then w is not in K so A(W) :t- O. Put c = A(w)/I W12 and let v = c*w. Then v E K.l, v :t- 0 and A(V) = A(C*W) = C*A(W) = c*c 1W12 = cc*(w, w) = (c*w, c*w) = (v, v). For any u E H put x = u - A(U)V/A(v). Then A(X) = A(U) A(U)A(V)/A(V) = 0 so X E K which implies (x, v) = O. Now (u, v) = (X+A(U)V/A(V),V) = (x, v) + (A(U)V/A(V), v) = 0 + [A(U)/A(V)](V, v) = [A(U)/A(V)]A(V) = A(U). Consequently, A(U) = (u, v) for each U E H.
To show that v is unique, let w be another vector in H such that A(U) = (u,w) for each U E H. Put z = v - w. Then (u, z) = (u, v - w) = (u, v) - (u, w) = A(U) - A(U) = 0 for each U E H. But then (z, z) = 0 which implies z = 0 so v = w.
-
Recall how a basis is defined in a vector space V. First we define a linear combination of vectors VI ..• Vn E V to be a vector sum of the fonn c 1 v 1 + ... + Cn Vn for some c 1 .•• Cn E C. We define the vectors VI' .• Vn to be linearly independentifclvl +",+cnvn=Oimpliesci=Oforanycj ... CnE C. A subset S of V is said to be linearly independent if every finite subset of S is linearly independent. The set V(S) of all linear combinations of finite subsets of S is clearly a vector space. In fact, it is easily seen to be the smallest subspace of V containing S. V(S) is called the span of S or the subspace spanned by S. If V(S) = V then S is called a spanning subset of V or we say S spans V. Finally,
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11. Spaces of Functions
we define a basis of a vector space V to be a linearly independent subset that spans V. PROPOSITION 11.8 A linearly independent subset of a vector space is a basis if and only if it is maximal. Proof: Let S be a maximal linearly independent subset of a vector space V. Assume V(S) '# V for otherwise S would be a basis for V. Then 0 E V(S) so there exists a U E V - V(S) with U '# O. Let v I . . . Vn E S and suppose C I v I + ... + CnVn + Cn+1 U = 0 for C I ••• Cn+1 E C. If Cn+1 '# 0 then U = Li'=1 CjVJCn+1 which implies U E V(S) which is a contradiction. Therefore, Cn+l = 0 which implies C I VI + ... + CnV n = 0 so Cj = 0 for each i = 1 ... n. But then Su{u} is linearly independent which implies S is not maximal which is a contradiction. Therefore, S is a basis for V.
Conversely, assume {u a} is a basis for V and suppose i U a I is not maximal. Then there exists a U E H with U '# 0 and {u a }u{ u} linearly independent. Since {u a} spans V, there exists a finite subcollection UI ... Un of {u a } and a finite collection CI ••• Cn E C with U = ClUJ + ... + CnU n. Then CIUI + ... + CnU n + (-I)u = 0 but (-1) '# 0, so {ua}u{u} is not linearly independent after all. • A set of vectors {u a} in a Hilbert space H is said to be orthogonal if (ua,u~)=Oifa'#~. {u a } is said to be normalized if Iual = 1 foreacha. If {u a} is both orthogonal and normalized it is said to be orthonormal. Clearly, {u a} is orthonormal if and only if (u a, u~) = 1 if ex = ~ and 0 otherwise. PROPOSITION 11.9 If UI ... Un is an orthonormal set and v = C I UI + ... + CnU n. then Ci = (v, ui)for each i = 1 ... n and Ivl 2 = Li'=I!c; 12. Proof: For each i = 1 ... n, (v, u;) = (c I UI + ... + CnU n' u;) = Ci(Ui, Ui) = Ci since UI ... Un is an orthonormal set. Also, Ivl 2 = (v, v) = (CIUI + ... + CnU n' C lUI + ... + CnU n) = C ICI *(u" ud + ... + CnC n*(u n, un) = Li'=,1 Ci 12 .• COROLLARY 1J.3 An orthonormal set is linearly independent. Proof: Let {u a} be an orthonormal set and let U! ... Un E {u a}. Suppose C 1 U1 + ... + CnU n = 0 for C I ... Cn E C. By Proposition 11.9, Ci = (0, u;) = 0 for each i = 1 ... n. Consequently, {u a} is linearly independent. • THEOREM 1 J.8 Each Hilbert space has an orthonormal basis. Proof: If H is a non-trivial Hilbert space, then there exists a U E H with IUI = 1. Then {u} is an orthonormal subset of H. Let P be the collection of all orthonormal subsets of H containing (u), partially ordered by set inclusion. Since {u} E P, P '# 0 so by the Hausdorff Maximal Principle (an equiValent
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fonn of the Axiom of Choice), there exists a maximal linearly ordered subcollection Q of P. Clearly {u} E Q, so Q =I- 0. Let R = uQ. If v, W E R then v E A E A and wEB E Q for some A, BE Q. Since Q is linearly ordered by inclusion, either v, W E A or v, wEB. Since both A and B are orthononnal subsets of H, (v, w) = 1 if v = wand 0 otherwise. Therefore, R is orthononnal. Suppose R is not a maximal orthononnal set. Then there exists an orthononnal set 5 containing R with 5 - R =I- 0. Now 5 is not in Q and 5 contains each member of Q so we can adjoin 5 to Q and still have a linearly ordered set with respect to inclusion which implies Q is not maximal which is a contradiction. Therefore, R is a maximal orthononnal set. By Corollary 11.3, R is linearly independent. Suppose R is not a maximal linearly independent set. Then there exists a linearly independent set T with ReT and T - R =I- 0. Let x E T - R. Let V be the subspace of H spanned by R. Then x is not in V which implies y = Pv.i(x) =I- 0 by Theorem 11.7, so (y, v) = 0 for each v E R. Put z = y/I y I which implies Izl = 1 and (z, v) = 0 for each v E R. Therefore, z can be adjoined to R to obtain an orthononnal set in H containing R as a proper subset which is a contradiction. Consequently, R is a maximal linearly independent set so R is a basis. Since R is orthononnal, R is an orthononnal basis. -
E
LEMMA 11.1 If V is a closed subspace of the Hilbert space H and H - V, then the subspace W spanned by Vu {u) is closed.
if u
Proof: Suppose v is a limit point of W. Then v = limn(v n + cnu) where {v n } C V and {c n } C C. Consequently, there exists a b < 00 such that IVn = Cnu I < b for each positive integer n. Assume {c n } has no convergent subsequence. Since closed and bounded subsets in C are compact, this implies limn I Cn I = 00. But then IV n + cnul/lcnl < b/lcnl for each n and limnbllcnl = O. Therefore, limn Ivn!cn + u I = 0, so limn(vn!c n} = -u which implies -u E V since V is closed which is a contradiction.
Hence we may assume {c n } has a convergent subsequence {c mn } that converges to some C E C. Now v = limn(v mn + cmn u) = limn vmn + cu which implies {v mn } converges to v - cu. Since {v mn } c V and V is closed, [v - cul E V and v = [v - cu + cul E W. Therefore, W IS c1osed.THEOREM 11.9 If u 1 . Un is an orthonormal set of vectors In the Hilbert space H and v E H, then 1 v - L;=! (v, Ui)Ui I :s; 1 v - Li'=! CiUi I for all c! .. Cn E C and equality holds if and only if Ci = (v, Ui) for each i = 1 ... n. The vector L;=! (v, Ui)Ui is the orthogonal projection of v onto the subspace W generated by {u! ... un} and if d is the distance from v to W then 1v 12 = d 2 + L;=! I (v, Ui) 12. Proof: The subspace {O} of H is obviously closed. By Lemma 11.1, the subspace of H spanned by {u I} is closed since it is the subspace spanned by {O} u {u 1 }. Proceeding inductively, it is clear that W is closed. Then by
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Theorem 11.6, Pw(v) is an element of W such that Iv - Pw(v) I ~ Iv - w I for each W E W and since the mappings Pw and pw:c are unique, Pw(v) is unique. Consequently, Pw(v) has the property that I v - Pw(v) I ~ I v - L?=1 CiUi I for each collection C 1 ... Cn in C. Let Pw(v) = L?=1 aiUi for some a 1 ... an E C. By Proposition 11.9, ai = (Pw(v), Ui) for each i = 1 ... n. Now v =Pw(v) + Ps.:c(v) so (v, Ui) = (Pw(v), uJ + (Pw:c(v), Ui) = ai + 0 since Pw:c(v) is orthogonal to all the Ui. Hence Pw(v) = L?=1 (v, UJUi' Therefore, Iv - L?=1 (v, UJUi I ~ Iv - L?=1 CiUi I for all C 1 ... cn E C and equality holds if and only if Ci = (v, Ui) for each i = 1 ... n. Finally, the distance d from v to W is the minimum value of Iv - w I such W E W. Therefore, d = I v - Pw(v) I so d 2 = (v - Pw(v), v - Pw(v» = (v-Pw(v),v) - (v - Pw(v), Pw(v». Now (v - Pw(v), Pw(v» = 0 so d 2 = (v - Pw(v), v). To see this, note that for each i = 1 ... n, (v - Pw(v), Ui) = (v - L?=1 (v, UJUi, Ui) = (v, Ui) - (v, Ui)(Ui, uJ = O. Therefore, (v - Pw(v), Pw(v» = (v - Pw(v), L?=1 (V,Ui)Ui) = L?=1 (v, Ui)(V - Pw(v), Ui) = O. Consequently, d 2 = (v - Pw(v), v) = (V-L'?=I(V,Ui)Ui, v) = Ivl 2 - L?=I(V, Ui)(Ui, v) = Ivl 2 + L?=II(v, UJI2 which implies Iv 12 = d 2 + L?=1 I(v, Ui) 12. that
Let {U a} be an orthonormal set in the Hilbert space H. For each vector v Va = (v, ua) for each a. By the symbol Lalval2 is meant the supremum of the set of finite sums of the form I VI 12 + ... + IVn 12 where Vi = (v, Ui) for each finite collection U1 ... Un E {U a}. With this notation we can state and prove the following classical result: E
H put
THEOREM 11.10 (Bessel's Inequality) La Iva 12 ~ IV 12. Proof: For any finite collection U1 . . . Un E {u a}, Theorem 11.9 gives LI=1 I(V,Ui) 12 = IV12 - d 2 where d is the distance from v to the subspace W spanned by the vectors U1 ... Un' Since d ~ 0, this means LI=1 I(v, Ui) 12 ~ IV12 and hence the supremum of such finite sums is less than or equal to Iv 12. -
Sums of the form LaC a where 0 ~ C a ~ for each a and where the summation is defined as the supremum of finite sums of the form C 1 + ... + Cn where C 1 ... Cn E {C a} are especially interesting in Hilbert spaces because they are used in the characterization of the structure of Hilbert spaces. We will now develop this characterization. Let X be a set. For each E E X put Il(E) = 00 if E is infinite and Il(E) = card(E) if E is finite. It is easily seen that 11 is a measure on the a-algebra of all subsets of X. The measure 11 is called the counting measure on X. Let f:X ~ C. Then it is easily seen that LxeX If(x) I, where the summation is the supremum of the finite sums of the form If(x 1) I + ... + If(Xn) I where Xl' .. Xn E X, is the Lebesgue integral of If I with respect to the counting measure on X. We use the notation [2(X) to denote the L2-space L 2(11) where 11 is the counting measure on X. 00
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In particular, if {u a I a E A} is an orthononnal basis in H and for some v E H, v':A ---+ C is defined by v'(a) == Va == (v, ua) for each a E A, then it is a consequence of Bessel's Inequality that v' E [2(A) since:
The importance of the Riesz-Fischer Theorem that we will prove next is that if { U a I a E A} is an orthononnal set of vectors in H then f E [2 (A) implies that f is of the fonn v':A ---+ C for some vector v E H. Before we prove the RieszFischer Theorem, we first observe that Bessel's Inequality implies an even stronger statement about a function v':A ---+ C defined by v'(a) == (v, ua) for each a E A. Bessel's Inequality implies the set of a E A for which v'(a) "# 0 is at most countable. To see this, suppose the set of a E A for which v'(a) "# 0 is uncountable. For each positive integer n put En == {a E AI I v'(a) I > lin}. Then for some positive integer m, Em must be uncountable. But then there exists a finite subset S of Em with LaeS I v'(a) 12 > IV12 which is a contradiction.
THEOREM 11.11 (Riesz-Fischer) If {Ua Ia E A} is an orthonormal set in a Hilbert space H and iff E [2(A), thenf == v' for some v E H. Proof: For each positive integer n put En == {a E A Ilf(a) I > lin}. Then each En must be finite, for otherwise there would exist a finite collection al ... an E A such that Li'=1 If(a;) 12 > If12' For each positive integer n let Vn == LaeEn If(a) IUa. Then each Vn is in H. For each n define v~:A ---+ C by v~(a) == (vn,ua)foreachaE A. ThenforagivenBE A,v~(B)==(LaeEnlf(a)lua,u~) == LaeEn If(a) I(u a , u~) == If(B) I if BEEn and 0 otherwise. Therefore, v~ == If I XE n' Consequently, If - v~ I ~ Ifl2 which implies If - v~ I ~ If I. Since Em C En if m < n, it is clear that limn v~ == f, so by Theorem 8.13, limn If - v~ I == 0 which implies lim" If - v~ I == 0 since we can choose an N large enough so that n > N implies [LaeEn If - v~ 12]112 < LaeEn If - v~ I and therefore If - V~ 12 == [LaeEn If - V~ 12]112 < LaeEn If - v~ I == If - v~ II for n > N. Then since lim" If - v~ 12 == 0, {v~} is a Cauchy sequence in [2(A). Now for each n, Vn == LaeEn If(a) IUa and since E" is finite we can apply Proposition 11.9 to get If(a) I == (v", ua) for each a E E". If m and n are positive integers with m < n, then Em C En. Put E(n, m) - En - Em. Then Iv~ - v~ I == [LaeE(",m) I(v", Ua) - (vm, Ua) 12]1/2 == [:EaeE(n.m) Ilf(a) I - 01 2]112 ==
(LaeE(n,m) If(a) IUa, :EaeE(n,m) Ifla) IUa )1!2 == (v n - vm, v" - vm)1I2 == Iv" - Vm
I,
so 1Vn - Vm I == I V~ - v~ 12 which implies {v n ) is Cauchy in H and therefore converges to some v E H.
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Then v':A ~ C defined by v'(a) = (v, ua) for each a is the desired function. To see this, for a fixed ~ E A, the function g(x) = (x, u~) for each x E H is uniformly continuous (Proposition 11.6). Therefore, since {v,,} converges to v, {g(v,,)} converges to g(v). Hence lim"(v,,, u~) = (v, u~). Then for each a E A, v'(a) = (v, Ua) = lim"(v,,, Ua) = lim" v~(a) =f(a) since {v~} converges to f. Consequently,! = v'. PROPOSITION 11.10 Let {u a} be an orthonormal set in the Hilbert space H. Then {u a} is a basis for H if and only if the set S of finite linear combinations of members of {u a} is dense in H. Proof: Assume {u a} is a basis for H. By Proposition 11.8, {u a} is a maximal linear independent subset of H. Suppose {u a} is not a maximal orthonormal set in H. Then there exists a U E H with IU I = 1 and {u a} U {u} orthonormal. By Corollary 11.3, {u a} U {u} is linearly independent which is a contradiction. Hence {u a} is a maximal orthonormal set in H. Now assume S is not dense in H. Then there exists a v E H - CI(S) which implies there exists ad> 0 such that Iv - U I > d for each U E CI(S). Now v = Ps(v) + Ps~(v) and by the remarks preceding Theorem 11.6, W = Ps~(v)/IPs~(v)1 is in S-L. Since Iwl = 1, { U a} U {w} is orthonormal which is a contradiction. Therefore, S is dense in H.
Conversely, assume S is dense in H and suppose {u a} is not a basis for H. Then {u a} is not a maximal orthonormal set in H, so there exists a U E H with IU I = 1 and U 1.. Ua for each a. Now there exists a sequence {u,,} c S such that lim"u" = u. Clearly U 1.. u" for each positive integer n. Since the function f:H ~ C defined by f(v) = (u, v) for each v E H is uniformly continuous, {f(u,,)} must converge to feu) = (u, u) = 1. But this is impossible since (u, u,,) = 0 for each n. Hence {u a} is a basis for H. The property that S is dense in H has some very interesting ramifications. In fact, it leads to a representation of all Hilbert spaces H as 12(A) where card(A) = card( {II a}) where {u a} is a basis for H. To show this is the case, we need the following lemma. LEMMA 11.2 Let {u a Ia E A} be an orthonormal set in the Hilbert space H and let S be the collection of finite linear combinations of members of {ua}. For each pair x, y E H let x', / E [2(A) be defined by x'(a) = (x, ua) and ira) = (y, ua)for each a E A. Then S is dense in H if and only if (x, y) = (x', y') for each pair x, y E H. Proof: First assume S is dense in H. Let v E H. Now choose e > O. Then there exists a finite col~ec~ion U1 . . . U" E {u a } and c 1 ••. C" E C such thalt w c 1 U1 + ... + C"U n IS wlthm e of v. By Theorem 11.9, Iv - L7;1 (v, uJu; I ~ v - wi < e. Putz=L7;I(V,U;)Uj. Then Iv-zl <e which implies Ivl < Izl +ewhichinturn implies (Ivl - e)2 < Izl2 = I(v, uI)12 + ... + I(v, U,,)12 ~ La Iv'(a)12 by Proposition 11.9. But La Iv'(a) 12 = Lav'(a)[v'(a)]* = (v', V'). Hence, for each v
=
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337
E H, (v, v) = IV 12 ~ (v', v'). By Bessel's Inequality, (v', v') ~ (v, v) so (v, v) (v', v') for each v E H.
=
Now let u, v E H. Then for each c E C we have (u + cv, u + cv) = (u' + cv' , U' + cv') which implies (u, cv) + (cv, u) = (u ' , cv') + (cv ' , u') or c*(u, v) + c(v, u) = c*(u', v') + c(v', u'). Since this holds for each c E C, it holds for c = 1 and c = i. When c = 1 we have (u, v) + (v, u) = (u', v') + (v', u'). When c = i we have (u,v) - (v, u) = (u', v') - (v', u'). Adding these two equations yields: 2(u, v) = 2(u',v') which implies (u, v) = (u', v'). Conversely, assume (x, y) = (x', y') for each pair x, y E H. Suppose S is not dense in H. Pick u E H - CI(S) and let W = Ps.l(u). Clearly W =I- 0 and (w, u a ) = o for each a. Put x = W = y. Then (x, y) = IW 12 > 0 but (x', y') = La(W,Ua)(w,u a )* = 0 so (x, y) =I- (x', y') which is a contradiction. Hence S must be dense in H.· Recall from Section 1.5 that two metric spaces X and M are isomorphic if there exists a uniform homeomorphism f:X ~ M that preserves distance. Two Hilbert spaces HI and H 2 are said to be isomorphic if there is an isomorphism f:H 1 ~ H 2 that is also a linear transformation, i.e., one that preserves sums and scalar products. Such a mapping is called a Hilbert space isomorphism. THEOREM 11.12 If {U a Ia E A} is an orthonormal basis for the Hilbert space Hand iffor each x E H, x' is the element of 12(A) defined by x'(a) = (x, u a ) for each a, then the mapping A:H ~ 12(A) defined by A(X) = x' for each x E H is a Hilbert space isomorphism of H onto [2(A). Proof: Let u, v E Hand c, dEC. Then A(CU + dv) = (cu + dv)'. Then for each a E A, (cu + dv)'(a) = (cu + dv, u a ) = c(u, u a ) + d(v, Ua ) = cu'(a) + dv'(a) = CA(V) + dA(V) so A is a linear transformation from H into 12(A). That A is onto follows from the Riesz-Fischer Theorem.
Suppose u =I- v but u' = v'. Then for each a E A, (u, u a ) = (v, u a ). Now (u-v)' E 12(A) and (u - v)'(a) = (u - v, u a ) = (u, u a ) - (v, u a ) = 0 for each a so (u-v), is the zero element of 12(A). Since u =I- v, Iu - vi > 0 which implies W = (u-v)/I u - vi is a unit vector in H such that w.1 Ua for each a, so {ua}u{w} is an orthonormal set in H. But then {u a} is not maximal which is a contradiction. Hence u - v = 0 so u = v. Therefore, A is one-to-one. Since (u, u) = (u', u') for each u E H, A preserves inner products and hence distance. Therefore, A is a metric space isomorphism between Hand [2(A). Since Ais a linear transformation, it is a Hilbert space isomorphism. •
EXERCISES
1. Show that the vector space C*(X) of all real valued continuous functions on
11. Spaces of Functions
338
x = [0, 1] is an inner product space with respect to (j, g) = Jxfg*dx (where dx denotes integration with respect to Lebesgue measure) but not a Hilbert space. 2. Show that if S is a closed subspace of the Hilbert space H then (S-L)-L
= S.
3. Show that a Hilbert space is separable if and only if it contains a countable orthonormal basis. 4. Let {un) be an orthonormal sequence in the Hilbert space H. Let S be the set of all v E H with v = L;;'=l CnU n where ICn I ~ lin. Then S is isomomorphic with the Hilbert cube and is an example of a closed and bounded subset of H that is not compact. 5. Show that no Hilbert space containing an orthonormal sequence is locally compact. 6. Show that for each pair of Hilbert spaces, one of them is isomorphic to a subspace of the other.
7. Let U be a member of the Hilbert space H and let S be a closed linear subspace of H. Show that min { IU - v II v E S) = maxI I(u, w) II W E S -L and Iwi = 1).
THE TRIGONOMETRIC SYSTEM Let T be the unit circle in the complex plane. If F:T ~ C is any function on T then the function f defined by f(x) = F(e ix) for each x E R is a periodic function of R of period 2n, i.e., f(x + 2n) = f(x) for each x E R. Conversely, if f:R ~ C is periodic with period 2n, it is easily seen that there is a function F:T ~ C such that f(x) = F(e ix ) for each x E R. Therefore, we can identify the complex valued functions on T with the complex valued 2n periodic functions on R. Define U (D, where 1 ~ p ~ 00, to be the class of all complex valued, Lebesgue measurable. 2n periodic functions on R for which
Iflp =
[1/2nfr-Jr.Jr) If(x) Ipdx] lip
= IPn - II = = 0, and hence for each IE C(n and e > 0 there exists a trigonometric polynomial P such that I/(x) - P(x) I < e for each x E R. 15. Show that {un a basis for L l{n.
In E
Z} where Un is defined by un{x) = e inx for each x
E
R is
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11.3 The Spaces U (11) and Banach Spaces Hilbert space norms are fairly specialized, being based on the concept of an inner product. This is what accounts for there being so few of them, essentially one for each cardinal. The LP -spaces are a special case of a more general class of spaces called the Banach spaces. These spaces derive their name from Stefan Banach who published a series of papers between 1922 and 1932 in several journals that culminated in the now famous Theorie des Operations lineaires (Monografje Matematyczne, Volume 1, Warsaw, 1932). In the special case p = 2, L 2 (11) is both a Hilbert space and a Banach space. As we shall see from the definition of Banach spaces, this is because all Hilbert spaces are Banach spaces. A normed linear space X is a vector space over the complex field C (i.e., an abelian group in which multiplication of group members by complex numbers, called scalar multiplication, is defined that satisfies the distributive laws) in which a non-negative real number Ixl (called the norm of x) is associated with each x E X that satisfies the following properties: Bl. Ixl = 0 if and only if x = O. B2. Ix + yl :0:; Ixl + Iyl for each pair x,y EX. B3.laxl = lallxl foreachxE XandaE C. Here, Ia I denotes the modulus of the complex number a. A metric can be defined in a normed linear space in the following way. Define d:X x X ~ [0,00) by d(x,y) = Ix - y I for each pair x, y E X. That d is actually a metric is left as an exercise. If the metric space (X, d) is complete, X is said to be a Banach space. The simplest Banach space is merely C itself, normed by absolute value, i.e., Ix I is simply the absolute value (modulus) of x for each x E C. One can also discuss real Banach spaces by restricting the field of scalars to R. The topology induced on X by d is called the norm topology and the set xl Ixl :0:; I} is the closed unit sphere in X. A mapping T of a normed linear space X into a normed linear space Y is said to be a linear transformation if T(x + y) = T(x) + T(y) and T(ax) = aT(x) for each pair x,y E X and a E C (linear transformations are sometimes called vector space homomorphisms - the definition does not depend on the norm, only on the vector spaces X and Y). Linear transformations are also commonly referred to as linear operators. In the special case the space Y is the Banach space C, Tis referred to as a linear functional. The kernel of a linear operator is the set of all elements in X that get mapped onto the zero element of Y, Le., Ker(1) = {x E X IT(x) = O}. The proof of the following proposition is left as an exercise. S(O,I) = {x E
PROPOSITION 11.11 The kernel of a linear operator T:X ~ Y from the linear space X into a linear space Y is a linear subspace ofX.
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A linear transfonnation T is said to be bounded if there exists a real number a such that IT(x) I ~ a Ix I for each x E X. The smallest a with this property is called the norm of T and is denoted by ITI. It is easily seen that ITI = sup { IT(x)l/lxl I x E X and x ::f:. O}. Since IT(ax) I = I aT(x) I = Ia II T(x) I for each x E X and a E C, we could restrict ourselves to unit vectors (i.e., x E X such that Ix I = I) in the definition of the nonn of T. In this case, we would have ITI = sup{ IT(x)l/lxl I x E X and Ixl = I}. A bounded linear transfonnation T maps the closed unit sphere in X into the closed sphere 5(0,ITI)= lYE ylly-Ol ~ ITI}. To see this,letx E S(O,l)inX. Then Ixl ~ 1. Since IT(x) I ~ ITllxl, we have IT(x)l/lxl ~ ITI. Since Ixl ~ I this implies I T(x) I ~ ITI which in turn implies IT(x) - 0 I ~ I TI so T(x) E 5(0, ITI). THEOREM J 1.13 Let T:X ~ Y be a linear operator from a normed linear space X into a normed linear space Y. Then the following statements are equivalent: ( 1) T is bounded. (2) T is uniformly continuous. (3) T is continuous at some point of x. Proof: If T is bounded, IT(x) - T(y) I = IT(x - y) I ~ I Til x - y I for each pair x,y EX. Thenif£>Oand Ix-yl O. Then there exists a 0 > 0 such that Ix - z I < 0 implies IT(x) - T(z) I < Eo Then Ix I ~ 0 implies I z + x - xl ~ 0 which in tum implies I T(z + x) - T(z) I < Eo But IT(z + x) - T(z) I = IT(x) I so Ix I < 0 which implies IT(x) I < £. Hence IT(x) 1/0 > Ix I/£ which implies I T(x) II Ix I > 0/£ which in tum implies IT(x) II Ix I < £/0. Therefore, Ix I < 0 implies IT(x) II Ix I O. Since {Tn} is Cauchy, there exists a positive integer N such that m, n > N implies I Tm - Tn I < E/lxo I. Now ITm(xo) - Tn(xo) I ~ ITm - Tn II Xo I < E so {Tn(xO)} is Cauchy in Yand therefore conver-
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ges to some xo' E Y. Define T:X ~ Y by T(x) = x' for each x converges to T(x) for each x E X.
E X.
Then {Tn(x) I
To show T E L(X, Y), we must show T is linear and bounded. To show Tis linear, let x, y E X and a E C. Then T(x + y) = (x + y)' and T(ax) = (ax)'. Let E > O. Since {Tn (x) I converges to T(x) for each x E X, there is a positive integer N such that n > N implies ITn(x) - x'i < E/2 and ITn(y) - y'l < E/2. Now ITn(x+y)-(x'+y')l = ITn(x)-x'+Tn(y)-y'l::; ITn(x)-x'l + ITn(y)-y'1 <E. Therefore, n > N implies I Tn(x + y) - (x' + y') I < E so {Tn(x + y) I converges to (x' + y'). Hence (x + y)' = (x' + y') so T(x + y) = T(x) + T(y). Also, there exists a positive integer M such that n > M implies ITn (x) - x'i < E/lal which in tum implies ITn(ax) - ax' I < E. Therefore, {Tn(ax) I converges to ax' so (ax)' = ax' which implies T(ax) = aT(x). Consequently, T is linear. To show that T is bounded, let E > O. Since {Tn I is Cauchy, there exists a positive integer M such that m, n > M implies ITm - Tn I < E. Pick m > M. Then ITml M, ITnl = ITn-ol::; ITn-Tml + ITm-ol N implies ITn(x)-T(x)1 <E. Now ITn (x) I ::; ITnl1xl O. Then there exists a positive integer M such that m, n > M implies ITm - Tn I < E/4. Suppose there is no n > M with ITn - TI < E/2. Then for each n > M, ITn - TI :::>: £/2. Let m > M. Then sup{ ITm(x) - T(x) I I Ixl = 1 I :::>: £/2 which implies there exists an x E X with Ixl = 1 such that ITm(x) - T(x) I :::>: £/2. Now, for each n > M, ITm(x) - Tn(x) I ::; ITm - Tn II X I < £/4. Therefore, for each n > M, 1 Tn (x) - T(x) I :::>: £/4 for otherwise, 1 Tm(x) - T(x) 1 ::; ITm(x) - Tn(x) I + ITn (x) - T(x) I < £/2 which is a contradiction. Hence {Tn (x) I does not converge to T(x) which is a contradiction. Therefore, there exists a k > M with Tk - TI < £/2. But then for eachn>M, ITn-TI::; 1Tn-Tkl + ITk-TI <Eso{TnlconvergestoT.1
L(X, Y), where Y is a Banach space, is our first example of a Banach space whose elements are functions. There are many more function spaces that are Banach spaces. It is easily shown that for each 1 ::; p ::; <Xl that U (I!) is a Banach space with respect to the LP -norm. If Z is a dense linear subspace of X and Y is a Banach space, then by Theorem 1.16, if T E L(Z, y), T has a unique extension to a member T of L(X, Y). THEOREM 11.15
ITI = ITI.
Proof: Clearly, 1 TI ::; 1 T I from the definition of the I I-norm. Suppose ITI < 1T I. Then there exists an x E X - Z with 1x 1 = 1 such that 1T(x) I > ITI. Since Z is dense in X, there exists a sequence {zn I c Z that converges to x. For
11.3 The Spaces U()l) and Banach Spaces
343
each positive integer n, put Wn = anzn where an = 111 Zn I. Then {w n } is a sequence of unit vectors in Z that also converges to x. This follows from the fact that Ix - Wn I ~ Ix - Zn I + IZn - Wn I and both terms on the right hand side of this inequality can be made arbitrarily small if n is large enough. IZn - Wn I = 1(1 - an)zn I = 11 - an II Zn I = I(IZn I - 1) I and IZn I ~ IZn - x I + Ix - 0 I = Izn-xl + 1. This last term converges to 1 as n ---7 so {I Zn I} converges to 1 which implies { IZn - Wn I} converges to O. 00
Therefore, IT'(w n ) I = IT(w n ) I ~ ITI for each positive integer n. Since 1 is continuous, {T'(w n )) converges to T'(x). But this is impossible since IT'(x) I > ITI and IT'(w n ) I ~ ITI for each n. We conclude that 111 = ITI. The complex numbers C, normed by their absolute value form a Banach space. L(X, C) is called the dual space of X and is denoted by X*. L(X, C) is a Banach space by Theorem 11.14. The interplay between X and its dual space X* is the basis of much of that field of mathematics known as functional analysis. To explore this interplay, one needs the Hahn-Banach Theorem. The Hahn-Banach Theorem essentially says that if Y = C, then we can drop the assumption that Z be dense in X. Linear transformations from Z into C can then be extended to X in such a way that Theorem 11.15 still holds. By a real linear functional, we mean a linear operator from a real vector space (vector space over the real field). Letfbe a complex linear functional on a linear space X. Then for each x E X, f(x) = u(x) + iv(x) for some real valued functions u and v on X. Since X is a vector space over the complex field, it is clearly a vector space over the real field as well. It is easily seen that the linearity of f implies the linearity of u and v, i.e., u and v are real linear functionals. PROPOSITION 11.12 If X is a linear space and f is a linear functional on X then: (1) Ifu is the real part offthenj(x) = u(x) - iu(ix)for each x EX. (2) Ifu is a real linear functional on X andf(x) = u(x) - iu(x)for each x E X, then f is a (complex) linear functional on X. (3)1fX is a normed linear space andf(x) = u(x) - iu(ix)for each x E X, then If I = Iu I. Proof: If a, ~ E R and 'Y = a + i~, then the real part of i'Y is -~. Therefore, y Re(y) - iRe(iy) for each y E C. Then (1) follows with Y=f(x). To show (2), it is clear that f(x + Y) = f(x) + f(y) and that f(a.x) = af(x) for each a E R. But we must also show this second equation for a E C. It will suffice to show it for a = i. For this note thatf(ix) = u(ix) - iu( -x) = u(ix) + iu(x) = if(x).
To show (3) note that Iu(x) I ~ If(x) I for each x E X which implies Iu I ~ E x. Put ~ = f(x)/lf(x) I. Then I ~ I = 1 and ~ If(x) I = f(x). Put a = Ial = 1 and af(x) = If(x) I. Hence If(x) I =f( ax) which implies
If I. Let x ~-l. Then
11. Spaces of Functions
344
f(ax) isreal sof(ax) = u(ax)-S: lullaxl. Then If I =sup{lf(x)lllxl =1}-S: Iu I. Therefore, If I = Iu I. One of the most important theorems in the theory of Banach spaces is the
Hahn-Banach Theorem. It allows us to extend bounded linear functionals on subspaces of a normed linear space in such a way that the norm is preserved. THEOREM 1I.16 (S. Banach, H Hahn, 1932) ffY is a subspace ofa normed linear space X and iff is a bounded linear functional on Y, thenf can be extended to a bounded linear functional F on X such that IF I = If I. Proof: We first prove the theorem assuming the field of scalars to be real, i.e.,
we assume X is a real normed linear space and f is a real bounded linear functional on Y eX. If If I = 0 then clearly, the desired extension is F = O. If If I > 0, we may assume, without loss of generality, that If I = 1 since if If I of. 1, there exists an a E R with Iallfl = 1 which implies Iafl = 1. We can then prove the theorem for af and simply divide the extension F by Ia I. Assuming If I = 1, pick z E X - Y and let N be the subspace of X spanned by z and Y (i.e., N = {x + azlx E Yand a E R}). DefinefN:N ~ R by fN(X + az) =f(x) + aA for any fixed A E R. It is left as an exercise to show that fN is a linear functional on N such that fN restricted to Y is f. We next show that it is possible to pick A such that IfN I = 1. For this, first note that by the definition of I I, that If I -s: IfN I. Also,
so that if If(x) + al..I -s: Ix + azl for each x E Y and a E R, then IfNI -s: 1. Hence If(x) + al..I + Ix + azl for each x E Yand a E R which implies IfN I = 1. Therefore, it suffices to show that A can be chosen such that If(x) + aA I -s: Ix+az I for each x E Yand a E R. This can be shown if we choose A such that If (x)+aA
Ia
I
I
-s:
I Haz I Ia I
for each x E Y and a E R.
But this is equivalent to showing that we can find a A such that If(x/a) + AI -s: Ix/a + zl which in tum is equivalent to finding aWE Y such that If(w) - AI -s: IW - z I in view of the substitution w = -x/a. For each W E Y put )'(w) = few) Iw-zl and P(w) =f(w) + Iw-zl. If),(w)-S:A-s:P(w),then If(w)-AI-s: Iw-zl. Hence IfN I = 1 if )'(w) -s: A -s: pew) for each W E Y To show this, let I(w) be the interval [)'(w), pew)] for each w E Y. Then IfN I = 1 if n{ I(w) IWE y} of. 0, or equivalently if )'(w) ~ P(v) for each pair W,v E Y. Now)'(w)-P(w)=f(w-v)-Iw-zl-Iz-vl. Sincef(w-v)~ If(w-v)l:5 If I Iw - v I = I w - v I ~ Iw - z I + Iz - v I, )'(w) - P(v) ~ 0 so )'(w) ~ ~(v) for each
11.3 The Spaces LP(Il) and Banach Spaces
345
pair w,v E Y. Hence n{I(w) IWE Y} # 0. Pick A E n{I(w) IWE Y}. Then y(w) : 0 such that IQ.g(x) I ~ 0 for each Q.g E G and every x E U. Let u E U. Since U is open there exists an e > 0 such that CI(S(u, e)) c U. Let b = (0 + bu)/e. To show If I ~ b for each f E F let f E F and x E X with Ixl = 1. Then I (ex + u) - ui = lexl = elxl = £. Hence (ex + u) E U. Therefore, If(x) I = If(e- 1 ex) I = e- 1 If(ex) I = e- i If(ex + u - u) I ~ c 1 (If(ex+u) I + If(u) I) ~ e-- 1 (0 + bu) = b. Consequently, If I = sup{ If(x) Illx = I} ~ b for eachfE F.· COROLLARY J J.8 (Banach-Steinhaus Theorem) If {fn} is a sequence of continuous linear operators from a Banach space X into a normed linear space Y that converges pointwise to an operator f:X --7 Y, then f is also a continuous linear operator.
Proof: Since limits are unique in Y, it is easily seen that f is a linear operator. Let x E X. Since {fn(x)} converges to f(x), it follows that b x = sup { Ifn(x) I} < 00. By the Uniform Boundedness Theorem, there exists abE R such that Ifn I ~ b for each n. Hence, for each x E X, If(x) I = limn Ifn(x) I ~ blxl since the distance function d(x, y) = Ix - y I is continuous in Y. But then f is bounded and therefore continuous. •
EXERCISES 1. Prove Proposition 11.11. 2. Prove Lemma 11.3. 3. Let I = [0, 1] and let f E £P (I) for p > 1. Show that there exists agE L q (I), where p and q are conjugate exponents, such that If Ip = I g Iq and f(
PROPOSITION 11.14 Let X be a space and let F = yX where (Y, v) is a uniform space. Let A c X and let 1t and 1tA denote the topologies of pointwise convergence on X and A respectively. Then (1) A subbase for 1t on F is the family of all sets of the form {fE Flf(x) E VI for some pOint x E X and open set V c y. (2) A subbase for 1tA on F is the family of all sets of the form {f E F If(x) E V} for some point x E A and open set V c y. (3) 1tA is the smallest topology for F that makes the restriction mapping p:F -t yA continuous. Proof: To prove (1) note that sets of the form p~~ (W I)" ... p~~ (Wn ) for some Wn and projections Pal' .. Pan (where for each i, Paj is the open sets WI projection of F onto the a;h coordinate subspace of yX = F), form a basis for 1t since 1t is the product topology on F. Now for each i,p~~(W;) = {fE FIPaj(f) E 0
••
11.4 Uniform Function Spaces
357
Wi) = (fE Flf(ai) E Wi)' Hence the sets of the form (fE Flf(x) E U) for some point x E X and open set U c Y forms a subbase for 1t on F. To prove (2), first note that by (l) above, sets of the form (f E yA If(y) E V} for some point YEA and open set V c Y form a subbase for the 1t topology on yA" Since 1tA is a uniformity that makes the restriction mapping p:F ~ yA uniformly continuous, each set of the form p~1 ({f E yA If(y) E V}) for some Y E A and open set V c Y belongs to the 1tA topology on F. Now p~I(lfE
yA
If(y) E V))
= (fE
Flf(y) E V},
so sets of the form (f E F If(y) E V} for some YEA and open V c Y belong to the 1tA topology on F. To show these sets form a subbase for the 1tA topology on F we need to show that for each open U in 1tA and g E U, there exists a finite collection of sets {f E F If(y dE V d ... (f E F If(yn) E Vn } for some points Y I ... Yn E A and open sets V I ... Vn contained in Y, such that g E n;=1 If E Flf(yi) E V;} C U. Since U E 1tA on F, there exists a U E 1tA (the uniformity) such that g E S(g, U) cU. Since 1tA is the smallest uniformity on F that makes p uniformly continuous, there exists a finite collection of open coverings V I ... Vn E 1t on yA such that W = n;=1 p~1 (V;) < U. Then g E S(g, W) cU. Now S(g, W) = If E Flf, g E W for some W E W} = (fE Flf, g E n;=IP~I(V;) for some VI ... Vn that are open in Y}. Therefore, S(g, W) = n;=1 (fE Flf, g E p~I(Vi)} for some VI ... Vn C Y which implies S(g, W) = n;=1 {fE Flf(Yi), g(Yi) E Vi} for some points Y I ... Yn E A and open sets V I ... Vn in Y. Therefore, sets of the form (fE Flf(y) E V} for some YEA and open V c Y form a subbase for the 1tA topology on F. To prove (3), note that since the 1tA topology on F is generated by the 1tA uniformity on F and the 1tA uniform is the smallest that makes p uniformly continuous, it is clear that the 1tA topology on F makes p continuous. Let 1 be another topology on F that makes p continuous. Then for each YEA and open set V c Y, the set (f E yA If(y) E V} is 1t-open in yA which implies
But then 1 contains a subbase for 1tA which implies 1tA is coarser than 1. Hence is the smallest topology for F that makes p continuous. -
1tA
From the definition of the product uniformity in Chapter 5 and by Exercise 5.2, it is seen that the coverings of the form 1tai(V) (where V E V and 1ta denotes the projection of F onto the alh coordinate subspace of yX = F) determine a subbase for the 1t uniformity on F. For a particular a E X and V E V we have 1t a (V) = 11ta'(V)I V EV} = {{fE Flf(a) E V} I V E VJ. Hence for g E F we have S(g, 1ta'(V» = (fE Flf(a), g(a) E V for some V E VJ.
358
11. Spaces of Functions
PROPOSITION J J.15 Let X be a space and leI F = yX where (Y, v) is a uniform space. Let A c X and leI TtA denote the uniformity of pOintwise convergence on A on F. Then (1) Coverings of the form Va = {If E F Ij(a) E VI I V E V I where a E A and V E V form a subbase for the TtA uniformity on F. (2) A net {f~ I c F is Cauchy with respect to TtA if and only if (f~(x)1 is Cauchy in Yfor eachx EA. (3) If (Y. v) is complete and p( F) is closed in yA with respect to the Tt topology on yA. then F is complete with respect 10 the TtA uniformity Proof: By the definition of the TtA unifonnity on F. sets of the form p-l (V) where V is a member of the Tt unifonnity on yA fonn a subbase for the Tt;>. unifonnity on F. For a particular V in the Tt unifonnity on yA we have that V is refined by some n£=i Tta~ (V;) for some a] .. an E A and VI .. Vn E v. Hence p-I (V) is refined by ni'=1 p-I (Tta; (V;), For each i = 1 " n,
Hence coverings of the fonn Va = {{f E F If( a) E V} IV and V E V fonn a subbase for the TtA uniformity on F.
E
V l where a
E
A
To prove (2) first assume {f~ I is Cauchy in F with respect to Tt;>.. Then for each covering of the fonn Va = {{f E F If( a) E V I IV E V I there exists a residual set R such that {f~ I ~ E RI C (fE Flf(a) E VI for some V E V. But then {f~(a)l~ E RI c V so {f~(a)1 is Cauchy in (Y, v). Hence {f~(x)1 is Cauchy in (Y, v) for each x E A. Conversely, assume (f~(x) I is Cauchy in (Y,v) for each x E A. Let Va be a subbasic member of TtA' For some V E V there exists a residual set R such that (f ~ (a) I ~ E RIc V since {f~ (a)l is Cauchy in (Y, v)o But then (f~ I ~ E RI c (f E Flf(a) E VI so tf~ i is Cauchy with respect to Tt;>. . To prove (3), assume (Y, v) is complete and p(F) is closed in yA with respect to Tt. Let {f13) be a Cauchy net with respect to TtAo Then {p(f~)1 is a Cauchy net in yA with respect to Tt since p is uniformly continuous. Since p(F) is closed in yA with respect to Tt and yA is complete with respect to Tt, since Tt is the product unifonnity on yA, {p(f~) i converges to some f E p(F) with respect to Tt. Let g E F such that peg) = f. Let V E TtAo Then there exists some al an E A and WI . Wn E Tt such that W = ni=IP-I(Tta~(W;) < Vo Now S(g,W) c S(g, V) and since {p(f~)} converges to f, there exists a residual set R with {p(f13) I ~ E R} c Tta~ (W;) for some Wi E W with f E Tta~ (W;) for each i = 1 no But then I p(f13) I ~ E R} c ni=ITta~ (W;) which implies If13 1 ~ E R} c ni'=l p-l (Tta~ (W;)o Moreover, g E ni'=l p-l (Tta~ (W;) c S(g, W)o Hence {f13 1 ~ 000
0
0
0
0
0
11.4 Uniform Function Spaces
E R} c S(g, V). Therefore, {fs} converges to g with respect to complete with respect to 1tA' •
359 1tA
so F is
While the 1t uniformity for F is one of the most simple, it is not the most interesting. In fact, the 1t topology, when Y = R, is so coarse that the characteristic functions of finite sets converge to the unit function. For each finite F c X let XF denote the characteristic function of F. Since the family of all finite subsets F of X is directed by set inclusion, the family {XF} is a net. Moreover, {XF} converges to the unit function 1 defined by 1(x) = 1 for each x E X. To see this, observe that for each x E X, {x I is a finite set so if F is a finite subset of X containing x then XF(X) = 1. Hence by Proposition 11.14.(1) {XF} is eventually in each subbasic open set containing 1. A topology for which the characteristic functions of finite sets get arbitrarily near the unit function is too coarse for investigating the problems we will be interested in throughout the remainder of this section. A more interesting uniformity for F is the uniformity of uniform convergence which we will denote as the uc uniformity. The uc uniformity is also independent of the topology of X, but it has the interesting (and useful) property that the family of continuous functions C from X to Y is closed in F. In other words, with the uc uniformity, the uniform limit of continuous functions is also continuous. The uc uniformity will be the finest uniformity we will consider for F. Let V E v. For each f E F define the ball about f of radius V, denoted B(f, V) to be the set {g E F I for each x E X, g(x), f(x) E V for some V E V}. Put V F = (B(f, V) If E F}. Let uc denote the family of all coverings of F that are refined by coverings of the form V F where V E v. If V F E uc then S(f, V F) = {g E Flf, g E B(h, V) for some h E F} = {g E FI for each x E X,f(x), hex) E VI and g(x), hex) E V 2 for some VI, V2 E V, for some h E F} = ig E Flfor each x E X, f(x), g(x) E S(h(x), V) for some h E F}. Since hex) ranges over all of Y as h ranges over F we have that S(f, V F ) = B(f, VLl) where V Ll = {S(y,v) Iy E Y} (i.e., V'" is the covering made from spheres with respect to V). If U F and V F are members of uc then clearly there exists aWE v such that W"'* < Un V (just pick W such that W** < Un V). Now
(Star({g E FI for each x E X, g(x),f(x) E W for some W E WI, W F ) if EO F} = { (h E F Ifor each x E X, hex), g(x) E W j and g(x),f(x) E W 2 forsomegEFandsomeW j ,W 2 E W}lfEF} = { (h E F Ifor each x E X, hex) E Star(S(f(x), W» If E Fl.
Since f(x) ranges over all of Y as f ranges over F we have:
360
w~
11. Spaces of Functions ({h E FI for each x E X, h(x),f(x) E W for some W E W il* lifE Fl.
Therefore, W~ = (B(f, Wil*) If E FI, Since W il* < UnV we have that B(f,Wil*) c B(f,U) and B(f, Wil*) c B(f, V) for each f E F, so W~ < U F and W~ < V F. Hence W~ < UFnV F , so uc is a unifonnity for F. We call uc the uniformity of uniform convergence and the topology generated by uc the topology of uniform convergence or simply the uc topology. PROPOSITION 11.16 uc is finer than 1t Proof: For a subbasic unifonn covering 1tul (V) in 1t and each f E F, the sphere S(f,1t ul (V» = (g E Flf(a), g(a) E V for some V E VI contains the set (g E F Ifor each x E X, f(x), g(x) E V for some V E V I = B(f, V). Therefore, VF < 1tul (V)il. Then for a subbasic covering 1tul (U) in 1t, there exists a V E v such that V* < U so 1tul (V) 0 and Sd(X, r) denotes the r-sphere about x with respect to a bounded member d of the family of pseudo-metrics that generates v, fonns a basis for v. Now V(d, r)F = (B(f,V(d, r» If E F} = {{g E Filar each x E X, g(X),f(X)E Sd(y,r)forsomeYE Y}lfE F}. Now d(f,g) = sup{d(f(x),g(x»lx EX} so if d(f, g) ~ r then for each x E X, f(x), g(x) E Sd(f(X), r) so V(d, r)F contains { (g E FI d(g,f) ~ r} If E FI = (Sd(f, r) If E F} Therefore, (Sd(f, r) If E F} < V(d, r)F so the family of pseudo-metrics of the form d(f, g) = sup{ d(f(x),g(x» Ix EX} where d is a bounded member of the family of pseudometrics that generates v, generates the uc unifonnity.
To prove (2) assume {fa} is a net in F that converges unifonnly to some f E F. Then for each V E v there exists a residual set R such that for each a E R, and x E X,f a (x),f(x) E Yea, x) for some Yea, x) E V. Therefore,fa E B(f, V) for each a E R. Since S(f, V F) = B(f, Vil ), fa E S(f, VF) for each a E R. Hence {fa} is eventuall y in VF for each V E V so {fa} is Cauchy with respect to uc. Moreover, since for each V E v, there exists a residual R such that for
11.4 Uniform Function Spaces
361
each a E R,fa(x),f(x) E V for some V E V, we have that {f a (x) ) is eventually in S(f(x), V) for each V E V so {fa(x») converges to f(x) for each x E X. Conversely, assume {fa) is Cauchy with respect to uc and that for each x E X, {fa(x») converges tof(x). Let V E v. Pick U E v such that U** < V. Since {fa) is Cauchy with respect to uc, there exists a residual set R such that {fa I a E R) c B(g, U) for some g E F. Then for each x E X and a E R,fa(x), g(x) E U(a, x) for some U(a, x) E U. Therefore,fa(x), g(x) E S(g(x), U) for each a E R and x E X. Now for each x E X, {fa(x)} converges to f(x) so there exists a Wx E U containing f(x) and a residual Rx such that {f a(x) Ia E Rx} c Wx' Now RxnR ~ 0 so pick ~ E RxnR. Then fj3(x) E Wx and fj3(x), g(x) E U(~, x). But f(x) E Wx and f a(x) E S(g(x), U) for each a E R and x E X. Therefore,fa (x),f(x) E S(g(x), U)uW x for each x E X which implies f a(x),f(x) E Star(S(g(x), U» for each x E X. But Star(S(g(x), U» c Vx for some Vx E V since U** < V. Then for each a E R and x E X,f(x),f a(x) E Vx for some Vx E V so {fa) converges uniformly to f To prove (3), let {fa) be a Cauchy net in F with respect to uc. Let V E v. Then VF is a basic open covering in uc, so there exists agE F and a residual set R such that {fa I a E R} c B(g, V). Then for each a E R and x E X, f a(x), g(x) E Yea, x) for some Yea, x) E V.
Fix y E X. Then f a(y) E S(g(y), V) which implies {fa(y)} is Cauchy in Yand therefore converges to some Px E Y for each x E X. Define f E F by f(x) = Px for each x E X. Then by (2) above, {fa) converges to f since {fa) is Cauchy with respect to uc and {fa(x») converges to f(x) for each x E X.· Notice that in (F, uc), convergence of a net {fa} to f E F implies uniform converge of {fa} to f This is because ~f {fa) converges to f with respect to uc then {fa} is Cauchy with respect to uc and since uc is finer than n by Proposition 11.16, {fa) converges to f with respect to n. But then {fa(x») converges to f(x) for each x E X so by Proposition 11.17.(2), {fa} converges uniformly to fwith respect to uc. THEOREM 11.24 Let X be a space and let F = yX where (Y, v) is a uniform space. Let C c F be the family of continuous functions from X to Y. Then C is closed in F with respect to the uc topology and hence (c, uc) is complete if (y, v) is complete.
Note that by the previous remarks, this theorem asserts that the uniform limit of a net of continuous functions from X to Y is itself continuous. Proof: Let G = F - C and let g E G. Then there exists an Xo E X such that g is not continuous at x o· Therefore, there exists a V E v such that g -I (S(g(x 0), V» is not a neighborhood of Xo· Choose U E v such that US < V (where US = U*****). Let f E S(g, Up) = B(g, U"i). Then for each x E X, I(x), g(x) E
11. Spaces of Functions
362
S(y,U) for some Y E Y. Hence f(x) E Star(S(g(x), U» and g(x) E Star(S(f(x),U» for each x E X. Consequently, if Z c X then for each z E Z, g(z) E S'tar(S(f(z),U» so z E g-l (Star(S(f(z), U), U) which implies
Z
C
g-l
(Star(fIZj, U*» for each Z c X.
Substituting r l (S(f(x), U» for Z we get
r
l
(S(f(x), U» c
g-I
(Star(fff-I (S(f(x),U»]. U*»
= g-I (Star(S(f(x), U), U*».
Since g(x) E Star(S(f(x), U» we have that S(f(x), U) c Star(S(g(x), U), U*). Hencerl(S(f(x), U» c g-I(Star(Star(S(g(x), U), U*, U*» c g-I(S(g(X), V» since US < V. But then r l (S(f(xo), U» cannot be a neighborhood of x 0 since g -I (S(g(x 0)' V) is not a neighborhood of x 0 so f is not continuous at x o. Consequently, S(g, Up) c G so G is open which implies C is closed. • Let 't be a topology for the function space F(X, Y) where (y, v) is a uniform space. Then we can consider the product space F x X and the mapping e:F x X ~ Y defined by elf, x) = f(x) E Y, for each (f, x) E F x X. The topology 't is said to be jointly continuous if the mapping e is continuous. Usually the topology 1t of pointwise convergence is not jointly continuous (see Exercise 1). However, if we restrict our attention to the continuous members of F that we denote by C, then the discrete topology is always jointly continuous, i.e., e:C x X ~ Y is continuous (see Exercise 2). Consequently, there is a finest topology on a family of functions that is jointly continuous. The opposite problem of finding a coarsest topology on a family of functions that is jointly continuous also has a solution sometimes as we will see a little later. However, in general, there is no coarsest topology for a family of functions that is jointly continuous. If A c X, a topology for F is jointly continuous on A if eA = e I A is continuous on F x A. If A is a family of subsets of X it may turn out that a topology for F is jointly continuous on each member of A E A. It will be shown that the uc topology is jointly continuous on C c F and that some interesting uniformities can be constructed on F by considering the uniformity of uniform convergence on each member A E A. These uniformities then generate topologies that are jointly continuous on each A E A. Joint continuity will play an important role in studying the so-called uniformity of uniform convergence on compacta on F. which we shall now begin. PROPOSITION 11.18 Let X be a space and let C be the family of continuous functions from X (0 a uniform space (y, v). Then the uc topology on C is jointly continuous. Proof: To show the continuity of e:C x X ~ Y at some point (f, x) E C x X, let V E V. It suffices to show there exists a basic open set W x U c C x X
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363
containing (j, x) such that e(W x U) c S(j(x), V). For this, first observe that e-1(S(j(x), V) = leg, z) E C x xlg(z) E S(j(x), V)}.
r
Pick U E v such that U** < Vo Put U = 1 (S(j(x), U». U is open in X since f is continuous and x E U. Also, W = B(j, U) is open in C and contains f. Therefore, W x U is a basic open set in C x X containing (j, x). Let (g, z) E W X U. Then g E B(j, U) and z E 1(S(j(x), U» so j(z),f(x) E U x for some U x E U and for each y E X,f(y), g(y) E U y for some U y E U. Hencef(z), g(z) E Uz for some Uz E U. Then f(x), g(z) E UxuU z so g(z) E Star(S(j(x), U» c S(j(x), V). Hence e(W x U) c S(j(x) , V) so e is continuous at (j, x). But then e is continuous at each (g, z) E C X X so uc is jointly continuous on C.·
r
Let us now consider some uniformities for F constructed by considering the uc uniformity restricted to some subsets A of X belonging to a family A. For each A E A let PA :F -t Y A be defined by PA if) = fA = f I A. Then define the uniformity uc IA to be the coarsest uniformity on F such that for each A E A, PA is uniformly continuous with respect to uc I A and the uc uniformity on FA = yA. PROPOSITION 11.19 A subbase for the uclA uniformity on F is the famity of all coverings oftheform VF(E) = {BE(j. V) IfE FE} where E E A and BE(f, V)={gEFlforeachxE E,g(X).j(X)E VforsomeVE V}andFE=yE.
The proof of Proposition 11.19 is left as an exercise. We are now in a position to state the central structural theorem of this section. Its proof is also left as an exercise. THEOREM 11.25 Let X be a space and let F be thefamity offunctions from X to a uniform space (Y. v). Let C A be thefamity of all functions in F that are continuous on each E E A where A is a family of subsets of X that covers X. Then (1) uc I A is finer than rt on F but coarser than uc. (2) A net Ifa} in F converges to f E F with respect to uc I A if and only if it is Cauchy with respect to uc IA and converges to f with respect to rt. (3) If (Y, v) is complete then (F. uc I A) is complete. (4) C A is closed in F with respect to uc IA. so if (Y. V) is complete then so is (CA , uc IA). (5) uc IA on CA is jointly continuous on each E E A.
If S is the family of singleton sets {x} such that x E X then uc IS = rt and the family C c F of continuous functions is generally not complete with respect to rt. Consequently, Theorem 11.25 does not allow us to say anything about C. However, if the continuity of each f E F on each member of A implies the con-
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tinuity of f on X then C = CA so by Theorem 11.25.(4) (C, uc IA) is complete if (Y, v) is complete. We now turn our attention to the collection K of compact subsets of X.
uc I K is called the uniformity of uniform convergence on compacta. We will denote uc IK by 1(. The topology generated by 1( is called the topology of compact convergence or simply the 1( topology. It turns out that on the family C of all continuous functions from X to Y that the 1( topology is identical to an important function space topology that is not dependent on the concept of a uniform structure on the function space. This topology is known as the compact-open topology which we now define. If K is a compact set in X and U is open in Y we write U(K) to denote the set {fE Flf(K) c U}. The family of all sets of the form U(K) where U is open in Y and K is compact in X is a subbase for the compact-open topology.
THEOREM 11.26 Let X be a space and let C be the family of all continuous functions from X to a uniform space (Y, v). Then 1( is the compactopen topology on C. Proof: Let f E C and let VcCK) be a subbasic covering in 1( for some compact set K c X and some V E v. Then VcCK) = {BKCg, V)lg E C K } where in this case CK denotes C(K, Y) [rather than the family of continuous functions with compact support]. ThenS(f, VcCK»= {gE Clg,jE BK(h, V)forsomehE C} = {g E C I for each x E K, g(x), hex) E V 1 and f(x), hex) E V2 for some V!, V2 E V and some h E C} = {g E C I for each x E K, g(x) E Star(S(f(x), V), V) I = {g E C Ig(K) c Star(S(f(x), V), V)}. Put W = Star(S(f(x), V), V). Then S(f,VcCK» = W(K) so S(f, Vc(K» is open with respect to the compact-open topology on C. Therefore, the compact-open topology is finer than the 1( topology. Conversely, if K is compact in X and U is open in Y then U(K) is a subbasic member of the compact-open topology on C: Let f E C such that f E U(K). Then f(K) is a compact subset of U so there exists a V E v such that Star(fIK], V) c U. Let U E v such that U* < V. If g E S(f, UcCK» thenf, g E BK(h, U) for some hECK' Then as we saw above, g(x) E Star(S(f(x), U» for each x E K so g[K] c Star(S(f(x), U), U) c S(f(x), V) for each x E K which implies g[K] c Star(fIK], V) cU. Therefore, S(f, UcCK» c U(K) so U(K) is open in the 1( topology. Hence, 1( is finer than the compact-open topology, so 1( is the compact-open topology on C.A topology for F is said to be jointly continuous on compacta if it is jointly continuous on each compact subset of X. In a paper titled On topologies for function spaces, published in the Bulletin of the American Mathematical Society in 1945(Vol. 51, pp. 429-432), R. H. Fox showed that the compact-open topology on C is coarser than each jointly continuous topology for C and that
11.4 Uniform Function Spaces
the compact-open topology on C compact.
365 IS
jointly continuous itself if X is locally
THEOREM 11.27 Let X be a space and let F = yX for a uniform space (Y, v). Each topology for F that is jointly continuous on compacta is finer than the compact-open topology on F. If X is regular or Hausdorff and CK is the family of functions f E F such that f is continuous on each compact subsec of X, then the compact-open topology on CK is jointly continuous on compacta. Proof: Let 't be a topology for F that is jointly continuous on compacta. Let U(K) be a subbasic member of the compact-open topology on F for some open U c Y and compact K c X. Let e be the mapping on F x X to Y defined by e(j;x) = f(x)o Put W = [F x X)ne- 1 (U). Then W is open in F x K since 't is jointly continuous on compacta. If f E U(K) then If} x K c W. Since {fi x K is compact, there exists a 't-open set V c F containing f such that V x K c e -1 (U) (see Exercise 5). If g E V then g[K] c U so g E U(K). Therefore, V c U(K) so for each f E U(K), there exists a V E 't such that f EVe U(K). Hence U(K) is 't-open so 't is finer than the compact-open topology on F.
Next assume X is either regular or Hausdorff and let K be a compact subset of X. Let (j, x) E CK X K. Then f E CK and x E K. Let U be an open set in Y containing f(x) = e(j, x). Since f is continuous on K, there exists an open set V in X containing x such that fey) c U. Since X is either regular or Hausdorff, there exists a compact neighborhood W of x withf(W) cU. Then U(W) x W is a neighborhood of (j, x) in C K x K and e(U(W) x W) c U. Hence e is continuous at x E K. But then e is continuous on K so the compact-open topology on CK is jointly continuous on K. Therefore, the compact-open topology on C K is jointly continuous on compacta.COROLLARY 11.9 If X is a regular locally compact space and C is the family of continuous functions from X to a uniform space (Y, v), then the K topology (compact-open topology) is the coarsest jointly continuous topology on C. THEOREM J 1.28 Let X be a regular of Hausdorff space and let CK be the family of all functions from X to a uniform space (Y, v) that are continuous on each compact subset of X. Let K and 1t denote the compact-open and the product topologies respectively. Then a subfamily E ofCK IS K-compact in CK if and only if (1) E is closed in CK with respect to K. (2) E[x] has a compact closure in Y for each x E X. (3) The 1t topology on the 1t-closure of E in F is jointly continuous on compacta. Proof: First assume E is K-compact in CK • Then E is closed with respect to K in
366
11. Spaces of Functions
C K so (1) holds. Let x E X. Then E[xj = (f(x) If E E} = (nx(f) If E E} = nAE). Since the Xlh projection is continuous with respect to the product topology, and hence continuous with respect to K. Efxj is compact so (2) holds. By the remarks following Theorem 11.23, since E is compact, the K and n topologies are identical on E. Consequently, E is closed in F with respect to n. Thus the n-closure of E is E. By Theorem 11.27, K is jointly continuous on compacta on E so n is jointly continuous on compacta on E. Hence n is jointly continuous on compacta on the n-closure of E so (3) holds.
Conversely, assume (1), (2) and (3) hold. Let E* denote the closure of E in F with respect to n. By (2), Cly(E[xJ) is compact for each x E X so DxExCly(E[x]) is compact. Since E* c DxExCly(E[x]), E* is compact with respect to n. By (3), n is jointly continuous on compacta on E* Let f E E* and let K be a compact subset of X. Then e:E* x K ~ Y is continuous at each point (f, x) such that x E K. Now e I !XK:{f} x K ~ Y is defined by e i !XK(f, x) = e(f,x) = f(x) so we can identify e I ! xK with f. Since e is continuous on If} x K,
f is continuous on K. Therefore, f is continuous on each compact subset of X, so E* C CK . By Theorem 11.27, since n is jointly continuous on compacta on E*, n is finer than K on E* so n is identical with K on E*. But then E* is compact with respect to K. By (1), E is closed in C K with respect to K so E is closed in C K with respect to n. Hence E = E* so E is compact with respect to K.
-
We now turn our attention to a concept that can replace condition (3) in Theorem 11.28, and which allows us to strengthen the conclusion of the theorem to the function space C rather than CK to get a uniform version of the classical Ascoli Theorem. This concept is called equicontinuity and it applies to afamily of mappings as does the concept of uniform convergence. A family of mappings E from a space X into a uniform space (Y, v) is said to be equicontinuous at a point x E X if for each y E v there exists an open set U c X containing x such thatf(U) c Star(f(x), V) for each fEE. In other words, for each V E v there exists a neighborhood U of x whose image under any fEE is V"'-small. E is said to be an equicontinuous family if E is equicontinuous at each point x E X. Clearly, equicontinuity implies continuity for each member of the family. We divide the proof of the Ascoli Theorem into three lemmas that are of significant interest themselves.
LEMMA 11.7 If E is equicontinuous at x, then the closure of E with respect to n in F is also equicontinuous at x. Hence if E is an equicontinuous family, then so is the n-closure in F. Proof: Let E* be the closure of E in F with respect to n. Let V E V and let U E v such that U* < V. Since E is equicontinuous at x, there exists an open U c X containing x such that feU) c S(f(x), U) for each fEE. Let H be the family of all members h E F such that h(U) c CI(S(h(x), U». Clearly E c H. Now
11.4 Uniform Function Spaces
H
=
{h
E
367
F I for each
"uEu{h
E
U E
V, h(u)
FI h(u) E
E
Cl(S(h(x), U»
I
Cl(S(h(x), U))}.
Since for each u E V, {h E F I h(u) E Cl(S(h(x). U» I is closed with respect to re, it is clear that H is closed in F with respect to re. But E c H so E* c H, so for each g E E*, g(V) c Cl(S(g(x), U» c S(g(x), V). Hence E* is equicontinuous atx. LEMMA 11.8 If E is equicontinuous then re is jointly continuous on E. Hence re is identical with K on E. Proof: First we want to show that re is jointly continuous on E. For this we need to show that e:E x X ~ Y defined by elf, x) = f(x) is continuous at each point of Ex X. Let (f, x) E E x X and let V E v. Pick U E v such that U* < V. Let W be an open set in X containing x such that g(W) c S(g(x), U) for each gEE. Put V(f, x) = {g EEl g(x) E S(f(x), U) I. Then V(t, x) is a subbasic open set in re. Now V(f, x) x W contains (f, x) in E x X and if (g, y) E V(f, x) x W then g(x) E S(f(x). U) since g E V(f, x) and g(y) E S(g(x), U) since YEW. But then g(y) E Star(S(f(x) , U), U) c S(f(x), V) so e(g, y) E S(e(f, x), V) for each (g, y) E V(f,x) X W. Hence re is jointly continuous on E. Then by Theorem 11.27. the re topology on E is finer than the compact-open topology on E so re is identical with the compact-open topology on E. Since the members of E are all continuous, by Theorem 11.26, re = K on E. COROLLARY 11.10 If E is an equicontinuous family offunctions then E is compact with respect to K if it is compact with respect to re. LEMMA 1 J.9 If E is a family of functions from a space X to a uniform space (Y,v) and E is compact with respect to some jointly continuous topology for E. then E is equicontinuous. E X and V E v. Pick U E v with U* < V. Since'! is jointly continuous on E, for each fEE there is a '!-open set U f containing f and an open set W c X containing x with e(U! x W) c S(f(x), U). If g E U f and y E W, then g(x) and g(y) are in S(f(x), U) which implies g(y) E Star(S(g(x),U),U) c S(g(x), V) for each YEW so g(W) C S(g(x), V) for each g E Ufo Since E is compact with respect to '!, there is a finite collection Uh ... Ufn E '! covering E and open sets WI' .. Wn C X, each containing x such that for each g E Uf" g(Wil c S(g(x), V). Put V = ,,7=1 Wi. Then V is open in X and contains x. Moreover, g(V) c S(g(x), V) for each gEE. Thus, E is equicontinuous. -
Proof: Let x
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THEOREM 11.29 (Ascoli) Let C be the family of all continuous functions from a locally compact space X to a uniform space (Y. v) and let K be the topology of uniform convergence on compacta. Then a subfamily E of C is compact with respect to K if and only if (1) E is closed in C with respect to K, (2) E[x] has compact closure in Y for each x E X and (3) E is equicontinuous. Proof: Since X is locally compact, X is uniformizable and hence regular. Also, local compactness implies C = CK (those members of F that are continuous on compact subsets). Then if E is compact in C with respect to K, (1) and (2) hold by Theorem 11.28, and the 1t topology on the 1t-closure of E in F is Jointly continuous on compacta. Then as shown in the last part of the proof of Lemma 11.8, 1t = K on E. Hence K is jointly continuous on compacta. But since X is locally compact, K is jointly continuous on E by Lemma 11.9, since E is compact with respect to K, E is equicontinuous. Hence (3) holds. Conversely, assume (1), (2) and (3) hold. By (3), E is equicontinuous so by Lemma 11.7, the closure E* of E with respect to 1t in F is equicontinuous. Then by Lemma 11.8, 1t is jointly continuous on E*. Hence 1t is jointly continuous on compacta on E* so (1), (2) and (3) of Theorem 11.28 hold which implies if E c C K then E is compact in CK with respect to K. Since X is locally compact, C = CK so E C CK and hence E is compact in C with respect to K. -
EXERCISES
°
1. Let X = [0, 1] = Y and let F = yX. Let f E F be defined by f(x) = if x is rational and f(x) = 1 otherwise. Then the function e:F x X --7 Y defined by e(f,x) = f(x) E Y is zero at (f, 1/2) E F x X. Show that e is not continuous at the point (f, 1/2). 2. Let X and Y be spaces and let G c yX. Let 0 be the discrete topology on G. Show that 0 is jointly continuous on G. 3. Prove Proposition 11.19. 4. Prove Theorem 11.25. 5. If X and Y are topological spaces and Hand K are subsets of X and Y respectively, and if U is open in X x Y and contains H x K, then there exists open sets V and W containing Hand K respectively such that V x W c U.
11.4 Uniform Function Spaces
369
THE 1t TOPOLOGY ON THE DUAL SPACE 6. Show that the weak* topology on the dual space X* of X in Exercise 8 of Section 11.3 is the topology 1t of pointwise convergence. 7. Show that the unit sphere in X* is compact in the weak* topology.
Chapter 12 UNIFORM DIFFERENTIATION
12.1 Complex Measures At the time of writing this chapter, very little is known about uniform differentiation. It is an area where all the good theorems remain to be proven. It is also an area where we know the approximate form we would like for some of these yet unproven theorems to take. This is because we expect the uniform derivative to be equivalent to the Radon-Nikodym derivative when it exists. In this chapter we develop both the concepts of the Radon-Nikodym derivative and the uniform derivative. However, the assumptions we make about the spaces on which the uniform derivatives are defined, and perhaps about the uniform derivative itself, are probably unnecessarily restrictive. This is because we do not know how to prove the equivalence of the uniform derivative and the Radon-Nikodym derivative at the present time without them. The concept of the Radon-Nikodym derivative is more general than the concept of the uniform derivative. The Radon-Nikodym derivative is established by a nonconstructive existence proof and consequently we know almost nothing about its structure. On the other hand, the uniform derivative is constructed in a more familiar manner as the limit of ratios of the measure of uniform neighborhoods of a point to the Haar measure of these uniform neighborhoods. Consequently, to even discuss the concept of uniform differentiation we need assumptions strong enough to guarantee the existence of Haar measure. The Radon-Nikodym derivative exists in a more general setting than that required to assure the existence of Haar measure. Both the Radon-Nikodym derivative and the uniform derivative of a measure can be defined for complex valued measures as well as for the positive measures that we have been considering up until now. In this section we will first introduce the concept of a complex valued measure and then develop some tools for dealing with them. For our purposes we define a complex measure to be a countably additive, complex valued function on a a-algebra. We will not develop the concept of complex valued measures on rings and a-rings. Let M be a a-algebra in a set X. A countable collection {En) eM is said to be a partition of E if EmnE n = (2) whenever m "#- n and if E = uEn • If J..L is a complex measure on M then J..L(E) = LnJ..L(En) for each partition {En} of E. Consequently, the series LnJ.!(En) must converge if J.! is a complex measure.
12.1 Complex Measures
371
Since the union uEn of the sequence {En} is not changed by a pennutation of the subscripts, each rearrangement of the series must also converge. From a well known result about infinite series that can be found in several advanced calculus texts, the series must converge absolutely Let A be a positive measure on M. A is said to dominate Il if IIl(E) I ~ A(E) for each E E M. Then for the partition {En I of E, A(E) = 'f,nA(En) ~ 'f,n IIl(E n) I so A(E) ~ sup {'f,n IIl(E n) II {En} is a partition of E}. We define a set function III I on M by III I (E) = sup {'f,n IIl(E n) II {En} is a partition of E}. This function III I :M ~ [0, 00) is called the total variation of Il. PROPOSITION 12.1 The total variation of a complex measure Il is a positive measure. It is the smallest positive measure that dominates Il. Proof: Let {En} be a partition of E E M. For each n let rn E R such that rn < III I (En)· Then each En has a partition {En m } such that 'f,m I J..l.(E nm ) I > rn' Since {En m In, m are positive integers} is a partition of E, we have 'f,nrn ~ 'f,n.m IIl(E nm ) I ~ IIlI (E). Then sup {'f,nrn I rn < III I (En)for each n) = 'f,n III I (En) and sUp{'f,nrn I rn < IIlI (En) for each n} ~ 'f,n,m IIl(E nm ) I so 'f,n IIlI (En) ~ IIlI (E).
To prove the opposite inequality, let {Am) be another partition of E. Then for a fixed n, {EnnA m ) is a partition of En and for a fixed m, {EnnA m } is a Therefore, 'f,m IIl(Am) I 'f,m I'f,nll(EnnA m) I ~ partItIOn of Am. 'f,m'f,n IIl(EnnA m) I ~ 'f,n III I (En). Since this holds for each partition {Am) of E, we have III I (E) ~ 'f,n III I (En). Consequently, 'f,n III I (En) = III I (E) so Il is countably additive. That III I satisfies the other properties of a positive measure are easily demonstrated. That III I is the smallest positive measure that dominates J..l. follows from the fact that if A is another positive measure that dominates j..I., then A(E) = 'f,nA(En) ~ 'f,n I j..I.(En) I for each partition {En) of E, so A(E) ~ 1j..I.l (E) for each E E M.· LEMMA 12.1 If C 1 that I'f,1 ESC I I ~ 1/6'f,;=1 ICI I.
. . . Cn E
C, there exists a subset of {I ... n} such
Proof: Put r = 'f,i=1 1CI I. C is the union of four "diagonal quadrants" bounded by the lines y = x and y = -x. Let Q J denote the one defined by I y 1 ~ x for each z = (x, y) in Q I. Now one of these quadrants, say Q, has the property that the sum of the ICI I 's for which Ci E Q is at least r/4. Assume first that Q = Q I. For CEQ 1, Re(c) ~ I C 1- 1/2 . If S is the set of all i such that Ci E Q I, it then follows that:
This proves the lemma for Q
= Q I.
If Q is one of the other quadrants, a similar
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372
argument will work by replacing the formula Re(c) appropriate formula for that quadrant.-
I c 1- 112 with the
::::0:
PROPOSITION 12.2 IfJl is a complex measure on X then IJlI (X)
0 there exists a partition {Xn} of X such that Ln IJl(X n) I > r. Now Jl(X) = c for some c E C so IJl(X) I < Put s = 6( IJl(X) I + 1). Then there exists a partition {En} of X with L;;'=l IJl(E n) I > s for some positive integer m. By Lemma 12.1, if we put Ci = Jl(E i ) for i = 1 ... m there exists a subset S of {I ... m} with I LnESJl(En) I ::::0: l/6L;;'=1 IJl(E n) I. Put H = unesEn and K = X - H. Then HnK = 0, X = HuK and IJl(H) I = IJl(unesEn) I = I LnesJl(En) I ::::0: 1/6L;;'=IIJl(En)1 > s/6 = IJl(X) I + 1 ::::0: 1, so IJl(H) I ::::0: l. (X).
(X).
For any pair a, b E C, it is easily seen that Ia - b I ::::0: Ib I - Ia I. Hence IJl(K) I = IJl(X) - Jl(H) I ::::0: IJl(H) I - IJl(X) I > s/6 - IJl(X) I = 1. Since IJlI (X) = IJlI(H) + IJlI(K) by Proposition 12.1, we have IJlI(H) = or IJlI(K) = Consequently, X is the union of two disjoint sets H and K such that for one of the sets, say H, IJlI (H) = 00 and for the other, IJl(K) I ::::0: 1. Put HI = H and K I = K. A similar argument can be used to show that H I can be partitioned into two disjoint sets H 2 and K 2 such that IJlI (H 2) = 00 and IJl(K 2) I ::::0: 1. Continuing inductively, we obtain a sequence {Kn} of subsets of X such that KmnK n = 0 if m of. n and I Jl(Kn) I ::::0: 1 for each n. Put Y = uKn. Then Jl(Y) = LnJl(Kn). But LnJl(Kn) cannot converge absolutely since IJl(K n) I ::::0: 1 for each n, which is a contradiction. Therefore, IJlI (X) < 00. (X)
(X).
Proposition 2.12 shows that complex measures are bounded in the sense that their range lies in a sphere of finite radius. To see this let Jl be a complex measure on a a-algebra M and E E M. Then IJl(E) I :s; IJlI (E) :s; IJlI (X). This fact has many interesting and useful consequences, among which is the the following: if Jl and A are complex measures on the same a-algebra M and if C E C, we define Jl + A and CJl on M as follows: for each E E M put [Jl + Al(E) = Jl(E) + A(E) and [cJll(E) = cJl(E). Clearly Jl + A and cJl are also complex measures on M. Therefore, the collection of complex measures on M is a vector space. In fact, if we put IJlI = IJlI (X), it is easily verified that the collection of complex measures on M is a normed linear space. Now assume that Jl is a positive measure on M and A is either a complex measure or a positive measure on M. A is said to be absolutely continuous with respect to Jl, denoted by A « Jl, if A(E) = 0 for each E E M for which Jl(E) = O. If there exists aYE M such that Jl(E) = Jl(En Y) for each E E M, we say Jl is concentrated on Y. In other words, Jl is concentrated on Y if and only if Jl(E) = o whenever En Y = 0. If Jl and f... are any measures on M such that there exists disjoint sets H and K with f... concentrated on H and Jl concentrated on K. then Jl and f... are said to be orthogonal or to be mutually singular. denoted Jl.l f....
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PROPOSITION i2.3 ifA. f.! and 0' are measures on a a-algebra M and if 0' is positive. then: (1) Iff.! is concentrated on E E M. then so is 1f.! I. (2)/ff.!.lAthen 1f.!ll.IAI. (3)lff.!«athen 1f.!1 «0'. (4) iff.!.l 0' and A .1 0'. then f.! + A .1 0'. (5) Iff.! «0' and A« 0'. then f.! + A« 0'. (6) Iff.!« 0' and A .1 0'. then f.! .1 A. (7) Iff.! «0' and f.! .1 0'. then f.! = O.
The proof of this proposition is left as an exercise.
EXERCISES 1. Prove Proposition 12.3.
2. Let f.! be a real valued measure on a a-algebra M and define f.!+(E) = 1/2[ If.!1 (E) + f.!(E)] and WeE) = l/2[ If.!1 (E) - f.!(E)] for each E in M. Show that both m+ and ware bounded positive measures on M. 3. Show that f.! = f.!+ - W and that If.! 1 = f.!+ + W· Expressing f.! as f.!+ - W is known as the Jordan decomposition of f.!. Also, f.!+ and W are said to be the positive and negative variations of f.! respectively. 4. Let A and f.! be measures on a a-algebra M such that f.! is positive and A is complex. Show that A is absolutely continuous with respect to f.! if and only if for each £ > 0 there exists a 8 > 0 such that 1A(E) 1 < £ for each E E M with f.!(E)
1 put Xn = En - u7~i Xi, Then {Xn} is a decomposition of X into disjoint measurable sets such that uXn = X and J.!(X n) < 00 for each positive integer n. Since A(X) < 00, A(Xn) < 00 for each n, so we can apply the result we just proved to each Xn to get a function fn ELI (J.!) such that fn = outside Xn and A(EnXn) = JEnXJndli. Now put f(x) = L';=dn. Then f:X -~ R is nonnegative and measurable. Now A(E) = A(EnX) = A(En[uXnD = A,(u[EnXnD = LnA{EnXn) = LnfEnXJnd!! = Lnfdnd!!. For each positive integer m, L~=lJdndJ.! = fE(L~=dn)dJ.! and by Theorem 8.9, JE(k~=dn)d!! converges to fEfdJ.! as m ~ 00. Hence L::'=l fEfndJ.! converges to fEfdJ.! as m ---7 00.
°
376
12. Uniform Differentiation
But L;:'=1 fElndll converges to LJElndll so LJElndll = fEldll. Therefore, A(E) = fEldll. Again, taking E = X shows that I ELI (11) since A(E) < 00. This concludes the proof for the case where 11 is a-finite and A is a positive bounded measure. Finally assume 11 is a-finite and that A is complex. Then A = Al + i~ for some real valued measures Al and ~ on M. We can then apply the above results to the positive and negative variations of Al and ~ (see Exercises 2 and 3 of Section 12.1) to obtain the desired result. To show that I is unique, let g be another member of L 1 (11) such that A(E) = fEgdll for each E E M. First assume I:X -7 [0,00) and g:X -7 [0,00). Now I - g is a measurable function. Let A = {x E xl [f - g](x) < O} and B = {x E xl [f-g](x) > OJ. Clearly both A and B are measurable and AuB = {x E XI/(x) "# g(x) J. Suppose A(A) "# O. For each positive integer n put An = {x E A l-l/2n : : : [f - g](x) < 1/2n J and let A 0 = {x E X I [f - g](x) < -1/2}. Then A = u;;'=QA n, AmnA n = 0 if m "# n, and each An is measurable. Therefore, there exists some nonnegative integer k such that A(A k) "# 0 which implies g(x) - I(x) > 1/2k for each x E Ak so fAJdA "# fAkgdA which is a contradiction. Hence A(A) = O. Similarly A(B) = O. Therefore, {x E X I/(x) "# g(x) J has zero measure, so I = g almost everywhere. Next assume f:X -7 C and g:X -7 C. Then 1=11 + ih for two real valued functions 11 and h and g = g 1 + ig 2 for two real valued functions g I and g 2· Then fEll dll = fa 1 dll and fEhdll = fa2d1l for each E E M. Now the argument above used to show I = g almost everywhere when I:X -7 [0,00) and g:X -7 [0,00) can easily be modified to show II = g I almost everywhere when 11 :X -7 Rand g I:X -7 R. Therefore, II = g I almost everywhere. Similarly h = g2 almost everywhere. Hence 1= g almost everywhere, so I is unique in L I (11).The existence of the Radon-Nikodym derivative is extremely significant and we will devote a good deal of space in this chapter to considering its consequences. For instance, it allows us to show that if p and q are conjugate exponents and 11 is a positive a-finite measure, then U(Il) is the dual space of U (11). It also allows us to extend the Reisz Representation Theorem to bounded linear functionals A on C ~(X) to obtain a representation of the form A(j) = fxldll for a unique regular, complex measure 11, for each lEe ~(X). Furthermore, it will be used to prove the Hahn Decomposition Theorem in the next section, However, as important as the Radon-Nikodym derivative is, its existence proof gives us no insight into the intrinsic nature or structure of this derivative. These results we have just mentioned only give us insight into its behavior. Later in this chapter we will use the Haar measure on an isogeneous uniform space to construct a derivative in the more familiar manner as a limiting process of ratios of a measure 11 with respect to the Haar measure m for suitably well behaved measures 11, to produce a uniform derivative dWdm. We
12.2 The Radon-Nikodym Derivative
377
will show that for these measures f.!, the uniform derivative is the RadonNikodym derivative. We now tum our attention to showing that for a positive a-finite measure f.! and conjugate exponents p and q that L q (f.!) is the dual space of U (f.!). If we let f.! be any positive measure, 1 ~ p < 00, and q is an exponent conjugate to p, then the Holder inequality (Theorem 11.1) implies that if gEL q (f.!) and if A is defined by A(j) = fxfgdf.! for each fEU (f.!) then A is a bounded linear functional on U(f.!) of norm IA I at most Ig Iq' To see this, note that Holder's inequality is simply Ifgl) ~ If I Iglq. Then IAI =sup{lfx/gdf.!l/lfl IfE U(f.!) andf;t O} ~ sup{fx Ifg Idf.!Jfflp IfE U(f.!) andf;t O} = sup { Ifg I dlfl p If E U(f.!) andf;t O} ~ sup { Iflp Iglq/lflp IfE U(f.!) andf;t O} = Iglq. Hence IAI ~ Iglq. Now if all bounded linear functionals on U(f.!) have a unique representation of this form, then we can define a one-to-one, onto mapping :U (f.!) * -7 L q (f.!) by (A) = g where g is the unique member of U (f.!) such that A(j) = fxfgdf.! for each f E U(f.!). Clearly, is a linear operator. If it were the case that IA I = Ig Iq where (A) = g then would be distance preserving (norm preserving) and consequently a Banach space isomorphism, so L q (f.!) would be the dual space U (f.!) * of U (f.!). This is indeed the case for 1 < p < 00. However, there are cases where it fails when p = 1 or p = 00. But if f.! is a a-finite, positive measure, then it holds for p = l.
THEOREM 12.2 If f.! is a a-finite, positive measure on X, 1 ~ P < 00, and A is a bounded linear functional on LP(f.!), then there exists a unique g E L q(f.!) , where p and q are conjugate exponents, such that A(f) = fx/gdf.!for each fE U(f.!) and such that IAI = Iglq.
Proof: The uniqueness of such a g is easily shown using the technique of Theorem 12.1, so it only remains to show that such a g exists and that IAI = Ig Iq' We first prove it for the case when f.! is bounded, i.e., f.!(X) < 00. As shown in the preceding discussion, HOlder's inequality implies that IA I ~ Ig Iq' If IA I = 0 then the theorem holds with g = 0 so suppose IA I > O. For any measurable set E c X put A(E) = A(XE) where XE is the characteristic function of E. Since A is linear, A is additive. Suppose {En} is a sequence of pairwise disjoint measurable sets and uEn = E. For each positive integer m put Fm = u':=lEn • Now IXE - XFm I = [f.!(E - F m»)lIp and the right side of this equation converges to zero as m -7 00. Since A is bounded, A is continuous and since [f.!(E - F m)] lip -7 0 as m -7 {XFm } converges to XE in the LP -norm, {A(XF m)} converges to A(XE) and hence {A(Fm)} converges to A(E). Consequently, A is a complex measure. If f.!(E) = 0 then A(E) = A(XE) = fxXEgdf.! = fEgdf.! = 0 so A is absolutely continuous with respect to f.!. Then the existence of the Radon-Nikodym derivative assures us of the existence of agE L 1 (f.!) such that for each measurable set E c X,
00,
378
12. Uniform Differentiation
Because of the linearity of A it is easily shown that whenever f is a simple measurable function that: (12.6)
A(j)
= fxfgdf.!.
If f E L ~(f.!) then there exists a b ~ 0 such that If I $; b almost everywhere on X. Let B = {x E X Ilf(x) I $; b). Then f.!(X - B) = 0 and If I is bounded on B. It is not difficult to show that the method of proof in Theorem 8.7 yields a uniformly convergent sequence {sn} of simple measurable functions on B that converges to If I. By Proposition 8.17 there exists a measurable function a on X such that Ia I = 1 and f = a If I. Since {sn} converges uniformly to If I, {as n } converges uniformly to f. Now the uniform convergence of {as n } implies Iasn-fl p converges to zero as n ~ 00. Since A is continuous on LP(f.!) and f.!(X) < 00 implies L ~(f.!) c U(f.!),f E U(f.!). But then {A(as n)} converges to A(j) on B as n ~ 00. Hence A(j) = limnA(asn) = limnixasngdf.! = limnfBasngdf.!. Now IaS n I(x) $; If I(x) for each x E Band fB If Idf.! < 00. Therefore, by Theorem 8.13, limnfBasngdf.! = fBfgdf.! = fxfgdf.!. Hence A(j) = fxfgdf.! for eachf E L ~(f.!).
Next we want to show that gEL q(f.!) and that IA I = Ig Iq. We can do this by showing that Ig Iq $; IA I because, as previously shown, the HOlder inequality implies IA I $; Ig Iq so we would then have IA I = Ig Iq which in turn implies gEL q(f.!) since A is bounded. For this there are two cases to consider. Case 1: (p = 1). Now p = 1 implies q = 00 and by (12.5), for each measurable set E, IfEgdf.! I = IA(XE) I $; IA II XE II $; IA If.!(E) = fE IA Idf.! which implies IA I = [l/f.!(E)]fE IA Idf.!. Then by Exercise 5 of Section 8.7, g(x) $; IA I almost everywhere on X. Hence Ig Iq = Ig I~ $; IA I. Case 2: (p > 1). By Proposition 8.7, there exists a measurable function a on X such that Ia I = 1 and g = a Ig I. For each positive integer n put En = {x E X II g(x) I $; n} and let hn = (l/a)XE n Ig Iq-I. Then for each x E En we have Ihn(x) Ip = I(1/a(X»XE n(x) Ig(x) Iq-I Ip = I(l/a(x» Ip Ig(x) Ipq-p. Since p and q are conjugate exponents, pq =p + q so Ihn(x) Ip = Ig(x) Iq. Hence Ih n Ip = Ig Iq on En. Clearly hn E L ~(f.!) for each n so (12.6) yields fEJ g Iqdf.! = fx Ig IXEJ g Iq-I df.! = fx(g/a)XEJ g Iq-I df.! = fxhngdf.! = A(hn) $; IA II hn Ip = IA I [fx Ihn IPdf.!] lip = IA I [fEn Ig Iqdf.!] lip .
k
Now Iglqdf.! $; IAI [fEn Iglqdf.!]"P so [fEn Iglqdf.!]I-IIP $; IAI which implies [fEn Ig Iqdf.!] IIq $; IA I so fEn Ig Iqdf.! $; IA Iq which implies IXXEn Ig Iqdf.! $; IA Iq. Then limnXEn Ig Iq = Ig Iq and XEm Ig Iq $; XEn Ig Iq if m < n so by Theorem 8.9, fx Ig Iqdf.! = limnfxXEn Ig Iqdf.! $; IA Iq so Ix Ig Iqdf.! $; IAI q. But then Ig Iq $; IA I. Consequently, g E U(f.!) and IA I = Ig Iq. It only remains to show that (12.6) holds for eachfE U(f.!).
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Both IA I (f) and fxlgd~ are continuous on U (~) since if lEU (~), g E Lq(~) implieslg E LI(~) and I/gll ~ I/lp Iglq (Proposition 11.2) so Ifxfgd~1 :5; fx I/g Id~:5; III pig Iq = IA IIII p = A(f). Hence fxfgd~ is a bounded linear functional on LP(~) and therefore continuous. Now L ~(~) is a dense subset of U(~). To see this recall that by Proposition 11.4, the class SM of simple measurable functions is dense in U(~). Clearly SM c L ~(~). Since ~(X) < 00 implies L ~(~) c U(~) we see that L ~(~) is dense in U(~). Since A(f) and fxlgd~ are continuous and agree on the dense subset L ~(~), they must agree on U (~). Therefore, (12.6) holds for each lEU (~). This concludes the proof for the case where ~(X) < 00. Next suppose ~ is a-finite and lEU (~). Then as previously observed in the proof of the Radon-Nikodym Theorem, there exists a sequence {En} of disjoint sets such that ~(En) < 00 for each n. For each positive integer n put An = u?=IEi . Nowforeachn, IXEJlp:5; I/lp sothemappingcI>n:U(~)---tCdefined by cI>n(f) = A(XEJ) for each IE U(~) is a bounded linear functional of norm at most IA I on X. Then since ~(En) < 00 for each n, we can apply the above proof to cI>n restricted to En to obtain a function gn on En with cI>n(XEnf) = JEJgnd~ for each/E U(~). Now extend each gn to X by setting gn(x) = 0 for each x not in En and put g = Lngn. Also, for each n let L~ and L~ denote U(~) and Lq(~) on An respectively. Since for each i, gi(X) = 0 for each x that is not in Ei , A(XAnf) = fAJ(g 1 + ... + gn)d~ for each I E U(~) and since ~(An) < 00 for each n, the theorem is true for each An> so (g 1 + ... + gn) I An is the unique member of L~ such that An(f) = fAJ(g 1 + ... + gn)d~ where An denotes A restricted to L~. Consequently, Ig 1 + ... + gn Iq = IAn I ~ IA I. For each positive integer n put hn = r.?=lgi. Then Ihn Iq = IAn I for each n. By Theorem 8.11, fx(lim inln~~ Ihn Iq)d~:5; lim inln~~fx I hn Iq d~ which implies fxlglqd~ ~ lim inln~~(IAnlq) ~ lim inln~~(IAlq) = IAlq. Hence [fxlglqd~]lIq:5; IAI SOgE Lq(~). Now let I E U(~). By Proposition 11.2,lg ELI (~) so Ix I/g Id~ < 00. Then by Theorem 8.13, since [fg](x) = limn[fhn](x) and Ifhn(x) I :5; I/g(x) I for each x E X, we have fxfgd~ = limnfxfhnd~ = limnIAnfhnd~ = limnAn(f) = limnA(XAnf). Since {XAJ} converges to I in U(~) and A is continuous, {A(XAnf)} converges to A(f). Therefore, A(f) = Ixfgd~. That IA I = Ig Iq follows from the fact that limnh n = g and Ihn Iq :5; IA I for each n. -
EXERCISES
1. Let A. denote Lebesgue measure on (0, 1) and let ~ be the counting measure on the a-algebra of Lebesgue measurable sets in (0, 1). Show that A is absolutely continuous with respect to ~ but there does not exist an I ELI (~)
380
12. Uniform Differentiation
such that A(E) = fEfd~ for each Lebesgue measurable set E. In other words, the Radon-Nikodym derivative may not exist if ~ is not a-finite. 2. Show that if I < P < 00, then Theorem 12.2 still holds even if ~ is not a-finite, i.e., if q is an exponent conjugate to p then L q (~) is still the dual space of U (~) even if ~ is not a-finite. 3. Let A, ~ be two measures on M with 1.« ~ such that the Radon-Nikodym derivative g exists. Show that for each E E M and measurable function f:X ~ R with respect to ~, fddA = fdg*. 4. Show that the Radon-Nikodym Theorem (12.1) can be extended to the case where both ~ and A are positive a-finite measures.
12.3 Decompositions of Measures and Complex Integration We have already encountered a decomposition of a real valued measure ~, namely, the Jordan decomposition of ~ defined in Exercise 2 and 3 of Section 12.1. This decomposition was helpful in proving the existence of the RadonNikodym derivative in the case where A was a complex measure, absolutely continuous with respect to ~ and ~ was positive and a-finite. In this section we introduce other useful decompositions, called the Lebesgue decomposition and the polar decomposition and also prove an important theorem about the Jordan decomposition called the Hahn Decomposition Theorem. All the major results in this section depend on the existence of the Radon-Nikodym derivative. The Lebesgue decomposition is introduced by means of the following theorem.
THEOREM 12.3 Let ~ be a positive a-finite measure on a a-algebra M in a space X and let A be a positive bounded measure or a complex measure on M that is absolutely continuous with respect to~. Then there exists a unique pair of measures Al and ~ on M such that A = Al + ~ and such that Al ..l~, ~ < < ~ and Al ..1 ~. Proof: We first prove the theorem for the case where ~ and A are positive bounded measures on M. We can use the first part of the proof of the RadonNikodym Theorem to obtain a unique function gEL 2(~) with g(x) E [0, 1] for each x such that equations (12.2) and (12.3) hold. Then, as in the proof of the Radon-Nikodym Theorem, let H = {x E X Ig(x) < I} and K = {x E X Ig(x) = I} and let Al (E) = A(KnE) and ~(E) = A(HnE) for each E E M. Then AI and ~ are both positive measures on M. If we take f = XK in (12.3) we get JK(1 - g)dA = fKgd~ which implies JKdA fKdA = fKd~ so ~(K) = O. Therefore, ~ is concentrated on H. But clearly AI is concentrated on K and HnK = 0. Hence AI ..l~. As shown in the proof of the
Radon-Nikodym Theorem, since g is bounded, (12.3) holds if we replace f by
12.3 Decompositions of Measures and Complex Integration
381
(l+g + ... + gn)XE for each positive integer n and each E E M, we get (12.4). At each point of K, g(x) = 1 so 1 - gn+l(x) = O. On the other hand, at each point x of H, {gn+l (x)} converges to zero monotonically. Consequently, the left hand side of (12.4) converges to A(HnE) = ~(E). Also, as shown in the proof of the Radon-Nikodym Theorem, the right side of (12.4) converges to fddJ.! and n ---+ 00 where f is the Radon-Nikodym derivative, so ~(E) = fddJ.! for each E E M. Therefore, J.!(E) = 0 implies ~ (E) = 0 so ~ «J.!. It remains to show that A) and ~ are unique since clearly Al ...L~.
But the uniqueness of A) and ~ is easily seen. If A~ and A; are measures such that A = A') + A; and such that A') ..1 J.! and A; « J.!, then A') - Ai = A; - ~ are measures such that A~ - A) ..1 J.! and A; - ~ «J.!. But then it is easily shown that A~ - Al = 0 and A; - ~ = O. Next we prove the case where J.! is a positive a-finite measure but A remains positive and bounded. Then, as shown in the proof of the RadonNikodym Theorem there is a disjoint sequence {En} c M with X = uEn such that J.!(E n ) < 00 for each n. Since A(X) < 00 we have A(En) < 00 for each n, so the above result holds for each n. Therefore, there exists sequences {A~} and {A~} of measures such that for each n, A~ and A~ are positive measures concentrated on En with A(EnnA) = A~(EnnA) + A~(EnnA) for each A E M, A~ ..1 J.!, A~ «J.! and A~ ..1 A~. Put A) = LnA~ and ~ = LnA~. Then for each A E M, Al (A) + ~(A) = LnA~(AnEn) + LnA~(AnEn) = LnA(AnEn) = A(A). If J.!(A) = 0 then A~(A) = 0 for each n so ~(A) = LnA~(A) = O. Therefore, ~ «J.!. Since A~ ..1 J.! for each n and J.! is concentrated on H, then A~ is concentrated on K for each n which implies A) is concentrated on K so Al ..1 J.!. Finally, since A~ is concentrated on K for each n, A~ is concentrated on H for each n which implies ~ is concentrated on H so A) ..l~. This proves the case where J.! is a positive a-finite measure but A remains positive and bounded since the uniqueness proof is the same as for the previous case. To complete the proof, assume A is complex and J.! is positive and a-finite. Then A = Y + ia for some real valued measures Y and 0'. Moreover, Y = y+ - Yand 0' = 0'+ - 0'- (see Exercises 2 and 3 of Section 12.1) where y+, y-, 0'+ and 0'are all positive bounded measures. Then applying the above result to these measures, we obtain the desired decomposition of A into a unique pair of measures Al and ~ on M such that A = Al +~, A) ..l~, Al ..1 J.! and ~ «J.!. The pair Al and ~ in Theorem 12.3 is known as the Lebesgue decomposition of A. Another useful decomposition of a complex measure is the so called polar decomposition. Like the Lebesgue decomposition we introduce it by means of a theorem. THEOREM 12.4 If J.! is a complex valued measure on a a-algebra M in X. then there exists a complex measurable function h on X with Ihex) I = J for each x E X and such that J.!(E) = JEhdl J.!I for each E E M.
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382
Proof: Clearly J..L « IJ..L I so there exists a Radon-Nikodym derivative f E Ll(IJ..LI) ofJ..L with respect to IJ..LI such thatJ..L(E) = fddlJ..L1 foreachEE M. We would like to show that If(x) I = 1 for each x E X. Let a E R and put A = {x E X Ilf(x) I < a). Then IJ..LI (A) = SUP{Ln IJ..L(En) II {En) is a partition of A). Let {An) be some partition of A. Then Ln IJ..L(An) I = Ln Ikfd IJ..L II ~ Lna IJ..L I(An) = a IJ..L I(A) so IJ..L I(A) ~ a IJ..L I(A). If a < 1 then IJ..L I(A) = 0 so If(x) I ~ 1 almost everywhere on X.
Now if E E M with IJ..LI(E) > 0 then I[1/1J..L1(E)lfddlJ..L1 I = I[1/ IJ..L I(E)]J..L(E) I = IJ..L(E) 1/1 J..L I(E) ~ 1. Then by Exercise 5 of Section 8.7, we have that If(x) I ~ 1 almost everywhere on X. But then If(x) I = 1 almost everywhere on X. If we define h:X ~ C by hex) = f(x) if If(x) I = 1 and hex) = 1 otherwise then fddlJ..L1 = fEhdlJ..L1 for each E EM, so h E L 1 (1J..L1), Ih(x)1 = 1 for each x E X and J..L(E) = fEhd IJ..L I for each E E M. The function h of Theorem 12.4 and the total variation IJ..L I are said to be the polar decomposition of J..L. This is because when we define integration with respect to a complex measure, which we will do next, we will have J..L(E) = fEdJ..L for each E E M so fEdJ..L = fEhd IJ..L I. We use the notation dJ..L = hd IJ..L I to denote fEdJ..L = fEhd IJ..L I for each E E M. It is this notation that gives intuitive meaning to the term polar decomposition. Notice that the polar decomposition is different in nature than the Jordan and Lebesgue decompositions where A = Al + Az. We cannot write J..L = hi J..L I meaningfully. It is only meaningful in the sense that dJ..L = hd IJ..L I. We defined integration with respect to a positive measure in Section 8.6 as the supremum of finite sums of products of complex numbers aj with measures of sets E j such that 0 ~ LjajMi ~ f. That definition of integration was extended to complex valued functions in Exercise 1 of Section 8.6. Since then we have been able to avoid the notion of integration with respect to a complex valued measure. But we now have a mechanism for defining integration with respect to a complex measure and a need for this idea in the next section. If J..L and Aare complex measures then so is J..L + A. Using our intuition about what integration with respect to a complex measure should mean and our experience with integration with respect to positive measures we would expect the following identities to hold for each E E M:
= fEdJ..L
(12.7)
J..L(E)
(12.8)
fdd(J..L + A) = fddJ..L + fddA
for each complex measurable function fEe ~(X). That (12.8) holds for the special case where J..L and A are positive measures is an easy exercise (Exercise 1).
By Theorem 12.4, there exists a complex measurable function h with Ih I = 1 such that J..L(E) = fEhd IJ..L I. Since we want (12.7) to hold, we expect our
12.3 Decompositions of Measures and Complex Integration
383
definition of integration with respect to a complex measure to yield 1l(E) = fE 1dll = fE Ihd 1111. If f = 1 (the unit function) then fddll = fEihd 1111. Using this formula for motivation, we define the integral of g with respect to 11 by (12.9) for each complex measurable function gEe ~(X) and for each E E M. Let us now see if the definition (12.9) implies (12.7) and (12.8). That (12.9) implies (12.7) is immediate since by Theorem 12.4, 1l(E) = fEhd 1111 so by (12.9), Il(E) = fEdll. That (12.9) implies (12.8) takes more work. To show (12.9) implies (12.8), first note that by (12.9), fxXEdll = fXXEhd 1111 = fEhd 1111 = 1l(E). Therefore, if 11 and 'A are complex measures and a = 11 + 'A, then for each E E M we have fxXEda
= a(E) = 1l(E) + 'A(E) = fXXEdll
+ fxXEd'A
so (12.8) holds if f = XE for some E E M. Next suppose f is a nonnegative bounded measurable function. By Theorem 8.7 there exists a sequence {sn} of simple measurable functions with 0 ~ Sn ~ f for each positive integer n and Sn ~ Sm if n < m such that {sn(x)} converges to f(x) for each x E X. For each Sn' Sn = L7'=l aiM; for some positive integer m and some nonnegative real numbers a 1 •• am and characteristic functions M 1 • • • XAm for measurable sets AI' .. Am. Let k be a complex measurable function with Ik I = 1 such that da = kd Ia I. Then fxsnda
= fxSnkdlal = fX L7'=laiM;kdlal = L7'=laJxXA;kdlal
=
So (12.8) holds for E = X and f a simple measurable function. Now let p be a complex measurable function with Ip I = 1 such that d'A = pd I'A I. Then if E E Mwehave
Since Sn is a simple measurable function, XESn is also a simple measurable function. Therefore, fXXESn da = ixXESndll + fXXESnd'A = fESnhd 1111 + JESnPd I'A I = JESndj..t + fESndA.. Consequently, (12.8) holds for the simple measurable functions. Define A on
12. Uniform Differentiation
384
C ~(X) by A(g) = Jxgda. By (12.9) A(g) = Jxgkdl a I so A is clearly a linear functional on C ~(X). Since Ig I~ < 00 and Ia I(X) < 00 we have IA(g)1
~
fxlgkldlal = fxlgldlal
~ Igl~lal(X)
so A is bounded and hence continuous by Theorem 11.13. Similarly define the continuous linear functionals .Q and on C ~(X) by .Q(g) = Jxgd/-! and (g) = JxgdA for each g E C ~(X). Since (sn I converges to f we have (.Q(sn) I converges to .Q(j), ((sn) I converges to (j) and (A(sn)) converges to A(j). Now for each positive integer n, A(sn) = fxsnda = fxsnd/-! + JXSndA so (A(sn) I = !fxsnd/-! + JXSndA I = (.Q(sn) + (sn) I· Clearly, (.Q(sn) + (sn) I converges to .Q(j) + (j) = Jxfd/-! + JxfdA and (A(sn) I converges to Jxfda. Since limits are unique, Jxfda = Jxfd/-! + JxfdA. Therefore, (12.8) holds whenfis a nonnegative bounded measurable function. It is a straightforward exercise (Exercise 2) to show that (12.8) holds for complex bounded measurable functions. Now if /-! is a complex measure then /-! = /-!l + i/-!2 for some pair /-!l and /-!2 of real measures. If E E M then i/-!2 (E) = iJEd/-!2' Consequently, JEd/-! = /-!(E) = /-!l (E) + i/-!2(E) = JEd/-!1 + iJEd/-!2 so
(12.10)
for each E E M. This concludes our discussion of integration with respect to complex measures. In the next section we will use these results to significantly generalize the Riesz Representation Theorem. PROPOSITION 12.4 If g E L 1(/-!), where /-! is a positive measure on a a-algebra MinX, and ifA(E) = JEgd/-!for each E EM, then IAI(E) = JElgld/Jfor each E E M. Proof: Let A = (x E X Ig(x) = 01 and B = (x E X Ig(x) oF 01. Then A(A) = 0 = JAlgld/J-. Since A(A) =Oimplies IAI(A)=Owehave IAI(A)=JAlgld/-!.
Next let E E M. Then IAI (EnB) = 0 implies A(EnB) = 0 so JEr>Bgd/-! = O. It can be shown that JEr>Bgd/-! = 0 implies /-!(EnB) = O. To see this note that fEr>Bgd/J- = fEr>Bgid/-! - fEr>Bgjd/-! + JEr>Bg2:d/J- - JEr>Bg2 dJ.l
forsomenonnegativefunctionsgi,gj,g2: andg 2. LetC I = {XE Xlgi(x»OJ, C 2 = (x E Xlgj(x) > OJ, C 3 = {x E Xlg2(x) > 01 and C 4 = (x E Xlg2(x) > 01. Clearly B = C 1 UC 2UC 3 UC 4 . Suppose J.l(EnB) > O. Then J.l(EnC i ) > 0 for some i = 1 ... 4. Assume J.l(EnC 1) > O. Then gj(x) = 0 for each x E EnC 1 so JC1 g IdJ.l = JclrEgidJ.l > O. But EnC 1 c EnB so 0 = A(EnC 1) = JEr.CIgdJ.l, and JEr.C1 gdJ.l oF 0 since JErCI g 1 dJ.l > 0 which is a contradiction. Hence J.l(EnC 1) =
O.
12.3 Decompositions of Measures and Complex Integration
385
Similarly ll(EnC 2) = 0 = ll(EnC 3) = ll(EnC 4). But then Il(BnE) = O. Therefore, IAI (EnB) = 0 which implies Il(EnB) = 0 so 11« IAI on B. Hence the Radon-Nikodym derivative dJ..LId IAI exists on B since IAI(X) < 00 by Proposition 12.3. By Theorem 12.4 there exists an h ELI (J.1) with Ihi = 1 such that A(E) = JEhd IAI for each E E M and by hypothesis, A(E) = JEgdll so JEhd IAI = JEgdll for each E E M. But then JEnBhdlAI = fEnBg[dJ..LIdIAI]dIAI for each E E M by Exercise 3 of Section 12.2. Thus h = gdJ..LId IAI almost everywhere with respect to IAI on B by Exercise 2 of Section 8.7. Therefore, IE nB Ihid IAI = JEnBlgl[dJ..LIdIAlldIAI since 11 and II.I are both positive, so IdJ..LIdl AI I = dJ..LIdIAI. But then IAI(EnB) = JEnBdlAI = JEnBlgldll by Exercise 3 of Section 12.2 and the fact that Ih I = 1. Consequently, IAI(E) = JE Ig IdJ.1 for each EE M. • We conclude this section with the so called Hahn Decomposition Theorem. This theorem is a statement about the Jordan decomposition of a real valued measure J.1 on a a-algebra M in a space X. It says that X can be decomposed into two disjoint sets A and B such that J.1+ is concentrated on A and W is concentrated on B.
THEOREM 12.5 (Hahn Decomposition Theorem) IfJ.1 is a real valued measure on a a-algebra M on a space X, there exists disjoint sets A and B with AuB = X such that J.1+(E) = J.1(EnA) and J.1-(E) = -J.1(EnB)for each E E M. Proof: By Theorem 12.4, there exists a complex measurable function f on X with If I = 1 and J.1(E) = Jdd IJ.11 for each E E M. Since J.1 is real f must be real almost everywhere. Define h on X by hex) = f(x) if f(x) is real and hex) = 1 otherwise. Then h is a real measurable function with Ih I = 1 and J.1(E) = JEhdlJ.11 foreachEE M. PutH={XE Xlh(x)=l}andK={xE Xlh(x)=-I). Clearly HuK = X and HnK = 0. Now J.1+ = (I J.11 + J.1)/2 and (1 + h)/2 = h on H and zero on K so for each E E M we have: J.1+(E) = fE(1/2)dlJ.11 +fE(1/2)dJ.1 = fd(1+h)/2]dlJ.11 =
fEnH[(1+h)/2]dlJ.11 + fEnd(1+h)/2]dlJ.11 = fEnHhdlJ.11 = J.1(EnH). Hence J.1+(E) = J.1(EnH). Now J.1(E) = J.1(EnH) + J.1(EnK) and J.1(E) = 1l+(E) WeE) so J.1(EnH) + J.1(EnK) = J.1+(E) - WeE). Since J.1+(E) = J.1(EnH) we have WeE) = -J.1(EnK). •
EXERCISES 1. Show that the identity (12.8) holds when J.1 and A are positive measures, without reference to complex measures.
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12. Uniform Differentiation
2. Show that the identity (12.8) holds when Il and ').., are complex measures, based on the fact that (12.8) holds when f is a nonnegative bounded measurable function. 3. Use a similar technique to the one used to prove the identity (12.8) to show that if c E C and Il is a complex measure, then for each f E C ~(X), JEfd(cll) = cJEfdll for each measurable set E.
12.4 The Riesz Representation Theorem One of the more important applications of the Radon-Nikodym Theorem is to generalize the version of the Riesz Representation Theorem presented as a series of exercises in Section 8.8, to include bounded linear functionals on C ~(X) as opposed to the positive linear functionals on CK(X) as given in Section 8.8. For this we will find the following lemma useful. LEMMA 12.2 Let Q be a bounded linear functional on C ~(X). Then there exists a positive linear functional A on CK(X) with IQ(f) I ~ A( If I) ~ If I~ for eachfE CdX). Proof: We first define A on the subset C1(X) consisting of all nonnegative real valued members of CK(X). For this put A(j) = sup{ IQ(g) II g E CK(X) and Ig I
~f} foreachfEC1(X). ThenA(j)~OforeachfE C1(X) and IQ(j) I ~A(lfl) ~ Ifl~. Moreover, if f, g E C1(X) withf~ g then A(j) ~ A(g) and if a > 0 then A(af) = aA(j). We want to show that if f, g E C1(X) that A(f + g) = A(j) + A(g).
For this letf, g E C1(X) and let e > O. Then there exists h, k E CK(X) with Ih I ~ f, Ik I ~ g and A(j) ~ IQ(h) I + e and A(g) ~ IQ(k) I + e. Moreover, there are complex numbers a and ~ with Ia I = 1 = I ~ I such that a.D.(h) = IQ(h) I and ~Q(k) = IQ(k) I. Then A(j) + A(g) ~ IQ(h) I + IQ(k) I + 2e = a.D.(h) + ~Q(k) + 2e = Q(ah + ~k) + 2e. Clearly Q(ah + ~k) ~ 0 so Q(ah + ~k) = IQ(ah+~k) I. Therefore, A(j) + A(g) ~ IQ(ah + ~k) I + 2e ~ A( Iah I + I ~k I) + 2e ~ A(f + g) + 2e. Since this inequality holds for each e > 0 we have A(j) + A(g) ~ A(f + g).
To show A(f + g) ~ A(j) + A(g), let h E CK(X) with Ih I ~ f + g. Put U = {x E Xlf(x) + g(x) > O} and F = X - U. Define s and t on X by s(x) = f(x)h(x) and t(x) = g(x)h(x)/[f + g](x) for each x E U and s(x) = 0 = t(x) for each x E F. Clearly s and t are continuous on U and h = s + t. Moreover, Is(x) I ~ Ih(x) I and It(x) I ~ Ih(x) I for each x E X so s and t are also continuous on F since h(F) = 0 and h is continuous on X. Therefore, s, t E CK(X). Since lsi ~fand It I ~gwehave: IQ(h) I = IQ(s) + Q(t) I ~ IO(s) I + 10(/)1 ~A(f) + A(g). Since this holds for each h E CK(X) with Ih I ~f + g we have A(J + g) ~ A(f) + A(g). Hence A(f + g) = A(f) + A(g) for each pair f, g E Ck(X).
12.4 The Riesz Representation Theorem
387
To extend A to the real valued members of CK(X) note that if I is a real valued member of CK(X) that = (III + /)/2 and 1- = (III - /)/2 so that both andr are in CKeX). Define A(j) = A(f) - A(J). To extend A to CK(X) let g E CK(X). Then g = h + ik for some real valued members h and k of CK(X). Define A(g) = A(h) + iA(k). Now A is defined for all g E CK(X).
r
r
It remains to show that A is linear on CK(X) and that IO(j) I ~ A( III) ~ III ~ for eachl E CK(X). We leave this as an exercise (Exercise 1). • A complex Borel (Baire) measure Il on X is said to be regular or almost regular if III I is regular or almost regular, respectively. The linear operator A defined on C ~(X) by A(j) = fEldll is clearly bounded as shown in the last section and therefore continuous. Moreover, IA I ~ III I (X) since IA I = sup{ Ifxldllll/E C~(X) and I/I~ = I} ~ sup{fxl/ldlllll/E C~(X) and I/I~ = I} = fxd III I = III I(X). Our generalization of the Riesz Representation Theorem shows that all bounded linear functionals on C ~ (X) are formed this way with respect to regular complex measures.
THEOREM 12.6 (Riesz Representation Theorem) Let 0 be a bounded linear lunctional on C ~(X) where X is locally compact and Hausdorff Then there exists a unique complex regular Borel measure Il on X such that Om = fxfdllior each I E C ~(X) and such that III I(X) = 101. Prool: Since CK(X) is a dense subspace of C ~(X) with respect to the supremum norm (Theorem 11.15) and each bounded linear functional A in CK(X) has a unique extension to a bounded linear functional on C ~(X) with the same norm (Theorem 11.6), it suffices to prove the Theorem for CK(X). For each I E CK(X) put (j) = O(j)/ I0 I. Then I I = 1 and is a linear functional on CdX). By Lemma 12.2, there exists a positive linear functional AonCK(X)with 1(j)1 ~A(I/I)~ 1/1~foreach/E CK(X). By Theorem 8.14 there exists a positive almost regular Borel measure 'A. such that A(j) = fxld'A. for each I E CK(X). From the proof of Theorem 8.14 (see Exercise I, Section 8.8) 'A.(X) = sup{A(j) II < X} = sup{A(j)I/E CK(X) and/(X) c [0, I]}. Since for each I E CK(X) with III ~ ~ 1 we have IA(j) I ~ 1 it is clear that 'A.(X) ~ 1. Hence 'A. is regular. Moreover, I(j) I ~ A( III) = fx III d'A. = IIII for each IE CK(X) where I II denotes the L I ('A.) norm. Now I II = sup{ I(j) III E CK(X) with IIII = I} ~ 1 so the norm of is at most 1 with respect to the L I ('A.) norm on CdX). By Theorem 11.4, CK(X) is dense in L I ('A.) and by Theorem 11.6 there exists a norm preserving extension to a linear functional ' on L I ('A.). By Theorem 12.2 there exists a unique Borel measurable function g with I g I~ = I' I such that (12.11)
'(j) = Jxlgd'A.
for each I ELI ('A.). Consequently, Ig I~ ~ 1 so Ig I ~ 1. Let lEeK(X) with
388
12. Uniform Differentiation
Ifl~ = 10 Then If(x) I $1 foreachxE X soJxlgldA~JxlfgldA~ IJxfgdAI for each f E CK(X) with If I~ = 1. Therefore, Ix Ig IdA ~ sup { IJxfgdAllf E CK(X) with If I~ = I} = sup{ I(f) lifE CK(X) with If I~ = I} = I I. Since I I = 1 (with respect to the 1 I~ norm on CK(X» we have Jx Ig IdA ~ 1. But since A(X) $ 1 and Ig I $ 1, this can only happen if A(X) = 1 and Ig I = 1 almost everywhere with respect to A. Redefine g so that I g(x) I = 1 for each x E X. Clearly we can do this without disturbing the results so far. Define~' on X by ~'(E) = JEgd IA1 for each Borel set E c X. Then~' is a complex measure and by Theorem 12.4, I~'I (E) = JE Ig 1d IAI = JEdA for each Borel set E. But then I~'I (X) = A(X) = 1 = I I. Now (12.11) and (12.9) yield (f) = fxfgdA = fxfgdl AI =
(12.12)
fxfd~'
for eachfE CK(X). Then n(f) = (f)lnl = Jxflnld~'. Put ~(E) = Inl~'(E) for each Borel set E c X. By Exercise 3 of Section 12.3 we have (12.13)
n(f) = Inl(f) = Inlfxf*' =
Inlfxfd(lnl~')
=
fxfd~
for each f E CK(X), so the bounded linear functional 0. on CK(X) is represented by the complex measure ~ in the sense that n(f) = Jxf* for each f E CdX). Moreover, I~I(X) = SUp{L;;'=ll~(En)11 {En} is a partition of X} = sup{ 10.1 L;;'=l I~'(En) II {En} is a partition of Xl = Inll~'1 (X) = 10.1. Hence 10.1 = I~I(X). Since the measure A was regular, it is not difficult to show that ~ is regular, but we leave that as an exercise (Exercise 2). It remains to show the uniqueness of~. For this let ~ and v be two Borel measures on X such that for each f E CK(X), Jxfd~ = n(f) = Jxfdv. Then Jxfda = 0 for each f E CK(X) where a = ~ v. Let h be a Borel measurable function on X with Ih I = 1 such that da = hdlal. Then h* E Ll(lal). Since CK(X) is dense in Ll(lal) we can find a sequence {fn} C CK(X) such that Ih* - fn I converges to 0 as n ~ 00. Now for each positive integer n, -fxfnda = 0 so -Jxfnhdl al = O. Therefore, lal(X) = fxlhldlal = fxh*hdlal - fxfnhdlal = fx(h*-fn)hdlal. Consequently, Ia I(X) = II a I (X) I $ fx Ih* - fn Ida = Ih* - fn 11' Since Ih* fn 11 converges to 0 as n ~ 00 we see that Ia I(X) = o. But then a(X) = 0 so a(E) = 0 for each Borel set E c X. Therefore, ~(E) = veE) for each Borel set E so~=v.-
EXERCISES 1. Show that the functional A constructed in Lemma 12.2 is linear and that In(f)1 $A(lfl)$lfl~ foreachfE CK(X).
12.5 Uniform Derivatives of Measures
389
2. Show that the measure constructed in Theorem 12.6 is regular.
12.5 Uniform Derivatives of Measures On the real line R there is a natural way to think about the derivative of a complex valued Borel measure f..l at a point x E R. Let m denote Lebesgue measure on R. Define f:R -7 C by f(x) = f..l« -00, x» for each x E R. Then it is possible to prove the following: THEOREM 12.7 f is differentiable at x with derivative DEC if and only iffor each E > 0 there exists a 0> 0 with I f..l(S)lm(S) - D I < E whenever S is an open interval containing x with mrS) < o. This theorem, whose proof we leave as an exercise, suggests defining the derivative of a complex Borel measure J.t on more general spaces, on which Haar measure m exists, as a limit of quotients of the form f..l(U)/m(U) for some suitably restricted class of open sets U, that contain some given point x, as these sets shrink uniformly to the poim x. In this more general setting, Haar measure replaces Lebesgue measure used in Theorem 12.7. Constructing such a generalization will be the program of this section. For this let (X, A) be a locally compact, isogeneous uniform space with isometric basis v. Let f..l be a complex Borel measure on X and let m denote Haar measure on X. If p E X and DEC such that whenever E > 0, there exists abE v with (12.14)
I f..l(S(p,a» m(S(p,a»
- DI
