The Joys of Haar Measure

Theorem 3.1. Let G be a topological group. If U is an open set containing

the identity e, then there is an open set V containing e such that e E V � V � U. Consequently, a To topological group is regular and so Hausdorff. Proof.

First things first: Let U be an open set that contains the identity

e. By continuity of multiplication, there is an open set W containing e such that W · W � U. If we set V = W n w-1 , then we have an open set that contains e, is symmetric (V = v- 1 ) , and satisfies V · V � U. We claim V � U. Take x E V. Then x V is an open set that contains x and so xV n V I- 0, so there must be v 1 , v 2 E V so that xvi = vz. But then X = v2 v! 1 E V · v - I = V · V � U.

The homogeneous structure of G now tells us that whenever x E G and U is an open set containing x , then there is an open set V such that x E V � V � U. Regularity follows from this and G's homogeneity. Again, homogeneity of G tells us that if x, y E G are distinct and there's a neighborhood of x that doesn't contain y, then we have a neighborhood of y that doesn't contain x. In other words, To topological groups satisfy the D Ti axiom. Henceforth, we assume that all topological groups are Hausdorff.

The surprising conclusion reached in Theorem 3 . 1 is a typical product of the mix of the algebra and topology in topological groups. Here's another: Proposition 3.2. Every open subgroup of a topological group is closed. Proof. Let H be an open subgroup of the topological group G. Take g E H. Every open set that contains g intersects H; g H is such an open set.

2. The Classical (Locally Compact) Groups

49

Therefore gH n H # 0. Since cosets are either the same or disjoint, gH = H. Thus g = ge E gH = H D and H � H. Exercise 3.3. Let G be a topological group, let F be a closed subset of G, and let K be a compact subset of G. Then FK and KF are closed in G. Exercise 3.4. If G is a connected topological group, then any neighborhood of the identity is a system of generators for G.

Let G be a topological group, and let N be the connected component of the identity in G. (i) N is a closed subgroup of G. (ii) N is an invariant (normal) subgroup of G. (Hint: Consider N', the set of inverses of N, which is connected, and so if x is a member of N, then x N' is connected and contains the identity, so x N' is contained in N for any x in N . ) Exercise 3.5.

Exercise 3.6. Let G be a locally compact group. Then G contains an open er-compact subgroup. (Hint: Choose an open symmetric subset U containing G's identity with compact closure, and look at the union of powers of U.)

2. The Classical (Locally Compact) Groups

JR.n and en will denote, as usual, real and complex n-spaces, respectively. Mn will denote the linear algebra of all n x n matrices with complex entries. We can associate with any ( aij) E Mn the point (b1 , . . . , bn2 ) E cn2 , where bi+ (j-l) n = aij · This establishes a bijective correspondence between Mn and cn2 , a corre­ spondence we use to equip Mn with the Euclidean topology of cn2 . Of course if a, (3 E Mn, then a · (3 E Mn, too, with n Cij = L aik bkj , k= l whenever a = ( aij ) , (3 = (bij ) , and a · (3 = ( Cij ) · It is easy to see that the operation (a, (3) f--t a · (3 is continuous from Mn x Mn to Mn. In addition to addition and multiplication, Mn is endowed by nature with

a couple of other natural operations-transposition and conjugation: if

3.

50

Introduction to Topological Groups

(aij ) E Mn , then at = (aj i ) and a = ( iiij ) , where a is the complex conjugate of the complex number a. Both of these operations are homeo­ a =

morphisms of Mn onto itself and each is of "order two" , that is, att = a and

a = a.

Some members of Mn have a multiplicative inverse; the collection of all such matrices will be denoted by GL(n,
iJ

_

Pij (a)

det( a ) ' where Pij (a) is a polynomial of the coordinates of a. It follows that the operation a ---+ a- 1 of GL(n,
Corollary 3. 7. GL(n,
After all,

z -:f. O}), and so GL (n ,
Inside GL( n,
GL(n, JR) : a E Mn by Mn (JR) so SL(n,
:

linear group.

is real if a = a; denote the set of real members of Mn

GL(n, JR) = GL(n,
the members of GL(n,
SL(n, JR) : SL(N,
3.

The Birkhoff-Kakutani Theorem

51

Again, SL(n, q , SL(n, �) , SO(n) , SU(n) are closed subgroups of GL(n, C) ) . Actually more can be said. Theorem 3.8. The groups U(n) , O (n) , SU(n) , and SO(n) are compact met­

ric topological groups.

Each of O(n) , SU(n) , SO(n) are closed subgroups of U(n) , so it's enough to establish that U(n) is compact. Now a E U(n) precisely when t a a = idic n . This in turn is the same as saying of a = ( aij ) that for each 1 ::; i, k ::; n, L aii ajk = 8ik · j Now the left side is a continuous function of a so U(n) is closed not just in GL(n, q but even in Mn. Moreover L aii aji = 1 j ensures that I aij I ::; 1 for 1 ::; i, j ::; n. So the entries of any a E U ( n) are bounded. But this ensures that U(n) is homeomorphic to a closed bounded D subset of cn2 • Proof.

3.

The Birkhoff-Kakutani Theorem

Our next result is technical, but it's an investment well worth the price. (Garrett Birkhoff and Shizuo Kakutani). Let (Un) be a se­ quence of symmetric open sets each containing the identity e of the topolog­ ical group G. Suppose that for each k E N,

Theorem 3.9

uk + i · uk + i � uk .

Let H = n k Uk · Then there is a left invariant pseudo-metric that (i) (]' is left uniformly continuous on G x G; (ii) (J' (x , y ) = 0 if and only if x E y H ; (iii) (J' (x, y) ::; 4 · 2 - k , if x E yUk ; (iv) 2- k ::; (J' (x, y) , if x (j. y Uk .

(]'

on G such

There is no harm in assuming that the sequence {Un} is decreasing. Otherwise we can replace Un with n�= l Uk , which will leave H unchanged and allow Un + l · Un + l � Un to hold for each n.

Proof.

3.

52

Introduction to Topological Groups

We start by reinterpreting the sequence's descending character with an eye toward defining O". For each k, set V2-k

=

Uk .

Next define v;. for r a dyadic rational number with 0 < r < 1 as follows. If r = 2-li + 2-! 2 + . . . + 2-ln , with 0 < li < · · · < ln , all positive integers, let Vr = lf2 -1 1 · lf2 -12 · · · V2-tn . For dyadic r's greater than or equal to 1, set v;. = G. Thus, r < s ==? v;. � Vs ; (4) further for any l E N, we have (5)

v;. · V2- 1 � l1r + 2-t+2 ·

We put off the somewhat tedious yet clever proofs of the indicated relation­ ships between the V's (equations (4) and (5) ) until the end of our general discussion. With these in hand we go forth to define O". First, for x E G, let

3.

The Birkhoff-Kakutani Theorem

53

So


From this we see that

2 O" (u, e) ::; r z + for u E V2- I ·

The third statement of the theorem follows from this and O"'s left invariance: if x E y Uk , then y- 1 x E Uk = V2-k so 2 O" (x, y) = O" (y - 1 x, e) ::; 2 - k + = 4 · 2- k . Next we deal with O"'s uniform continuity. Suppose x, y, x, and f) satisfy y - 1 x E V2-1-1 and f) - 1 x E V2-1-1 . Then - - 1 X E v -_1_ · T/"v2-z-1 = T/"v2-z-1 · T/"v2-z-1 C /" X- - 1 YY _ Tv2-1 , 2 1 1 and so IO" (x, y) - O" (x, f)) I

IO" (y - 1 x, e) - O" ( fj - 1 x, e) I < I O" (y - 1 x, f) - 1 x) I = IO" (x - 1 f)y - 1 x, e) I < 2 - 1 + 2 (by (iii)) ,

and this is what we mean by O" is left uniformly continuous. For (iv), suppose y- 1 x ¢ U1 = V2-z . Then ¢ (y- 1 x) 2:: 2- l and so O" (y - 1 x, e) 2::
=

Finally, (ii) is an easy consequence of (iii) and (iv) .

D

The hard work of Birkhoff and Kakutani pays off in a couple of fundamen­ tal consequences-consequences which underscore the special character of topological groups. Corollary 3. 10. Le t G be a topological group. If G has a countable neigh­

borhood base at { e}, then G is metrizable. In this case the metric can be taken to be left invariant.

3.

54

Suppose {Vn

:

Introduction to Topological Groups

E N} is a countable open base at e. Let U1 =

v1- 1 , and let U2 be a symmetric open neighborhood of e such that U2 � U n Vi and U2 · U2 � U1 . Continuing, let Un be a symmetric open

Proof. Vi n

n

1

neighborhood of e such that

Un � U1 n U2 n · · · n Un - 1 n Vn ,

The family {Uk k E N} satisfies the conditions set forth in the Birkhoff­ Kakutani theorem. Further, H = n n Un = { e} . Let a be the left pseudo­ metric introduced in the theorem. In fact, a is a true metric: a(x, y) = 0 if and only if x = y, since after all, H = { e} ! Since :

the· topology defined by a coincides with the given topology of G. Corollary 3 . 1 1 . Let G be a topological group, let a E G, and let

0

F be a

closed subset of G such that a � F. Then there is a continuous real function x on G such that x(a) = 0 and x (x) = 1 for all x E F. Consequently, every topological group is completely regular.

Let U1 be a symmetric neighborhood of e such that (aU1 ) n F = 0. Choose a sequence (Un n 2: 2) of open neighborhoods of e such that each Un is symmetric, Un + l · Un+ l � Un , and let H = nn Un . Apply the Birkhoff­ Kakutani theorem to the Un 's. For x E G, define x(x) by x(x) = min{l, 2a(a, x) } , where a is the left-invariant, uniformly continuous pseudo-metric produced in the Birkhoff-Kakutani theorem. Now x is continuous by (i) of our the­ orem, and x (a) = 0. If x E F, then a- 1 x E a- 1 F, a set disjoint from U1 ; consequently, a- 1 x � U1 and a(a, x) 2: 2 - 1 by (iv) of our theorem. It follows that x (x) = 1 for all x E F. 0 Proof.

:

We'll be spending much of our time in a locally compact setting, and again there's more than meets the eye because of the group's structure. Theorem 3 . 1 2 . Any locally compact topological group is paracompac t, hence

normal.

3.

The Birkhoff-Kakutani Theorem

55

Proof. Let G be a locally compact topological group, and let V be an open set in G containing the identity of G and having a compact closure V. By Theorem 3 . 1 there exists a symmetric open set U such that

Look at H ( hence

= Un un ' where un = � . Then H is an open subgroup n

closed ) of G. Also H is a-compact since

H = LJ .___,_. U · · · U � H' u . . . u .. � LJ .___,_. .. -

n 2n

-

n n

a countable union of compact subsets. Hence H is Lindelof and regular, so H is paracompact [85] . Let U be an open cover of G. Each coset xH of H is also a-compact and so there is a countable subfamily {V;� : n E N} of U that covers xH. Naturally, {Vx� n xH n E N} is an open cover of xH and since xH is paracompact, there is a locally finite open cover of xH that refines {v;� n xH : n E N} , call it {W��}�= l · Note that for each n = 1 , 2, . . . , w�� � xH. For each n = 1, 2, . . . , let :

U w<xHn) •

xHEG/H

so {Wn }�= l is a locally finite open cover. Therefore W = U n wn is an open cover of G that's plainly a-locally finite and refines U. Thus each open cover U of G admits a a-locally finite open cover W that refines U so G is paracompact and therefore normal ( [66] , Theorem 5.28 and Corollary 5 . 32) . D Warning. Not every topological group is normal as exhibited by the fallowing classical example. Example 3. 13. Let m be an uncountable cardinal number. Then zm is a To group, hence completely regular. But zm is not normal.

3.

56

Introduction to Topological Groups

We'll write zm as rri E J '7!,i where each '7!,i is '7J, and I I I = m. For the sake of the present efforts, let A, B � zm be given as follows A = { (xi ) E zm : for any n -j. 0, there is at most one index i for which Xi = n } and B = { ( Xi ) E zm : for any n i- 1, there is at most one index i for which Xi = n } . Then A and B are disjoint. If (xi ) i E J ¢ A, then there are io, ii E J, io i- ii such that for some n E Z, n i- 0, Xio = Xi 1 = n. The set { (Yi) i EI E zm : Yio = Yi 1 = n } is an open set containing (xi ) i E J but no point of A, so A is a closed set. Similarly B is a closed set. Let U, V be open subsets of zm such that A � U and B � V. We claim that U n V i- 0, which will show that zm is not normal. Let (x� i ) ) i E J E zm be defined by x� i ) = 0 for each i E J. Since (x� i ) ) E A � U, there are distinct indices ii , . . . , im 1 E I such that (x? ) ) i EI E { (xi ) i E J E zm : Xi 1 = Xi2 = . . . = Xim1 = O} � u. Let (x� 2) ) i E J be defined by for i = i k , k = 1, . . . , m i , x � 2 ) = Ok otherwise. Since (x? ) ) E A � U there exists im 1 +i , . . . , im2 E J, distinct indices from each other and from ii , . . . , im 1 such that (x� 2) ) i E J E {(xi ) i E J E zm : Xi 1 = 1, Xi2 = 2, · · · , Xim1 = m i , Xim1 +i = 0, • • • , Xim2 = O} � U, Continue in this manner. Define ( Yi)i E J E zm as follows: Yik = k for any k and Yi 1 if i i- i k . Plainly, ( Yi)i E J E B. Hence for some finite subset J of I, { (xi ) i E J E zm : Xi = Yi , for i E J } � V. But J is finite so there is an no such that i k ¢ J whenever k > mn0 • Look at ( zi ) i E J E zm , where if i = i k , k :S ffin0 , if i = i k , mn0 + 1 :'.S k :'.S mn0+l , otherwise. Proof.

i

{

3.

The Birkhoff-Kakutani Theorem

57

So So ( zi ) iEI E { (xi ) iE/ E zm : Xi i = l , Xi2 = 2 , . . . , Ximo = mno ,

Ximno + l = Ximno +2 = . . . = Ximno+l = O} c - u. So ( zi ) iEI E V, ( zi ) iEI E U, and U n V -/= 0. Therefore zm is not normal. D After Thoughts. The missing steps of the Birkhoff-Kakutani theorem. To

prove (4)

r<s

Assume that r < s < 1 , where

=>

Vr � Vs.

r = 2 -l i + 2 -! 2 + . . . + 2 - ln , 8 = 2 - m1 + 2 - m2 + . . . + 2 - mp , and where 0 < li < l 2 < · · · < ln and 0 < m 1 < m2 < · · · < mp are integers. Since r < s, there is a positive integer k such that li = m 1 , l2 = m2 , . . . , lk - l = mk - 1 , but l k > mk . Let W = V2 -11 V2 - 1 2 · · · v;-1 k _ 1 . Then VTtn- 2 v;- ln - 1 V2- ln \1r = W 112 -lk v;-l k + I •

•

•

� W 112- lk V2 - lk + I •

•

� W V2 - lk VTt k +l •

•

•

•

•

•

•

•

•

•

•

•

•

V2 - l n - 2 . V2 - ln-l +I � v; - l n- 2 v; - ln- 2 , •

'-.,.-"'

where the first and third inclusions are due to the decreasing nature of the Un 's, and the second inclusion uses the fact that Uin- i · Uin- i C Uin- i - l · In this way, in these successive inclusions, we can "clip off the ends" one at a time. Continuing in this way, we have that \1r �

W · 112 -l k · V2 - lk +I · · · V2 -l n- 2 · v; - l n- 2

W · V2 -i k V2 -i w . 112-i k +l W · V2 -mk W · V2 - m k · v; -m k +l · · · 112-mp = Vs , � � � �

•

k

where the last inclusion uses the fact that the Un 's each contain the identity

e.

3.

58

Introduction to Topological Groups

With the same representation of r in mind, we set off to prove formula (5) v;. · V2 -1 � l1r+ 2 -1 + 2 .

We suppose that r + 2 - 1 + 2 < 1 . If l > ln , then

v;. · V2 - 1 = l1r +2 -1 ,

and all is well. So we look to the case that l :::; ln . Let k be the positive integer such that Let r i be given by

r i = 2- l + l - 2 - lk - 2 - lk +l - . . . - 2- ln

and It's plain that so

v;. · V2 -1 � ll;.2 · V2 -1 = ll;.2 + 2 -1 � l1r +2 - i+ i +2 -1 � l1r+ 2 -1+ 2 ,

and that's that.

D

( i ) Let G be a locally compact group, let K be a compact subset of G, and let U be an open set containing the identity. Then there is a open set V containing the identity so that for each x E K, Vx � xU. ( ii ) Deduce from ( i ) that if G is a compact metrizable group, then G has a bi-invariant metric generating its topology.

Exercise 3. 14.

( i ) Let G be a locally compact metrizable group with a left invariant metric d generating G's topology. Then ( G, d) is a complete metric space. ( ii ) Deduce from ( i ) and the Birkhoff-Kakutani theorem that every locally compact separable metrizable group is homeomorphic to a complete separable metric space. ( iii ) Suppose H is a continuous homomorphism of the locally compact separable metrizable group G onto the locally compact separable metrizable group G' . Then H takes open sets to open sets.

Exercise 3 . 1 5 .

4.

Products of Topological Spaces

59

4. Products of Topological Spaces

Of considerable importance in many topological affairs is the notion of a product space. Suppose (Sa , TabE A is a nonempty family of nonvoid topo­ logical spaces, and let

IJ Sa aE A be the Cartesian product of the sets {Sa : a E A} . We equip S with the weakest (i.e., smallest) topology T such that for each a E A, the projection Pa : S-+Sa is continuous. So we ask of T that P/; (Ua) E T for each Ua E Ta and each a E A. The collection {Pi; (Ua) : Ua E Ta , a E A} is a subbase for the product topology T . A base T is the family of all finite S=

intersections of the subbase members, i.e., all sets of the form

n

aEF

Pi; (Ua ) ,

where F is a finite subset of A and Ua is an open set in (Sa , Ta) . Here are some easily established facts and features (also known as "exercises" if you please) of the product topology. • Each Pa : S-+Sa is an open map, i.e., Pa (U) is open in Sa for each open set U � (S, T) . • For any topological space T, a function f : T-+S is continuous precisely when Pa o f : T-+S is continuous for each a E A. • If (xd)dEV is a generalized sequence (i.e., a net) in S, then (xd)dEV is convergent to xo E S if and only if for each a E A, we have Paxd = xd, o = Paxo . lim Xd, a = lim v v Of course we are naturally interested in the stability of various topological invariants. Here is a basic result. Theorem 3. 16. The produc t of Hausdorff spaces is a Hausdorff space.

If x, y are distinct members of S = TI A Sa , then there is an a E A so that xa =f. Ya i since Sa is a Hausdorff space there are open sets U and V in Sa so xa E U, Ya E V and U n V =f. 0. The sets Pi; (U) and Pi; (V) D are disjoint open sets in S with x E U and y E V Proof.

3.

60

Introduction to Topological Groups

A result of great importance throughout mathematics is the following famous result of Tychonoff. Theorem 3 . 1 7

(Tychonoff's Product Theorem). The product of compact

spaces is compact.

Proof. (Chevalley and Frink [17] ) . Let F be a family of subsets of the product S of the compact spaces Sa, a E A, and suppose that any finite subfamily of F has a nonempty intersection. Our aim is to show

n F tf 0 . F

Let M be a maximal system of subsets of S with the finite intersection property (FIP) containing F. Consider

Ma = {Fa = PaF: F E M}. Then each Ma has the FIP. Each Sa is compact so for each a E A there is a point Pa common to n Fa. FEM Let p be the point of S with Pap = Pa · We want to show that p E n F. FEF Now the very way we chose Pa ensures that if Ua is any open set containing Pa in Sa, then Ua has a point in common with any member of Ma. Hence, in S, the open set Wa = P/;(Ua) containing p has a point in common with each member of M. But M is maximal so Wa must belong to M. Similarly each and every finite inter­ section of open sets such as Pi;(Ua) must belong to M. But M is still maximal so M contains all such finite intersections. But sets of this sort form a basis for the open sets containing p: each open set in S that contains p is a member of M. Therefore every open set containing p has a point in common with each member of M. Hence p is in M for each M E M, and so D p E n F. FEF Exercise 3 . 18. Let S be a topological product of the spaces (Sa, ra). If each Sa is connected, then so is S. If each Sa is T1 , then so is S. •

•

5. Notes and Remarks

61

•

The product of countably many spaces, each of which has a count­ able basis for its topology, has a countable basis for its topology. • The product of locally compact spaces is locally compact if (and only if) each coordinate is locally compact and all but finitely many are compact. What about topological groups? Well, it's easy to see that if (Ga , Ta) is a topological group for each a E A and G = IT Ga A

is the Cartesian product, then G is a group with coordinate-wise operations and a topological space. It is, in fact, a topological group as well. To see this, let x = ( xa ), y = (Ya) be points in G, and let U be any open set in G that contains xy. Then, by definition of the product topology, there are open sets Ua in Ga containing XaYa for all a such that, except for finitely many a 's, Ua = Ga. But each Ga is a topological group, and so for each a we can find open sets Va and Wa such that x a E Va, Ya E Wa and Va · Wa � Ua . Moreover if we choose Va = Wa = Ua whenever Ua = Ga , then V = IT va A is open in G, contains x, and W = IT wa A is open in G, contains y and V · W � U. Exercise 3. 19.

Modify this argument to show that x t-+ x - 1 is also contin­

uous on G. Of course, in tandem with Tychonoff's theorem, this provides us with a plethora of compact groups. If we take into account the last statement of Exercise 3 . 18 we also have a boat load of locally compact, noncompact groups. 5.

Notes and Remarks

The definition of topological group, as we use it, is due to F. Leja [69] ; soon after a large number of mathematicians were uncovering interesting and important features of the subject. The early development can be traced through the papers of D. van Dantzig [129] , [130] , H. Freudenthal [40] and J. von Neumann [133] . The monograph of L. Pontrjagin [104] organized

62

3.

Introduction to Topological Groups

much of what was known and presented it in a palatable form. Often the hy­ pothesis that the topology satisfy the second axiom of countability (thereby ensuring that the group's topology is metrizable) was added. A. Weil's monograph [136] rid the added hypothesis of second countability wherever possible. A. Weil also gave us the theory of uniform spaces [135] isolated with the express purpose of providing a common ground for studying metric spaces and topological groups. In particular, the realization that every topological group is completely regular is due to Weil. Our presentation of the classical locally compact groups was greatly influ­ enced by H. Weyl's [137] monograph, The Classical Groups. The Birkhoff-Kakutani theorem is due to G. Birkhoff [11] and S. Kakutani [60] . Kakutani noted that his metric could be chosen to be left invariant. That locally compact topological groups are normal was pointed out (to E. Hewitt and K. Ross [55] ) , by E. A. Michael. It seems like the paracompact­ ness was first noted by Hewitt and Ross in their first volume on Abstract

Harmonic Analysis. That zm. is not normal if

m is an uncountable cardinal is due to A. H. Stone [122] who also noted that a product rri EZ Gi of topological groups being normal implies that all but a countable number of the groups are countably compact. The subject of topological groups holds many fascinating avenues of inves­ tigation and has been given correspondingly fine expositions. In addition to the monographs already mentioned, the books of P. J. Higgins [56] , K. H. Hoffman and S. A. Morris [57] , D. Montgomery and L. Zippin [84] and S. A. Morris [86] should keep the reader busy and entertained.

Chapter 4

Banach and Measure

Like most abstract analysts of his day, Banach took a keen interest in de­ velopments related to invariant measures. In this chapter, we present some of Banach's work in this direction. We open with his discovery of Banach limits, a remarkable consequence of the Hahn-Banach theorem ( which we also prove ) . We follow this with Banach's proof of the existence of a Borel measure on a compact metric space invariant under "congruence" ; in par­ ticular, it is to be noted that there exists a left-invariant probability Borel measure on any compact metrizable topological group. After this we take a slight detour to present S. Saks' proof that the positive linear functionals in C(K)* , the dual of the space C(K ) of continuous real-valued functions defined on a compact metric space K, consist of regular Borel measures on K.

The result on the dual of C(K) , when K is a compact metric space, was first derived by Banach, and we present his proof in Section 4. We have included Banach's Daniell-like description of the positive linear functionals in C(K)* for its elegance. To see that his abstract Lebesgue integral does indeed apply to the case of positive linear functionals on C(K) , where K is a compact metric space, requires us to present some of the most over­ looked results appearing in his Theorie des Operations Lineaires [7] , results that demonstrate his complete understanding of life with bounded, finitely additive measures. Particular attention ought to be paid to Banach's approach to Lebesgue integrals, an approach taken to describe C(K)* in terms of ( regular ) Borel measures on K. Like his work on invariant measures, this appeared in the appendices of Saks' beautiful book Theory of the Integral [1 1 1 , 1 1 3] . -

63

4. Banach and Measure

64

We follow Banach's view of the abstract Lebesgue integral with his solution of the so-called easy problem of measure for the real line and the plane. To our knowledge, Banach's proof has slipped through the cracks of mathe­ matical knowledge. We follow with a very brief discussion of the Hausdorff paradox, and its role in solving the easy problem of measure negatively for Euclidean spaces of dimension greater than two. An all-too-brief discussion of amenability follows. 1 . Banach Limits

(Hahn-Banach Theorem). Let X be a real linear space, and let S be a linear subspace of X. Suppose that p X --+ JR is a subadditive, positively homogeneous functional and f S --+ JR is a linear functional with f(s) :S p(s) for all s E S. Then there is a linear functional F defined on all of X such that F(x) :S p(x) for all x E X, and F(s) = f(s) for all s E S. Theorem 4.1

:

:

Proof. Our first task is to see how to extend a functional like f one dimen­ sion at a time while preserving the domination by p. With this in mind, let x E X\S, and notice that for any linear combination s + ax of a vector in S and x, whatever the linear extension F's value at x (say that value is "c" ) , we must have

F(s + ax) = F(s) + aF(x) = f(s) + ac.

So it must be (if we are to have F dominated by p) that f(s) + ac :S p(s + ax)

holds for all s E S and all real a 's. We'll follow where this leads us. For all a E JR and s E S we have to have ac :S p(s + ax) - f(s) .

For a > 0 this tells us that c <

� (p(s + ax) - f(s) ) p ( ; + x) - f ( ; ) .

While if a < 0, then ( - a > 0 and) - ac 2:: f(s) - p(s + ax) ,

1.

Banach Limits

65

or since - a > 0, c >

ax) - -a f (s) - -p(s -a + 1

f

1

( :a ) - P ( :a - x) .

Taking into account the linearity of S, we see that what we seek is a c E � (which will be F ( x)) so that regardless of s , s ' E S satisfies f (s 1 ) - p(s ' - x) ::; c ::; p(s + x) - f (s) . Does such a c exist? You bet! After all if s , s ' E S, then f(s) + f (s 1 ) = f ( s + s 1 ) ::; p(s + s1 ) ::; p(s - x) + p(s1 + x). So for all s , s ' E S f (s) - p(s - x ) ::; p( s1 + x) - f (s' ), and we can chose c in an appropriate manner. Now we know that we can extend linear functionals one dimension at a time while preserving p's domination. It's time for some transfinite hijinks. We consider the collection of all linear functionals 9 defined on a linear subspace Y of X such that S � Y, 9 l s = f, and, on Y, 9 ::; p. We partially order this collection by saying that "91 ::; 92 " if 92 is an extension of 91 (so 9 1 is defined on a linear subspace Y1 that is contained in 92 's domain). Hausdorff's maximal principle ensures us that there is a maximal linearly ordered subfamily {ga } of linear extensions of f so that on 9a 's domain Ya (which contains S) , 9o: ::; p. We define F on the linear space that's the union of the domains of the 9a 's as one might expect! So F(x) = 9a (x), if x is in 9a 's domain. Because of the ordering described above, the domain of F is a linear subspace of X and on that domain F is linear and dominated by p. Of course, F is a linear extension of f and the domain of F must be all of D X -this is assured us by the opening salvo. We put the Hahn-Banach theorem to immediate use by establishing the existence of "generalized limits" or Banach limits, as we'll refer to them henceforth. We will call on two spaces: l 00 , the space of bounded, real­ valued sequences; and the linear subspace c of l 00 , consisting of all the convergent sequences. Typically if x E l 00 , then l l x l l oo = sup{(xn) : x E N} . Theorem 4.2

that

(Banach). There exists a linear func tional LIM on l 00 such

(i) LIM(x) 2: 0 if x = (xn ) E l 00 and Xn 2: 0 for all n;

4. Banach and Measure

66

( ii ) J LIM(x) J :S J lxJ J 00 , for all x E Z 00 ; ( iii ) If x E z oo and Tx = (x 2 , x3, . . . ) for x = (x 1 , x 2 , . . . ) then LIM(Tx) = LIM(x) ; ,

( iv) For any x E z oo

lim sup x . Xn :S LIM(x) :S lim n--+inf n--+ oo n oo

Let p z oo -T � be given by . sup X1 + · · · + xn . p(x) = hm n n --+oo It is easy to see that p is subadditive and positively homogeneous. Next, let f c-T � be the linear functional J(x) = lim n xn . Since f and p agree on c, f(x) :S p(x) is trivially satisfied. LIM is any Hahn-Banach extension of f to all of z oo . LIM is a linear functional on z oo such that for any x E z oo LIM(x) :S p ( x ) . To see item ( i ) , take x E z oo and suppose Xn � 0 for all n. Then Proof.

:

:

- LIM(x)

so LIM(x)

>

=

LIM ( - x) :S p( - x) ,

sup - x 1 - x 2 n- · · · - xn - lim n -TOO + X 1. . 1nf I X2 +n · · · + Xn >- O , nim-Too -p(x)

=

because Xn � 0 for all n . Item ( ii ) is plain since p(x) :S l l x J J 00 for all x E Z 00 • To see item ( iii ) , take x E Z00, then . sup X I - Xn+l = 0, p(x - Tx) = hm n n-TOO since x is bounded. It follows that LIM(x - Tx) :S p(x - Tx)

=

0.

=

- LIM(Tx) ,

So for any x E z oo (6)

LIM(x) :S LIM(Tx) .

This applies as well to - x so - LIM(x)

=

LIM (-x) :S LIM( - Tx)

2.

Banach and Haar Measure

67

and so LIM(x) 2: LIM(Tx) as well. LIM(x) = LIM(Tx) is the only conclusion that can be drawn from (6) and (7) . For item (iv) , let E > 0. Find N E N so inf n Xn � x N � inf n Xn + E . Then Xn + E - XN 2: inf n Xn + E - XN 2: 0 . Hence (7)

by (i). Hence

inf n Xn � XN � LIM(x) + E . Since E > 0 was arbitrary, inf n Xn � LIM(x). For any k but

limninf Xn = sup kinf n ?:_ n Xk � sup n LIM(x) = LIM(x). Again using LIM's linearity, we see that lim sup xn = - lim inf( -xn) 2: -LIM(-x) = LIM(x) . n-+oo n

The proof of the existence of Banach limits appears.

D

2. Banach and Haar Measure

Let Q be a fixed compact metric space. Definition 4.3. We suppose that for subsets of Q the notion of congruence is defined to satisfy the following conditions (here A � B means A is congruent to B) : (i) A � A. (ii) A � B {:} B � A. (iii) A � B, B � C ==:;. A � C. (iv) If A is an open set, then so is any set congruent to A.

4. Banach and Measure

68

(v) If A is congruent to B and A can be covered by a sequence ( An) of open sets, then B can be covered by a sequence (Bn ) so that Bn � An for each n. (vi) For any open set A the collection of sets congruent to A cover Q. (vii) If (Sn) is a sequence of open concentric balls with radii tending to zero, and if Gn � Sn and an , bn E Gn with limn an and limn bn existing, then these limits coincide. Proposition 4.4. If G is a compact metrizable group, then A

A = gB for some g E G is a congruence.

� B whenever

Proof. Let ( Sn ) be a sequence of concentric open balls (with common center c) with radii tending to zero. Suppose ( An ) satisfies An � Sn for each n and

suppose an , bn E An have limn an = a, limn bn = b. We claim that a = b. Since An � Sn , there exists gn E G such that An = gn Sn . Hence an = gnSn and bn = gn sn for some Sn, s � E Sn . Plainly sn -+c, s� -+c, an -+a, and bn -+b so gn = an s n- 1 -+ac- 1 and gn = bn ( Sn ) - 1 --+ bc - 1 . D Limit one to a customer, so ac 1 = bc- 1 and a = b as we claimed. I

I

Proposition 4.5. Let Q be a compact metric space, and let G be a transitive

equicontinuous group of homeomorphisms of Q onto itself. If A, B E Q, then A � B whenever A = g(B) for some g E G is a congruence. Proof. Transitivity of G is simply the feature that if qi , q2 E Q, then there is a g E G so that g (q1 ) = q2 . Because members of G are homeomorphisms of Q onto Q, (iv) and (v) of Definition 4.3 are so. Transitivity of G assures us of (vi). Equicontinuity just says that given an E > 0, there is a o > 0 so if x , y E Q and o( x , y ) ::; o, then d ( f ( x ) , f ( y)) ::; dor all f E G. Because each g E G is a homeomorphism of Q onto itself, each such g is a uniformly continuous map on the compact set Q as is g- 1 . So G is an equiuniformly continuous group of uniform isomorphisms of G onto G. Thus (vii) follows from this. D

Given two nonempty open sets A, B, by Definition 4.3(vi), the collection of sets congruent to A covers B, a compact set. Hence there is a finite

69

2. Banach and Haar Measure

collection of sets congruent to A that still cover B. This motivates Haar's covering function h( B, A) : h( B, A) = the least number of sets congruent to A needed to cover B. Proposition 4.6. Suppose A, B, and C are nonempty open subsets of

Then

(i) (ii) (iii) (iv) (v)

Q.

C � B => h(C, A) :S h(B, A) . h(B U C, A) :S h(B, A) + h(C, A) . B � C => h(B, A) = h(C, A) . h(B, A) :S h(B, C)h(C, A) . If d(A, B) = distance from A to B is positive (so A n fJ = 0) and (Sn) is sequence of open concentric balls with radii tending to zero, then there is a number N so that for n 2: N h(A u B, Sn) = h(A, Sn) + h(B, Sn) ·

Item (v) requires some serious and careful attention. Suppose (v) fails. Then there is ( n k) so that Proof.

h(A u B, Snk )

< h(A, Snk ) + h(B, Snk )

for each k. We can plainly suppose that the nk's are chosen so large that Snk n A i= 0 and Snk n B i= 0 cannot both occur. It follows that there is a sequence (Gk) such that Gk � Snk and Gk n A i= 0 and Gk n B i= 0. Why? We'll fix nk momentarily and imagine that any G that is congruent to Snk could meet at most one of A and B. If we cover A U B by h(A U B, Snk ) many sets congruent to Snk , then this cover (call it C) would be the disjoint union of the collection A (respectively B) where A (respectively B) consists of the members of C that meet only A (respectively B) . Consequently, h(A U B, Snk ) = I C I = I A I + I B I 2: h(A, Snk ) + h(B, Snk ) ,

which is not an option. So we get a sequence (Gk) of sets with Gk � Snk and Gk n A i= 0, Gk n B i= 0. From each of the sets Gk n A, pick a point ak and from each Gk n B, pick a point bk. The sequences (a k ) , (bk) lie inside Q so there is a .lJ E P (N) so that a = lim a) ' b = lim b jEJI J jEJI 00

·

·

both exist. Of course, a E A, b E fJ . But now we're in precisely the position to which Definition 4.3(vii) is applicable: ak, bk E Gk, Gk � Snk · Hence a = b. But A n B = 0. OOPS! The denial of item (v) leads to unnecessary 0 chaos.

70

4. Banach and Measure

Let (Sn ) be a sequence of open concentric balls with radii tending to zero. For any open set A � Q, define ln (A) = h (A, SSn )) ' h(Q, n Then h(A, Sn ) � h(A, Q) . h(Q, Sn ) and h(Q, Sn ) � h(Q, A) . h(A, Sn ) tell us that 1 � ln (A) � h(A, Q) . h(Q, A) Therefore (ln (A)) is a bounded sequence of real numbers, each of whose terms exceeds the fixed positive number 1 / h( Q, A) . Let LIM be a Banach limit, that is, LIM E B1* , and LIM satisfies lim inf x � LIM(x) � lim sup x (8) for each and every x E l00 • Let l(A) = LIM((ln (A))) for any open set A � Q. D Note. We did not use the shift-invariance of LIM; rather we just used the fact that it satisfied (8) . CXl

Proposition 4. 7. If A and

(i) (ii) (iii) (iv) (v)

B are open sets, then 0 < l(A) < oo, as long as A -/= 0. A � B =? l(A) � l(B). l(A U B) � l(A) + l(B) . A � B =? l(A) = l(B) . If d(A, B) > 0, then l(A U B) = l(A) + l(B) .

To see Proposition 4.7(v) , note that by Proposition 4.6(v), there exists an N such that for all n � N, h(A U B, Sn ) = h(A, Sn ) + h(B, Sn ) · From this we easily see that for all n � N , ln (A U B) = ln (A) + ln (B) , and (v) follows. D Proof.

71

2. Banach and Haar Measure

Let X � Q. Define >.( X )

as

{

follows:

}

>.( X ) = inf L l ( An) : X � LJ An , An open . n n Here are the fundamental properties of >.. Theorem 4.8. Let Q be a compact metric space. Then

( i ) 0 s >.( X ) . ( ii ) If X is a nonempty open subset of Q, then 0 ( iii ) X � Y � Q ==> >.( X ) s >.( Y ).

< >.( X ) <

oo.

( iv ) X � U n Xn ==> >.( X ) S L: n >.( Xn) · ( v ) X � Y ==> >.( X ) = >.( Y ). ( vi ) d ( X, Y ) > 0 ==> >.( X U Y ) = >.( X ) + >.( Y ) .

Proof. Items ( i ) - ( iv ) tell us that >. is an outer measure, item ( v ) assures us that >. respects congruence, and item ( vi ) says that >. is a "metric outer measure" . It is a known consequence of A's metric outer measure character that every Borel set B � Q is >.-measurable. Item ( i ) deserves comment. If X is open, then >.( X ) s l ( X ) < oo. If, in addition, X =I- 0, then for any E > 0, we can find a sequence ( An) of open sets so that X � Un An and

n Now if S is an open ball centered at a point in X and a subset of X, then only finitely many of the An 's, say A i , A2 , . . . AN, are needed to cover S. Then 0 < l (S) S l (S) S l ( A 1 U · · · U AN ) S L l ( An) < >.( X ) + E , n and 0 < >.( X ) follows. Item ( vi ) , too, deserves proof-after all, it's item ( v ) that ensures that >. is a metric outer measure. By Proposition 4.7 ( iv ) , it suffices to show that >.( X ) + >.( Y ) :::; >.( X U Y ) . If d ( X, Y ) > 0, then there are disjoint open sets U, V such that d ( U, V ) > 0 with X � U and Y � V; this is so thanks to

4. Banach and Measure

72

normality of metric spaces, if you please. Let E > 0. Pick a sequence of open sets such that

(An)

n

and

Ln l(An) :::; A(X U Y) + E . Now each of the sets An n U, An n V are open and d(An n U, An n V) 2: d(U, V) > 0.

Hence by Proposition 4.7(v)

l((An n U) U (An n V)) = l(An n U) + l(An n V) :::; l(An ), where the last inequality follows since An n U and An n V are disjoint open sets whose union is a subset of An . Further So

n

n A(X) :::; L l(An n U),

n

This in turns ensures that A(X) + A(Y) < < <

A(Y) :::; L l(An n V) . n

Ln l(An n U) + Ln l(An n Y) Ln (z(An n U) + l(An n V) ) Ln l(An) ·

It follows that A(X) + A ( Y ) :::; A(X U Y) .

D

From this we know that A is a metric outer measure on Q which assigns the same value to any pair of congruent subsets of Q. Hence the collection of A-measurable sets is a a-field of subsets of Q which contains the Borel a-field, and on this a-field, A is countably additive and assigns congruent measurable sets the same measure. Exercise 4.9. Notice that the full strength of Banach limits was not used in the above proof. Indeed, what was used was that LIM satisfies that it is a linear functional on the space of all bounded real-valued sequences such that for any bounded real sequence x lim inf x :::; LIM(x) :::; lim sup x.

3.

Saks ' Proof of C( Q)* , Q a Compact Metric Space

(9)

73

(i) If F is a positive linear functional on the Banach space of bounded real-valued sequences, then F satisfies lim inf x � F(x) � lim sup x. (ii) Denote by ,BN the Cech-Stone compactification of the (discrete) space N of natural numbers. The defining characteristic of the compact space ,BN is that it contains N as a dense subspace and every bounded real-valued function defined on N has a unique continuous real-valued extension to all of ,BN. So C(,BN) is isomet­ rically isomorphic to £00 • If we take a point z E ,BN that's not in N, then it defines a contin­ uous linear functional Fz on C(,BN) (or what's the same £00 ) via the formula Fz ( J ) = f ( z ) . The functional Fz satisfies condition (9) .

The two parts of this exercise combine to establish the beautiful result of Nakamura and Kakutani [93] to the effect that there are 2c many generalized Banach limits-after all, there are 2c many points in Remark 4.10.

,BN.

The opening steps of Banach's proof of the existence of a Haar measure on a compact metrizable group follow Haar's argument. One should be aware that both Haar and Banach were interested in "bigger fish" , and each dealt with the case of locally compact, separable metrizable underlying space. We opted to follow Banach's approach because it broaches the subject of homogeneous spaces (and the groups that act on them), and because of our functional analytic prejudices. 3.

Saks' Proof of C( Q)* , Q a Compact Metric Space

Soon after the appearance of Saks' monograph [1 1 1] , Saks published an alternative proof of the theorem of Banach [1 12] regarding positive linear functionals on C(Q) , Q a compact metric space. This proof relies on the theory of metric outer measures to ensure that the resulting measure is a Borel measure. We set our notation. Let Q be a compact metric space (with metric d) , C(Q) is the Banach space of continuous real-valued functions defined on Q, and (to be consistent with Saks) q> is a positive linear functional on C(Q) . Recall that l q. (x) I � q. (1)

74

4. Banach and Measure

whenever x E Bc ( Q) • so is a member of C(Q) * with norm (l). For any q E Q, r > 0, denote by Ur(q) and Br (q) the sets Ur (q) = { y E Q : d(q, y) < r } , Br(q) = { y E Q : d(q, y) :::; r } . Stage I. For E � Q, define >.(E) by >.(E) := inf{(x) : x E C(Q), x(q) � XE(q), for all q E Q } . Here's what's so about >.: (i) if A � B, then >.(A) :::; >. (B) ; (ii) if A, B � Q, then >.(A U B) :::; >.(A) + >. (B) ; (iii) if d(A, B) > 0 (which is the same as A n B = 0) , then >.(A U B) = >.(A) + >.(B). Item (iii) demands comment and proof, even. Let E > 0. Pick x E C(Q) so x(q) � XAus (q) for all q E Q (where A, B � Q satisfy d(A, B) > 0) and so that (x) :::; >.(A U B) + t: . Next chose h E C(Q) so that 0 :::; h(q) :::; 1 for all q, with h(q) = 1 for q E B, h( q) = 0 for q E A. Let x A = ( 1 - h )x and x B = hx. Both xA , x B E C ( Q) ; also x(q) if q E A > XAus (x) if q E A X A (q) 0 if q E B 0 if q E B =

and

{ { 01

-

{

if q E A if q E B = XA ( q)

xB (q) =

{ x(q) 0

if q E A if q E B -> XB ( q )

for all x E Q. It follows that >.(A) + >.(B) < (x A ) + (x s ) = (x A + x s ) = (x) < >.(A U B) + t: . Stage II. Let E � Q, and define µ(E) by f'(E) '= inf

{�

A(Gn) ' G. is open, E <:;

� Gn } .

3.

Saks ' Proof of C(Q)*, Q a Compact Metric Space

75

Then µ is an outer measure on Q, a metric outer measure, with the added property that for any closed subset F of Q, µ(F) = >.(F). We'll establish this last claim. Suppose Gn is a sequence of open sets so that F � LJ Gn . n Since F is closed, Q's compactness is inherited by F, and so we can find N such that It follows that It follows that

N

>.(F) :S L >.(Gi ) :S L >.(Gn) · n i= l

>.(F) :S µ(F) . Now let E > 0 be given. Pick x E C(Q) so that x ( q ) 2:: XF(q) for all q and (x) :S >.(F) + E . Look at the open set G = {q E Q : x ( q) > ( 1 + t:) } . So F � G since if q E F, then x ( q ) 2:: XF (x) = 1 . So µ(F) < >.(G) (since G is open and F � G) < (x) (by definition of >.(G))

<

(x)

(>.(F) + E ) . Let E \.0. Then µ(F) :::; >. (F) follows. Stage III. Like all metric outer measures, µ has among its µ-measurable sets each and every Borel subset of Q. In particular, each x E C(Q) is µ­ measurable, and of course, bounded. Hence each x E C(Q) is µ-integrable. Let's check (x) vis-a-vis J xdµ. We'll show that (x) ::;

j x dµ

for every x E C(Q) . Take x E C(Q). Let E > 0. Realizing that µ(Q) = (l) , we assume that x ( q ) > 0 for all q E Q-just add an appropriate constant to x and note that this has the exact same effect on (x) and J xdµ, leaving their relationship unchanged. Choose rJ > 0 so that if d ( q , q1 ) :S rJ, then l x(q) - x(q' I :S E . Cover Q by open balls Ur 1 ( q1 ) , . . . , Urn (qn ) , of radii r 1 , . . . , Tn , with each < rJ/2 centered at q1 , . . . Qn , respectively, with the added feature that µ( {y E Q : d(y, Qi ) = ri } ) = 0.

4.

76

Banach and Measure

This last feature can be assumed since µ(Q) < oo, and so for a fixed qo E Q, only countably many of the sets { y E Q d(y, qi) = r } can have positive µ-measure. Now that the U's are in place, set :

Ei Ur 1 (q1 ) E2 = Ur2 (q2 )\E1 E3 = Ur3 (q3 )\(E1 U E2 )

Each set Ei , . . . , En is closed with diameter :S 'f} and Q = Ei U · · · U En .

What's more-and this is crucial-the Ei's overlap only in a set of µ-measure O! Let mi = min{x(q) : q E Ei } , i = 1 , 2 , . . . , n . Consider n

a(q) = L miXEJ q) ; i=l notice that x(q) ;:::: a(q) with µ-almost everywhere. Hence n

n

L miµ( Ei) = L mi>.( Ei) · i 1 i=l For each k = 1 , . . . , n , pick Xk E C(Q) so that for all q E Q, xk(q) ;:::: XEk (q) and -

Put Since the oscillation of x on Ek is no more than E , then n

u(q) = L mi xi(q) + E ;:::: x(q). i=l

77

4. The Lebesgue Integral on Abstract Spaces

It follows that

Jxd

i=l > tmi( � (xi) - : ) n i i=l � (tmi Xi) - E i=l �(u)(u - E) -(E)E E � -� > � (x) - E(� ( l ) Let E 0 and, be happy, don't worry-after all, J xd � �(x) for all x C(Q). µ

>

.

+ 1).

�

µ

D

E

4. The Lebesgue Integral on Abstract Spaces

In this section we present Banach's approach to the Lebesgue integral in abstract spaces. Banach's approach is a Daniell-like construction built using his clear and deep understanding of lim sup's and lim inf's. His starting point is a positive linear functional f acting on a vector lattice C of real­ valued functions defined on some set K; we suppose (with Banach) that the functional satisfies a kind of Bounded Convergence theorem (BCO) on C. It is important to note that in the previous section we presented Banach's famous result characterizing weakly convergent sequences in spaces C Q) , Q a compact metric space; it follows from this that should the initial vector lattice C be such a C(Q) , then every positive linear functional satisfies the (BCO) hypothesis. After an initial discussion of technical consequences of the (BCO) hypothesis involving lim sup's and lim inf's of functions in C, Banach introduces an upper and a lower integral. Were we doing measure theory, this piece of the puzzle would be concerned with properties of outer and inner measures generated from an initial set function. Next the class of integrable functions is isolated; it is identified as those real­ valued functions for which the upper and lower integrals coincide and are simultaneously finite. The classical Monotone and Dominated Convergence Theorems are derived, and all is well with the world.

(

4. Banach and Measure

78

We follow with a discussion of what the construction does in the all-impor­ tant case that the initial vector lattice C = C(Q) , with Q a compact metric space. It is noteworthy that this construction of Banach led him to a description of C(Q)* . It is impossible to tell for certain (but easy to imagine) what Saks thought of Banach's "construction" of C(Q)*. It appeared after all, as an appendix to Saks' classic monograph Theory of the Integral [1 1 1] , yet uses practically none of the material from the monograph! Whatever Saks thought, he soon presented an elegant proof of Banach's result about C(Q)* in the Duke Mathematics Journal [1 12] . We presented Saks' proof in Section 3. (The numbering and notation throughout this section is consistent with Banach's [1 1 1] . ) Let C denote a vector lattice of real-valued functions defined on a set Q (i.e., if x, y E C, then so is x V y = inf { x, y} and x /\ y = sup{ x, y}). A linear functional f on C is a positive linear functional if f(x) � 0 for any x E C, x � 0. Throughout this section, we will suppose f is a positive linear functional on C satisfying if (xn ) � C, M E C, with l xnl ::; M and limn Xn (t) = 0 (BCO) for all t E Q, then limn f(xn ) = 0. If C = C(Q) , Q a compact metric space, then any positive linear functional f on C satisfies (BCO) thanks to Theorem 4. 24. This is an example well worth keeping in mind. 1° If (xn ) � C, m E C, z � 0, xn � m, and lim infn Xn � z, then lim n xn - lxnl = 0. Now lim infn Xn � z ensures lim infn xn (t) � 0 for each t E Q; that is, for each t E Q, nlim -+oo inf{xn (t), Xn+ i (t) , . . . } � 0. So, given 'f/ > 0 there's n = n ( 'f/) so that for all k � n, 4 . 1 . A Start.

{

(10)

Look at Xn - l xnl : (xn - l xnl )(t) =

{

if Xn (t) � 0 if Xn (t) < 0. Naturally, (xn - l xn l )(t) ::; 0 for all t. So, if limn (Xn - l xnl ) exists, it must be less than or equal to 0.

��n (t) ,

4. The Lebesgue Integral on Abstract Spaces

79

Let's check on contrary possibilities. Can it be that limninf(xn - l xnl ) (to) < 0 for some to E Q? Let's suppose that this is possible. Notice since m E C limninf(xn - l xnl ) (to) 2: inf{2m(to), O} > - oo. If limninf(xn - l xnl )(to) < 0, it's because there's a subsequence (x� ) of Xn so that for some Eo > 0 2x� (to) = x� (to) - l x� (to) I < - Eo for all n. It follows that , - Eo xn (to) < 2 for all n. But if n is BIG we can arrange , - Eo xn (to) > 4 (that's what our opening words (10) of wisdom guarantee). Aha! While we also have

- EQ · xn (to) < 2 Drawing the conclusion from this that � < � is easy and leaves us with a clear-cut contradiction to all that is right in our world. We conclude that limninf(xn - l xn l ) 2: 0. We know that limnsup(xn - l xn l ) :S 0, D and so 1° follows. 2° If (xn ) � C, m E C, Xn 2: m, and lim infn Xn 2: 0, then limninf f(xn ) 2: 0. Again start with a look at Xn - l xn l · As before since 2m E C, - oo < inf {2m, O} :S Xn - l xnl :S 0. By 1°, we know that lim(x n n - l xn l ) (t) = 0 for each t E Q, and since C is a lattice, we can use the the (BCO) condition on f to conclude lim n f(xn - l xnl ) = 0. I

4. Banach and Measure

80

Can lim infn f(xn ) < O? If so, then it's because there is a subsequence (x� ) of (xn ) and an Eo > 0 so that f(x� ) < - Eo for all n. We can (and do) assume that lim n f(x� ) exists as well. But now lim n f(x� ), lim n f(x� - l x� I) both exist, and so lim n f( l x� I) exists too with lim lim(f( n f( l x� I) n l x� I - x� ) + f(x� )) lim n f(x� ) :S - Eo, 0 which is not possible, and so we have 2°. Let £* denote the set of all real-valued functions z on Q for which there exist two sequences (xn ), ( Yn ) � C such that lim sup Yn :S z :S limninf Xn . n C* is a linear space containing C since C is a vector lattice. Given z E £*, the upper integral of z, denoted by f (z), is defined by inf {limninf f ( Xn ) : there exists m E C, ( Xn ) � C, Xn 2:: m, limninf Xn 2:: z}; the lower integral of z, denoted by l(z), is defined by sup{lim sup f(xn ) : there exists M E C, (xn ) � C, xn :S M, lim sup xn :S z}. n n Obviously 4.2. Upper and lower integrals.

jz = - 1(-z).

Note. In each of the above definitions we can suppose that limn f(xn ) exists and is real valued since lim inf's and lim sup's are taken over sequences which are eventually finite. So we replace lim infn f (xn ) and lim supn f (xn ) with limn f(xn ) throughout. From the definitions we have 3° If z E £,* , z 2:: 0, and f (z) < P < oo, then we can find (xn ) � C, Xn 2:: 0, lim inf Xn 2:: z with f(xn ) < P for all n.

4. The Lebesgue Integral on A bstract Spaces

81

The value of 3° is found in the accessibility it affords us to epsilonics; since ( with Banach) we frequently call on computing lim sup's and lim inf's. This is a critical aid. Lemma 4. 11. For any x E C, J (x)

= f(x) .

On the one hand, we can let Xn = x for all n and m = x. This done, we plainly have limninf Xn ;::: x, and Xn ;::: m; hence (x) ::; li�inf f(xn ) = f(x). On the other hand, if (xn ) � C and m E C with limninf Xn ;::: x, and Xn ;::: m, then limninf ( Xn - x) ;::: 0 and Xn - x ;::: m - x. 2° steps in to say "O :S limninf J (xn - x) = limninf J (xn ) - f(x)" ; we see that f(x) :S limninf f(xn ) , and with this we conclude 0 f(x) :S (x). Lemma 4.12. If z1 , z2 E £,* with f( z1 ), f ( z2 ) < oo, then f( z1 + z2 ) < f ( z1 ) + f ( z2 ). Proof.

J

J

Proof.

Suppose P1 , P2 are numbers such that

J(z1 ) < P1 and J(z2 ) < P2 .

There are sequences (x�1 ) ), (x�2 ) ) � C and functions m 1 , m2 E C such that x�1 ) ;::: m 1 for all n, limninf x�2 ) ;::: z2 , x�2 ) ;::: m2 for all n, and 2) 1) lim n f(x� ) < P1 , lim n f(x� ) < P2 . Letting Xn = x�l ) + x�) and m = m 1 + m2 , we see that limninf Xn ;::: z1 + z2 , Xn ;::: m.

4. Banach and Measure

82

It follows that

J ( z1 -

+ z2 )

2) 1) ::::; lim n f(xn ) = lim n f(x� ) < P1 + P2 . n f(x� ) + lim

Enough said.

D

z E £* , J (z) ::::; J(z) . There is nothing to prove if J ( z) = If J ( -z) = (z) = -f(-z), so again there is nothing to prove. If f(z), J(-z)

Lemma 4.13. For any

+oo.

Proof.

f

+oo,

then Lemma 4.12 kicks in to give 0 = f(O) =

so that Lemma 4. 14.

If

then

< oo,

J(o) = J(z - z) ::::; J(z) J(-z) +

z E £* and f(z) then J ( z � lzl ) J(z) = J ( z �l z l ) J ( z � l z l ) .

D

< oo,

< oo

and

+

f(z) z,

m l2 l - m -2 l m l.'

m m,

Suppose < P < oo . Find E C and (xn ) � C so that Xn 2: for all n, and lim infn Xn 2: limn f(xn ) < P. Notice that if Xn 2: then Xn - xn > Proof.

this can be seen by a simple analysis of cases. Hence f Xn f Xn Xn

( � l xnl )

( ( � l xnl )) x J (xn ) f ( Xn � l n l ) f(xn ) - 1 ( m � l m l ) _

_

<

4. The Lebesgue Integral on Abstract Spaces

and

J ( z � lzl )

83

( � l xnl ) li[? f(xn ) - f ( m � l m l )

< limninf f Xn <

00 .

<

Now P > lim n f(xn ) > limninf f Xn � l xn l

(

>

J ( z � lzl ) J ( +

)

(

limjnf f Xn � l xnl z � lzl +

}

)

it follows from P's arbitrary nature among members > f ( z ) that z lzl + z � lzl . (z) �

J J( � ) J(

)

Lemma 4. 12 tells the rest of this tale.

D

Two more lemmas are plain and worth mentioning. Lemma 4.15. If z1 , z2 E £* satisfy z1 < ular if z E £* and z � 0, then f ( z ) � 0. Lemma 4.16. If z E £,* , then 4.3. The Integral.

£, = { z

z2 , then f ( z1 ) ::; f ( z2 ); in partic­

J(>, z ) = >.. f ( z ) for any real number >.. � 0.

Let £, be the set E

£,* :

J(z) j (z) , with both finite}. =

£, is a linear space and J is a linear functional on £. More­ over C � £, and J extends f .

Lemma 4.17.

Lemma 4.18. Proof.

Since

If z E £ , then lzl E £ , that is, £, is a vector lattice.

( � ) cz i ; z ) ,

lzl = z lzl

+

it's enough (thanks to L's linearity) to show that �l £, if z E £. z

l

z , z

-;l l both belong to z

4.

84

Banach and Measure

( )<

We recall that Lemma 4 . 14 ensures that if z E £, then J z-i;l z l J z-;l z l > - oo, as well as z lzl + z lzl . (z) =

( )

oo

and

J J( � ) J( � )

Symmetry (applying Lemma 4. 14 to -z and using J - z = -J z ) shows that z-i;lzl < oo and J z-;l z l > - oo as well as J z lzl + z lzl . (z) =

( ) j J( � ) J( � )

( )

But z E [, says that Jz = j z , and so

[/ ( � ) 1 ( � ) ] + [! ( z � lzl ) 1 ( z � lzl ) ] = [/ ( z � lzl ) + J ( z � lzl ) ] [f ( z � lzl ) + 1 ( z � lzl ) ] = j z - j z = O. -

z lzl

z lzl

-

_

_

-

-

_

Now Lemma 4. 13 kicks in to say that the finite quantities z lzl z lzl . must, in fact, be equal, and the finite quantities z lzl , z lzl

J( � ) 1( � )

J( � ) 1 ( � )

follow suit.

D

An old friend is next on the agenda. (Monotone Convergence Theorem). if ( zn ) � C, Zn � Zn+I for all n and z = limn Zn with (MCO) lim n J( zn ) < oo, then z E C and J( z ) = limn J( zn ) ·

Lemma 4.19

{

We can, and do, assume z1 = O; otherwise subtract z1 from each of the Zn 's. Next note that z 2: Zn for all n and so since f ( zn ) = -f ( zn ) , we have Proof.

(1 1)

4. The Lebesgue Integral on Abstract Spaces

Let € > 0. Let Wn = Zn+ l - Zn 2 0. For each wk(n) 2 0 , lI" mk wk(n) 2 Wn and f(wkn) ) � (wn ) + 2: , say. Write Yn = wn( l )) + . . . + wn(n) '. so

85 n

find (wkn) )

� C

J

Yn = w�l ) + w�2) + · · · + w�n) E C . Since Yn = w�l ) + w�2 ) + · · · + w�n) 2 w 1 + w2 + · · · Wn , limninf Yn 2 limninf ( w 1 + W2 + · · · + Wn) = '"" = z. L.., Wn n At the same time, f( Yn ) f(w�1 ) ) + · · · + f(w�n) ) < (w 1 ) + � + (w2 ) + ;2 + · · · + (wn ) + 2: < =

J J J J(WI ) + + J(Wn ) + € J(z2 ) + J(z3 - z2 ) + + J(zn+l - Zn ) + € J(z2 + (z3 - z2 ) + + (zn+l - Zn )) + € J(zn+ l ) + € � li;ri J(zn ) + € (by Lemma 4.15). ·· ·

·· ·

·· ·

=

It follows that

J(z) � limninf f(Yn ) � li;ri J(zn) +

and by e's arbitrary nature,

0 < < <

J(z) � li;ri J(zn) J (z) by (11) J(z ) <

oo.

E,

with

4. Banach and Measure

86

Therefore D

Another old friend: Lemma 4.20

C and lznl

::;

(Dominated Convergence Theorem). Suppose ( zn ) � £, M E

M. Then

g = limninf Zn , h = limnsup zn E C

with Consequently, (DCO)

{

Suppose ( zn ) � £, M E C satisfy lznl ::; M. If z (t) � limn Zn t for each t E Q, then z E C and

()

f( z ) = limn J ( zn ) ·

For each i and for each j � i, write 9ij = min{ zi, Zi+l , . . . , Zj}· Then the sequence (9ij ) 'J°=i is decreasing, each member belongs to £ , and so the sequence (M - 9ij ) 'J°=i is an increasing sequence of members of £. (MCO) guarantees that if 9i = liJ:µi 9ij then Proof.

'

and that is, 9i E C and

M - gi E C

J(gi ) = liyi !(9ij )·

Applying (MCO) again reveals g = limninf Zn = liJ:µi 9ij E C with D J(g) = lif1 J(gi) ::; li�inf (zn ) · Lemma 4.21. If z E £, z � 0 and Jz = 0, then whenever the function x satisfies lx l ::; z , we have that x E C and Jx = 0. This is an immediate consequence of Lemma 4. 15.

J

4. The Lebesgue Integral on Abstract Spaces

87

4.4. We have the integral. Now we turn to Banach's treatment of C( Q)*, Q a compact metric space. We begin by recalling a few results from his classic book Theorie des Operations Lineaires [7] . Let X be a normed linear space. A linear functional on X is bounded if there is a c > 0 so that l f(x) :::; c l l x l l for all x E X. The appellation "bounded"

pertains to f's boundedness on the closed unit ball Bx = { x E X : l l x l l :S

1}

of X. It is easy to show that a linear functional's boundedness is tantamount to its continuity, and this, in turn, is assured by f's continuity at the origin. Then 1 1 1 1 1 = sup l f (x) I xE Bx

is a norm on X* and is in fact "complete" . So Cauchy sequences in X*, with respect to this just-defined norm, converge. The Hahn-Banach theorem provides any normed linear space with lots of linear functionals in its dual. • Let S be a linear subspace of the real normed linear space X, and let s * be a continuous linear functional on S . Then there is an extension x* of s* to all of X so that l ls * l l = l l x* l l · We simply let p(x) = l l s* l l l l x l l and apply the Hahn-Banach theorem to find a linear functional F defined on all of X with F(x) :::; p(x) for all x E X. Because F( - x) :S p( - x) = p(x) ,

we see that

•

F(x ) :S p(x) = l l s* l l l l x l l for all x E X, and so F = x* is in X* and l l F l l :S l l s* l l · Since F extends s* , l l s* l l :S l l F l l also. Let x be a member of the (real) normed linear space X . Then there is an x* E X* with l l x* l l = 1 so that x* (x) = l l x l l . Indeed,

let

S = {ax : a E �}, define s * (ax) = a l l x l l , apply the Hahn-Banach theorem, and get x* , the norm-preserving extension of s * .

It is frequently the case that for a particular normed linear space the norm­ topology is too coarse to uncover special and delicate phenomena peculiar to that space. Oft times such phenomena are best described using the weak topology.

88

4. Banach and Measure

base for the weak topology of the normed linear space X is given by sets of the form U(x; x ! , x2 , . . . , x� ) = { y E X : jxi (x - y ) j < E, i = 1, . . . , n}, where x E X, xi, . . . , x� E X*, and E > 0. Definition 4.22. A

The topology generated by this base is easily seen to be a Hausdorff linear topology so the operations of (x, y)-tx + y and (.A, x)-t.Xx of X x X to X and JR x X to X, respectively, are continuous. This topology is never metrizable and never complete for the infinite-dimen­ sional normed linear spaces! Usually closures are not determined sequen­ tially. Nevertheless, weak sequential convergence in special spaces often holds secrets regarding the inner nature of such spaces. To be sure, if X is a normed linear space and (xn ) is a sequence of vectors in X, then we say that (xn ) converges weakly to x E X (sometimes denoted by x = weaklimn xn ) if for each x * E X*, lim n x * (xn ) = x * (x). It is an integral part of basic functional analysis to compute the duals of special normed linear spaces and using the specific character of the spaces in­ volved to characterize when sequences are weakly null (tend to zero weakly). We now turn our attention to life inside spaces of the form l00(Q) , where Q is a set and l00(Q) denotes the normed linear space of all bounded real-valued functions x defined on Q, where x's norm is given by l l x l l oo = sup { l x(q) I : q E Q}. Now to bring tools such as the Hahn-Banach theorem to bear on the study of l00(Q), we need to know something about l00(Q)*. Banach knew a great deal about this (as did F. Riesz before him); he didn't formulate an exact description of l 00 (Q) * but, nevertheless, understood the basics. To begin, if x* E l00(Q)*, then x* is entirely determined by its values at members of l00(Q) of the form X E where E � Q; after all, simple functions are dense in l00(Q) . Now if F(E) = x* ( XE), then F is a bounded finitely additive real-valued measure on 2 Q , the collection of all subsets of Q. If we define I F I by IF l (E) = sup{F(S) : S � E} for E � Q, then IFI is a nonnegative real-valued function defined on 2 Q and IF(E) I :::; IFl (E) for each E � Q. (Remember: X0 = 0 so F(0) = x * (x0) = x * (O) = 0.) What's more, IFI is also finitely additive! If G � Ei U E2 where

4.

The Lebesgue Integral on A bstract Spaces

89

E1 and E2 are disjoint subsets of Q, then G = (G n E1 ) U (G n E2 ), and so F(G)

F((G n E1 ) u (G n E2 )) F(G n E1 ) + F(F n E2 ) � I F l (E1 ) + IFl (E2 ).

It follows that On the other hand, if E1 and E2 are disjoint subsets of Q, then for any E > 0 we can pick G 1 � E1 and G2 � E2 so

But now < =

<

F(G 1 ) + "2E + F(G2 ) + "2E F(G 1 ) + F(G2 ) + E F(G 1 U G2 ) + E (since F is finitely additive) I F l (E1 u E2 ) + E j

since E > 0 was arbitrary, we see that whenever E1 and E2 are disjoint subsets of Q. Here's the punch line: if x* E l 00 (Q) * , then x* defines a bounded finitely additive measure on 2Q -call this measure F. From F we generate I FI , all of whose values are nonnegative; I FI - F is also a nonnegative bounded real-valued finitely additive map on 2 Q and F = IFI - (IFI - F) . So F is the difference of nonnegative bounded finitely additive maps on 2 Q . In turn, such nonnegative additive bounded maps on 2 Q define positive linear functionals on l 00 (Q) , functionals that are necessarily bounded linear functionals. Why is this last statement so? Well suppose G : 2 Q -t[O, oo ) is finitely additive. If A � B � Q, then B = A U (B\A) so G(B) = G(A u (B\A)) = G(A) + G(B\A) � G(A); it follows that for any E � Q, G(E) s G( Q ), and G is bounded by G(Q) . Moreover if E1 , . . . , Ek are pairwise disjoint subsets of Q and ai , . . . , an E JR,

4 . Banach and Measure

90

then

I ?= is n

ai G (Ei )

I

<

l � •iXE,t e� G

< GQ ( J

l l � ·•x& L;

E;

)

and so G determines a linear functional

L aiXEi-1- L aiG(Ei)

on the simple functions which is "bounded" there on. This bounded linear functional extends to l 00 ( Q) in a bounded linear fashion, a positive functional to be sure. Now x E Bzoo ( Q ) means I x( q) I ::::; 1 for all q E Q; it follows that - 1 ::::; x( q) ::::; 1 for all q E Q. If f is a positive linear functional defined on l 00 ( Q ) , so f(x) ;:::: 0 whenever x(q) ;:::: 0 for all q E Q, then f( - 1) :S f(x) ::::; f(l) , or l f(x) I ::::; f(l), whenever x E Bzoo ( Q ) " Positive linear functionals on l 00 (Q) are bounded linear functionals. Whenever Q is infinite, £00 (Q) is ( highly) nonseparable. Nevertheless, using his mastery of lim sup's and lim inf's, Banach was able to establish the fol­ lowing stunning characterization of weakly convergent sequences in £00 ( Q) , a characterization which is sequential in all regards. Banach limits play a huge role. ( Banach ) . Let Q be a (nonempty) set, and let (xn) be a (uniformly} bounded sequence in l 00 (Q) . Then (xn ) is weakly null if and only if lim (12) n limkinf l xn (qk ) I = 0

Theorem 4.23

for each sequence (qk ) of points in Q.

Suppose to the contrary that there is a sequence of points (qk ) in Q such that lim sup limkinf l xn (qk ) I > a > 0 n for some a . Then unraveling the meaning of lim sup's, we can find a strictly increasing sequence ( nj) of positive integers such that limkinf I Xn . ( qk ) I > a > 0 Proof. Necessity.

J

4 . The Lebesgue Integral on Abstract Spaces

91

for each j. Now turning to the meaning of lim inf, we find a subsequence (qkm ) of (qk) such that for each j. Let x * E l 00 (Q) be given by x * (x) = LIM((x(qkm ))m) where LIM E l 00 (N) * is a Banach limit. Then for each j, l x * (xni ) I > a and so lim sup l x * (xn ) I > a > 0. n It follows that (xn ) is not weakly null in l00(Q) . Sufficiency. By remarks preceding the statement of this theorem, to test (xn ) 's weak nullity, it suffices to check the action of x * E l 00 (Q) * , for x * a positive linear functional of norm 1 , on the sequence (xn ) · So suppose x* is such a functional, with lim sup x* (xn ) > a > 0, n

where (12) holds

lim n limkinf l xn (qk) I = 0 for each sequence (qk) of points in Q. Let Sn be the sequence if Xn (q) 2: 0, Sn ( q) = 0Xn (q) otherwise, and let tn = Xn - Sn · One of lim supn x * (sn ) and lim supn x * (tn ) must exceed � ; after all, Xn = Sn + tn , so x * (xn ) = x * ( sn ) + x * (tn ), ensuring that lim sup x* (xn ) = lim sup(x*(qn ) + x *(tn )) S lim sup x * ( sn ) + lim sup x* (tn ) · n n n n If we clip off the "bottoms" of Sn by defining S if Sn (q) 2: � ' Yn ( q) = 0n (q) otherwise, then

{

{

4. Banach and Measure

92

What's more, lim sup x* ( Yn ) n so that

lim sup x * (sn - (sn - Yn )) n lim sup(x * ( sn ) - x * (sn - Yn )) n

im:up X * (Yn ) �

1.

Let

a

a

a

- = 2 6 3.

and look at XSn . Since l l Ynl l oo ::; l lsn l l oo ::; l l xn l l oo ::; M, say, we see that for any q E Q Yn (q) X Sn ( q ) > - IVl ' so that From this we see that lim sup x* (xsn) � 3� = : f3 > 0. n For E � Q, let F(E) = x * (xE)i of course F E l 00 (Q)* and lim sup F(Sn) > (3. n Let ni be the smallest positive integer such that lim sup F(Sn n Sn1 ) > 0. n

Such an n exists by the way! This is the crux of the matter! In fact, other-

wise,

lim n F(Sn n Sk) = O for each k so (because F is additive) li� F

(u

J=l

(Sn n Sj)

for each k as well. Let ki = 1. Pick m 1 > ki so large that F(Sm 1 ) > /3 and

)

=

0,

4. The Lebesgue Integral on A bstract Spaces

93

Let k2 > m1 . Pick m2 so large that F(Sm2 ) > /3 and F(Sm2 n (81 U · · · U Sk2 )) < 2/3 · Continuing in this fashion, produce k1 < m1 < k2 < m2 < · · · , with and

F(Smj n (81 U · . . U Ski )) < 2/3 · Now disjointivity: let Tj be given by

By construction F(Tj )

>

/3 2

and this is a "no-no" since F takes disjoint sequences to 0. So n l does indeed exist such that lim sup F(Sn n Sn1 ) > 0. n Believe it or not. Once faith has been established for n l we see that there are n2 < n3 < · · · < nk < · · · so that limnsup F(Sn n Sn1 n · · · n Snk ) > 0 for each k . The all-important point here is that for each k there is at least one point qk so qk E Bn1 n Sn 2 n · · · n Snk . Of course if j 2".: k, then qj E Snk ' and so by how the Sn 's were defined, l xnk (qj ) I

It soon follows that

2".:

a

5·

a

limnsup lim_inf l xn(qj ) I 2".: -6 . J This contradicts (12) and thus, the sufficiency is proven.

D

4. Banach and Measure

94

This result is remarkable because it characterizes weak convergence in a highly nonseparable, nonmetrizable situation in terms of sequences (qk) in

Q.

Suppose Q is a nonvoid compact metric space. Then the space C(Q) of all continuous real-valued functions defined on Q, equipped with the norm l lx l l oo = sup{ l x(q) I : q E Q } is a closed linear subspace of l00(Q) (the space of all bounded real-valued functions on Q) . We can give a much more succinct characterization of when a bounded sequence (xn ) is weakly null in C(Q) than just what Banach has above. Indeed, and again this was observed by Banach, the following is so. Theorem 4.24

(Banach).

weakly null if and only if for each q E Q.

A (uniformly) bounded sequence

(xn ) in C(Q) is

lim n xn (q) = 0

Proof. If (xn ) � C(Q) is weakly null and q E Q, then using the point evaluation Oq E C(K) * , where Oq(x) = x(q) we see that 0 = lim n xn (q) . n Oq(Xn ) = lim Now assume that (xn ) is a (uniformly) bounded sequence in C (Q) for which limn Xn (q) = 0 for each q E Q and imagine that (xn ) is not weakly null in C(Q) . Of course the Hahn-Banach theorem tells us that (xn ) is not weakly null in l 00 (Q) either. So by Theorem 4.23 there must be a subsequence (x�) of (xn ) and a sequence of points (qk) in Q and an a > 0 so that for each n limkinf l x� (qk) I -> a > 0. (13) But (qk) is a sequence in the compact (hence sequentially compact) metric space Q and so (qn ) has a subsequence (q� ) that converges to some qo E Q. By (13) it must be that for all n l x� (qo) I 2: a > 0. OOPS. 0

This theorem of Banach actually applies to any compact Hausdorff space O; that is to say, if n is a compact Hausdorff space and (xn ) is a uniformly bounded sequence in C(O) such that limn Xn (w) = 0 for each w E n, then (xn ) is weakly null in C(O) . The usual proof of this relies on the extension

95

4. The Lebesgue Integral on A bstract Spaces

to general O's of the representation of C(O)* as the space of regular Borel measures on n. However using a device by S. Eilenberg, this result can be deduced directly from Banach's result. How? Well, suppose n is a compact Hausdorff space and (xn) is a uniformly bounded sequence in C(O) for which limn Xn (w) = 0 for each w E 0. Introduce an equivalence relation on points in 0 as follows: w w ' if Xn (w) = Xn (w 1 ) for all n. If we define the distance between the equivalence classes [w] and [w ' ] containing w and w ' , respectively, by ,..._,

n then D is a metric on the space n of equivalence classes modulo ",....," . What's more, the map w ---+ [w] that takes a point w into the equivalence class [w] that contains w is a continuous surjection. By the way, we defined each Xn lifts to an x;;, E C ( fi ) ; moreover, the range of Xn is precisely that of Xn · If we call q t�e map taking w E 0 to [w] E fi, then q is a �ontinuous surjection of n onto O; consequently the linear operation q0 : C(D)---+ C (O) given by ,..._, ,

q 0 (f) (w) = f(q(w) )

is an isometric embedding. The point is just this: for any w E 0 , lim n X'-;;, ( [w] ) = lim n xn (w) = 0.

Hence by Banach's theorem, (X'-;;, ) is weakly null in C ( fi ) . Since Xn = q0(xn ) , the sequence (xn) is also weakly null. Now let's turn to Banach's approach to integration. Start with C = C(Q), Q a compact metric space. If f is a positive linear functional on C, then l f(x) I is bounded by f(l) so long as x E Be. Why? Well if x E Be then Jx l :::; 1 and l x(q) I :::; 1 for all q E Q. So x E Be means - 1 :::; x(q) :::; 1 ; from this it follows that - f(l) = f( - 1) :::; f(x) :::; f(l) . Okay? Any positive linear functional f on C(Q) is a bounded linear functional with norm f(l) . By Banach's theorem (Theorem 4.24) , if (xn) � Be and l xn l :::; M E C with limn Xn( q) = 0 for all q E Q, then (xn ) is weakly null in C(Q), hence limn f(xn ) = 0. This is (BCO) in Banach 's integration theory! So we can extend f to a vector lattice £ of real-valued functions defined on Q in a linear fashion so that the extension, denoted by J df, enjoys the fruits of (MCO) and (DCB) : (MCO) If ( zn ) � £, ( zn ) an ascending sequence with z = limn Zn , then z E £ whenever limn J Zn df < oo. In this case J z = limn J Zn .

4 . Banach and Measure

96

(DCB) Suppose ( zn ) � .C and M E .C satisfy l zn l S M. Then lim inf Zn , lim sup Zn E .C, and lim inf Zn S lim inf Zn S lim sup Zn S lim sup zn .

J

J J

J

So should ( zn ) � .C, M E .C satisfy lznl S M, if z (q) = limn Zn (q) for each q E Q, then z E .C and J z = limn J Zn . What does this provide us with? To start, let F be a closed subset of Q. Consider the continuous function F ( q) = d( q; F) ; notice that q E F precisely when p(q) = 0. If we consider
{

n (q) = inf p(q) , �

},

then limn
so xun En E .c also-here you can appeal to (MCB) or (DCB) , take your choice. Therefore { E E Q : XE E .C} is a O"-algebra of subsets of Q containing each and every closed subset of Q. Therefore for any Borel set B � Q, XB E .C. Thus we have proved Theorem 4.25 . .C contains all of the indicator functions on Borel sets.

Suppose (Bn ) is a sequence of pairwise disjoint Borel subsets of Q. Then XBn ' XUn Bn E .C and XLJn Bn = L XBn ; n

97

5. Notes and Remarks

an appeal to (MCO) or (DCB) soon reveals that

f XLJn Bn dj

=

j n

L Xn dj.

J df acts in a countably additive fashion on Bo(Q) ! It is a measure on Bo(Q) ;

i.e.,

J

J

µf = df, where µJ(E) = XEdf, whenever XE E .C is a countably additive measure defined on some o--field that contains the Borel o--field. The total mass of µ! is f(l) . We initially thought to include the above construction as an appendix; that's how it appeared originally, as an appendix to Saks' book [1 11] Theory of the Integral. But it is too beautiful to be perpetually relegated to appendices. 5.

Notes and Remarks

5.1. The Problem of Measure. Classically there were two such prob­ lems: the difficult problem of measure and the easy problem of measure. The reader should be aware that these adjectives are arbitrary and some­ what inaccurate! The "difficult" problem of measure asks if one can assign to every set E in �n a nonnegative real number µ(E) , its measure, so that (i) if E = [O, l] n , then µ(E) = 1; (ii) if p : �n -+�n is a surjective isometry, then µ(E) = µ(pE) ; (iii) if E is the union of any sequence (Ek) of pairwise disjoint sets, then µ(E) = L µ(Ek) · k We have seen that Vitali's construction of a nonmeasurable set answers this question in the negative in �; in the Notes and Remarks to Chapter 7, Vitali's construction is modified to show that in any nondiscrete locally compact topological group, if :E is a o--field of sets containing the Borel sets and µ is a translation-invariant nonnegative, countably additive extension of Haar measure defined on :E, then there is a non-µ-measurable set. The "easy" problem of measure asks if one can assign to every set E in �n a nonnegative real number µ(E), its measure, so that (i) if E = [O, l] n , then µ(E) = 1; (ii) if p : �n -+�n is a (surjective) isometry, then µ(E) = µ(pE) ;

4.

98

Banach and Measure

(iii) if E is the union of the pairwise disjoint sets Ei , . . . , Ek, then µ(E) = µ(E1 ) + · · · + µ(Ek) · The lessening of the third condition seriously changes the nature of the problem, and the solution to this problem now depends on the dimension n of the space ffi.n on which the measures reside. For n = 1, 2, there is a measure µ solving the easy problem of measure; this is due to S. Banach, and it is not easy. For n � 3, there is no such measure and, again, the solution due to F, Hausdorff is anything but easy. Since it was proved first, we take up Hausdorff's negative solution for ffi.3 as a start. This solution revolves around the paradoxical theorem known as Hausdorff 's paradox. For a beautiful exposition of this and related top­ ics, we refer the reader to the paper of Karl Stromberg in the American Mathematical Monthly [123] . Theorem 4.26 (Hausdorff's Paradox) . The unit sphere 82 of ffi.3 can be

decomposed into the disjoint union 82 = Q U R U S U T of four sets: Q is countable, R, S, T are pairwise congruent, and R is con­ gruent to S U T. Now suppose that µ is a solution to the easy problem of measure in ffi.3 . Normalize µ so µ(BH:f.3 ) = 1. With Hausdorff's paradox in hand, write Bo = BH:f.3 \ {O} in the form Bo = Q o u Ro U So u To ,

where the subscript "O" indicates that the set is the union of rays emanating from the origin and ending on the sphere at a point of the subscript set (e.g., Qo = LJ q E Q (O, q]) . We first show that µ( Q0) = 0. Introduce spherical coordinates and assume (a rotation of 82 may be needed) that Oz passes through the none of the points of Q. A point in 82 is determined by two angles: "latitude" which measures N S and "longitude" which measures E W The set of longitudes of points of Q is countable as is the set of differences thereof. So we can choose an angle A distinct from all the angles of longitude of Q and different from all the differences. Rotate O z through the angle Aj Q moves to a (countable) set Q 1 which is disjoint from Q. Again we can find an angle A1 distinct from all the points of Q U Q1 and from all the differences thereof. Rotate Oz through the angle A1 to get a set Q11 which is the image of Q 1 ; Q, Q1 and Q 11 are congruent by construction. Continue in this vein. -

-

99

5. Notes and Remarks

For any n, we get pairwise disjoint sets Q ' QI ' Q l ' . . . ' Q (n- l ) and accompanying them are the pairwise disjoint sets Q o, Q I Q l Q (n - l ) Of course all these sets are pairwise congruent as well. Then µ( Bo) > µ(Qo U · · · U Q �n - l ) ) = µ(Qo ) + µ(Q� ) + µ(Q � ) + . . . + µ (Q�n - l ) ) = nµ(Qo) . But µ (Bo) is finite, and we can do the above for each n E N. It must be that µ(Qo) = O. We also have µ( Bo) = µ(Qo U Ro U So u To) = µ(Qo) + µ(Ro) + µ( So ) + µ( To) (because Qo, Ro , So, To are pairwise disjoint) µ(Ro) + µ( So ) + µ (To) (because µ(Q o) = 0) 3µ(Ro) (because Ro , So, To are congruent) . At the same time µ( Bo) = µ(Ro) + µ( So u To) = 2 µ(Ro) because Ro and So U To are congruent. Hence 2 µ(Ro) = 3 µ(Ro) = µ( Bo) . Since µ(Bo) > 0, chaos is the only explanation. One can easily piggyback this negative solution of the easy problem of mea­ sure in JR3 to a negative solution in !Rn for any n > 3 . After all, if v were to be a positive solution to the easy problem of measure for JR4 , then µ( E ) = v( [O, 1] x E ) would be a positive solution for JR3 , etc. Banach's positive solution for JR (and JR 2 ) is a delicious mix of his natural cleverness and ideas from functional analysis-ideas similar to what goes into the Hahn-Banach theorem. A few words on Banach's proof seem in order. In what follows we will be considering bounded real-valued functions defined on JR which are periodic of period 1. If f is such a function, then f can be viewed as a bounded real-valued function defined on the circle C of radius 2� having center at the origin. The value f(x) corresponds to f being evaluated '

0,

0, . . . '

0

.

4.

100

Banach and Measure

at an arc x on C where we proceed x units in a counterclockwise fashion starting at the point ( 2� , 0) . Let f be such a function. Definition 4.27. We say that f ,...., 0 if given an E > 0, there exist ai , . . . , a n E � such that for all real x

l � t.

f (x + a • )

I

«

We say that f ,...., g (for two bounded functions defined on C) if ( ! - g) ,...., 0, that is, if given an E > 0 there are ai , . . . , a n E � such that for all real x

Proposition 4.28. Some basic facts:

(i) (ii) (iii) (iv) (v)

0 and g 0, then f + g 0. If f ,...., 0 and C E �, then C · f ,...., 0. If f(x) 2:: c > O for all x, then it is not so that f ,...., 0. If f ,...., 0 and a E � ' then fa(x) := f(x + a) also has fa ,...., 0. If f

rv

rv

rv

If f ,...., g and g ,...., h, then f ,...., h.

Hence ",....," is an equivalence relationship among the bounded real-valued functions on C. If f is a bounded real-valued function defined on C, then the equivalence class [f] of f will be denoted by F, and we shall refer to F as a hyperfunction, keeping Banach's terminology. Some natural conventions will be followed: • If F is the hyperfunction [f] and a is a real-valued constant func­ tion, then A · F will be the hyperfunction [a · f] , where A = [a] . • If F = [ !] and G = [g] , then F + G will be [f + g] . • If F = [f] , then - F will be [ - f] . One easily realizes much of the usual relationships hold in the space of hyperfunctions. A fundamental question to be entertained is, What about order? It is easiest to discuss the order by working with (bounded real-valued) functions f on C.

101

5. Notes and Remarks

We say f >-- 0 if there exists c > 0 and a l , . . . , an E JR so that for all real x, 1 n -n L f (x + ak) 2:: c. Definition 4.29.

k=l

If - f >-- 0, then we write f -< 0. We say that f >-- g if (f - g) >-- 0. If F1 = [fi] and F2 = [h] are hyperfunctions, we'll say that F1 > F2 if for any f E [f1 - h ] we have f >-- 0. Proposition 4.30. More basic facts:

(i) ( ii ) ( iii ) ( iv )

If f >-- 0 and a E JR, then fa >-- 0 where fa(x) := f(x + a ) . If f >-- 0 and g >-- 0, then (f + g) >-- 0. If f(x) > 0 for all x, then it is not so that f -< 0. At most one of the following holds for any f : f )>- 0, f 0, f -< 0. (v ) If f >-- 0 and g 0, then (f + g) >-- 0. ( vi ) If f >-- g, f Ji , and g 91 , then Ji >-- 91 · ( vii ) Let F = [f] be a given hyperfunction. Then there are hyperfunc­ tions C1 = [c 1 ] and C2 = hl where c 1 , c2 E JR are constant func­ tions such that C1 > F > C2 . rv

rv

rv

rv

We turn now to the study of linear spaces of hyperfunctions with special attention being paid to O ( R.) and 0(£) . Here O(R.) is the linear space of all hyperfunctions F = [f] where some f E F is Riemann integrable (keep in mind that Riemann integrable functions are necessarily bounded) , and O(L) is the linear space of hyperfunctions each of whose members contain a bounded Lebesgue integrable function. Generally if n is a linear space of hyperfunctions and � = [¢] is a hyper­ function that does not belong to n, then n( � ) denotes the smallest linear space of hyperfunctions containing n and �- Now O(�) consists of all hy­ perfunctions of the form F + C · � where c is a real constant and F E n. Naturally C = [c] . We'll attempt to avoid being too pedantic by using c1 for both the constant real number and the hyperfunction [c 1 ] . With this in mind, notice that for any IJt E 0( �) , IJt can be represented in one and only one way in the form IJt = F + c�. In fact, if we also have 'ljJ

= F + c �' I

I

102

4.

Banach and Measure

then

(c - c' ) = F1 - F. If c = c1 , then c - c1 = 0 and F1 = F. If c f. c' , then

1

I

1

= --, F , F, c-c c-c --

and so E 0, which is assumed not to be. Either way the uniqueness of the representation W = F + cc/> is established. We further take note that if '11 1 , '11 2 , W 3 E O() and W3 = a 1 W 1 + a2 W 2 (a 1 , a2 E JR ) , then on representing Wi , '11 2 , W 3 as '11 1 = F1 + c 1 , '11 2 = F2 + c2 , W3 = F3 + c3 , we know that F3 = a 1 F1 + a2 F2 and c3 = a l c1 + a2 c2 , thanks to the uniqueness of the representation of '11 1 , '11 2 , and W 3. Now let 0 be a linear space of hyperfunctions. Definition 4.31. A map

A : 0---+JR is called linear if for any constants c1 , c2 and any F1 , F2 E 0 we have

A(c 1 F1 + c2 F2) = c1 A(F1 ) + c2 A(F2 ) . We say that A is a positive linear functional if in addition to being linear, A ( F ) 2:: 0 whenever F E 0 satisfies F 2:: 0. Theorem 4.32. Let be a hyperfunction not belonging to the linear space

0 of hyperfunctions, and suppose that there exist F1 , F2 E 0 such that

F1 > > F2 . Let A : 0---+ JR be a positive linear functional. Then there exists a positive linear functional A : O()-+JR that extends A to O () .

The hypothesis of Theorem 4.32 tells us that a = inf { A ( F ) : F E O, F > } is real. Now if '11 E 0 ( ) is written in the form '11 = F + c where c E JR, F E 0, then A(w) = A ( F ) + ca is a positive linear functional on 0 ( ) that extends A. Banach now calls on transfinite induction and with this firmly under control proves the basic:

103

5. Notes and Remarks

(!t a ) a<,B be a well ordered (ascending) family of linear spaces of hyperfunctions (so !t a � n'Y if a < 'Y < /3) , and suppose that for each a < /3 , Aa is a positive linear functional on !t a such that A-y l na = Aa whenever a < 'Y < (3. Let !1 = u n a , a<,B and for F E n set A(F) = A a ( F ) (F) , where a (F) denotes the first ordinal so that F E !ta . Then A is a positive linear functional on n. Proposition 4.33. Let

Proposition 4.34. Suppose n is a linear space of hyperfunctions containing the hyperfunction l . Assume A is a positive linear functional defined on n.

Then A can be extended to a positive linear functional A defined on the space of all hyperfunctions.

Now we're prepared for the main course of Banach's offerings. First an appetizer. Theorem 4.35. Suppose f is Riemann integrable. Then

f01 f(x) dx .

Proof.

f

,....,

c

where c =

Suppose f is Riemann integrable, and define fn on C by n n 1L 1" fn(x) = - f x + - = - h (x) . n k=l L.....t

( k) n

n k=l

n

Then Un) converges uniformly to J� f(x) dx. Hence if we choose n very big for all x; i.e. , for n big

,....,

for all x. This is tantamount to f - c 0 or f

,....,

c.

D

Before the next delightful tidbit we recall the following feature of JR: The real line JR can be written as the disjoint union of two sets and F, where N is a set of Lebesgue measure zero, and F is a set of the first Baire category.

Fact 4.36.

N

4.

104

Banach and Measure

(

To see this, let qn) be a listing of the rationals, and let Iij be an open interval with center qi of length 2 - i -j . Set Gj = LJij Iij and N = nj Gj. If E > 0 is given and 2 -j < E , then N � nJij, and L l(Iij) = L 2i1+j = 21J < E . Proof.

i

i

Therefore N is a set of Lebesgue measure zero. Of course each Gj is open and contains Q so N is a dense gi5 set. On the other hand, each Gj is nowhere dense, and so Nc = ( nj Gj r = Uj (Gj ) D is of the first category. p p(x) d x = 0 such that if with f01 ¢ is Riemann integrable and ¢ >- p, then f01 ¢ (x) dx 1.

Proposition 4.37. There exists a bounded Lebesgue integrable function >

If N is a set of Lebesgue measure zero with Nc a set of the first Baire category, then p = XN , the indicator function of N, is bounded, Lebesgue integrable, and has d = 0. This p fits the bill.

f� p(x) x

Remark 4.38.

If p is the function of Proposition 4.37, then

{ fo 1 ¢ (x) dx : ¢ >- p and ¢ is Riemann integrable } = 1. After all, the constant functions ¢(x) = 1 + E are Riemann integrable and satisfy ¢ >- p. inf

Moreover the p in Proposition 4.37 is not equivalent to any Riemann in­ tegrable function in the sense of Indeed suppose p """ 'I/; where 'I/; is so p c """ 0. But p c """ 0 Riemann integrable. Then p """ c = means that given E > 0 we can find . . . , an E JR so that for all real

a1,f01 'lj;(x) dx x (14) l��[p(x+ a•) - c] I < E But we've chosen p so that there are points where � p(x + ak) takes the value one and points where � p(x+ak) is zero. There is absolutely no way for (14) to hold in light of this. ",_.... " .

-

I:�= l

-

I:�= l

Theorem 4.39. There is a positive linear functional H defined on the vec­

tor space of all bounded real-valued functions on the circle C satisfying the following: (i) H c = c for each real constant c;

()

105

5. Notes and Remarks

(ii) for any a E JR. and <J = ±1, if f% ( x ) := f( <Jx + a) , then H (f) = H ( J: ) ; (iii) if f is a bounded Lebesgue integrable function, then

H (f) =

fo 1 f(x) dx.

Denote by O(.C) the linear space of hyperfunctions, each containing a bounded Lebesgue integrable function on C. For any X E O(.C), we let A be the operation A(X) = L - ( x ) dx, where is a bounded Lebesgue integrable function belonging to X. The operator A(X) is well defined: Proof.

fo1

fo1

L - f(x) dx = 0 for each and every f that's Lebesgue integrable and satisfies f ,..., 0. Why is this so? Suppose f 0, and let E > 0 be given. Then one can find reals a , . . . , an so that for all real x f(x + ak ) «

1

,...,

It follows that

I I � t. L f1 I � t f(x + ak) I dx lo k=l I L - lf1o � k=lt f(x + ak) dx l

<

and so But

E,

<

E.

1o 1 f(x) dx = L - 1o 1 f(x + ak) dx = -1 k=ll::n L - 1o 1 f(x + ak ) dx. So 1 I L fo f(x) dx l Since 0 was arbitrary, we're left with the conclusion that L - fol f(x) dx = 0, L

n

-

E >

whenever f ,..., 0.

<

E.

106

4. Banach and Measure

The operation A is linear: if X1 , X2 and X are in n(.C) with X = c1 X1 + c2 X2 (c1 , c2 E JR) , and if ¢ 1 , ¢2 are bounded Lebesgue integrable functions in the hyperfunctions X1 , X2 , respectively, then

fo c1 ¢1 (x) + c2¢2 (x) dx b b c1 L - 1

A(X) = L

=

l

-

c1 A(X1 ) + c2 A(X2 ) . The operation A is positive: Suppose X 2:: 0 , and let

- 0 or

seen that if 0, then rv

fo 1
and so

What if

- O? Well, in this case there are reals a1 , . . . , an and c that for all real x, It follows that

>

0 so

1

1 1 n L c. -n L o k=l

But as we've seen (using the invariance of the integral under translation(!)), this is tantamount to 1 L-

fo
>

c,

and so A(X) > c > 0. Regardless, A(X) 2:: 0 if X 2:: 0. Because n(.C) contains the constant hyperfunction 1, Proposition 4.34 is applicable. On using Proposition 4.34 we find a positive linear functional A defined on the linear space of all hyperfunctions such that A l n (.C) = A. Now we define an operation G with domain the linear space of all bounded real-valued functions that are periodic of period 1 (hence bounded function­ als defined on C) . Let f be such a function, and suppose f is a member of the hyperfunction F. As you might expect, set G(f) = A ( f ) . Then G is well defined because F is the only hyperfunction containing f, G is linear (this is straightforward) , and should f 2:: 0 (i.e. , f(x) 2:: 0 for all x) then

5. Notes and Remarks

107

G(f) � 0 as well. To see that this last assertion is so, for any E 0, let (x) := f(x) + c, and realize that >- 0 so the hyperfunction determined by verifies 0 and so A ( ) � 0. In turn, this tells us that G ( ) � 0 so G(E + J ) � 0. But G is additive so G(f) � -G(E) = -E, which because E 0 was arbitrary tells us that G(f) � 0. Since fa ,...., f (recall fa(x) : = f(x + a)), f and fa determine the same >

>

>

hyperfunction F, and so

G(fa) = G(f) = .A (F) .

fo1

It is plain that

G(f) = L- f(x) dx if f is Lebesgue integrable-one need only recall the opening arguments of this argument. Now we'll define H. Let f be a bounded real-valued function, periodic with period 1. Denote by f ( + l ) the function f(x) and f ( - l ) the function f ( - l ) (x) : = f(-x). Then

( + l ) + Gf ( - l ) ) . H f -- �(Gf 2 The functional H does all that we have claimed: it's a positive, linear func­ tional. If f ( x) = c is a constant function, then A(!) = "21 (G(c) + G(c)) = c, H (f�a ) ) = � (Gg + Gf� - a ) ) = � (G r + Gf ( - a ) ) = H (f) ,

and finally if f is Lebesgue integrable, then (since it's assumed that f is periodic with period one) ,

( + l ) + Gf) �2 (Gf ( + l ) + Gf ( - l ) ) = �(Gf 2 = � (Lf(x) dx + L- f(x) dx) = L where f(x) = f(l -x). H (f)

fo1

fo 1

-

fo1 f(x) x

d ,

Banach makes special note that the above extension of Lebesgue measure is 0 noteworthy in that it satisfies conditions (i) , (ii) , and (iii) . In fact there is a positive linear functional defined on C that satisfies con­ ditions (i) and (ii) of Theorem 4.39 but is not identical to the Lebesgue integral for all bounded Lebesgue integrable functions.

4. Banach and Measure

108

Indeed, consider the linear space O(R) consisting of hyperfunctions that contain a bounded Riemann integrable function, and let p be the bounded Lebesgue integrable function of Proposition 4.37. Suppose p generates the hyperfunction R. Define the functional A : n (R)---+ JR by

fo

A(qi) =

l

(x) dx,

where E qi is a Riemann integrable function. Then A is a positive linear functional on O(R) . We extend A to a positive linear functional A i on the linear span of O(R) U { R } by A i ( qi + cR) = A(qi) + c. By Theorem 4.32 and by Remark 4.38, if a = inf{A(qi) : qi E O (JR) ; qi > R }, then a = 1 . In particular A i (R) = 1 . Now we can apply Proposition 4.34 to extend A i to a positive linear func­ tional A on the space of all hyperfunctions. If we now proceed as in the proof of Theorem 4.39, we find a positive linear functional H defined on the space of all bounded real-valued functions on C such that H(p) = 1 , where we keep in mind that L

fo

l

p(x) dx = 0.

We make special mention of the fact that H(f) = H(g) whenever f g. To see this, it suffices to show that H(f) = 0 whenever f 0 (this, thanks to H's being a positive linear functional) . Suppose f 0. Then regardless of E > 0 we can find reals a i , . . . , an so that for all x n 1 - E � - 2: f ( x + ai ) � E . n rv

rv

rv

k=l

It follows that

(

H n1 and so

n

�

fa;

)

-E

=

1

n

n

�

H(fa;) = H(f) ,

� H(f) � E .

1 09

5. Notes and Remarks

Now E > 0 being arbitrary gives rise to the realization that H (f) = 0. Of course, if f is Riemann integrable, then 1 f c = f(x) dx,

fo fo

rv

and so

1 H (f) = c = f(x) dx. One consequence of Theorem 4.39 worthy of note is the following solution of the problem of measure for � (we use Banach's notation): Theorem 4.40.

One can assign to each subset E of points of the unit circle

C a number H ( E ) in such a way that (i) H ( E) 2: O; (ii) H ( C) = 1; (iii) H ( E1 U E2) = H ( E1 ) + H ( E2) whenever E1 n E2 = 0;

(iv) if E1 and E2 are congruent (there is an isometry of � into � that takes E1 onto E2 }, then H ( E1 ) = H ( E2) ; (v) H ( E ) = Lebesgue measure of E if E is Lebesgue measurable. Banach turns his attention to � 2 . Here's what he does. Again, C will denote the (true unit) circle with center the origin of the plane and the radius 1/27r . Let f (x, y) be a bounded real-valued function defined on a (nonempty) bounded domain E � �2 , and let H be the functional whose existence was the point of Theorem 4.39. Let e be an angle (point on C) . Rotate the OX, OY axes system through the angle e (in a counterclockwise manner, if you please), and let (x' , y1 ) be the new coordinates of (x, y) in the 0 x ' , OY 1 system. Recall from your days of analytic geometry that

x = x cos e - y sin e and y = x sin e + y cos e (so the transformations (x, y) -+ (x' y1 ) and (x ' , y1 ) -+ (x, y) are linear isome­ tries). Look at l( x ' ' y ' ) = f ( X1 cos e - y 1 sin e ' x' sin e + y ' cos 0 defined on the set E of points (x' , y1 ) which one obtains by rotating E through the angle -e. I

I

I

,

I

1 10

4. Banach and Measure

We define g(x 1 ) as follows: Let a ' E JR be given. Define ' · line x ' = a ' intersects E, 9 ( a ' ) = 0H ( f( a , ) ifif not. Let ( � ) = H (g) . We define h(g 1 ) as follows: Let b1 E JR be given, and set H ( f(· b1 ) if line y' = b1 intersects E, h( b' ) if not. 0 Let x (�) = H (h) . Then

{

¢

=

{

,

H (f) = H

( ¢ ; X ).

Now H is quite a functional. Here are the properties of H noted specially by Banach (again we use his notation).

One can associate to each bounded function f defined on a (nonempty) bounded domain in JR2 a real number H (f) so that the following hold: (i) if Ji and f2 are bounded real-valued functions with common domain and c 1 , c2 E JR, then

Theorem 4.41 .

(ii) if f � 0, then H (f) � O; (iii) if f = c on a domain of area 1, then H (f) = c; (iv) if f and g are congruent {i. e., there is an isometry p of JR2 onto JR2 that takes the domain of f onto the domain of g with g(p(x, y)) f(x, y) for all (x, y ) in the domain on !), then H (f) = H (g) ; (v) if f is Lebesgue integrable {with domain E a measurable set}, then =

H (f) =

j l f(x, y) dxdy.

Alternatively we have

One can assign to each bounded subset E of the plane a number H ( E ) in such a way that (i) H ( E ) � O; (ii) H ( E ) 1 whenever E has Lebesgue measure 1 ; (iii) H ( E1 U E2) H ( E1 ) + H ( E2) if E1 n E2 = 0;

Theorem 4.42.

=

=

111

5. Notes and Remarks

( iv ) H ( E1 ) = H ( E2) if E1 and E2 are isometric under an isometry of

IR2 onto IR2 · ( v ) H ( E ) coincides with the Lebesgue measure of E, if E is Lebesgue measurable with finite Lebesgue measure. What is remarkable is that there is a common explanation covering all n: it explains simultaneously why the easy problem of measure has a positive solution for n = 1, 2 and a negative solution in case n 2: 3. This approach relies on a keen understanding of the group Gn of isometries of !Rn onto itself. A few words about Gn are in order. Any isometry of !Rn is affine, and so is the composition of a translation and a linear isometry. We denote by Tn the group of translations and by On the group of linear isometries of !Rn . Of course Tn is abelian ( it's isomorphic to !Rn ) . The map taking a member of Gn to its linear component is a homomorphism of Gn onto On with kernel Tn so Gn /Tn is isomorphic to On . Members of On have determinant ±1 and so the subgroup of 80n of On con­ sisting of those linear transformations having determinant + 1 is an invariant subgroup of On with On / 80n isomorphic to Z2 . In the case of n = 1, the only orthogonal linear maps of IR 1 are the identity and its negative. In case n = 2, if we denote by PB the counterclockwise rotation about the origin, then PB is represented by the matrix cos () - sin () sin () cos () ' and so 802 is plainly isomorphic to the circle group PB +-+ () E [O, 27r ) . In the notation of group theory (where H1 :'.SlH2 means that H1 is an invariant subgroup of H2 and H2 / H1 is abelian) , we have for n = 1 { id } :sl Ti :sl Gi , and for n = 2 { id } :sl T2 :sl 8G2 :sl G2 , where 8G2 denotes the group of members of G2 whose linear component is a member of 802 . Thus G 1 and G2 are solvable. The notion of an amenable group is due to J. von Neumann. A group G is amenable if there is a finitely additive, left-invariant probability defined on all the subsets of G. '

(

)

112

4.

Banach and Measure

He showed that solvable groups are amenable and noted that Hausdorff's paradox leads to the existence of a free subgroup of Gn ( n 2: 3) of rank 2, something forbidden for solvable groups. If one proceeds like what was done following the statement of Hausdorff's paradox, then it is easy to see that the amenability ( or nonamenability ) of Gn holds the answer to the easy problem of measure in �n .

Results related to the problem of measure often lead to paradoxical state­ ments, and so attract the attention of many from different areas of mathe­ matical endeavors. I. P. Natanson's Theory of Functions of a Real Variable [94] relates the problem to classical real variables. K. Stromberg's paper [123] The Banach-Tarski Paradox presents the Hausdorff paradox, and its more famous companion, the Banach-Tarski paradox, in a clear and com­ plete way. And S. Wagon's volume "The Banach-Tarski Paradox" in Cam­ bridge University's Encyclopedia of Mathematics series gives an overview of this fascinating area of mathematics. Banach's work, as outlined above, solves ( negatively ) the case for n = 1 of the Banach-Ruziewicz problem, which asks if the ( normalized) Lebesgue measure is the unique finitely additive rotation-invariant probability defined on the Lebesgue measurable subsets of the n-sphere. For n > 1, the answers to this problem are all of a positive nature. For n = 2 and n = 3, this uniqueness of normalized Lebesgue measure among finitely additive rotation invariant probabilities was established by V. G. Drinfeld [33] ; while for all n > 3 this was established independently by G. Margulis [77] and D. Sullivan [125] , based in part on earlier work by J. Rosenblatt [106] and D. Kazdan [65] . For the Euclidean spaces, Banach [4] showed that there was more than one rotation invariant finitely additive measure, normalized so that the measure of the unit ball is one, on n-space in case n = 1 or n = 2. For n > 2, G. Margulis [78] has shown that Lebesgue measure is the unique finitely additive, suitably normalized, rotation invariant measure on the o--field of all Lebesgue measurable sets. A remarkable construction due to S. Kakutani and J. Oxtoby [64] shows that Lebesgue measure in n-space can be extended in a countable addi­ tive, nonnegative, translation-invariant manner to very large o--fields. The interested reader will find a discussion of this work in their Appendix 2.

Chapter

5

Compact Groups Have a Haar Measure

1. The Arzehi-Ascoli Theorem

When we first study analysis, one of the basic results encountered is the fact that if f is a continuous real-valued function defined on a closed, bounded interval, then f is uniformly continuous. This result has a natural extension to the setting of topological groups. The proof of this extension is easy but nevertheless worthy of close attention.

Let G be a topological group, and let M be a nonempty com­ pact subset of G. Then any continuous function f : G ---+ � is left uniformly continuous on M; i. e., given E > 0, there is an open set V containing the identity of G so that if x, y E M and x E yV, then l f(x) - f(y) I :S E . Theorem 5 . 1 .

Proof. Let E > 0 be given. For each a E M, there is an open set Va that contains the identity such that if x E M and x E a Va , then l f(x) - f(a) I :S � · Since e · e = e, there is an open set Wa that contains the identity e and satisfies Wa · Wa � Va . Now if a E M, then a E a · Wa so {a · Wa : a E M} covers M; we can find a i , . . . , an E M so that

Look at V

= Wa 1 n · · · n Wan .

Then V is open and contains e . Let x, y E M with x E yV. To see that l f(x) - f(y) I :S E , note that if y E M, then y E ai · Wai for some i = 1, . . . , n.

-

1 13

114

5. Compact Groups Have a Haar Measure

It follows from this and Wa � Va that I f (y) - f(ai ) I � � · Also and so l f(x) - f(ai ) I � � too. In sum, if x, y E M with x E yV, then E E l f(x) - f(y) I � l f(x) - f(ai ) I + l f(ai) - f(y) I � 2 + 2 = E .

D

Note. Of course, we can also show that the continuous function of Theorem 5.1 is right uniformly continuous. If S is a compact Hausdorff space, then we denote by C(S) the Banach space of all continuous real-valued functions on S equipped with the sup norm 1 1 · l l oo : l l f l l oo = sup { l f ( s ) I : s E S} . The task before us is to identify the subsets of C(S) that are relatively norm compact. Let K be a nonempty collection of members of C(S) . We say K is equicon­ tinuous at a point s E S if, given E > 0, there is an open set U which contains s so that whenever t E U, then l f(t) - f(s) I � E for all f E K. We say that K is equicontinuous if K is equicontinuous at each point of S. Our first result shows that equicontinuity allows us to arrive at uniform conclusions from pointwise hypotheses. If K is equicontinuous, then the topology of pointwise convergence on K and the norm topology coincide. Fact 5.2.

We need to show that if Jo E K and E > collection s1 , . . . , Sn E S so that should Proof.

for i = 1, . . . , n, f E K, then

0,

then there is a finite

I l l - fo l l � E .

Since K is equicontinuous, for each point s E S we can find an open set Us in S containing s so that if t E Us , then E l f(t) - f(s) I � 4

1.

115

The Arzela-Ascoli Theorem

for all f E / C Now

S=

LJ Us

s ES . , Sn

is compact so there are points s 1 , . . E S so that each t E S belongs to a Us3 , 1 :S j :S nj ; i.e., there are points s1 , . . . , Sn E S so that for any t E S there is an Sj , 1 :S j :S n, for which E (15) l f( t ) - f(sj ) I :::; 4 for all f E / C We have our sought after points s1 , . . . , Sn · Assume f E /(, and E (16) l f (sj ) - fo (sj ) I :S 2 for j = 1, . . . , n. Then given t E S there is j , 1 :S j :S n, so that (15) applies. It follows that l f(t) - fo (t) I < l f(t) - f(sj ) I + l f(sj ) - fo (sj ) I + l fo (sj ) - fo (t) I < 4E + 2E + 4E = E

by (15), (16) , and (15) , if you please.

D

Equicontinuity is contagious. Fact 5.3. If /(, � C(S) is equicontinuous, then K's pointwise closure (in JR8 ) is also equicontinuous. Proof. Let s E S if t E Us , then

and E > 0. There is an open set

Us

containing s so that

E l f ( t) - f(s) I < 3

for all f E / C Let g E J(P , JC's pointwise closure in JR8 . Then given s , t E S there is f E /(, so that E l f( s ) - g(s) I , l f( t ) - g (t) I < 3 ' since this is what it means for g to be in /(P . If t E Us , then jg(s ) - g ( t ) I < j g(s ) - f(s) I + l f( s ) - f(t) I + l f(t) - g( t ) I E E E D < 3 + 3 + 3 = E. Now we are ready for the centerpiece of this section. (Arzela and Ascoli). A nonempty subset /(, of C(S) is rela­ tively norm compact if an d only if JC is uniformly bounded and equicontinu­ ous.

Theorem 5.4

116

5. Compact Groups Have a Haar Measure

Proof. Assume /(, is uniformly bounded and equicontinuous. By Fact 5.3 the pointwise closure, K,P of /(, is also equicontinuous. But /(, is bounded in each coordinate by its uniform bound, and so /(,P is also bounded in each coordinate by the same uniform bound. Tychonoff's theorem tells us that K,P is compact in JR8 , the topology of pointwise convergence. But now Fact 5.2 tells us that K,P is norm compact so K is relatively norm compact. On the other hand, if /(, is relatively norm compact in C(S) , then /(, is uniformly bounded. Further, if c > 0 is given then there are Ji , . . . , fn E K so that each f E /(, is within i of one of the f/ s, 1 ::; j ::; n. Suppose s E S. For each j, 1 ::; j ::; n, there is an open set Uj in S containing s so that € l fi (t) - fj (s) I ::; 3

for all t E Uj . It follows that if f E /(, and t E U1 n · · · n Un , then judicious choice of j allows us to estimate: I f (s) - f (t) I < I f (s) - fj (s) I + l h (s) - fj (t) I + l fj (t) - f (t) I € € € < 3 + 3 + 3 = E, D and so /(, is indeed equicontinuous. An added bonus.

G be a topological group, and let M be a nonempty compact subset of G. Suppose K � C ( G ) is equicontinuous on M . Then given c > 0 there is an open set V containing the identity of G so that if x, y E M and x E yV, then l f(x) - f(y) I ::; c for all f E K . Corollary 5.5. Let

Proof. (It is useful to take a close look at the proof of Theorem 5.1 before looking at this proof.) Let c > 0 be given. For each a E M there is an open set Va that contains the identity e of G such that if x E M and x E a Va, then l f(x) - f(a) I ::; � for all f E /(,, Since e · e = e there is an open set Wa that contains e and satisfies

Wa · Wa � Va. Now if a E M, then a E a · Wa. Hence {a · Wa : a E M} forms an open cover of M. Therefore we can find a 1 , . . . , an E M so

Clearly,

117

2. Existence and Uniqueness of an Invariant Mean

is open and contains e. Let x, y E M and suppose x E yV. Since y E M, y E ai · Wai for some i, 1 S i S n. It follows that E I J (y) - J(ai) I s 2 for all J E J C After all, Wa � Va . Further, and so

E l f(x) - f(ai ) I S 2

for all f E JC as well. In sum, if x, y E M and x E yV, then, regardless of f E JC,

E E l f(x) - f(y) I S l f(x) - f(ai) I + l f(ai ) - f(y) I S 2 + 2 = E .

D

2. Existence and Uniqueness of an Invariant Mean

In this section, we'll show how to ascribe to each f E C ( G) a mean M( J) , which is at one and the same time linear in f, nonnegative when f is, and is a true average with the values at f and any right translate of f, identical. Let G be a compact topological group. Denote by F( G) the collection of nonempty finite subsets of G and by C ( G) the Banach space of all continuous real-valued functions defined on G, equipped as usual with the supremum norm. Throughout this section, if F1 , F2 E F(G) , then by F1 F2 , we mean all words a · b, where a E F1 and b E F2 ; in particular, if ai · b 1 = a2 · b 2 but ai =/= a2 , then we distinguish ai · b1 and a2 · b 2 . ·

Definition 5.6.

If F E F(G) and f E C ( G ) , then RAveF f (x) : = 1 f(xa) , x E G.

Naturally RAveFf E C ( G ) .

L

fFT a E F

(i) If f E C ( G ) , then min f, max f, and Osc f max f - min f all exist. (ii) If f E C ( G ) and F E F(G), then Osc RAveFf S Osc f. In fact, min f S min RAveFf S max RAveFf S max f.

Lemma 5.7.

118

5. Compact Groups Have a Haar Measure

(iii) If f E C(G) and F1 , F2 E F(G), then

Proof.

To see (iii), if x E G, then

and we have proved (iii) . Lemma 5.8.

that

D

If f E C(G) is not constant, then there is an F E F(G) such Osc RAveFf < Osc f.

Proof. After all, f's not being constant ensures that there is an a such that min f < a < max f. Set

U = [! < a] = {x E G : f(x) < a}.

Since min f < a, U is a nonempty open set in G and G = U aEG ua- 1 . (If x E G, then for any y E U, x = y(y- 1 x) E U(y- 1 x) � U aEG ua- 1 . ) Now U is open (since f E C(G)), and U #- 0 so ua- 1 is also a nonempty open set for each a E G. Therefore the ua- 1 ' s cover the compact G. There is F E F( G) such that

119

2. Existence and Uniqueness of an Invariant Mean

Therefore for any x E G there exists ax E F such that x E Ua;;; 1 ; i.e., for any x E G there exists ax E F such that f(xax) < a. Thus RAvep f(x) = 1

L f(xa) l � I ( L f (xa) + f (xax) ) 1 f(xa) + a IFI L

TFf

aE F

a E F,a#ax

<

a E F,a#ax

F < ( I l - l) max f + a IFI ( I F I - 1) max f + max f < IFI

max f.

Therefore Osc RAvep f ::; Osc f .

D

C(G), and define JC = {RAvepf : F E F(G)}. Then JC is uniformly bounded, equicontinuous family in C(G).

Lemma 5.9. Let f E

Proof. The key to this precious fact is that f is of course uniformly contin­ uous. So given an E > 0, there is an open set V in G containing G's identity such that if xy- l E V, then l f(x) - f(y) I ::; E. Notice that if a E G and xy- 1 E V, then (xa) (ya) - 1 = xaa- 1 y- 1 = xy- 1 E V. So once xy- 1 E V,

l f(xa) - f(ya) I ::; E

for all a E G. But now if F E F(G), then whenever xy- 1 E V we have IRAvep f(x) - RAvep f(y) I =

� I L f(xa) - aLE F f(ya)

l l aE F

1

< TFf <

L l f(xa) - f(ya) I

aE F

1 F E = E. TFf l I

I

120

5. Compact Groups Have a Haar Measure

Note that /(, is uniformly bounded since

IL L

I

I RAveF f(x) I = l l � aE F f(xa) 1 < TFf l f(xa) I aE F 1 < TFf IFI · l l fl l = l l f l l oo·

Thus K is uniformly bounded, equicontinuous family in C ( G) .

0

Lemma 5.9 brings the Arzela-Ascoli theorem into play. With Lemmas 5.8 and 5.9 in hand, the plan of attack is clear. We want an averaging technique which will give a true average, assigning values in a uniformly distributed manner. If the function f is constant, then we will plainly want to assign that value of constancy to f. With the aforemen­ tioned lemmas in hand, we handle nonconstant functions thusly: If f is not constant, then we can find F1 E F( G) so that OscRAveF1 < Oscf. If RAveF1 (!) is constant, then its value of constancy is the natural value to ascribe to f. If RAveF1 (!) is not constant, then we appeal to Lemma 5.9 again to find F2 E F ( G ) so that OscRAveF2 RAveF1 < OscRAveF1 (!) . Continuing in this vein, we see that in the worst case, we can find a sequence (Fn ) � F ( G ) so that for each n OscRAveFn+ 1 RAveFn (f) < OscRAveFn (f). Appealing to Messrs. Arzela and Ascoli, we can pass to a sequence (F�) c F( G ) so (RAveF'n (!) ) is uniformly convergent. The point is that because our averages were taken with respect to right translates, in the long run, judicious choices of the Fn 's ought to produce an average that is right invariant. Remarkably enough the wisdom needed has already been provided by von Neumann. Lemma 5. 10. Let

Proof.

f E C ( G) and K = {RAveFf : F E F ( G ) }. Then inf Osc g = 0. gE/C

Let s

= gE/C inf Osc g = inf{Osc RAveFf : F E F ( G ) }.

121

2. Existence and Uniqueness of an Invariant Mean

Therefore there exists (Fn) in F ( G ) such that Osc(RAveFn ) ( J ) � s. Thanks to Arzela and Ascoli we can, by passing to subsequences if necessary, assume that RAveFn f --+ g E C ( G ) uniformly. It's plain that on assuming the uniform convergence of (RAveFn ) that min RAveFn f --+ min g and max RAveFn f --+ max g, and so OscRAveFn f --+ Oscg. Thus Oscg = s . Here's the point: g is constant! Indeed if g were not constant there would be an Fo E F( G) such that so = OscRAveF09 < Oscg = s thanks to Lemma 5.8. Since (RAveFn f) is uniformly convergent, there exists N such that s so l l RAveFN f - gl l oo < -3 - ; i.e., for any x E G, s - so I RAveFN f(x) - g(x) I :::; 3 - . But this is quickly seen to mean s - so I RAveF0 RAveFN f(x) - RAveF0g(x) I :::; 3 for all x E G as well. It follows that s - so I OscRAveF0 RAveFN f - OscRAveFo91 < 2 -3- ; -

-

-

(

( s so )

i.e.,

)

I OscRAveF0 RAveFN f - so l < 2 --3- · But this in turn means that s - so 2 1 OscRAveF0 RAveFN f < so + 2 3- = 3 s + 3 so < s. But

(

and

)

s = F infG OscRAveFf.

EF( ) This should elicit an "OOPS" because RAveF0 RAveFN f = RAveFo· FN f E / C

122

5. Compact Groups Have a Haar Measure

Therefore g is constant and s = O ; i.e., inf Oscg = 0. g E/(

D

We say the real number p is a right mean of f if for each E > 0, there is an F E F( G) such that I RAvep f(x) - P l < E for all x E G; i.e., l l RAvep f - P l l oo < E . Theorem 5 . 1 1 . Every f

E C(G) has a right mean.

By the techniques used in Lemma 5. 10, there is a constant function h (say h(x) = p) and a sequence (Fn ) � F(G) such that lim l l RAvepnf - h l l oo = O ;

Proof.

n

i.e., as n --+ oo . Plainly p is a right mean of f.

D

It's plain that each f E C(G) has a left mean as well, that is, there is a q E ffi. so that if E > 0 is given, there exists an F E F( G) so that

for all x E G. For obvious reasons, we define 1 LAvep f(x) = TFf f(ax) .

L

aE F

E C(G) . Let p be a right mean of f, and let left mean of f . Then p = q .

Theorem 5.12. Let f

Proof.

Let E > 0. Find A, B E F( G) so that

q

be a

123

2. Existence and Uniqueness of an Invariant Mean

Now

1

RAveA TBT L J(bx)

RAveALAvesf(x)

1 1 TJiT TBT 1 1

bEB

LA S(xa)

�

(where S(x) =

L J(bx))

�B

L L f(bxa)

TAT TBT aE A bEB

1 1

L L f(bxa)

TBT TAT bEB a EA

1

1

TBT L TAT L f (bxa)

1

bEB

aE A

TBT L RAveAf(bx) bEB

LAvesRAveA f.

Further, and

RAveA(LAvesf - q) = RAveA LAvesf - q

LAves (RAveAf - p) = LAvesRAveAf - p. So for any x E G, IP - RAveALAvesf(x) + RAveALAvesf(x) - q i IP - q i < IP - RAveALAvesf(x) I + I RAveALAvesf(x) - q i IP - LAvesRAveAf(x) I + I RAveALAvesf(x) - q i I LAves (P - RAveAf(x)) I + I RAveA (LAvesf(x) - q) I < IP - RAveAf(x) I + I LAvesf(x) - q i (since I LAvesf l ::; I i i and RAvesf l :S I i i ) E E < 2 + 2 = E,

and p = q. Go figure.

Corollary 5. 13. For any f

both a right and left mean.

D

E C(G) there is a unique number M(f) that is

Theorem 5. 14. The functional M on C(G) satisfies the following:

(i) M is linear. (ii) M f 2'. 0 if f 2'. 0.

5.

124

Compact Groups Have a Haar Measure

(iii) M(l) = 1. (iv) M(af) = M(f) = M(fa) for each a E G, where af(x) = f(ax) and fa(x) = f(xa). (v) M(f) > 0 if f ?:. 0 but f =/= 0. (vi) M(f) = M( J ) where f(x) = f(x - 1 ) for each x E G. We start by showing M(RAvepf) = M(f) (17) for each f E C ( G) and each F E F ( G ) . Suppose that M( J ) = p. If E > 0 is given to us, then we can find Fo E F ( G ) such that l l LAvep0f - P l loo :S c; i.e., f(bx) - p Proof.

l�'

1 1;.1 ,t;,

for all x E G. It follows that for any x E G, I RAvepLAvep0 f(x) - Pl :S € . Since RAvepLAvep0 f = LAvep0 RAvep f, p is a left mean of RAvep f. Hence by our previous result, M(RAvep !) = p and M(RAvep !) = M(f). To see that M is linear, let M(f) = p and M(h) = q. Pick Ho E F ( F ) so that i.e., for all x E G,

I 1 �01 � ,

,

h(xb) - q

i � <;

i.e., if E E F ( G) and x E G, then I RAveE -Ho h(x) qi = I RAveERAveH0h(x) - qi < E . Now p = M(f) = M(RAvep !) for any F E F. Therefore p is the right mean of RAveH0 f. Hence there exists Fo E F ( G ) so that l l RAvep0 RAveH0f - P l l oo :S c; -

2. Existence and Uniqueness of an Invariant Mean

1 25

i.e., for all x E G, I RAvep0.H0f(x) - Pl = I RAvep0 RAveH0f(x) - Pl S E . Since we already know that for all x E G and each E E F(G) ,

I

i

RAvee.H0h(x) - q S E ,

it follows that by taking E = Fo we get for each x E G, I RAvep0.H0 (f + h) (x) - (p + q) I S 2 E . Thus M(f + h) = M(f) + M(h) . It follows from this and the easily established fact that M(kf) = kM(f) that M is linear, and we have shown (i). Parts (ii) and (iii) are clear. To see (iv) , since (18) RAvepf(xa) = RAvea·Ff(x), M(fa)

Similarly, (19)

and so M(af)

and thus

M(RAvep fa(x)) (by (17)) M(RAvep f(xa)) M(RAvea · F f(x)) (by (18)) M(f) (by ( 1 7) ) .

LAvep f(ax) = LAvep .af(x), M(LAvep( af(x))) (by (17)) actually it's equivalent with left averages) M(LAvep f(ax)) M(LAvep.a f(x)) (by ( 19) ) M(f) (by ( 1 7) actually it's equivalent with left averages) ,

M(af) = M(f) = M(fa) · For (v), suppose that f E C ( G ) , f ;:::: 0, f ¢. 0. Then there is a > 0 such that U = [f > a] is nonempty and open. It's easy to see that {ua - 1 : a E G} is an open cover of the compact G. It follows that for some a i , . . . , am E G G = Ua! 1 u Ua2 1 U · · · U Ua;;/ . Let's check to see how this plays out.

5.

126

Compact Groups Have a Haar Measure

If x E G, then x E U ak, 1 for some 1 f (xak) > a. It follows that

:::;

k

:::;

m. Hence, xak E U and thus

for all x E G. Therefore 0 < ma :S M(RAve {a i , . .. ,am} f) = M(f). -

Almost done: we have but to show that M(f) and M(f) agree. To establish this, define N(f) = M(f o inv) , where inv : G --+ G is given by inv(x) = x - 1 . N is a linear functional on C(G), N(f) 2'. 0 if f 2'. 0, and N(l) = 1. Moreover M(fa o inv) N(af) M( a - 1 /) (since fa o inv(x) = fa(x - 1 ) = ](a - 1 x) = a - 1 f(x)) M(f) (by (iv)) N(f). But by Corollary 5.13, there is only one invariant mean on C(G) so N(f) = 0 M(f). 3.

The Dual of C(K)

Theorem 5.14 tells us that if G is a compact topological group, then there is a positive linear functional x * on C(G) such that for any x E C(G) and any a E G x* (ax) = x*(x) = x * (xa), where ax(t) = x(at) and Xa(t) = x(ta) for t E G and x* (l) = 1. In this section we will realize x* as a measure acting on the Borel sets in G. In fact, we will show that for any compact Hausdorff space K every positive linear functional x * of norm one on C ( K) can be realized as integration vis-a-vis a regular Borel measure on K. Our plan of attack goes as follows. First, we show that positive linear functionals on C(K) extend to positive linear functionals of the same norm on l00(K) . This calls on Kantorovich's extension theorem (a natural corollary of the Hahn-Banach Theorem 4.1). Next, denoting by A the algebra of subsets of K generated by K's topology, we study how the extension of a positive linear functional on C(K) acts on B(A) , the uniform closure of the space of A-simple functions on K. The

3.

The

Dual of C(K)

127

action is that of integration against a nonnegative finitely additive measure defined on A. To be sure we gain enough of the mass of the original function to compute its norm (as a functional on C(K)), we need to ask more of the nonnegative finitely additive measures-weak regularity is what we need. We then show how to replace a finitely additive probability ,\ on A by a weakly regular one, µ, in such a way that for x E C(K) , and so that sup

J x d,\ = J x dµ, { I J x dµ I : x E C(K) , l l x l l oo :S 1} = l l x* l l c(K)* ·

Theorem 5.15 (L. Kantorovitch). Let x* be a positive linear functional on C ( K) . Then there is a positive linear functional y* on l 00 ( K) which is an extension of x* without increasing its norm. The proof relies on the fact that C ( K) * "majorizes" z oo ( K) in the sense that given any b E l 00 (K) there is an x E C(K) such that x - b � 0. This allows us to build a sublinear functional perfectly suited to applying the Hahn-Banach theorem with a happy ending (the conclusion) . Let x* be a positive linear functional on C(K) . For b E l00(K) define p(b) by p(b) = inf {x* (x) : x E C(K) , x � b}. Then p is a sublinear functional on l00(K) and in fact, for x E C(K) we have p(x) = x* (x) . The Hahn-Banach theorem provides us with a linear extension y* of x* to all of l00(K) such that y* (b) :S p(b) for all b E l 00 (K) . Notice that if b E l 00 (K), b � 0, then - b :S 0 E C(K) so - y * ( b ) = y * ( - b) :S p( - x) :S x* (O) = 0. It follows that y* (b) � 0. Suppose that K is a compact Hausdorff space and A is the algebra generated by K's topology. Let ,\ : A-+[0, 1] be a finitely additive probability. We can define J f d,\ for any A-simple f : K-+�: if f = Z:::� 1 ai XAi where ai , . . . , an E � and Ai , . . . , An are pairwise disjoint members of A, then

j f d..\ = t ai .A (Ai )· i= l

It is easy to see that J f d,\ is linear in f, J f d,\ � 0 if f � 0 and if If ( k) I :S M, then J I f I d,\ :S M . Consequently, J f d,\ has a unique linear extension to the uniform closure B(A) of the space of simple functions

128

5. Compact Groups Have a Haar Measure

defined on K. This extension, whose value at f E B(A) will still be denoted by J f .A, satisfies f d.X � l f l d.X � I I f I l oo · Moreover, if f 2: 0, then J f d.X 2: 0. It is worth noting that C(K) � B(A) . In fact, if E > 0 and x E C(K), then we can approximate x by an A-simple function X E as follows. Cover x(K) by open sets Gi , . . . , Gn each of diameter less than E . Let Ai = G 1 , A2 = G2 \ G 1 , . . . , Aj = Gi \ (G 1 U · · · U Gj - 1 ), . . . If Ai =1 0, let x E ( k ) = x ( k ) for all k E Aj, and x E ( k ) = 0 if k f/. Uj=1 Gj . It's plain that

IJ I J

l x(k) - x E (k) I � E

for all k E K and X E is A-simple. Of course, J d.X acts as continuous positive linear functional on C(K) with norm (as a functional) less than or equal to 1, which is .X(K). In order to assume more about .A, namely, we need to know that A is weakly regular in the sense that given E > 0 and A E A there is a closed set F � A and an open set U 2 A so that .X(U \ F) < E. The result follows: If A is a weakly regular finitely additive probability on A, then d.X is a norm one functional on C ( K) . f f-+ J

Fact 5 . 16.

Let E > 0 be given. Let Ei , . . . , En be pairwise disjoint member A so that K = Ei U · · · U En . For each j, 1 � j � n, pick Fj closed, Fj � Ej with .X(Ej \ Fj) � �- Using K's normality, let G 1 , . . . , Gn be open sets that are disjoint and so that Fj � Gj. Shrink the G/s if necessary so that a (Gj \ Fj) < � -A's weak regularity comes in handy again. Let xi , . . . , Xn E C(K) be chosen so that 0 � Xj � 1 and Xj = 0 outside of Gj but Xj = 1 inside of Fj. Set n xo = L xi E C(K). j= l Then l l xo l l oo = 1 and Proof.

It follows from this and the usual epsilonics that 0

3. The Dual of C(K)

129

Now let x* be a positive linear functional on C(K). Because x E C(K) has l l x l l oo S 1 means -1 S x S + 1 (as functions on K), -x * (l) = x * (-1) :::; x* (x) :::; x * (l) , and so l l x* l l = x * (l). Applying Kantorovich's theorem to extend x* to a positive linear functional y* on l00(K) we see that l lY * l l = y* (l) as well. Now for any A E A, y* (xA ) = .X(A) is a nonnegative finitely additive measures on A. If we assume l lY * l l = 1(= l l x* l l ) , then .A is a finitely additive probability and for any f E B(A) , y * (f) =

J 1 d.X.

If .A is weakly regular, then the functional x --+ J x d).. already achieves x*'s norm on C(K). Otherwise? Otherwise, we will show how to modify .A to get a finer representing measure. Generically, F is a closed subset of K, G is open subset of K, and E is any subset of K. Set µ i (F) = inf{).. ( G) : F � G}, µ2 (E) = sup{µ 1 (F) : F � E}. Both µ i , µ2 take nonnegative values and each is ascending on its domain. Suppose F1 \G 1 � G. Then Fi � G 1 U G, and so ).. ( G 1 U G) s .X(G 1 ) + .X(G) implies Therefore

µ i (F1 ) S ).. ( G 1 ) + µ(F1 \G 1 ), after all, F1 \G 1 is a closed set contained in G. Since G 1 is any open set, we let G 1 range over the open sets that contain a given F n F1 , and we see µ i (F1 ) S µ i (F n F1 ) + µ i ( Fi \G 1 ) . But F1 \F = F1 \(F1 n F) which contains F1 \G 1 so µ i (F1 ) S µ i (F n F1 ) + µ2 ( Fi \F). If we now let E � K and F1 range over the closed sets contained in E the result is (20) where E and F are unrestricted within their generic classes. We plan to show the reverse inequality stated in (20) holds as well.

130

5.

Compact Groups Have a Haar Measure

Suppose F1 and F2 are disjoint closed sets. Envelop F1 and F2 in disjoint open sets G 1 and G2 , respectively. Let G contain F1 U F2 . Then ,\(G) � ,\(G n G 1 ) + ,\(G n G2 ) � µ i ( Fi ) + µ i (F2 ); hence µ i (F1 U F2 ) � µ(F1 ) + µ(F2 ). Now look to generic E, F, and let F1 range over the closed subsets of E n F, and let F2 range over the closed sets contained in E\F. The result is µ2 (E) � µ i (F1 U F2 ) � µ2 (E n F) + µ2 (E\F) , and so (21) Look familiar? It should. Were µ2 an outer measure, then (21) would say that each closed set F is µ2 -measurable. Now µ2 is not necessarily an outer measure but it's not far from being such. We say that a set E � K is a µ2-set if µ2 (A) = µ2 (A n E) + µ2 (A n Ec) for each A E A. What we've just seen is that each closed set is a µ2 set. Moreover, following the reasoning encountered in the Caratheodory procedure, the collection of µ2 -sets is an algebra of sets. Let's see that this is so. Of course, 0, K are both µ2 -sets and E is a µ2-set if and only if Ec is. Suppose Ei and E2 are µ2 -sets. Take A E A. Since Ei is a µ2 -set (22) µ2 (A n E2 ) = µ(A n E2 n Ei ) + µ2 (A n E2 n El). Since E2 is a µ2 -set, (23) Also µ2 (A n (E1 n E2 )c) = µ 2 (A n (E1 n E2 )c n E2 ) + µ2 (A n (E1 n E2 )c n E2) . But and so (24) µ2 (A n (E1 n E2 )c) = µ(A n E2 n El) + µ2 (A n E2). It follows that µ2 (A n E2 ) + µ2 (A n E2) (by (23)) µ2 (A) µ2 (A n (E1 n E2 )) + µ2 (A n E2 n El) + µ2 (A n E2) (by (22)) = µ2 (A n (E1 n E2 )) + µ2 (A n (E1 n E2 )c) (by (24)),

131

3. The Dual of C (K)

and E1 n E2 is also a µ2-set. The collection of µ2 -sets is an algebra of sets containing the topology of K. If E1 and E2 are disjoint µ2 -sets and A E A, then µ2 (A n (E1 u E2 )) = µ2 (A n (E1 u E2 ) n E1 ) + µ2 (A n (E1 u E2 ) n El) (since E is a µ2 -set) = µ2 (A n E1 ) + µ(A n E2 ), and so µ2 is finitely additive on the algebra of µ2 -sets. If we restrict µ2 to A we get a finitely additive probability µ. Moreover for any closed set F � K, as

careful inspection of the respective definitions will reveal. Let A E A. By µ2 's very definition µ(A) = µ2 (A) = sup{µ 1 (F) : F is closed and F � A} sup{µ(F) : F is closed, F � A}; further, µ(A) = µ( K ) - µ(Ac) = µ(K) - sup{µ(F) : F is closed, F � A} = µ( K ) + inf {µ(Fe) - µ(K) : F is closed, F � Ac} inf{µ(Fc) : F is closed, F � A } sup{µ( G) : G is open A � G}. It remains to show that x dµ

j x d).. = j

for each and every x E C ( K ) . But again we need only do this for nonnegative real-valued x's. Since each x E C(K) is bounded, we'll turn our attention to x E C(K) so that 0 :S x :S 1. Let E > 0. Partition K into disjoint members E1 , . . . , En of A in such a way that +< x

J dµ, ,;, t. ( .��/(k) ) µ,(E;)

Since µ is regular, we can find closed sets Fi , . . . , Fn with Fj � Ej such that

5. Compact Groups Have a Haar Measure

132

But K is normal and x E C(K) so we can envelop the F/ s within disjoint open sets Gj where Fj � Gj for j = 1 , . . . , n with inf

kEGj

This being done,

j

inf

x (k) 2::

x dµ

kEEj

x (k) -

:S t. (.�it

But remember that µi (F)

n

(µ�K)) .

)

x(k) µ(G; ) + 3 ' .

= µ2 (F) = µ(F)

for closed sets F � K, and if G is open with F � G, then µ(F) follows that µ( G) :::; .A ( G) and

� inf x(k) � kEG ·

j=l

(

J

)

� kEG inf · x(k) �

µ(Gj ) :::;

j=l

(

J

)

.A(Gj ) :::;

j

:::; .A(G) .

It

x d.A.

So

j

x dµ

< < <

� inf x(k) � kEG ·

( ) � ( inf ) J j=l

J

� j=l

kEGJ·

µ(Gj ) + 3c:

x(k) .A(Gj ) + 3c:

x d.A + 3E.

Since E > 0 was arbitrary, we have that for any x E C(K), with 0 :::; x :::; 1 ,

J :::; J x dµ

If we apply this to

(1 - x)

x d.A.

and keep in mind that µ(K)

= .A (K) ,

then the reverse inequality follows from

j

(1 - x)dµ :::;

j

(1 - x)d.A,

and that is all she wrote! Our next task is to relate positive linear functionals on C(K) to countably additive regular Borel measures on the Borel o--field B0 (K) . Here's a safe first step.

1 33

3. The Dual of C(K)

(Alexandroff) . Let µ be a regular finitely additive nonneg­ ative measure defined on the Borel field A {the field generated by the open sets) of the compact Hausdorff space K. Then µ is countably additive on A.

Theorem 5 . 1 7

Proof. Let E > 0. Suppose (En) is a sequence of pairwise disjoint members of A with E = Un En E A. There is an F E A, F closed, F � E, with

µ (E \ F) < E.

Moreover for each n we can find a Gn that is open with En � Gn and

µ (Gn \En) < 2En .

Now

n

n

so K's compactness, which is inherited by the closed set F ensures us that there is an N so It follows that

Ln µ (En)

> > >

>

Ln l µ l(Gn) - E L lµl (Gn) - E n� N µ (F) - E µ (E) - 2 E.

So E > 0 being arbitrary soon reveals that

( )

Ln µ (En) � µ (E) = µ LJn En .

On the other hand, for any M E N, so

( ) � µ ( n�LJM En) = n�LM µ(En) , µ ( LJ(En ) ) � L µ (En) n n

µ (E) = µ LJ En n

too. So µ is countably additive.

0

Now we note that having a nonnegative real-valued countably additive mea­ sure µ on the field A, if A � K and we define µ* ( A) by

µ' ( A) = inf

{ � µ(En) } ,

134

5.

Compact Groups Have a Haar Measure

where the infimum is taken over all sequence (En) of members of A such that A � U n En , then the result that µ* is an outer measure on K. Two points are worth making here: First, Caratheodory's Theorem (see Chapter 2) tells us that the µ*-measurable sets constitute a a-field of subsets of K on which µ* is countably additive. Second, if E E A, then µ* (E) = µ(E) . We need to establish this second feature of µ*, so let E E A. Of course E � E tells us immediately that µ * (E) :S µ(E). On the other hand, if E E A and E � Un En where each En E A, then the disjointification of En

j-1 Di = Ei, D2 = E2\E1 , . . . , Dj = Ej\ LJ Ek, . . . , k=l is a disjoint sequence of members of A with Un En = Un Dn. It follows that µ(E) = µ (E n ( LJ Dn )) n µ ( LJ(E n Dn) ) n µ(E n Dn ) (µ is countably additive on A) L n < L µ (Dn) n < L µ (En) · n It follows that µ(E) :S µ* (E) too. So if E E A, then µ(E) = µ* (E) . Finally, we note that each E E A is µ*-measurable. To show this, we need

to test vis-a-vis arbitrary A � K. Since µ* is an outer measure and then

µ * (A n E) + µ * (A n Ee) ::::; µ* (A). Now let E > 0 be given and find (En) � A such that and

A � LJ En n Ln µ (En) :S µ* (A) +

E.

1 35

3. The Dual of C ( K)

Notice that A n E � LJ ( En n E ) n Therefore

and

n

n n n

< µ* ( A ) + E .

But E > 0 was arbitrary so we have µ* ( A n E ) + µ ( A n Ee) :::; µ* ( A ) . We have shown that for every E E A and for every A � K, µ * ( A) = µ * ( A n E ) + µ * ( A n Ee ) , and so each member of A is µ*-measurable. Alas when we consider the extensive µ* of µ to the Borel a-field Bo(K) , we get a countably additive regular Borel probability. Let's see why µ* is regular. Let E E Bo(K) , and let E > 0 be given. We can find a sequence ( En) in A such that E � Un En and µ* LJ En \ E < n But µ is a regular on A so for every n there exists an open set Gn such that En � Gn and µ ( Gn \ En) < 2n€ Now E � LJ En � LJ Gn = G, an open set. Further

(

)

�·

·

( )

µ* G \ E

( ( n )) + µ* ( ( un En) \ E) µ* ( y ( Gn \ En )) + µ * ( ( y En ) \ E) Ln µ* (Gn \ En ) + µ* ( ( LJn En ) \ E) <

:::; µ* G \ u En

= =

and so µ* is outer regular on Bo(K) and inner regularity follows.

E,

5.

136

Compact Groups Have a Haar Measure

4. Notes and Remarks

We have followed Pontrjagin's [104] classic in our discussion of von Neu­ mann's proof of the existence of Haar measure (or invariant mean) for com­ pact groups. A careful look at von Neumann's proof will indicate why he was so quick to see the possibility of an abstract presentation of solutions to the easy problem of measure; of course, as any good Aristotelian under­ stands, it's a long way between possibility and execution. Our presentation of the computation of C ( K) * follows that of Dunford and Schwartz, as found scattered throughout Part I of their three-part bible [34] . 4. 1 . Other Proofs of the Existence of Haar Measure in Compact Groups.

4. 1.1. Marriage and Haar Measure. Can one produce an invariant measure "abstractly" ? There are several ways of doing so; presently we will follow a path indicated by Maak, as expressed in the notes of V. Milman and G. Schechtman [83] . It calls on basic tools of functional analysis and a combinatorial solution to the "marriage problem" . It is well motivated by a simple example. Let II denote the unit circle, II = {z E
Then II is a compact metric topological group, a prototype of such, if you please. It also possesses a natural invariant probability Borel measure, nor­ malized Lebesgue measure >. , a measure that requires construction, to be sure. However, >. can be realized by an abstract limiting procedure if we but think in a Riemannian manner. For each n, let {z1 , . . . , zn} be the n-primitive roots of unity, and let µn be n 1 µn = -n L bz ' k=I where bz is the (continuous linear) functional on C(II) of evaluation at z, &z (f) = f( z ). The sequence (µn ) lies in the dual unit ball Bc ( IT ) * of C(II)* and so has a weak* limit point. (This is Banach's proof of the Banach-Alaoglu theorem.) Actually, more is so. The sequence (µn ) is itself weak* -convergent, that is, for any f E C (II), limn µn (!) exists and n 1 lim L f(zk ) = O f(t ) d>.(t ). n µn (f) : = lim n -n k=l k

12�

137

4. Notes and Remarks

After all, the n primitive roots of unity are uniformly distributed about II, and so an appeal to the basic theory of the Riemann integral tells us the limiting procedure above is valid. We want to modify this procedure to a more general setting and so call on functional analysis for tools to do so. The basic tool will be the Banach­ Alaoglu theorem. Recall that if X is a Banach space with dual X * , then the weak* -topology of X* is the topology of pointwise convergence on members of X. Salient features of this topology include • the closed unit ball Bx* of X* is always weak * -compact; • Bx· is metrizable in the weak * -topology if and only if X is separable. Our setting will be C(S) where S is a compact metric space; in this case, C(S) is a separable Banach space and so Bc( S) * is weak* -compact and metrizable. In particular, any sequence (µn) in Bc( S) * has a weak* -conver­ gent subsequence. We now present a combinatorial lemma, sometimes referred to as the mar­ riage problem. Its proof goes back to P. Halmos and H. E. Vaughan [5 1] . Suppose for each boy bi in the set B of n boys, B = { bi, . . . , bn }, there is a collection G ( bi ) of girls with whom bi is acquainted.

With bigamy an anathema, for each boy in B to be able to marry a girl with whom he is acquainted it is both necessary and sufficient that regardless of C � B

Theorem 5 . 1 8 .

(25)

holds. Proof. Since it is plain that (25) is necessary, we'll concentrate on proving sufficiency of (25) . So we suppose (25) to be in effect and prove the possi­ bility of a wise matchmaker by an induction on the number of boys in B. Though not universally associated with marriage, we nevertheless introduce the notion of perfect cliques. Definition 5 . 19.

A clique C � B is perfect if

I U G ( b) I = 1 ° 1 · bEC

138

5. Compact Groups Have a Haar Measure

Lemma 5.20. Proof.

If C and c' are perfect, then so is C U c' .

By (25) we know that

I

l i

,

LJ G ( b ) � c u c' .

bECUC'

Further

I C u c' i = I C I + 1 c ' 1 - 1 c n c' i , and, since it's too much to expect all the G ( b ) ' s to be singletons,

I

u G ( b)

bECUC1

'

�

I u G ( b) ' + I u G ( b ) l - 1 I C I + 1 c' 1 - I u G ( b ) ' bEC'

bEC

u G(b)

bECnC'

'

bECnC'

So C u c' is perfect.

< I C I + i c' l - I C n c' i (by (25)) = I C U c' 1 . D

So for now we suppose that for any collection of fewer than n boys, (25) is sufficient to warm the cockles of the matchmaker's cold heart. Let B be a collection of n boys which given any C � B, we have

IU

bEC

I

G ( b) � 1 c 1 .

We'll take a good look at G ( bn) · If g E G (Bn), then looking at G ( b 1 ) \ {g} , G ( b2) \ {g} , . . . , G(bn - 1 ) \ {g} , we see two possibilities: MAYBE for some g E G ( bn) G ( b 1 ) \ {g } , G(b2) \ {g } , . . . , G(bn - 1 ) \ {g} satisfies (25) and so we can effect a marriage of each b 1 , b2 , . . . , bn - l to a girl of his acquaintances and still have the lovely g left to marry bn when g E G ( bn) · Of course in this case we're happy onlookers and the proof is done. MAYBE for each g E G (bn), the collection G ( b 1 ) \ {g } , G ( b2) \ {g } , . . . , G(bn - 1 ) \ {g}

139

4. Notes and Remarks

fails ( 25 ) ; i.e., for each g E G ( bn) there is a Cg � {b 1 , b2 , . . . , bn - 1 } such that

Iu

bi EC9

I

G ( bi) \ { g } < I Cg l ·

But deleting g from G ( bi) can change the number of girls in this collection by at most 1 and { G ( bi) : bi E Cg } satisfies ( 25 ) by our inductive hypothesis so each of the G ( bi) 's, with bi E Cg must contain g! It follows that

Iu

bi EC9

I

G ( bi) \ { g }

=

Iu

biEC9

I

G ( bi) - 1 < I Cg l ·

Since { G ( bi) : bi E Cg } satisfies ( 25 ) , it must be that

Iu

bi EC9

G ( bi)

I

= I Cg I ;

i.e., {G ( bi) : bi E Cg } is perfect for each g E G ( bn) · Our lemma now assures us that {G ( bi) : bi E ug EG(bn) Cg } is also perfect and so LJ{G ( bi) : bi E u Cg } = u Cg ·

I

Keep in mind that

'

g EG(bn)

I

l

g EG(bn)

G ( bn) � LJ G ( bi) · biEC9

So if we add bn to LJ{ Cg : g E G ( bn) } , we have

I

G ( bn) U {G ( bi) : bi E LJ

Cg }

g EG(bn)

' I I l { { } =

LJ{G ( bi) : bi E LJ

Cg }

g EG(bn)

' }j

'

LJ{Cg : g E G ( bn) }

<

and we've found a collection C � B,

C = {bn } u

LJ

{ bn} u

g EG(bn)

LJ

g EG(bn)

Cg

,

Cg ,

which violates ( 25 ) . We see then that the second of the possible outcomes is defined to fail, and D so we deduce that the first "MAYBE" is so.

5.

140

Compact Groups Have a Haar Measure

We will apply our result to proving the existence of a probability measure µ on a compact metric space S that is G-invariant, where G is the group of isometries of S onto itself. The use of the solution to the marriage problem in this context is due to Maak. Let S be a compact metric space, and let G be the group of isome­ tries of S onto S. Let E > 0. Let NE be a minimal E-net; i.e, 8 = u {y E 8 : d ( x , y) < E }, x ENE and I NEI is minimal. For f E C (S), define Proof.

Clearly, µNE is a probability measure on S and as such belongs to Bc(S) * , the unit ball of C ( S)*. Since C ( S) is a separable Banach space, Bc (S) * is compact and metrizable in the weak *-topology. Hence for a suitably chosen sequence En \.i 0, we have that (µNEJ is weak *-convergent to a probability µ E BC(S) * · Our first task will be to show that µ is indep�dent of the choices of NE made, E-by- E . To this end, !!.e will show that if NE is another minimal E-net, then not �ly are NE and NE equinumerous but we can find a bijection
(u

{s ' E S ; d ( t , s1 ) < E }

tE NE Were L to have fewer members than K, then the set L U (NE \ K)

)

"#

0.

141

4. Notes and Remarks

would have cardinality I L u (NE \ K) I

I L i + l (NE \ K) I < I K I + l (NE \ K) I = I NE I · Yet for any x E S, x within E of some point of NE , maybe x is within E of a point of L and maybe not! But if x is not within E of a point of L, it is not within E of a point of K-after all, that's how L was defined! So x is within E of a point (NE \ K) . Then L U (NE \ K) is an E-net for S with fewer members than NE . Oops! Our claim is verified. Now our marriage counselor steps in to advise us that there is a bijection ¢> of NE onto NE such that for each t E NE , ef>(t) is acquainted with t; i.e., there is a bijection ¢> : Nc-tNE such that for each t E NE , d(t, ef>(t)) < 2E. But this entails the following: if f E C ( S) , then <

< <

where

Ll I �E I tEN E Wf (2E) ,

f(t) - J (ef>(t)) I

WJ('fJ) : = sup{ lf(t) - f(s) I : d(s, t ) � 'fJ}

is the modulus of (uniform) continuity of f E C(S) . It follows from this that weak * lim µNEn , µNEn , weak * lim n n coexist and coincide. What of it? Well if g E G, then NE = {g(t) : t E NE } is a minimal E-net if NE is, so for any f E C(S) """" µ( J o 9) = h.� N.1 L.J J (g(t)) I E I tENE h.� 1 """" f(s) I NEI s� ENE lim µN-En ( j ) = µ(!) , n and µ is G-invariant.

0

142

5. Compact Groups Have a Haar Measure

4.1.2. Fixed Points in Haar Measure. We start with the following classical result. (Markov-Kakutani Fixed Point Theorem). Suppose K is a compact convex subset of the locally convex {Hausdorff) linear topological space E, and suppose that T is a continuous linear operator T : E-+E such that T(K) � K. Then there is an element k E K so that T(k) = k. Theorem 5.21

Fix ko E K, and set 1 n kn = n +-l (ko + T(ko) + · · · + T (ko)). Each kn E K. Because K is compact the sequence ( kn) has a limit point k. Let 0 and n o E N there is an n > no so that l ( ( kn) + ( kn) - 0 and no E N are arbitrary, it follows that
-----

Here's an easy consequence of the Markov-Kakutani Fixed Point Theorem using some transfinite hijinks.

4. Notes and Remarks

143

Suppose A is a commuting family of continuous linear operators on the locally convex {Hausdorff) linear topological space, K is a compact convex subset of E, and T(K ) � K for each T E A. Then there is a k E K such that T(k) = k for each T E A. Corollary 5.22.

First we notice that if T1 , . . . , Tn E A, then there is a k E K so that T1 (k) = k, T2( k) = k, . . . , Tn( k ) = k. This follows by induction on n . The Markov-Kakutani theorem shows it so for n = 1. Suppose we know it to be so for n > 1, and we let Kn = {k E K : T1 (k) = k, T2(k) = k, . . . , Tn(k) = k}. Then Kn is a compact, convex subset of E, and if k E Kn , then for j = 1, . . . , n we see that Tn+ 1 k = Tn+i ( Tj ( k)) = Tj (Tn+ 1 ( k)) since members of A commute. So Tn+i ( k) E Kn too. Applying the Markov­ Kakutani theorem again tells us that Tn+l has a fixed point in Kn . So for any finite set F � A, if KF = {k E K : T(k) = k for T E F}, then KF is a nonempty closed convex subset of K and the family {KF : F � A, F finite} has the finite intersection property. By K's compactness n { KF : F � A, F finite } is nonempty. Of course any point in the intersection is a common fixed point for all T E A. D Proof.

A consequence of the above corollary is the existence of a Haar measure on a compact abelian group. Since the main idea of this proof will soon be revisited to deduce the existence of a Haar measure on arbitrary compact group topological groups, we present the proof in outline only. The key is the fact the dual C (G) * of the Banach space C(G ) of continuous scalar-valued functions on G is the space M(G ) of regular Borel measures on G. Inside M(G) we concentrate on the weak *-compact convex set P (G) of probabilities in M (G) . For f E C(G) and g E G, define T9 : C(G)-+C(G)

5. Compac t Groups Have a Haar Measure

144 by

Tg (f)( h ) = f(gh) for h E G. The collection {Tg : g E G} is a commuting family of linear isometries of C(G) onto itself. What's more, each T; takes P(G) into itself:

J

J

J

(T; ( µ) ( J )) = (Tg f) dµ = f(gh) dµ( h ) = f(h) dµg ( h ) , where µg ( A ) = µ(g - 1 A ) . This shows T;µ = µg . The map T; is weak * -weak * continuous on M(G) to M(G ) . By the Markov­ Kakutani theorem and its corollary, there is a common fixed point in P(G) . It's plain that the common fixed point is a Haar measure. For general compact topological groups we need the following. Theorem 5.23

( Ryll-Nardzewski ) . Suppose S is a semigroup of continuous

linear operators on a Hausdorff locally convex topological linear topological space E, and let K be a compact convex subset of E. Assume that S ( K ) � K for each S E S. Suppose further that whenever k l and k2 are distinct points of K, then 0 ¢ {S(k 1 - k2) S E S } . Then there exists k E K such that for each S E S Sk = k. :

Remark 5.24. It must be mentioned that this result of Ryll-Nardzewski [109, 1 10] is a consequence of the Markov-Kakutani theorem and martin­

gale argumentation.

Let G be a compact topological group. Then there exists a regular Borel probability µ on G such that µ(g - 1 A ) = µ ( A ) for each g E G and each Borel set A � G. Corollary 5.25.

Proof. The proof proceeds along the same path as that outlined in the case when G is abelian. This time we see that the collection {Tg : g E G} is a group of isometries of C(G ) onto C(G) and, as before, T; (µ) = µg , ensuring that each T; is a weak *-weak * continuous linear operator taking P (G) into P(G) . What remains to be shown is that the added condition of

145

4. Notes and Remarks

Ryll-Nardzewski is met. To this end, suppose p 1 , P2 are distinct members of P (G) . There must be an f E C(G) , l l f l loo = 1 so f dp 1 - f dp2 = € > 0. Then {T9 f g E G} is equicontinuous and norm-bounded, and so (thanks to Arzela and Ascoli ) it is relatively norm compact in C(G) . Hence there are 91 , . . . , 9n E G such that if g E G there is an i, 1 ::; i ::; n so that (26) l l T9- 1 (f) - Tgi (f) l loo < � · If we let

J

J

:

then V is a weak * open set containing zero. Then if g E G and 9i satisfies (26) , l (T;p1 - T;p2)(T9J ) I 2". l (T;p 1 - r;p2)(T9- 1 ! ) I - l (T;p 1 (T9- 1 f - T9J )) l - l (T;p2(T9- 1 f - T9J )) I € € € -> € - 4 4 = 2 It follows that r; (P1 - P2) tf_ V The hypotheses of the Ryll-Nardzewski theorem are fulfilled. The common fixed point µ of all the r; 's satisfies µ ( A ) = µ (g - 1 ( A )) , 0 for all g E G and all Borel sets A � G. -

-

-

- .

The elder of the authors was shown this proof by D. J. H. "Ben" Garling, whose Cambridge University notes, The Martingale Convergence Theorem and its Applications, are a treasure trove of fascinating proofs and mathe­ matical connections. Of course we specially direct the ambitious reader to the original papers of Ryll-Nardzewski ( [109] , [110] ) , which use ergodic theoretic ideas to prove his fixed-point result.

Chapter

t

Applications

In this chapter we present a few applications of Haar measure on compact groups. We open with a discussion of homogeneous spaces, that is, compact Haus­ dorff spaces on which a compact group acts as a group of homeomorphisms under composition. We show how Haar measure naturally induces an image measure on the space via a natural quotient operator. Assuming the group acts transitively, this invariant measure is also unique. These results go back to A. Weil's classic book [136] . Next, we show how the existence of a normalized Haar measure on a com­ pact group can be used to prove the famous Peter-Weyl theorem on the existence of a complete system of irreducible unitary finite-dimensional rep­ resentations of the group. In the third section we introduce the class of absolutely p-summing oper­ ators between Banach spaces, indicate the remarkable Pietsch domination theorem ( wherein measure theory makes a surprise appearance ) , and then show that if a p-summing operator acts on a Banach space of functions on a compact group that possesses suitable translation invariance and the operator acts in a respectful manner from the space to itself, the so-called "Pietsch measure" is the normalized Haar measure. We close, as usual, with a section of Notes and Remarks.

-

147

148

6.

Applications

1. Homogeneous Spaces

Let G be a compact topological group, and let K be a compact Hausdorff space. We say that G acts transitively on K if there is a continuous map G x K-+K : (g, k) i-+ g(k) such that (i) e(k) = k for all k E K (e is the identity of G) ; (ii) (gig2 ) (k) = gi (g2 (k)) for all gi , 92 E G, h E K; (iii) given ki , k2 E K, there is a g E G so that g(ki) = k2 . It is noteworthy that each g E G may be viewed as a homeomorphism of K onto itself; after all, the map k-+g(k) is continuous and has k-+g - i (k) as an inverse. Condition (iii) says, in particular, that the space K is homogeneous; i.e., we can move points of K around K via homeomorphisms (members of G, in fact) of K onto itself. If µ is the unique translation invariant Borel probability on G, then µ induces a G-invariant Borel probability on K. This is an important construction, one worth understanding in general as well as in special cases. Suppose H is a closed subgroup of the compact topological group G. Con­ sider the set G/ H (of left cosets of H in G) with the so-called "quotient topology" , that is, the strongest topology that makes the natural map qH : G-tG/H (taking g E G to gH E G/H) continuous. So U � G/H is open precisely when qif (U) is open in G. In other words, a typical open set in G/ H is of the form { xH : x E V} when V is open in G. Because H is supposed to be closed, this topology is Hausdorff; because qH is continuous and surjective, G / H is compact. More is so. G acts transitively on G/H. The map (g, g1 H)-+gg1 H fits the bill in the definition. In fact, any transitive action of a compact group on a compact space is of the sort just described. To be sure we need to tell when seemingly different spaces are the same under G's action. Let G act transitively on each of the compact Hausdorff spaces Ki , K2 . We say that Ki and K2 are isomorphic under G 's action if there is a homeomorphism

1.

Homogeneous Spaces

149

Fix ko E K. Look at H = {g E G : g(ko) = ko}. H is called an isotopy subgroup. It is plain that H is a closed subgroup of G. A natural candidate for the isomorphism of G/ H and K is at hand:

Nate that because of this isomorphism theorem we can consider any G/ H where H is the isotopy subgroup associated with any ko E K. Now we're ready for the main course. (Weil). Suppose the compact group G acts transitively on the compact Hausdorff space K. Then there is a unique G-invariant regular Borel probability measure on K.

Theorem 6.2

We identify K with G / H where H is an isotopy subgroup of G as in our previous theorem. Let qH : G --+G/H be the natural quotient map. Suppose µ is the normalized Haar measure on G and define µG/ H on G/H by µG/ H ( B ) = µ( qjj (B)) for any Borel set B � G/ H. Proof.

150

6. Applications

If 9 E G and B is Borel subset of G/ H, then {9X : xH E B} 9(qj{ (B)) {9x : 9xH E 9B} qj{ (9B) . Therefore µG/ H (9B) = µ(qj{ (9B)) µ(9(qj{ (B)) µ(qj{ (B)) = µG/ H( B), and µG/ H is G-invariant. Uniqueness is a touchier issue, as is always the case it seems. We take a close look at how regular countably additive Borel measures, members of rca( B0(G)) = C(K)* act on C(G). Take
¢(9) =

L
defines a member if> of C ( G). Suppose 91 H = 92 H. Then 91 1 92 E H,

¢(91 ) = = =

=

L cPg1 (h) dµH(h) L
Therefore ¢ is constant on the left cosets of H so we can lift ¢ to a continuous function if> on G/ H : (i> (9 H ) = ¢(9) . To summarize: if

E C(G) and from this we get if> E C(G/H).

1.

151

Homogeneous Spaces

Remarkably, each member of C ( G/ H ) comes about from this procedure. In fact, if f E C ( G/ H ) , then f o qH E C ( G) and for any g E G ----

( ! o qH) (g H )

=

-

( ! o qH)(g) ( ! o qH )9(h) dµH(h)

l l ( ! qH ) (gh) dµH(h) l f(ghH) dµH(h) l f(gH )dµH(h) o

=

f(gH) µH( H ) = f(gH) .

In other words, f = ( ! o qH) . Now we look at G's action. Take any G-invariant regular Borel probability measure v on G/ H. For
>-. (¢) = l{ if; (gH) dv(gH). a/ H Then ).. is a probability measure in C ( G ) * . Moreover, ).. is translation in­ variant. Indeed if x E G

>-. (¢x )

=

=

l{a/ H Jx (gH) dv(gH) l{a/ H ef>x (9) dv(gH) l{a/ H -. (¢).

So ).. is nothing else but normalized Haar measure on G.

152

6. Applications

If v1 and v2 are G-invariant regular Borel probabilities on G/ H and if x = ( x o QH ) E C ( G/ H ) , then V1 ( x ) = Jr x ( gH ) dv1 ( gH ) -----

c/H

f

� ) ( gH ) dv1 (gH )

f

� ) (gH ) dv2(gH )

}G/H ( .A ( x o QH ) jG/H (

v2( x ) ; D in other words, v1 and v2 are the same. The worth of an abstract construction lies, at least in part, it its applicability to concrete cases. Our first application is classical and was well known before Weil's general theorem. It is, nonetheless, interesting and important. Consider O (n ) the orthogonal group of order n, a compact group, and sn - l , the unit sphere in !Rn , which is our compact Hausdorff space. The action of O ( n) on sn - l is given, naturally by (u, x ) -tu (x ) . If we consider the rotations of ( n + l ) -dimensional Euclidean space, then they act as isometries on the n-sphere. Since Hausdorff measures are invariant under all such maps on any given metric space, the Hausdorff n-measure is invariant under the action of the natural rotations ( that generated the homogeneous space in the first place! ) Exercise 6.3.

Verify that O ( n ) acts transitivity on sn - l in a suitable fash­

ion. Acknowledging the descriptions of members of O ( n ) as rotations of !Rn , a direct application of Weil's theorem says, "There is a unique rotation­ invariant regular Borel probability measure on sn - 1 . ''

Let p denote the unique rotation invariant Borel probability on the sphere of !Rn and for y E !Rn denote by fy the ( continuous ) function defined on sn - l by fy (x) = ( x, y) , where ( ·, · ) is the inner product in !Rn . Show that if l lYl l = 1, then Exercise 6.4.

3n - l

l l fy l l L1 (p)

=

Jl

xn l

dp(x) ,

1.

153

Homogeneous Spaces

and so the map where Mn = I J fy J J V (µ) ' is a linear isometry of E into L 1 (p) .

Geometry is replete with examples of compact Hausdorff spaces that are homogeneous spaces on which various compact groups act transitively. Here are a few more. Again our group is 0( n) but this time our underlying compact space is L (n) = { (x, y) E sn- l x sn- l : x J_ y}, where x 1- y means x is perpendicular to y. Note that (x, y) E l:(n) precisely when for any real numbers a, b, J J ax + by J J 2 = a 2 + b2 . It is easy to see from this that l:(n) is a closed subset of sn - l x sn - 1 , hence, is compact. The action of O(n) is natural enough, too: ( u , (x, y))-+( ux, uy) .

Verify that O(n) acts transitively on l:(n) . Let 1 :S m :S n. Denote by E(m ) (n) the set Exercise 6.5.

( m)

L:: ( n)

{

= (x1 , . . . , Xm ) E 3n - l

X

}

· · · X 3n - l : { x 1 , . . . , Xm } is orthonormal ·

m times Note that (x 1 , . . . , Xm ) E E(m ) (n) precisely when, regardless of the real numbers ai , . . . , am , we have 2 aJ . aj Xj = J=l J =l With this in mind, E(m) (n) is a compact set of (sn - l) m is easy to see.

If

Exercise 6.6.

ll f

Verify that the action of O(n) on L(m) (n) given by ( u , (x 1 , . . . , Xm ))-+(ux1 , ux 2 , . . . , uxm )

is a transitive one, establishing, with a modicum of tender love and care, that O(n) acts transitively on E(m) (n) .

154

6. Applications

Finally, let 9m( n ) denote the m-dimensional Grassmanian manifold, that is, 9m( n ) is the space of all m-dimensional linear subspaces of �n . There is a natural surjection of L: (m ) ( n ) onto 9m( n ) that takes (x 1 , . . . , Xm) E L: (m) (n ) to the linear span of {x 1 , . . . , xm} E 9m( n ) . If we equip 9m( n ) with the natural quotient topology, the result is a compact Hausdorff space. Verify that O ( n ) acts transitively on 9m( n ) . The map re­ flecting the action of O(n ) on 9m is plain: if {x 1 , . . . , Xm } is an orthonormal set in �n , then (u, span{x 1 , . . . , Xm}) = span{ux 1 , . . . , uxm} · Here we interject that the geometry imparted above on 9m( n ) is such that if E = span{x 1 , . . . , x m } and E1 = span{x� , . . . , x:n} are members of 9m( n ) and if each X k is close to x� in �n , then E is close to E1 in 9m( n ) . Exercise 6.7.

In this way we find that there is a unique rotation invariant probability Borel measure on the m-dimensional Grassmanian manifold 9m( n ) . 2. Unitary Representations: The Peter-Weyl Theorem

We enter here with a discussion of the wonderful theorem of F. Peter and H. Weyl which asserts that every compact topological group G has a complete system of irreducible unitary representations. What's this all about? While topological groups have very regular features, they remain quite general. To pursue a deeper understanding of them, it is necessary to hypothesize some added structure, such as assuming the groups are locally compact or even compact. A worthy goal is to be able to aptly compare such groups with classically studied, well-understood groups, such as groups of matrices with complex entries and multiplication being that of matrix multiplication. The Peter-Weyl theorem gives secondary information about compact groups in terms of groups of matrices Our approach is modeled after that presented in Pontrjagin's classic [104] , with a bit of help from more modern mathematical technology. A unitary representation of a compact group G is a homomorphism x -+ Ux of G into the group of unitary operators on some Hilbert space H with the added feature that for each h E H, x -+ Ux (h) is a continuous complex-valued function on G. The representation x -+ Ux is irreducible if the only closed linear subspaces of H left invariant by every one of the operators Ux are {O } and H.

2. Unitary Representations: The Peter-Wey] Theorem

155

We say that G admits a complete system of irreducible unitary representa­ tions if given any g E G with g # e (the identity of G), then there is an irreducible unitary representation X -*Ux of G over the Hilbert space H such that U9 # idH. We open with the classical Schur's Lemma, a purely algebraic tool that is indispensable to our needs. We follow Schur's Lemma with a proof that for compact groups, irreducible unitary representations are necessarily finite dimensional. We follow this with a proof of the so-called "orthogonality relations" calling on Schur's Lemma to sort out the differences in the entries of nonequivalent irreducible representations of a compact group. We next pass to a description of integral operators on a compact group, con­ centrating on highly symmetric such operators. Particular attention is paid to convolution operators. The principal feature of such operators is their approximative power. It is essential, too, that the eigenspaces of these op­ erators are translation invariant, and so give rise to unitary representations, albeit not necessarily irreducible. Finally we find out that the entries of irreducible unitary representations are sufficiently rich and diverse so as to establish the completeness of the system of all irreducible unitary representations of a compact group. To get us started, we call in a classical piece of algebra. (Schur's Lemma) . Let E and F be finite-dimensional complex linear spaces, and suppose R and S are irreducible families of linear trans­ formations on E and F, respectively. Assume that T E --* F is a linear transformation such that given any B E S there is an A E R such that B T = TA . Then precisely one of the following holds: (i) T = 0 or (ii) dim E = dim F, and T is invertible. Lemma 6.8

:

Step One. Suppose dim E = m , dim F = n, and { ei , . . . , em } , {/1 , . . . , fn} are a bases for E and F, respectively. Let bj = Tej , and L = span{b 1 , . . . , bm } = T(E) . Then L is invariant under the action of S. To see this, let B E S. Pick A E R so that Proof.

BT = TA.

156

6. Applications

If Aei = �'}= 1 ai; ej then Bbi = BTei = TAei =

r( jf=l ai; ej ) = jf=l ai; bj E L;

so B leaves L invariant. Step Two. We apply the irreducibility of S and conclude that either L = {O} or L = F. In case L = {O} , we have TE = L = {O} , and T = 0. What happens if L = F? Well, TE = F, and so dim TE = dim F = n , so m = dim E 2: dim T(E) = n . Step Three. Let E' and F' denote the duals of E and F, respectively, and R' and S' denote the collections of adjoint operators of R and S, respectively. Suppose T f:. 0 (so m 2: n ) . Then T' f:. 0 as well. Since R' and S' are also irreducible families and for any B' E S' we can find an A' E R' so that T'B' = A'T', we can mimic the argument of Step Two but armed now with T' f:. 0. The conclusion is T'F' = E', and so n = dim F = dim F' 2: dim T'F' = dim E' = dim E = m . Therefore m = n , and T is surjective between two vector spaces of the same D finite number of dimensions so T is invertible. Corollary 6.9. Let E be a finite-dimensional complex linear space, and let R be an irreducible family of linear transformations on E. Suppose A is a linear trans/ormation on E such that AB = BA for each B E R. Then A = a idE for the some a E
If A f:. 0 and a is a nonzero eigenvalue of A, then A - a idE commutes with each B E R. Schur's Lemma tells us either A - a idE = 0 or A - a idE is invertible. The latter is not in the cards, otherwise what does an eigenvalue mean? So A - a idE = 0 is what is left, and that is as it should be. D Proof.

2. Unitary Representations: The Peter-Wey] Theorem

157

To put Schur's Lemma into play, we establish the following stunning feature of irreducible unitary representation of compact groups.

Let G be a compact topological group, and let x -+ Vx be an irreducible unitary representation of G over the Hilbert space H. Then H is finite dimensional. Theorem 6. 10.

Let µ be normalized Haar measure on G; keep in mind that µ is right-, left-, and inversion-invariant. For h E H, look at the sesquilinear form on H x H given by Proof.

( u, v ) =

la

(Vx h, v) (Vxh, u)

dµ (x) .

By the F. Riesz theorem, there is a bounded linear operator so that for any u , v E H (Bhu, v )

= ( u , v) =

Notice that for any y E G After all, if u, v E H, then (BhVyu, v)

l = la = la = la =

la

(Vx h, v) (Vxh, u)dµ (x) .

(Vx h, v) (Vxh, Vy u ) dµ (x) (Vxh, v) (VJ Vxh, u )

dµ (x)

(Vxh, v) (Vy- 1 Vxh, u )

dµ (x)

(Vxh, v) ( Vy - 1 Vxh, u)

dµ (x) .

Thanks to the unitary nature of Vy- I , (Bh Vyu, v)

l = la =

(Vy- I Vxh, Vy - 1 v) (Vy- I Vxh, u)dµ (x) (Vy- ix h, Vy - 1 v) (Vy- 1x h, u )

dµ (x) .

6. Applications

158

Now using µ's left invariance, we have (Bh Vyu, v) =

l (Vxh, Vy- 1 v) (Vxh, u) dµ(x)

= (Bhu, Vy -1 v) = (Bhu, vy- 1 v) = (Bhu, Vy*v) = (VyBh, v) ,

and we conclude that (Bh Vyu, v) = (VyBhu, v)

for each u, v E H. It follows then that BhVy = VyBh. But x --+ Vx is irreducible so there is a(h) E C so Bh = a(h)idH ;

that's for Schur. So a(h) (u, v) = (Bhu, v) =

for any u, v E H. It follows that

l (Vxh, v) (Vxh, u) dµ(x)

fc 1 (Vxh, u) l 2dµ(x) = a(h) (u, u) = a(h) l l u l l 2 .

Now we take apt advantage of the unimodular character of G: If f E £ 1 (µ) and f * is defined by f * (x) = f(x- 1 ) , then f0 f * dµ = fc f dµ. It follows that

fc 1 (Vxh, u) l 2 dµ(x) = fc i (u, Vxh) l 2 dµ(x) = fc 1 (V; u, h) l 2 dµ(x) = fc 1 (Vx- 1 u, h) l 2 dµ(x) = fc 1 (Vx- 1 u, h) l 2 dµ(x) = fc 1 (Vxu, h) l 2 dµ(x)

a(h) l l u l l 2 =

= a(u) l l h l l ·

Thus regardless of u E H,

a(u) l l ul l 2

a(h) l lhl l2 '

2. Unitary Representations: The Peter-Wey] Theorem

159

so a(h) = c l l h l l 2 for some constant c. In turn this says

la l (Vxh, h) l 2 dµ(x) = a(h) l l h l l 2 = c l l h l l 4 .

If we try h with l l h l l = 1 , it follows that c > 0. Now suppose (¢n) is an orthonormal sequence in H. Set h = ¢ 1 and u = ¢k for k > 1 . Then

la l (Vx¢k , ¢1 ) 1 2 dµ(x) = a(¢k ) l l¢1 l l 2 = c.

But (Vx
t, la

l ( V, ¢. , ¢i ) l 2 dµ(x) =

j l l¢1 l l 2dµ(x) =

la t, l (Vx ¢i ) l 2 dµ(x)

1.

Since c > 0, this means nc ::::; 1 severely limits the length of any orthonormal 0 sequence in H, and with it the dimension of H. Suppose x --+ Vx and x --+ Ux are two unitary representations, the first over the Hilbert space H (V) and the second over the Hilbert space H(U) . We say these representations are equivalent if there is a linear isometry T of H (V) into H(U) so that for any x E G, TVx = UxT. Our next result shows that inequivalent irreducible unitary representatives of a compact group carry considerable information by their very inequivalence. Again Schur's Lemma plays a central role. G be a compact topological group with normalized Haar measure µ. Suppose x --+ Ux = (uij (x)) and x --+ Vx = ( v k1 (x)) are nonequivalent irreducible unitary representations of G on the finite­ dimensional Hilbert spaces E and F, respectively. Here {el , . . . , em} is an orthonormal basis of E, {!1 , . . . , fn} is an orthonormal basis of F and (uij (x)) and ( vk1 (x)) are the matrix representatives of x --+ Ux and x --+ Vx on E and F with respect to these orthonormal systems. Then

Theorem 6 . 1 1 . Let

la Uij (x)vk1 (x) dµ(x) = 0.

Let T : F --+ E be a linear transformation. For any x E G, let Proof.

160

6. Applications

Each Ax is a linear transformation from F to E, and as x varies through G, Ax varies continuously through the finite-dimensional space L( F, E ) of all ( bounded ) linear maps from F to E. Let A : F --+ E be the average of the Ax's, that is, A = Ax dµ(x) . Then A is a linear transformation from F to E. Note for any a E G

l

Indeed, UaAVa-1 = Ua

l UaAxVa-1 dµ(x) = l UaUxTVx -1 Vx-1 dµ(x) = l UaxTV(ax) -1 dµ(x) ( by left invariance of µ ) = l UxTVx -1 dµ(x) = A. =

So for each a E G

l Axdµ(x)Va-1

UaA = AVa. But x --+ Ux and x --+ Vx are not equivalent, so Schur's Lemma tells us that A = 0. In other words, UxTVx - 1 dµ(x) = 0, and this is so for any linear transformation T : F --+ E. If T is the linear map that takes !1 to ej but annihilates

l

Ji , · · · , fz - 1 , f1 + 1 , · · · ' fn , then in terms of the entries ( Uii ( x) ) of Ux and ( Vkl ( x)) of Vx, this says ( since Vx -1 = (vk1 (x)) ) D Uij (x)vk1 (x) dµ(x) = 0.

l

A fixed irreducible unitary representation also packs a lot of information. If X --t Ux = ( Uij ( X) ) is such a representation over the finite-dimensional Hilbert space E, then we will follow a path similar to the one followed in the above proof to deduce the pertinent information.

161

2. Unitary Representations: The Peter-Wey] Theorem

Let S : E --+ E be any linear transformation. For any x E G, let Bx = UxSU; 1 = UxSUx -1 ; each Bx is a linear transformation in F and, as before, x --+ Bx is a contin­ uous action taking G into the space L(E, E) of all (bounded) linear trans­ formations in E. The average B, given by B=

l Bx dµ (x) ,

is a (bounded) linear transformation on E , and for each a E G BUa = UaB. Irreducibility tells us (in the voice of Schur) that there is an a E B = a idE, that is, Hence

C,

so

( l UxSUx-1 dµ(x) ) = l tr(UxSUx -1 ) dµ ( x ) = l tr(UxUx -1 8) dµ ( x ) = l trS dµ ( x )

ma = tr( a idE) = tr

= tr(S) . So regardless of the linear transformation S : E --+ E tr(S) = a ,

-m

and so

--

. UxSUx -1 dµ ( x) = tr(ms ) IdE. G Picking S judiciously will reveal that

1

1

G

--

oikoil Uii ( x ) ukz ( x )dµ ( x ) = --. m

Theorem 6.11, and the above information are often referred to as the "or­ thogonality relations" of irreducible unitary representations of a compact group. To bring them to bear on the study of compact groups, we make a brief visit to the theory of integral equations on a compact group.

162

6. Applications

Throughout this next discussion, G is a fixed compact topological group and µ is the normalized Haar measure on G. We let C(G) denote the Banach space of continuous complex-valued functions on G, equipped with the supremum norm 1 1/l l oo = sup { l / ( x ) I : x E G} , and for 1 � p < oo , LP ( G ) denotes the Banach space of ( equivalence classes of ) pth power integrable functions with respect to the Haar measure µ. Let k : G x G --t C be a continuous function, and suppose k(s, t) = k(t, s) for all s , t E G. Define the integral operator K on L 1 (G) by Kf(s) =

l k(s, t)J(t) dµ(t) .

Notice that K takes its values in C(G) , thanks to k's uniform continuity on G x G. In fact, if f E Bu ( G) and E > 0, then there is an open set V in G that contains the identity so that if s - 1 s' E V, then l k(s, t) - k (s ' , t) I � E .

It follows that

fa 1 k(s, t) - k(s' , t) l l f(t) I dµ(t) � fa 1! (t) I dµ(t) �

I K f(s) - K f(s' ) I �

E

E.

So in truth K takes the closed unit ball of £ 1 ( G ) into a uniformly bounded ( by l l k l l 00 ) , equicontinuous set in C(G) . Thus K acts as a compact linear operator from £ 1 (G) into C(G ) . Also K acts as a bounded linear operator from the Hilbert space £2 (G) to itself. Viewing K thusly, we take note that if ( , ) denotes the inner product in L 2 (G) so -

( !, g) =

then

·

l J(t)g(t) dµ(t) ,

ll k(s, t)f(t) dµ(t)g(s) dµ(s) = l l k(s, t)f(t) g(S'j dµ(t) dµ(s) = l f(t) ( l k (t, s)g (s) dµ(s) ) dµ(t) = l f(t)Kg(t)dµ(t) = ( ! , Kg) ,

(Kf, g) =

2. Unitary Representations: The Peter-Wey] Theorem

163

with the obligatory incantation of Fubini at the appropriate juncture. Thus K acts in L 2 (G) as a self-adjoint operator. Its action on L2 (G) is also compact. The spectral theorem for compact self-adjoint operators gives very precise information about K. To wit: • l l K l l L 2 G -+£2 G ' the operator norm of K, is given by () () llKll L2 (G) ---+L2 (0) = sup{ l (Kf, !) I : 1 1!1 1 2� 1 }. •

K's nonzero eigenvalues (>.n ) are all real and can be listed as a null sequence (perhaps only finitely nonzero) in descending order of absolute value:

•

If >. is a nonzero eigenvalue of K, then the eigenspace E(>.) = {f E L 2 (G) : Kf = >.!} corresponding to >. is a finite-dimensional (closed) linear subspace of L2 (G) . • Moreover, since K's values are actually in C(G), each eigenspace E(>.) has a (finite) orthonormal basis consisting of members of C(G). Of critical interest will be those integral operators generated by members ,.., E C(G) for which K(s) = K(s - 1 ) for all s E G. Here if k(s, t) = K(C 1 s) , then the operator K is just Kf(s) = k(s, t)f(t)dµ(t) = K(C 1 s)f(t)dµ(t) (also known as) = (f * K) (s) ,

la

la

a "convolution operator" . Convolution operators, such as described above, serve many purposes. With­ out further comment, they are a smoothing process; more to the present point, they help approximate. Here's how. Still in the setting of a compact group G with normalized Haar measure µ, if U is an open set in G that contains G's identity e, then let Ku E C( G) be a nonnegative function with support contained in U such that • K u ( e) = 1, • fa K U dµ = 1 , • K'{; (s) = Ku (s- 1 ) = Ku(s) for all s.

6. Applications

164

Order these open sets by reverse inclusion: U � V if U � V. Theorem 6.12. The net (,.., u * J)u converges to f in

C(G) for each f .

Let f E C(G) and E > 0 be given. By f's uniform continuity, there is an open set W containing e such that regardless of s E G if t E W, then Proof.

l f(C 1 s) - f(s) I � E .

Let U be any open set containing c such that U � W ( that is, U � W ) . Then l ,..,u * f(s) - f(s ) I = �

= �

I la la l la

,..,u (t) f(C 1 s) dµ(t) - f(s )

,..,u (t ) l f(C 1 s) - f(s) I dµ(t)

I

,.., u (t) l f(C 1 s) - f(s) I dµ(t)

K,u(t) · E dµ(t) = E,

and, like a man falling off a steep cliff, this proof has a sudden end.

D

Next, we take advantage of the symmetry possessed by the ,..,u ' s, ,.., u (z) = ku(z- 1 ) . As a result we see that ( ! * * ,.., u ) (s)

=

la la la k la

f* (t),..,u (C 1 s) dµ(t) f(t- 1 ),..,u (t - 1 s) dµ(t) f(t- 1 ),..,u (s - 1 t ) dµ(t) .

Let T = s- 1 t so C 1 = r- 1 s- 1 and dµ(t) = dµ(T) . Then f(T - 1 s - 1 ),..,u (T) dµ(T)

( ! * * ,.., u )(s)

=

,.., u (T) f(T- 1 s - 1 ) dµ(T)

= (,..,u * J) * (s) .

It follows from this, the fact that f ** = f, and the previous theorem that for any f E C ( G), the net ( ! * ,..,u ) u converges in C ( G) to f.

2. Unitary Representations: The Peter-Weyl Theorem

165

A simple, but important, computation involving Asf( t ) := f( s- 1 t ): Asf * K ( x ) = Asf(t)K(t - 1 x ) dµ( t )

la la f(s - 1 t) K (C 1 x) dµ(t) .

Again, if we let T = s- 1 t ( so t = sT) , then the left invariance of µ gives us Asf *l'r, ( x ) = f(T) K (T - 1 s - 1 x ) dµ (T)

la

As

(1 f(T) K (T- 1 x) ) dµ(T)

= As(( ! * K ) (x)) . If f is an eigenvector of the operator f --+ f * K with eigenvalue 1J, then hence Asf is also an eigenvector of f --+ f * K with eigenvalue 7J . Therefore the eigenspaces of the operator f --+ f * K are invariant under translation by As. We have seen that the eigenspaces E (A) of an eigenvalue A of a convolution operator f --+ f * K, for suitable symmetric K, is left invariant by the operator Asf(x) = f( s- 1 x ). Now As l E ( >-) : E (A) --+ E (A) is unitary for each s E G. If {91 , . . . , 9d J is an orthonormal basis of E (A) , then the functions Xki( s ) = (As9i , 9k) belong to C(G), and because As9i E E ( A) for each s E G, .

d;..

9i( s - 1 t ) = As(9i)( t ) = L ( As9i, 9k )9k ( t ) k=i holds for almost all t E G ( denoting by d.>, the dimension of E (A)). But all in sight is continuous, so d>.

holds for all s, t E G. Let Us = As l E( >-) > so

9i( s - 1 t ) = L ( As9i, 9k)9k( t ) k=l

Us = (Xij ( s )) = ( As9i, 9j ) · Since As is unitary, Xij ( s ) = Xji( s- 1 ) or u; = Us- i .

166

6. Applications

Further,

(>.t gi , >.s- 1 gj ) d>.

L (gk , As-l gi) (>.t gi , gk)

k=l d>.

L Xk; ( s )xkj (t )

k=l

or Ust = UsUt . Therefore s --+ Us is a unitary representation of G over E ( >. ) . Return now to the operators of the sort K (f) = f * K, when K is a continuous function on G satisfying K ( s ) = k(s- 1 ) for each s E G. The operator K has a sequence (>.n) of eigenvalues which we list in order of their descending module: l >. 1 1 � l >.2 1 � · · · � l >.n l � · · · '\i 0; for each n, the eigenspace E ( >.n ) is a finite-dimensional subspace of L 2 ( G ) , and we can list the orthonormal bases of the E ( >.n ) 's thusly, gl( 1) ' · · · ' gd(1) ' g l( 2 ) ' · · · ' gd( 22 ) ' · · · ' gl(n ) ' · · · ' gd(nn) ' · · ' ' 1 ...._,_, "'-.,.-' ...._,_, E (>.1 )

E (>.2)

E (>.n)

where dn = dim E ( >.n ) and gin) , . . . , g�:) is an orthonormal basis of E ( >.n ) , consisting of members of C ( G) . For any f E L 2 ( G ) , we know that f = L Cn gn + go, n

where Cn = ( f, gn) , gn in the list above, and Kgo = 0. It follows that

n 2 where the sum is taken in L ( G ) . But Ln cngn converges in L 2 ( G ) so given E > 0 there is NE E N so This ensures that

167

2. Unitary Representations: The Peter-Wey] Theorem

Of course K9m = µm9m so K

( f,_ Cm9m ) f,_ Cmµm9m N,

N,

=

is a linear combination of the 9m ' s, m = 1, . . . , Nf . Alas,

ll KJ - mt=l Cmµm9m l l ll K (mL>n, Cm9m ) ll ::; l l K ( L Cm9m ) l l 2 l l K l l E, m >n, =

00

oo

<

and K f is, in fact, the uniform limit of linear combinations of the 9m 's. Suppose x -t Ux is a unitary representation of the compact group G over the finite-dimensional Hilbert H. Then there are (closed) linear subspaces H1 , . . . , Hn of H such that H = H1 EB H2 EB · · · EB Hn , and the restriction x -t Ux l Hi is an irreducible unitary representation of G over Hi for each i = 1, 2, . . . , n. After all, if S is a (closed) linear subspace of H that is left invariant under all the Ux's (x E G) , then so too is 81- : suppose s E S, s ' E 31- , x E G; then ( Ux s1 , s) = (s ' , u;s) = (s1 , U; 1 s) = (s1 , Ux -1 s) . Since Ux -1 E S, the last quantity is 0. It follows that Ux s E 31- if s ' E 31- . Continue this idea until you run out of room. To summarize, we start with f E C(G). Then f can be approximated by f * Ku for suitable U. If K(f) = f * Ku , then K(f) is the uniform limit of linear combinations of orthonormal bases of the (finite-dimensional) eigenspaces of K. These eigenspaces, in turn, are the underlying Hilbert spaces of unitary representations of G. If we look closely at an eigenspace of K, say E, and monitor what the rep­ resentation s -t Us is doing on E, we see that this representation is a direct sum of irreducible unitary representations of G. So E has an orthonor­ mal basis consisting of blocks, each an orthonormal basis of an underlying Hilbert space of an irreducible unitary representations of G. It follows from this that the system of irreducible unitary representations of G is a complete system. Indeed, if not, then there would be a g E G, g =/:= e

168

6. Applications

so that for every irreducible unitary representation x-+Ux of G over H, U9 = idH = Ue. This in turn implies that f(g) = J (e) . But we have already noted that G is a completely regular Hausdorff space. 3.

Pietsch Measures

In 1967 in [100] , A. Pietsch introduced the absolutely p-summing operators. Soon after his introduction of this class of operators, J. Lindenstrauss and A. Pelczynski [71] , used these operators to clarify points of Grothendieck's mysterious Resume [45] . In particular, they used Pietsch's p-summing op­ erators and Grothendieck's inequality to inspire a generation of functional analysts and effect a rebirth of interest in Banach space theory. Much of the power of Pietsch's p-summing operators lies in the surprising role measure theory plays in their study. A linear operator u : X -+ Y between the Banach spaces X and Y is absolutely p-summing (here 1 :::; p < oo ) if there exists K > 0 such that for any x 1 , . . . , xn E X we have L l l u(x k) l l P :::; KP sup L l x*xk l P : x* E Bx * · ( 27) k �n k�n Here Bx * denotes the closed unit ball of dual X*of X, Bx * = {x * E X * : l l x* l l :::; l}. It is not difficult to see that u : X -+ Y is absolutely p-summing precisely when given a sequence ( xn) of members of X for which En l x*(xn) IP < oo for each x* E X*, I: I l uxn l IP < oo . Thanks to the closed graph theorem of Banach, it follows that there is a K > 0 so that L l l xn l l P P :::; K sup I: l x*xn l P P : x * E Bx· (28) for any sequence ( xn) � X. The constants K in (27) and (28) are inter­ changeable, and the p-summing norm 7rp( u) of u is defined to be the infimum of all such K's. The collection of all p-summing linear operators from X to Y is denoted by IIp(X; Y) ; it is a linear space and with 7rp as a norm, (IIp(X; Y) , 1rp) is a Banach space. Suppose K � Bx· is weak *-closed (closed in the topology of pointwise convergence on X, a topology in which Bx· is compact, by the way) and norms X, so for any x E X l l x l l = sup { jx * (x) I : x * E K}.

{

(

)

1

{(

}

)

1

}

3. Pietsch Measures

169

Then u : X -+ Y is p-summing precisely when l:n J J uxn J JP < oo whenever l:n J x*(xn) JP < oo for each x* E K. Along with this we get that

{(

)

1

(

)

1

}

1rp (u) = sup L J Juxn J JP i> : L J x* (xn) JP i> :S 1 , all x* E K · The principal example of the above phenomenon is when F is a closed linear subspace of a C(K)-space. In this case, R = {ok : k E K} � C ( K ) * is a weak *-closed, norming subset of BF* , where ok (f) = f( k ) is the usual point change. Now one can stare at the definition of an operator being p-summing and not realize that measure theory is in play. That this is so is a profound result of Pietsch, often called Pietsch's domination theorem. Here's what it says: (Pietsch) . Let u : X -+ Y be a bounded linear operator be­ tween Banach spaces X and Y. Let K � Bx• be a weak* -compact, norming set. Then u is an absolutely p-summing operator if and only if there exists a regular Borel probability measure µ on (K, weak* ) such that for each x E X

Theorem 6 . 1 3

J JuxJ J :S 1rp (u)

(L J x* (x) JP dµ(x* ) )

1

i> .

Pietsch's original proof was somewhat complicated, but in the early seventies a shorter proof was offered by B. Maurey [81] . We present an outline of Maurey's proof of the existence of the "dominating probability" . For the finite set F = {x 1 , . . . , xn } � X, we define the function ef>F : K -+ IR, ef>p (x * ) = L ( J Juxk J JP - 7r� (u) J x*(xn) JP) . k5,n The collection K = { ef>F : F is a finite collection in X} is a convex subset of C(K) . Because u is absolutely p-summing with p-summing norm 1rp (u), K is disjoint from the positive cone P = { ! E C(K) : f (x*) > 0 for all x* E K}, an open convex set. Mazur's separation version of the Hahn-Banach theo­ rem tells us there is a continuous linear functional µ E C(K)*, the space of regular Borel measures on K, such that for some real constant c (29) ef>F dµ ::; c < f dµ

L

L

for all finite sets F � X and all f E P. Since 0 E K, c � 0. Since c < JK f dµ for every f E P, c :S 0. So c = 0, and µ is a nonzero, nonnegative measure. The defining inequality (29) is impervious to dividing by positive constants,

170

6. Applications

so we can assume that µ is a probability. Testing (29) against ¢{x} E K gives the result. Incidentally, if there is a probability µ on K so that

l l ux l l ::; G then < < <

( [ l x*(x) IP dµ(x* ) )

1

p'

1

GP L l x* ( xj ) I P dµ( x* ) j'S_ n K GP L l x* (xj ) IP dµ( x* ) K j'S_ n GP jr sup L l x* ( xj ) IP dµ K x *EK j'S_ n

1

(because µ is a probability), and so u is absolutely p-summing with 11'p (u) ::;

c.

Since µ comes about as a result of the tandem of the Hahn-Banach theorem and the Riesz representation of G ( K )* , it appears that little can be said of a personal nature about µ. This is not so.

Let G be a compact topological group, and let F be a closed translation invariant linear subspace of G ( G ) that separates the points of G. Suppose that u : F -+ X is an absolutely p-summing linear operator, and suppose that for each f E F and g E G l l u( J ) l l = l l u( Jg ) l l , where f9 ( x ) = f (gx) . Then the normalized Haar measure on G is a Pietsch measure for u .

Theorem 6. 14.

Let µ be a Pietsch measure for u; this says µ is a regular Borel probability measure on G such that for any f E F Proof.

l l uf l l P :S 7r� (u)

fa i f(x) IP dµ(x) .

For g E G let µ9 E C ( G)* be given by µg ( J ) = µ(Jg ) ·

171

3. Pietsch Measures

Here f9 ( x ) = f(gx) for all x E G. Then µ9 is a regular Borel probability measure on G and is, in fact, a Pietsch measure for u. In fact, for any f E F, f9 E F

l l u(f) l l P

l l u(f9 ) l l P 7r� (u)µ( l fgl P) 11'� ( u ) µg ( l f IP) 11'� ( u) I f (x) IP dµ9( x ) .

<

la

The map taking g E G to µ9 E C(G)* is a continuous function from G onto (Bc ( G) * • weak*) . Hence the Gelfand integral I/

=

la

µg dµ

la

µg dµ

exists and lies in Bc ( G) * · Here dg indicates we are integrating with respect to the normalized Haar measure on G. Of course, v is the Gelfand integral I/

=

and means that for any f E C ( G)

It's easy to see that v is nonnegative, since f 2: 0 implies µ9 ( ! ) 2: 0 for all g E G. It's also plain that v is a probability: v(l)

=

la

µ9 ( 1 ) dg =

la 1 dg = 1.

Therefore v is a regular Borel probability on G. It is also true that v is a Pietsch measure for u since for each f E F, g E G,

we have that as well.

6. Applications

1 72

Finally v is translation invariant: if h o E G and f E C(G), then

v(fh0) = =

la µ (fh0) dg = lala fh0 (x) dµ9 (x) dg

la la la la la la la la la la la la la

fh0 ( gx ) dµ( x ) dg

f(hogx) dµ( x) dg f( hogx ) dg dµ (x) f(gx) dg dµ( x) f(gx) dµ( x) dg f(x) dµ9( x ) dg

v(f)dg = v(f).

Therefore v is the normalized Haar measure on G.

0

4. Notes and Remarks

The material from Section 1 of this chapter traces its origin in the form presented herein to A. Weil's [136] treatise. Another excellent source is Nachbin's delightful book, The Haar Integral [89]. We have cited but a few of the homogeneous spaces that admit invariant probabilities. It would be hard to study modern geometry or analysis with­ out encountering examples of invariant measures on homogeneous spaces. The power inherent in the use of these techniques can be appreciated, for ex­ ample, in the study of A. Dvoretzky's celebrated spherical sections theorem [35] . Roughly speaking, this theorem says that in any infinite-dimensional normed linear space, one can (theoretically) find finite-dimensional sub­ spaces of arbitrarily high dimension that are but small perturbation of Eu­ clidean spaces. In his original exposition of this result, Dvoretzky calls on the invariant probabilities living in the Grassman manifold Q(n) . Other proofs ( simpler in execution ) have since appeared, but each makes its fair share of use of invariant measures on homogeneous spaces. We refer the bravest readers to the paper of T. Figiel, J. Lindenstrauss, and V. Milman [39] for a proof calling on the isoperimetric inequality and invariant mea­ sures on homogeneous spaces and to the survey papers of Lindenstrauss [70]

4. Notes and Remarks

173

and V. Milman [82] . Once hooked on the beauty of the subject, the V. Mil­ man and G. Schectman volume, Asymptotic Theory of Finite Dimensional Normed Spaces [83] , will be a natural. Our presentation of the Peter-Weyl theorem follows the path laid by L. Pontrjagin set forth in Chapter IV of Topological Gro ups [104] . Relying as it does on the existence of a normalized Haar measure on a compact group, the Pontrjagin presentation carries over to any compact group. Schur's Lemma (Lemma 6.8) it to due to I. Schur [114] and is present in virtually all presentations of representations of groups. Theorem 6.10 was uncovered by A. Gurevic [46] in case the compact group satisfies the second axiom of countability. See also P. Koosis [68] and L. Nachbin [88] . K. Shiga [119] modified the Gurevic construction to apply it to irreducible representations of compact groups over general (complex) Banach spaces. The notion of an absolutely p-summing operator is due to A. Pietsch [100] who is responsible for the magical theorem wherein "Pietsch measures" make their appearance. Though A. Grothendieck isolated the absolutely 1-sum­ ming operators and the absolutely 2-summing operators in different guises, it was the identification by Pietsch of the norm 1fp that was to play a critical role in the development and application of these operator ideals. Theorem 6.14 appears in J. Diestel, H. Jarchow, and A. Tonge [25] , where it is admittedly inspired by earlier works of Y. Gordon, S Kwapien and A. Pelczynski. Pietsch's isolation of the p-summing operators (and his definition of their norm) was recognized by J. Lindenstrauss and A. Pelczynski [71] as an appropriate vehicle to reformulate many of the most mysterious of Grothen­ dieck's results from his Resume [45] . In addition, Lindenstrauss and Pelczynski recast Grothendieck's "fundamental theorem of the metric theory of tensor products" in a more user-friendly version; they went on to solve a number of long-standing problems in Banach spaces. The literature related to Grothendieck's inequality (an alias for his fun­ damental theorem) is enormous. It is probably wisest to recommend to interested readers that they go to the master and study the amazing survey paper of G. Pisier [102] . For an exposition of Grothendieck's original argu­ ment for his fundamental inequality in its tensorial home, a look at Diestel, Fourie, and Swart [26] might help. Alternative views of the inequality, as well as varied applications, can be found in [24] , [25] , and [101] .

1 74

6. Applications

The use of compact groups and their Haar measure has been a frequent tool in the local theory of Banach spaces. We recommend the papers of P. Mankiewicz [75] , P. Mankiewicz and N. Tomczak-Jaegermann [76] , as well as the book of N. Tomczak-Jaegermann [127] .

Chapter

7

Haar Measure on Locally Compact Groups

1 . Positive Linear Functionals

1.1. Introduction. Let S be a nonempty locally compact Hausdorff space. Denote by IC(S) . the linear space of continuous real valued func­ tions f : S ---+ � such that {s E S : f( s ) =I= O} is compact. JC + (S) will denote the collection of f ' s in /C(S) such that f(s) 2: 0 for all s and f( s ) is not identically zero. Let : JC(S) ---+ � be a linear functional for which

( l f l ) Of course,

( g ) if f, g E /C(S) and f 2: g. It follows that ( - 1 ! 1 ) = - ( l f l ) .

Fact

7.2. If K is a compact subset of S , then there is a f3K > 0 so that if f E /C(S) and {s E S : f( s ) # 0} � K, then l U ) I � /3K l l f l loo·

Fact

-

1 75

7. Haar Measure on Locally Compact Groups

1 76

7.3. Let 1J be a nonempty subset of K + (S) , and suppose 1J is directed downward (so given f, g E 1J there is an h E 1J so h :::; f, h :::; g) and i nf{ f (s) : f E 1J} = 0 for all s E S. Then given an E > 0, there is an fE E 1J so that fc( s ) < E for all s E S. Consequently, inf { (!) : f E 1J} = 0. Fact

The first assertion of Fact 7.3 can be viewed as a directed version of Dini's theorem. Let E > 0 be given. For f E 1J define B1 = {s E S : f( s ) � t:}. Each Bf is compact and, by hypothesis, n B1 = 0 f Hence there is Ji , f2 , . . , fn E 1J so that B1i n · · · n B1n = 0 Let ft: E 1J be chosen so fc :::; Ji, for all 1 :::; i :::; n . If g E 1J, then there is an h E 1J so h :::; g and h :::; ft:· If K = {s E S : g(s) > O}, then Fact 7 . 2 says there is f3K > 0 so that l (h) I :'.S l
.

.

.

0

A function f : S --+ [O , oo] is lower semicontinuous if for each so E S the following holds: • if f( s o) < oo , then given E > 0 there is an open set U containing so so that for any s E U, f(s) > f( so ) - t: ; •

if f(so) = oo, then for any P > 0 there is an open set U containing so so if s E U , then f(s) > P.

1.

Positive Linear Functionals

177

We denote by £ + the collection of all lower semicontinuous functions f : S --+ [O, oo] . A basic feature of members of £ + -a motivating factor for their introduction-is the following. If f E £+ , then f ( s) = sup{g(s) : g E J(, + ( s), g :::; !}. The truth of Fact 7.4 is plain if f (so) = 0. Suppose then that f( so) > 0. Then for any P > 0 with 0 < P < f (so), there is an open set U containing so so if s E U, then f( s ) > P. We can (and do) assume U is compact. Now Urysohn's lemma is available and on call; it provides us with g E J(, + (S) so that {s : g ( s ) =!= O} � U with g (so) = P and g(s) :::; P for all s E S. Of course, 0 :::; g :::; f and P is any number with 0 :S P :S f ( so ) , so Fact 7.4 is established. Fact 7.4.

Basic facts about the set £ + : (i) If f E £ + , a > 0, then a f E £+ . (ii) If Ji , . . . , fn E £+ , then min{f1 , . . . , fn } E £ + . (iii) If 'D is a nonempty subset of £ + , then sup{! : f E 'D} E £ + . (iv) If Ji , . . . , Jn E £+ , then Ji + . . · + fn E £ + . Let f E £+ . Then we know from Fact 7.4 that f (s) = sup {g(s) : g E K + ( s ) , g :::; !}. In light of this, we define 1 (!) by 1(!) = sup { (g) : g E K + ( s) , g :::; !}. Keep in mind that 1(!) = + oo is a distinct possibility. Fact 7.5.

Basics about 1: (i) If f E J(, + (S) , then f E £+ and 1(!) = (!). (ii) If Ji , h E £+ and Ji :::; h , then 1(!1 ) :S I( f2 ). (iii) If f E £+ and a 2: 0, then I( a f) = al (!). Now we modify Fact 7.3 to the context of 1. Fact 7.6.

Let 'D be a nonempty subset of £+ that is directed upward (so given f, g E 'D, there is an h E 'D so h 2: f and h 2: g). Then I(sup{f : f E 'D}) = sup{I(f) : f E 'D}.

Theorem 7.7.

178

7. Haar Measure on Locally Compact Groups

Consequently, if Un ) � c+ is an ascending sequence, then I (lim n I Un ) · n fn ) = lim Keep in mind that c + is permanent under suprema. Maybe 1J and Jo = sup{f : f E 1J} belong inside JC + (S) ; in this case, {Jo - f : f E 1J} falls prey to Fact 7.3, and so inf {

(go) = sup{ (go) < sup{
D

1 79

1 . Positive Linear Functionals

Since go was any member of JC, + (S) with go :::; Jo , it follows that l (fo) = sup { ( go) : go E K + ( s ), go :::; Jo} :::; sup l (f) . D

This proof is now complete.

0

If Ji , h E .c + , then 1(!1 + h ) = 1(!1 ) + l ( f2 ). This is an easy consequence of 's additivity on JC + (S), and the fact that Ji + h = sup{g1 + g2 : gl , g2 E JC + , g 1 :S Ji , g2 :S f2 }. Piggybacking, we get Corollary 7.8.

Corollary 7.9.

For any nonempty V � .c + ,

Next we will extend 1 to the set of all nonnegative extended real-valued functions defined on S. Let f : S-t [O, oo] be any function with nonnegative, extended real values defined on S. Denote by VJ the set VJ = {g : S-t[O, oo] : g E .c+ , J (s) :::; g(s) for all s E S}. Trivially g ( s ) = + oo belongs to VJ, regardless of f. Definition 7.10.

We define the upper integral of f, Jf, by

J

f

= inf { 1 (g) : g E VJ}.

Take note that if J E .c+ , then JJ = 1 (f) . Basic facts about J : (i) If Ji , h : S-t[O, oo] and Ji :::; h , then JJi :::; Jf2 . (ii) If c 2:: 0 and f : S-t [O, oo] , then JcJ = cff. (iii) If fi , h : S-t[O, oo] , then f (!1 + h ) :S JJi + Jf2 .

Fact 7 . 1 1 .

Theorem 7.12.

S-t[O, oo] , then

If Un ) is an ascending sequence of functions with fn

7. Haar Measure on Locally Compact Group�

180

Proof.

It's plain from Fact 7.ll ( iii ) that

J s�p fn 2: s�p Jfn ,

so we will establish the reverse inequality, namely,

J J s�p fn ·

s p fn 2: �

If supn Jfn = + oo , then there is nothing to be proved. So we suppose for the sake of argument that supn Jfn < oo . Of course this supposition gives us access to epsilonics. Let E > 0. For each n, choose 9n E c+ so that fn ::; 9n and Set

h i = gi , h2 = max g 1 , g2 , . . , hn = max gi , . . . , 9n } · Then the sequence ( hn ) is ascending, hn 2: 9n and

{

}

{

.

max{ hn , 9n+ I } + min{ hn , 9n+ I } hn + 9n + I · So

I ( hn+ I ) + I ( min { hn , 9n + i}) = I ( hn ) + l ( 9n+ I ) ·

It follows that

I ( hn+ I )

I ( hn ) + I (9n + I ) - J (min{ hn , 9n + I })

< J ( hn ) + I (gn+ I )

J

- Jfn

< I ( hn ) + fn+ I + 2n� l

alternatively, for n = 1 , . . . , k - 1 ( say ) ,

- Jfni

We add these inequalities up from n = 1 until n = k - 1 , and we find that

181

1 . Positive Linear Functionals

Hence

Jfk + I(h1 ) JJi + � Jfk + (I (91 ) JJi ) + � Jfk + � + � = Jfk + E .

I(hk) <

-

-

< It follows that

sup k I (hk) 1(sup k hk)

J S�Pfk,

>

with the critical equality of supk I (hk) and I (sup k hk), a result of Theorem D 7.7. Now E > 0 was arbitrary, so Theorem 7 . 1 2 is proved. Corollary 7. 13.

S-t[O, oo] , we have

Given any co untable collection { !j } of functions fj :

f I; fj � I; ffj · J

J

Corollary 7.13 is crucial in our generation of outer measures. 1 . 2 . The outer measure associated with a positive linear func­ tional. For any A � S the o uter meas ure µ ; (A) (shortened to µ(A)) is

given by

J

µ(A) = XA · This µ is an outer measure; i.e., µ(A) � 0 for any A � S. If A � B � S, then µ(A) � µ(B). If (An ) is a sequence of subsets of S, then

Fact 7. 14. • • •

(

)

µ LJ An � L µ(An ) · n n Fact 7. 14 follows from the basic facts (announced earlier) about J ; µ respects the topology of S and behaves accordingly.

7. Haar Measure on Locally Compact Groups

182

Theorem 7.15.

then

(i) If U is a family of pairwise disjoint open sets, µ

( LJu) = L µ(U). u

u

(ii) !J U is a family of open sets that is directed upward (given Ui , U2 E U there exists U E U so that U1 U U2 � U }, then µ

( � u) = s�p µ(U) .

(iii) If K is a compact subset of S, then µ(K) < oo . (iv) For any A � S, µ(A) = inf{µ(U) : U is open, A � U}. ( v) If U is open, then µ(U) sup{µ(K) : K is compact, K � U} = sup{µ(V) : V is open, V is compact, V � U}. open, then xu E £ + . Indeed, if so E U, for any s E U; i.e., Xu (s) > h. If s o ¢ U, then xu( s o ) = 0 so xu( s o) > h means h is negative, and so xs (S) 2 0 > h always holds. Hence µ(U) = xu = I (xu ) , and so (i) follows from Corollary 7.13. Item (ii) follows from the observation that xu belongs to £+ and Theorem Proof. To see item (i) : if U is then 1 = xu (so) > h also holds

J

7.7.

For item (iii) : if K is compact, then we can find an open set V in S so K � V and V is compact. Hence there is f E K, + (S) so that XK :S f :S XV· It follows that µ(K) = XK :S f = l (f) = ¢(!) < oo . We need only prove item (iv) in the case µ(A) < oo, so epsilonics are avail­ able. Accordingly let E > 0. Suppose 0 < E < 1 . By definition there is f E £+ so that XA :::; f and f :::; µ(A) + E .

J

J

J

1.

183

Positive Linear Functionals

-

Let U = [f > 1 E] . Since f E .c + , it follows that U is open. Since XA S f, it follows that A � U. On the other hand, f 2: (1 E ) X u , so E ) X u s J (f) S µ(A) + E . ( 1 - E )µ(U) s Hence ( 1 - E )µ(U) s µ(A) + € . For the last part of the theorem, suppose t < µ(U) . There is an f E J(, + (S) so that f S X u and ¢(!) > t. Again, xu E .c + so

J(l -

-

J

µ(U) = xu = I (xu) = sup{¢(!) : f E K + ( s ), f s xu} . Let 0 < E < 1 , and let Fa = [f 2: a] , Va = [f > a] , V = [f > OJ . Plainly V � U so µ(V) s µ(U) and ¢(!) S µ(V) , because all of f's values are between 0 and 1 (f S xu) , and whenever f(s) > 0, s E V so xu (s) = 1 2: f(s) . But now t <
!- lim J- lim

a '-,.0

<

a'-,.0

J J

X Fa

limO µ(Fa) = a\,,

X Ua

lim µ(Ua) · = a'-,.0

So we can choose a small enough that both µ(Fa ) and µ(Va) are larger than

t.

D

7.16. A subset E of S is µ-measurable if and only if µ(U) 2: µ(U n E) + µ(U n Ee) holds for each open subset U of S with µ(U) < oo . If suffices, of course, to show the above condition is sufficient for E's µ­ measurability. With this in mind, let A be an arbitrary subset of S. If µ(A) = oo , then µ(A) 2: µ(A n E) + µ(A n Ee) is plain and trivial. If µ(A) < oo and U is an open set containing A with µ(U) < oo, then µ(U) 2: µ(U n E) + µ(U n Ee) . Fact

184

7. Haar Measure on Locally Compact Groups

But we know from Theorem 7.15(iv) that µ(A) = inf {µ(U) : U is open, A � U} > µ(U n E) + µ(U n Ee) > µ(A n E) + µ(A n Ee) . Theorem

7.17. All Borel subsets of S are µ -measurable.

Suppose V is an open subset of S. Let U be an open subset in S with µ(U) < oo, and let E > 0 be given. Use Theorem 7.15 to find an open subset 0 of S so that u n vc � 0 and (31) µ(0) < µ(U n vc) + 2E . Proof.

Now appeal to Theorem 7.15 to find an open set W such that W is compact, w � u n v, and (32) µ(W) + 2E > µ(U n V) . Let Wo = u n o n (wr Then W and Wo are disjoint open sets. Now U u vc � Wo � 0, so (31) tells us that E l µ(Wo) - µ(U n vc) I < 2 · In tandem with (32), we see that (33) J µ(W) + µ(Wo) - (µ(U n V) - µ(U n vc)) I < E. But the disjointness of W and W0 ensures (by Theorem 7.15) that µ(W U Wo) = µ(W) + µ(Wo) . So since both W � U and Wo � U, µ(U) � µ(W U Wo) = µ(W) + µ(Wo) � µ(U n V) + µ(U n vc) - E by (33). All in sight is finite and E is arbitrary, so µ(U) � µ(U n V) + µ(U n vc) holds whenever U is open in S and µ(U) < oo . Therefore V is µ-measurable.

1.

Positive Linear Functionals

185

Of course, we know that the collection of µ-measurable sets is a O"-algebra, and since it contains every open subset of S, it must contain every Borel set D as well. Here's a basic regularity feature of this construction. Fact 7. 18.

Let A be a µ-measurable set, and suppose µ(A) < oo . Then µ(A) = sup{µ(K) : K is compact, K � A}.

Let E > 0. Choose an open set U so that A c U and µ(U n Ac ) < £ . Now appeal to Theorem 7.15 to find a compact set K � U so µ(UnKc) < � · Now, let W be an open set so that U 2 W 2 U n A and µ(W) < � · Let F = K n we . Then F is compact, and F � A. What's more, µ(A n Fe) µ((A n Kc) u (A n W)) € € ::::; µ(U n Kc) + µ(W) < "2 + "2 , and so (since F � A ensures µ(A n Fe ) = µ(A) - µ(F)), it follows that D µ(F) > µ(A) - E . Proof.

Our next task is to show that the measure µ generated by positive linear functionals

(µ) so useful in uncovering the structure of the space of integrable functions. To start, we call on some epsilonics to simplify the establishment of a set's measurability. Lemma 7. 19. A subset A of S is µ -measurable precisely when either of the fallowing conditions is satisfied: (i) A n U is µ -measurable for each open set U of finite measure; (ii) A n K is µ -measurable for each compact set K � S. Naturally, A's µ-measurability entails the measurability of A n U and A n K for every open U and compact K since all Borel sets are µ-measurable. Now suppose A n U is µ-measurable for each open set U of finite µ-measure. Assume µ(B) < oo and E > 0 is given. Pick an open set U 2 B with µ(U) ::::; µ(B) + E . Since A n U is µ-measurable, µ(U) = µ(U n A) + µ(U n Ac) , and so µ(B) + E 2:: µ(U) 2:: µ(U n A) + µ(U n Ac) 2:: µ(B n A) + µ(B n Ac),

186

7. Haar Measure on Locally Compact Groups

after all, B � U. Thanks to E 's arbitrariness, µ(B) � µ(B n A) + µ(B n Ac) , and so A is measurable. To include (ii) we bootstrap on the result of (i). Suppose A n K is µ­ measurable for each compact set K. Assume U is an open subset of S with µ(U) < oo, and appealing to Theorem 7.15(v) , find an ascending sequence (Kn) of compact subsets of U so that µ(U \ Kn) '\i 0. So A n Kn is µ­ measurable for each n and A n (U \ Un Kn) satisfies

So A n (U \ Un Kn) is µ-measurable. But AnU

�

( (

An U\

� Kn)) U (y (A n Kn) ) ,

and so A n U is the countable union of µ-measurable sets and as such is entitled to call itself µ-measurable. Now we come to the principal technical step we need to make: to overcome the possible non-o--finiteness of the measure µ. This calls on an exhaustion argument, more or less covering virtually all of S with the well-disposed disjoint compact sets. Theorem 7.20 (Exhaustion Principle) . There exists a family C of compact subsets of S such that

(i) (ii) (iii) (iv)

distinct members of C are disjoint; if K E C, then µ(K) > O; if U is open and K E C satisfies K n U i- 0, then µ( K n U) > O; if A is measurable with µ(A) < oo , then A n K i- 0 for at most a countable collection of K 's in C and

µ(A) = L µ(A n K) ;

c (v) A is measurable if and only if A n K is µ-measurable for each K E C; (vi) f : S--t� is µ-measurable if and only if f X K is for each K E C. Proof. Let F be the collection of all families K of compact subsets of S having the properties:

1.

187

Positive Linear Functionals

(a) distinct members of }(, are disjoint; (b) any member of }(, has positive µ-measure; ( c) if U is open and U n K =/= 0 for K E JC, then U n K has positive µ-measure. Partially order F by inclusion. Note that F contains the empty set by default, and so F is itself nonempty(!). Linearly ordered subfamilies of F have an upper bound (their union) and so Zorn's lemma is available for use. On Zornicating, we find a maximal element C in F. We will check C for the properties (i)-(vi) . Properties (i), (ii), and (iii) are all part and parcel of C's membership in F. Suppose A is µ-measurable with µ(A) < oo. By Theorem 7.15(iv) there is an open set U, A � U, with µ(U) < oo. Look at Cu = {K E c : µ(U n K) > 0}. For each n E N, let

{

Cu,n = K E Cu : µ(U n K)

>

� }·

Because µ(U) < oo, the disjointness of distinct members of C ensure that each Cu, n is finite and so

Cu = LJ Cu,n n

is countable. Now each member of C meeting A a fortiori meets U so belongs to Cu by (ii). It follows that there are only countably many members of C that meet A. List these in a sequence (Kn) · Since A is µ-measurable and the Kn ' s are pairwise disjoint, µ(A)

�

( (y Kn)) ( (y Kn) ') ' � Kn ( (y Kn ) )

µ An

µ(A n

+µ An

)+µ An

.

So we want to show that

Since A is µ-measurable (as are all Borel sets) , the set A n ( Un Kn ) c ) is plainly µ-measurable and, sitting as it does inside A, has finite µ-measure. Suppose C is a compact subset of A n ( Un Kn ) c and imagine that µ( C ) > 0. Now µ's restriction to the µ-measurable subsets of C is a regular measure on C. The support of this restriction (call it Cµ ) is compact, µ( Cµ) = µ( C) > 0,

188

7. Haar Measure on Locally Compact Groups

and if U is an open subset of S so that UnCµ =f. 0, then µ(UnCµ) > O; after all, Cµ is the support of µ inside C. But C (and hence Cµ) meets none of the members of C and so C U { C} satisfies (a)-( c), and is therefore a member of F, larger than the maximal member C of F. Our supposition on C must be suppressed with µ(C) = 0, the only option left. This establishes (iv) . Let's establish (v) . Suppose A n K is µ-measurable for each K E C, and let C be any compact subset of S. By what we just proved in establishing (iv) , there is a sequence (Kn ) in C so that µ(C) = L µ(C n Kn ), n and so But A n Kn is µ-measurable for each n, and so A n C n Kn is, too. Thus the set A n c n (Un Kn ) c satisfies and A n C n (Un Kn ) c is therefore µ-measurable. Hence A n C is the union of the A n Kn 's and the µ-null set A n C n (Un Kn ) c is µ-measurable, and this is so for any compact set C in S. An appeal to Lemma 7.19(ii) now tells us that the proof of (v) is complete. Alas (vi) follows from (v) since for any Borel subset E of � and any K E C, D r- ( E ) n K = (f XK)+-- ( E) n K. Before presenting the proof that L 1 (µ) * is L00 (µ), we make mention of the fact that VX)(µ) always sits comfortably inside L 1 (µ) * via the integration of any f E L 1 (µ) against an element of L00(µ) . Moreover this inclusion is a linear isometry. The issue is whether or not there is anything else in L 1 (µ)*. Now we're ready to establish the duality of L 1 (µ) and L00(µ). It proceeds from the basic fact (a consequence of the Radon-Nikodym theorem for finite measures) that if v is a finite (positive) measure, then L 1 (v)* is identifiable with L00(v) via the formula y * ( f ) = f g dv, f E L 1 (v), relating y * 's in L 1 (v) * and g 's in L00(v) . ·

J

What we have in hand is that for any compact set K � S (with µ(K) > 0) , if we restrict µ to the µ-measurable subsets of K, we get L 1 (K, µ) where

1.

Positive Linear Functionals

189

(K, µ) serves as a v to which we can apply the comforting duality of L 1 and £00 • To be sure we can describe a functional on all of L 1 (µ) : we call on our exhaustion principle. To represent x* E L 1 (µ) * we use the exhaustion principle to "localize" the action of x*. Take a typical f E L 1 (µ). As is always the case for L 1 -functions, f has a-µ-finite support, that is, there is a sequence ( An) of µ-measurable sets of finite µ-measure such that f(k) = 0 if k
l f(k)gK(k) dµ(k) ,

and

l l 9K l l L00 ( K,µ) = l l x* I L 1 ( K,µ) l l :S l l x* l l · We now define g on S by g(s) =

{ 9KO (s)

if s
By the disjoint nature of distinct members of C, g is well defined. By (iv) of the exhaustion principle, g is µ-measurable, and, thanks to the constraints on l l 9K l lL= ( K,µ) • we have that g E L00 (µ) with l l 9 l loo :S l l x * l l ·

7. Haar Measure on Locally Compact Groups

190

Finally if f E £ 1 (µ) , then we locate the sequence (Km) in C so that f vanishes outside Um Km. We see that

( ) ( ) � lm f(k)9Km (k) dµ(k) � lm f(k)9(k) dµ(k) 1Um Km f(k)9(k) dµ(k) J f(k)9(k) dµ(k) ,

x* (f) = x* fxum Km = l: x* f XKm m

=

=

and the duality of £ 1 (µ) and VXJ (µ) is established. 2. Weil's Proof of Existence

In this short section we present Andre Weil's proof of the existence of a left Haar measure for any topological group. The initial stages of the proof are combinatorial averaging properties comparing continuous functions having compact support. Though these properties are uncomplicated and, once formulated, easily established, they are nonetheless basic to establishing the existence of a Haar measure in a general locally compact group. Throughout G is a locally compact topological group. Let K ( G) denote the linear space of continuous functions f : G-+IR with compact support, and let J(,+ (G) denote those nonzero f E K(G) with nonnegative values. Fix, once and for the rest of this discussion, w E JC+ ( G) . Note that if f,

91 , . . . , 9n E G so that (34)

n

f � L Ci g;

i=l 1 where 9; ¢(9) = ¢(9; 9) denotes a left translation of ¢. To see why (34) holds, let K be a compact subset of G such that f(9) = 0 if 9 ¢ K, and let M > max{f(9) : 9 E K}. Since

O; by (�o ) for all u E Uo. For each h E G, hUo is an open set and K � LJ hUo . hEG

191

2. Weil 's Proof of Existence

K's compactness leads us to hi , . . . , hn E G so K = hiUo U · · · U hnUo. Letting h ¢(g) = rp (h- i g) , we see that if g E K, then g E hm Uo for some 1 S m S n and h;;?g E Uo so

;

( o) hm rp (g) = rp (h-;;/ g ) > rp ;

hence

2M

h rp (g) > M > f(g) , rp (go) m

for all g E G and (34) follows. With this in mind we define (f : rjJ) = inf

{ t,

Ci : f S

t,

}

Ci gi ¢, for some ci , . . . , Cn 2: 0, gi , . . . , 9n E G ,

and (f : rjJ) is the functional covering number of f via translates of ¢ . Lemma 7.21. For any f, g, ¢ , 'lj; E K,+ (G) , we have

(i) 0 < (f : ¢) < oo ; (ii) if h E G, then (f : ¢) = (hf : ¢) ; (iii) ( f : rjJ) is subadditive and positively homogeneous in f ; i. e., ( f + g : ¢) S (f : ¢ ) + (g : ¢) and ( >-. f : ¢) = >-. ( ! : ¢ ) for any ).. 2: O; (iv) if f S g, then (f : ¢) S (g : ¢ ) ; (v) (f : ¢) S (f : 'lf; ) ('lf; : ¢) .

Proof. All of the statements follow easily using the definition of (f : rjJ) . We comment on (v): to show (f : ¢) S (f : 'lf; ) ('l/J : ¢) , fix E > 0, and let ci , . . . , Cn 2: 0 and gi , . . . , 9n E G be chosen so that

n

f s L Ci gi 'l/J, i=i

with (f : 'l/J) S L� i Ci S (f : 'lf; ) + E and di , . . . , dm 2: 0 and hi , . . . , hm E G be chosen so that m

j= i

with ('lj; : ¢) S I:j=i dj S ('l/J : ¢ ) + E .

7. Haar Measure on Locally Compact Group1

192

It follows that i= l

n

m

L Ci L dj gi h/ P , i= l j= l

which, since says

n

m

(f : ¢) � L Ci L dj � ( (f : 'lj;) + E ) ( ( 'lj; : ¢) + E ).

Let E \.i 0. Then

i= l j= l

(f : ¢) � (f : 'lf; )( 'lf; : ¢).

D

For any f, ¢ E K,+ ( G) we define µ ( ! ) =

(f : ¢) (w : ¢) '

We'll see that µ ( ! ) is an approximation to our ultimate J f dµ. As we squeeze the support of ¢ to ever smaller sets, we'll get better and better approximations. At least, that is the game plan. 7.22. For any f, g, ¢ E K,+ (G) ( i ) µ ( ! ) > O ; ( ii ) µ ( g f) = µ ( ! ) for all g E G; ( iii ) µ is subadditive and positively homogeneous; i. e. , µ ( ! + g) � µ ( ! ) + µ<1> (9) and µ (>.. ! ) = >.. µ ( ! ) for all >.. � O ; ( iv ) µ is monotone nondecreasing;

Lemma

(v)

µ ( ) (f ) (w : ! ) � ! � : w . 1

We comment on (v ) : w , f and ¢ are all nonzero, so quantities in­ volved in the inequalities from ( v ) of Lemma 7.21 are positive. Therefore (f : ¢) � (f : w ) (w : ¢) Proof.

193

2. Weil 's Proof of Existence

and

(w : ¢) � (w : J) (f : ¢) ,

and (v) of Lemma 7.22 is immediate from these.

D

Recall that any f E K,+ (G) is both left- and right-uniformly continuous. and fix E > 0. Then there exists a neighborhood V of the identity so that if cp E J(,+ ( G) with the support of cp contained in V, then

Lemma 7 .23. Let

Ji , h E

J(,+ ( G) ,

In effect (by Lemma 7.22), as we squeeze the support of cp toward { e } , the function µ
Proof.

XK � f � 1 .

Fix {J > 0 and consider Define h i , h 2 by

{ bf{Ji ((g)g)

f{J = fi + h + f> f .

{

f2(g) if g E K, if g ¢ K, h 2 (g) = ffJ (g) if g ¢ K. if g E K, o Then h i , h 2 E JC+ (G) and each has its support inside that of f{J · So Ji = h if{J , h = h 2 f{J , h i + h 2 � 1 . Because any member of J(,+ ( G) is (left) uniformly continuous on G given 1 i E > 0, there is an open set V containing the identity so that if x - y E V, then 1 l h i ( x ) - h i ( Y) I � E and l h 2( x) - h 2(Y) I � / . Suppose cp E J(,+ ( G) with the support of cp contained in V, and suppose

h i (g) =

i= i for some c i , . . . , Cn 2". 0 and gi , . . . , 9n E V. Then for any g with 9i i 9 E V, n n i fi (g) = f{J (g)hi (g) � L Ci c/J (g:; g)hi (g ) � L Ci c/J (g:; i g ) ( h i (9i) + / ) i= i i= i

194

7. Haar Measure on Locally Compact Groups

and

n n 1 L L h (g) = f8 (g) h 2(g) ::; Ci(!1 ) + µ ( h ) < ( 1 + 2 c1 )µ<1>(!8 ) = ( 1 + 2E1 )µ<1> ( !1 + h + 8 ! ) < (1 + 2 / )( µ<1> ( !1 + h ) + 8µ¢(!)) µ(fl + f2 ) + 2 €1 µ( JI + f2 ) + ( 1 + 2E1 ) 8µ¢(!) . An elementary application of epsilonics will finish the proof of Lemma 7.2 3 . D

All the ingredients are present to proceed to a proof of the existence of Haar measure on G if we follow A. Weil and call on transfinite methods. In fact we can use J(+ ( G) as an indexing set and keep in mind that for any ¢, f E J(+ (G) , 1 :w ( 3 5) ( w : f) ::; µ¢(!) ::; ( ! ) .

With this in mind, we let I(!) be the closed bounded interval I ( !) =

[ ( � !) ' (! : )] w

w

in � and call on Tychonoff's theorem (in its original form) to conclude that I = IT I(!) JC+ (G)

is a compact Hausdorff space. For a fixed

3. A Remarkable Approximation Theorem of Henri Cartan

195

thanks to (35) . For any open set V in G that contains the identity, let Fv denote the set of all points ( µ¢(!)) / E K + ( G) • where cp's support is contained in V. Each Fv is nonempty. If Vi , Vz and V are open sets each of which contains the identity with V � Vi n Vz , then Fv � FVi n FVi · By I's compactness there is a point (µ(f)) jE K + ( G) E n Fv . What does this mean? Well given Ji , . . . fn E x:+ ( G ) , E > 0 and an open set V containing G's identity, there is a ¢> E x:+ ( G) with the support of ¢> contained in V so l µ( f1 ) - µ¢( !i) I , · · · ' l µ (fn) - µ

.. ! ) - >.µ( ! ) , µ( f) 2:: (w:f) > 0, whenever Ji , f2 E !(,+ ( G) and >. > 0. If we define µ( O) = 0, then we have what's needed to define an invariant measure on G. ,

{

3.

A Remarkable Approximation Theorem of Henri Cartan

(Bochner and Dieudonne, specially tailored to our pur­ poses). Let G be a locally compact topological group, let K be a nonempty compact subset of G, and suppose Vi , . . . , Vn are open subsets of G each with compact closure so that K � Vi U · · · U Vn . Then there are continuous func­ tions fl , . . . , fn : G---+ [ 0, 1] such that the support of each fi is contained in Vi and I.:�=l fi ( k) = 1 for each k E K. The Ji , . . , fn are referred to as a continuous partition of unity subordinate to the cover {Vi , . . . , Vn } of K.

Theorem 7.24

.

Proof. In Chapter 3 the "Introduction to Topological Groups" we proved that G is paracompact; hence, normal. By G's normality we can find another open cover {W1 , . . . , Wn } of K with Wi � Vi, i = 1 , . . . , n. (Induction: n = 1 follows by the very definition of normality. If we assume that the statement holds for any compact subset of G and any open cover of the set by m-open sets for any m < n, then let's see what we can do if K is a compact set and K � Vi U · · · U Vn . Look at the compact set C = K n (Vi n · · · n Vn) c � Vi .

196

7. Haar Measure on Locally Compact Groups

Normality gives us a W1 so that C � W1 � W 1 C Vi . Our induction hypothesis applies to K n Wf � Vz U · · · U Vn , providing us with the rest of the Wi's, W2 , . . . , Wn .) Now suppose K � Vi U · · · U Vn ; look for W1 , . . . , Wn so Wi � Vi as in the above. Appeal to Mr. Urysohn, and he generously provides us with continuous functions gi , . . . , 9n : G -+ [0 , 1] such that Xwi :'.S 9i :'.S XVi . Set . Ji = � D �i�1 9i

f E J(,+ (G) , let K be a compact set in G, and let E > 0 be given. Then there exist k l , k2 , . . . , kn E K, h 1 , . . . , hn E !(,+ ( G) so that if g E G, k E K, then Lemma 7.25. Let

Proof.

then

Let V be a neighborhood of the identity chosen so that if g0 E Vg,

l f(g) - f(go) I :'.S E , remembering that f E JC+ (G) must be (right) uniformly continuous on G. Now cover K by a finite collection k l V, . . . , kn V of translates of V where k l , . . . , kn E K. Next, choose a partition of unity h 1 , . . . , hn E JC+ (G) sub­ ordinate to the covering {k l V, . . . , kn V} of K so that • each hi = 0 outside of ki V, and • l: �=l h i( k ) = 1 for each and every k E K. This we can do thanks to the Bochner-Dieudonne partition of unity theorem (Theorem 7.24). Notice that if k E V and k E ki V, then (ki l g) (k - l g) - 1 = ki l gg - l k = ki l k E V, and so by choice of V, l f(ki 1 9) - f(k - 1 g) I :'.S E . If k E V but k
3. A Remarkable Approximation Theorem of Henri Cartan

197

and this is so for i = 1 , . . . , n. Adding these inequalities up and recalling that l:�=l hi( k) = 1 for each k E K, we get

I t.

'

hi( k) kJ (g) - k f(g) ::; E

for all k E K and all g E G, as advertised.

D

Let Ji , h , . . . , fn E K,+ (G) , 0 :S A l , A2 , . . . , An , and E > 0 be given. Then there is an open set V containing the identity of G so that if 'ljJ E !(,+ ( G) with the support of 'ljJ contained in V, then Lemma 7.26.

t.

AiµVi ( Ji ) ::; µVi

( t. Adi) +

E.

Let K be a compact set that contains the support of Ji , . . . , fn · Choose f E !(,+ ( G) so that XK :'.S f :'.S 1 . Let fJ > 0 . Look at ffJ = Aif1 + · · · + Anfn + flf E K + ( G) . Define hi , 1 ::; i ::; n by Adi (g) hi(g ) = ffJ (g) 1'f g E K' 0 if g rt K. Notice Adi = ffJ hi and l:�=l hi :S 1 . Moreover hi E K,+ (G) . The uniform continuity of members of !(,+ ( G) ensures that if E1 > 0 is given, then there is an open set V that contains the identity so that if y E x V, then 1 l h 1 (x) - h l ( Y ) I , . . . , l hn( x ) - hn(Y) I :S E • Suppose 'ljJ E !(,+ ( G) has its support contained in V and suppose c 1 , . . . , Cm > 0 and gl , . . . , gm E V are chosen so that n Aif1 + " · + Anfn + flf = ffJ :S L Cj gj 'l/J. j=l If g E gj V, then for each i = 1 , . . . , n m Adi( g) = ffJ hi (g) :S L cj 'l/J(gj 1 g ) hi( g) j=l m < L Cj 'l/J(gj 1 g) ( h i (gj ) + / ) . j=l Proof.

{

7. Haar Measure on Locally Compact Groups

198

It follows that for i = 1, . . . , n,

m

>.i( fi : '¢ ) = (>.di : '¢ ) � 'L cj ( hi( gj ) + E' ) , j=l and so keeping in mind the fact that L:�= l hi � 1, m n n 'L >.i( fi : '¢ ) = L (>.di : '¢ ) � 'L cj (l + n/ ) . i= l

i= l

j=l

By the arbitrariness of c 1 , . . . , Cm > 0 and g1 , . . . , 9m E V vis-a-vis f0 , we see that n L Ai( fi : '¢ ) � (1 + n/ ) Uo : '¢ ) . i= l

On dividing everything in plain sight by (w : 'ljJ), we get n L >.iµ1fJ ( fi ) < (1 + nc=.1 ) (µ1/J ( JJ )) i= l

(1 + n/ ) (µ1/J (>.if1 + · · · + >.nfn + of)) < (1 + nE1 ) (µ1/J (>.if1 + · · · + >.nfn) + µ1/J (of))

µ1/J (>.if1 + · · · + >.nfn) + m' µ1/J (>.if1 + · · · + >.nfn) + o ( l + nc=.1 )µ1/J (f) . The standard dose of epsilonics finishes the proof.

D

A key element in the proof of the existence (and uniqueness) of Haar measure is the following stunning approximation theorem of H. Cartan. Remarkably, there is no measure theory in the statement of the theorem. 7.27 (Cartan). Let f E K,+ (G) , and let 0 < E be given. Then there is an open set V in G that contains the identity and has compact closure such that if ¢ E J(,+ ( G) with the support of ¢ contained in V, then there exist g 1 , . . . , 9n E supp( !) , c 1 , . . . , Cn � 0 so that

Theorem

Proof. To start, let 0 < c=.1 < E, and choose an open set e E V so that l f(g) - f(go) I � c=. 1 whenever g - 1 go E V.

V with V compact,

Let K be a compact set containing the support of f. Let 8 > 0. Apply Lemma 7.25 to find g 1 , . . . , 9n E K and h i , . . . , hn E K,+ (G) so that if

3. A Remarkable Approximation Theorem of Henri Cartan

9 E G, k E K, then

199

I t, hi( k)g/P (9) - kc/> (9) 1 :::; 8 .

Multiplying by f(k), we see that for all 9 E G, k E K, we have (36)

but the support of f is contained in K, so if k 's support lies in V, so if k- 1 9 E V, we see If (9) - f ( k) I :::; c' , and so 1 lkc/> (9)f(9) - k ¢(9)f(k) I :::; E k c/> (9) . On the other hand if k- 1 9 (9) = c/> (k - 1 9) = 0. Regardless of where k - 1 9 is located, we have that for every k E G and 9 E G, (37) l k c/> (9)f(k) - k ¢(9)f(9) I :::; c' k c/> (9) . In tandem, (36) and (37) tell us that for any k, 9 E G

(38)

I t, 9i c/>(9) hi( k)f(k) - kc/>(9)f(9) I

:::; I t,

I

9i c/> (9) hi( k)f(k ) - kc/> (9)f( k) + lkc/> (9)f(k) - k ¢(9)f(9) I

D f(k) + E1 k c/> (9) . Notice that if (/> (9) = ¢(9 - 1 ) , then kc/> (9) = c/> (k - 1 9) = (/> (9 - 1 k ) = g (/> (k) . If we now replace k c/> (9) by 9 (/> ( k ) in (38) , we get that for all k, 9 E G, :S

( 39 )

I t,_

., (g) h; (k)f(k) - , ,P ( k)f(g)

I

<;

O f(k) + ,' , ,P (k) .

Now viewing g as fixed and letting k be free, we get the functional inequality

I t,

,,. (g ) h; f - ,
I

<; /j

f + ",¢.

We now call on the averaging machinery developed earlier.

7. Haar Measure on Locally Compact Groupt

200

With the inequality

in hand, note that for 'ljJ E tells us that

!(,+ ( G) ,

the subadditivity of the functional µ'l/J

Now ef> =/= 0 and by Lemma 7. 22, µ'1/1 (9 (/>) = µ'l/J ((/>) > 0, so division by µ'1/1 (9(/>) is legal (if not ethical). Do it! The result: (40)

This is so for any 'ljJ E !(,+ ( G) . But Lemma 7.25 says that given rJ > 0 there is an open set W containing the identity of G so that if 'ljJ E J(,+ ( G) and the support of 'ljJ is inside W, . "i = 9iµ,µ(¢) (g) ) then (settmg \

(41)

Note. W acts as a catalyst ensuring near additivity of µ1/J, but also providing

a term

to compare with both n

f(g) and a sum of the sort I:>i gi ef>( g ) . i= l

We're in business! Let Ci =

µ'l/J ( hd ) , µ1{J ((/>)

201

4. Cartan 's Proof of Existence of a Left Haar Integral

then each Ci > 0 and for any g E G

' f (g)

-

n f= Cig//>(g) =l •

'

=

<

+

' f (g) ti=l µµ1/J1/J(h(
/

o µ1f1 ( {) + / + 'f/ ( by µ1/J (
(40)

and

(41 ) ) .

D

4. Cartan's Proof of Existence of a Left Haar Integral

Before presenting Cartan's proof of the existence of a left invariant Haar integral, we'll state a crucial lemma. We delay the proof of the lemma because it's also critical to Cartan's proof of uniqueness, and the pattern of its proof is worthy of close inspection. Lemma 7.28. Let f E K, + ( G ) and E > 0 be given. Then there is an open set U containing G 's identity for which, given any h E K, + ( G ) with the support of h contained in U, there is c 2: 0 and an open set V = Vu containing the identity so that l µU ) - cµ ( h ) I :::; € + whenever E !(, ( G) with the support of contained in V . The existence of a left Haar integral follows from our next result. Theorem 7.29. Let f E K, + ( G ) , E > 0 . Then there is an open set V containing G 's identity for which l µU ) - µ1/J ( J ) I < € + whenever 'I/; E !(, ( G ) with the support of and the support of 'I/; contained in V .





Proof. Suppose 0 < 7.28 that there is an

o :::; 1/7. Then according to the dictates of Lemma open set U containing G's identity so that for any h E K, + ( G ) with the support of h contained in U, there is a CJ 2: 0 and another open set Vu containing the identity so that (42)



whenever E !(, + ( G ) with the support of contained in Vu .

202

7. Haar Measure on Locally Compact Groups

If we apply our lemma again (Lemma 7.28) with_our beloved w replacing f and the same o in place, we find an open set U containing the identity so that if h E J(, + ( G) with _! he support of h contained in fJ, then for some Cw � 0 and some open set Vu containing the identity, we have (43)

-

-

-

-

whenever E J(, + (G) with the support of contained in Vu · On replacing U, U by U n U and Vu, Vu by Vu n Vu , we can and do suppose that the same U, V work in both (42) and (43). Notice that o < 1. So (43) ensures us that Cw > 0, making division by Cw a legal operation. Let (44) R = CJw . C If

(f) - RI = l µ<1>(f) - CJµ( h ) + CJµ( h ) - R I < l µ<1>( f) - CJµ( h ) I + CJ µ( h ) c < 0 + CJ l cw µ ( h ) - l l Cw < o + Ro = (1 + R)o. Appealing again to (42) , we see that

,

�I

(45)

because (f : ) :::; (f : w ) (w : ¢) . By (43) we have (46) Cw µ ( h ) � 1 - 0 . Since µ( h ) > 0 we see (using O :::; 1/2) that O R = CJw = cwh µ ( h ) -< (f 1: w ) 0+ -< 2(! : w ) + 1. C C µ ( h) Investing this knowledge wisely we find l µ<1>(f ) - RI :::; (1 + R)o :::; (1 + 2(! : w ) + l)o = 2(1 + (f : w))o so long as E J(, + (G) with the support of contained in V. Cutting to the chase: given f E K, + ( G) and E > 0 there is an R � 0 and an open set V containing G's identity so that

20 3

5. Cartan 's Proof of Uniqueness

whenever ¢ E

K,+ (G)

with the support of ¢ contained in V. Of course if

'lj; E K,+ ( G) has its support inside V, then

also.

1 µ1/J (f) - R I :S E

Conclusion. Given f E K,+ ( G) and E > 0 there is an open set V containing the identity so that if ¢, 'lj; E K,+ ( G) have their support inside V, then l µ¢ ( f ) µ1/J ( f ) I :S 2E. D Among friends, this is good enough to finish the proof. -

Our hard work is done. It's time to reap the rewards. Existence of a Left Invariant Integral. We now look at all pairs (V, ¢) where V is an open set in G containing G's identity and ¢ E K,+ ( G) with the support of ¢ contained in V. We order the pairs (V, ¢) by (Vi, ¢ 1 ) :S (Vi, ¢2 ) if Vi � Vi ; so convergence with respect to this direction measures what happens as that V coordinates shrinks to { e}. Fixing f E K,+ ( G), we see that our last theorem tells us that µ( f ) = lim µ¢ ( !) (V,¢)

exists. By Lemma 7.2 3 , the functional µ is additive on K,+ (G) ; Lemma 7.22 tells us that µ is positively homogeneous and invariant under left translates. What's more, for any ¢ E K,+ (G), : ¢) = 1, µcl> ( w) = (w (w : ¢) so µ(w) = 1. Therefore µ is a left invariant Haar measure on G. 5.

Cartan's Proof of Uniqueness

We restate Lemma 7.28 and include its proof. E K,+ (G) and E > 0 be given. Then there is an open set U containing G 's identity for which, given any h E K,+ (G) with the support of h contained in U, there is c 2: 0 and an open set V = Vu containing the

Lemma 7.30. Let f

identity so that

l µ¢ ( f ) - cµ¢ ( h ) I :S E

whenever ¢ E K,+ ( G) with the support of ¢ contained in V.

204

7. Haar Measure on Locally Compact Groups

Let W be an open set containing the identity with W compact, and let K = {x : f(x) =/:- O} be the support of f. Fix j' E K, + (G) so that Proof.

J'

?: X K . W ·

For any rJ > 0, we know (by Cartan's approximation theorem) that there is an open set U containing the identity (we can suppose that U � W) so that if E J(,+ ( G) with the support of contained in U, then there are ki , . . . , kn E K and c1 , . . . , Cn ?: 0 so that

h

h

h

ki h

Notice that vanishes outside of U and so outside of W. It soon follows that each of the functions vanish outside ki W and so outside K W. But f itself vanishes outside K which lies inside K · W. So f and both vanish outside K · W. Using this and the behaviour of j' we see that we can upgrade the above inequality to

ki h

·

I f - t Ci ki h l :S ry/ , i=l

a functional inequality involving members of K, + (G) . Now if ¢ E K, + (G) , we see that

(the last inequality follows by Lemma 7.22(v)). An appeal to our new additive lemma (Lemma 7.26) tells us that there is an open set V = Vu containing the identity so that if we further stipulate that

205

5. Cartan 's Proof of Uniqueness

But µ¢ ( ki h ) = µ¢ ( h ) , so combining (47) and (48) and letting c = 'L:�= l Ci , l µ

l "·ui - "·( t, "' "' h) + µ. ( t,"'

( t, ) I l ( t, ) t, I 1

) t, I

h -

c; µ , ( h )

Ci ki h

� µ¢ ( ! ) - µ¢ + µ
..

Ci ki h -

Ciµ¢( h )

� T/ ( / : w) + 'f/ . Epsilonics now rules!

D

Uniqueness of µ . To prove the uniqueness of left invariant integrals on J(, + ( G) , we consider a functional v on JC + ( G) which satisfies v(f) > 0 for each f E J(, + (G) , v is additive and positively homogeneous, and v is left invariant so v(f) = v(9 J ) for any g E G. We further assume that v(w) = 1 , where w is our chosen one in J(, + (G) . What do we want? We want to show

that v is in fact, lim 0 there is an open set U containing the identity (contained, if we wish-and we do-inside W) so that if h E J(, + ( G) with the support of h contained in U, then there are ki , . . . , kn E K and c 1 , . . . , Cn 2:'.: 0 so that As before, analysis of the supports of f and h soon tell us that Now the inequality ( 47 )

1

µ¢ ( !) - µ
I

f-

ti= l Ci ki h l � 'f// .

( t, Ci ki h) I � 'f/µ¢(/ ) � T/ (/ : w)

is replaced by

Since v is additive and positively homogeneous, the inequality (48)

is replaced by the identity

The left invariance of v tells us that v(kih) = v(h) , so letting c = L:�= l Ci, Iv(!) - cv(h) I

�

< <

I v (!) - t, c;v(h) I I v (!) - v ( t, c;., h) I

'IJV(/' ) .

Note. The "c" is the same here as it was in the case of µ or at least, it can

be chosen to be so. If we now turn to the proof of the existence of a left invariant integral µ (Theorem 7.29) , we can easily see that the constant Cw found therein also works with v replacing µ, this time without any codicils about ¢'s support. As a result, if R = : as in that proof (see (44) ) , we can conclude the same truism: there is can open set U containing the identity so that for any h E K, + (G) there is a Cw > 0 so that l l - Cwv(h) I = l v(w) - Cwv(h) I � & so long as h's support is contained in U. The point, once again, is that Cw can be chosen within the realm of epsilonics to be the same for v as for µ; again the roots are in Cartan's approximation theorem. The resulting R for v is the same as for µ and so ultimately, Iv(!) - µ(!) I � 2 € just as in the proof of existence. The upshot: for any E > 0

207

6. Notes and Remarks

6. Notes and Remarks

Our discussion of positive linear functionals follows M. A Naimark's Normed Rings [91] with occasional help found in D. Cohn's treasured book [20] . The proof in §2 is due to A. Weil [136] . Our presentation of Weil's proof (and Cartan's, for that matter) is based on positive linear functionals. Again our functional analytic bias entered our decision. For those who would prefer a more purely measure theoretic approach, we recommend the treatment found in D. Cohn [20] or P. Halmos [50] . H. Cartan's wonderful proof in [15] of the existence and uniqueness (achieved more or less simultaneously) makes no call on the Axiom of Choice. The elegance of his approach far outweighs the added complexity. There do not appear to be any essentially different proofs for general locally compact groups, other than those of Cartan (with a simplified variation due to Alfsen [1] ) and Weil; the generalization due to Steinlage is itself a mix of Cartan and Weil with a touch of Banach. We went with Cartan's original proof for historical reasons. Let G be a nondiscrete locally compact topological group, and let µ be a left Haar measure defined on G. Here's how to find a nonmeasurable subset of G. Start with an open set U that contains G's identity e and has compact closure. Of course 0 < µ(U) < oo . Let V be an open set containing E such that V = v - I and V · V V � U. Let D be a countably infinite subset of V. Let H be the subgroup of G generated by D. Then H is also countably infinite. Look at the collection {Hx a : a E A} of all right cosets of G (modulo H) and denote by Ao the set Ao = { a E A : H x a n V =!= 0} . For each a E Ao, pick exactly one point Ya such that Ya E HX a n v = HYa n v. Our nominee is E = {ya : a E Ao}. 6.1. Existence of a non-Haar measurable set

·

a

la Vitali.

208

7. Haar Measure on Locally Compact Groups

We observe that H n (V · V) is countably infinite. It's countable since it's a subset of the countable set H, and it's infinite since D s;;; V s;;; V · V, and D s;;; H. Consider ( H n V · V) · E. It's a disjoint union of sets of the form xE, where x E H n V · V; each of the countably infinite many sets xE, where x E ( H n V · V) · E has the same measure µ(E). It follows that the measure of ( H n V · V) · E has two choices: µ((H n V · V) · E) = 0 if µ(E) = 0 or µ((H n V · V) · E) = oo if µ(E) > 0. But V s;;; ( H n V · V) · E s;;; U. Indeed, if v E V, then v E HxCL for some CL E A. Therefore v E V n HxCL, and so CL E Ao, which means that v E HYCL · Thus v = hyCL for some h E H and CL E Ao. Therefore h = vy(;/ E V · v - 1 = V · V, and so v E ( H n V · V) · E. On the other side of the ledger, ( H n V · V) · E s;;; V · V · E s;;; V · V · V s;;; U. From this it follows that µ(V) :::; µ((H n V · V) · E) :::; µ(U) , and this is impossible. A close look at what has just been done is warranted. Suppose E is a o--field of subsets of G that contains all the µ-measurable sets, gE E E if g E G and E E E, and µ is a countably additive left invariant extension of µ to E. Reasoning as before, we still cannot admit E to E without running afoul of the rules of logic, so E just does not gain membership in :E. After all, if E E E, then ( H n V · V) · E E :E too. As before il(( H n V · V) · E) = 0 or il(( H n V · V) · E) = oo. D But V s;;; ( H n V · V) · E s;;; U makes either possibility untenable.

6. Notes and Remarks

209

P. Erdos and R. Daniel Mauldin [36] showed that if G is an uncountable group, then there does not exist a nontrivial o--finite countably additive measure on the field of all subsets of G that is left invariant. Their proof relies on Ulam's proof (using the Axiom of Choice) that the first uncountable cardinal is not a measurable cardinal. 6.2. Von Neumann-Kakutani-Weil Proof of Uniqueness. There is

another well-known argument for the uniqueness of Haar measure in the setting of a general locally compact group. This proof revolves around the judicious use of Fubini's theorem. The proof was given independently by J. von Neumann [1936] and S. Kakutani [59] . Each needed an added hypothesis on the group to ensure the availability of the appropriate Fubini theorem. As was his wont, A. Weil [ 13 5] realized the availability of a suitable version of the Fubini scheme and gave a general proof of uniqueness. As a result, this might be called the Von Neumann-Kakutani-Weil proof. How's that for pedigree? We'll be dealing with Haar measures µ and v on a locally compact topological group G. Here's the utility grade Fubini theorem that we will need. (Fubini). Let h E K ( G x G ) . Then (i) for each x, y E G the functions f(x, · ) , f( -, y) belong to K ( G) ; (ii) the functions

Theorem 7.31

x--+

l f(x, y)dv(y)

y--+

l f(x, y)dµ(x)

and belong to K ( G ) ; (iii)

J J f(x, y)dv(y)dµ(x) = J J f(x, y)dµ(x)dv(y) .

For a clear exposition of what's involved, we suggest the beautiful book Measure Theory by Donald L. Cohn [2 0] .

Let µ and v be left Haar measures on the locally compact group G. Then there is a positive real number c such that v = cµ. Theorem 7.32.

210

7. Haar Measure on Locally Compact Groups

Proof.

ratio

In fact, let g E K + ( G) , g -/= 0, and let f E K . Since J gdµ -=!= 0 the J fdµ J gdµ

is legal. We will show that this ratio depends only on f and g. Now if h E K ( G x G ) , then Fubini's theorem in its utility grade version tells us that both integrals

J J h (x, y) dµ(x)dv(y) , J J h (x, y)dv(y)dµ (x) exist and are equal. We look at the right-hand integral and apply the Fubini theorem, substituting y - 1 x for x to get

j j h (x, y)dv(y)dµ(x) = j j h (y- 1 x, y)dµ(x)dv(y) ,

which in reapplying the Fubini theorem and substituting xy for y gives

J J h (x, y)dv(y)dµ (x) = J J h (y- 1 x, y)dµ(x) dv(y) = J J h(y - 1 , xy)dv(y ) dµ( x ) .

(49)

Now a judicious choice of h comes to the fore: Let g (yx) h (x, y ) = J f(x) g ( z x )dv( z ) · We have that h E K + ( G ) and the function x-+ J g ( zx ) dv( z ) is continuous and never vanishes. Moreover if K = support( ! ) and K = support(g), then h's support is contained in the compact set K x K K- 1 . Further we see that -1 h(y - 1 , xy) = f gf (yzy- l) g (x) z ' ( )dv( ) and so looking to ( 49) we get f(x) g(yx ) dv(y)d (x) = J (y - l ) g (x) dv(y ) dµ( x ) . µ J g ( z x ) dv( z ) J g ( zy- 1 )dv( z ) Since g(yx ) f(x) g(yx ) J g ( zx )dv( z ) dv(y)dµ (x) = f(x) J g(zx )dv( z ) dµ (x) = J (x)dx,

JJ

JJ

jj

j j

j

211

6. Notes and Remarks

we have

f f(x)dµ(x) ff J g(f(yzy-- 1 l)g()dv(x) z) dv(y)dµ(x) f g(x) f J g(zJy-(y-1 )d1 ) v(z) dv(y) dµ (x) f g(x) dµ(x) f J g(zf(yy--1 )1dv() z) dv(y) . =

In other words,

f

f(y - 1 ) dv(y) , J f(x)dµ( x) = J g( zy- 1 )dv( z ) J g(x)dµ (x) regardless of v and regardless of µ.

D

6.3. Form of Invariant Measures on Some Classical Groups. Fol­ lowing the directions of Hewitt and Ross [55] we "compute" the Haar mea­

sure in case G is a topological group with easily checked qualifications: ( i ) G is an open set in �n for some n, so if x = (x 1 , . . . , Xn ) , Y = ( YI , . . . , Yn ) E G, then X · y = F(x 1 , . . . , Xn , 1 , . . . , Yn ) E G for F : G x G-+G. d ( ii ) � F, � F exist and are continuous throughout G. dXk dYk Denote by la , Ta the left and right translations by a E G, la ( X) = ax, Ta ( x) =

xa.

( iii ) The Jacobians of la and Ta are constant, that is, depend only on a.

If J( r) denotes the Jacobian of the transformation r , let L(a) = I J( la) I , and R(a) = I J( Ta) I . The nature of Jacobians ensures that J( la o h ) = J( la) · J( h )

( since la o h = lab ) , and so

L(a)L(b) = L(ab) .

Similarly, R(a)R(b) = R(ab) .

212

7. Haar Measure on Locally Compact Groups

Moreover le = ida = re so

L (e) = 1 = R(e) . Each of the maps L, R : G--t(O, oo) is a continuous homomorphism of G onto the multiplicative group (0, oo) . Left Haar integration? Take ¢ : G--t!K continuous with compact support, and denote by cp(x) dx the Riemann integral of ¢ ; this makes sense because ¢ has compact support. Set (x) I(!) = {a LJ (x) dx l so I(!) is a left Haar integral. Indeed using the classical change of variables formulas, we see that for any a E G (a - 1 f)(x) I(a- 1 ! ) =

l

l L�:) l (a-l f la) (y) (L �a ) (y) dy O

0

r f(y) dy = I(!) . Ja L(y )

So I is a left invariant functional on the collection of continuous functions with compact support. Similarly J ( J ) = f(x) is a right invariant integral. Let's see this Haar measure machine in action. We start with Q.C(2, JR) , the general linear group on JR2 . Here

l R�:)

la (x) =

So

213

6. Notes and Remarks

and
an
(�

n

�22 � 0

12

)

+ a12
(�

21

11

� �

12

)

0 a21 a22 a21 a22 an (a11a22a22 - a12a21a22) + a12 (a12a21a21 - aua21a22) a� 1 a�2 - 2 an a12a21a22 + a�2 a� 1 (aua22 - a12a21 ) 2 = (det x) 2 ,

and thus the left invariant acts like -+ (x) Ldx (X) =

1Q.C(2,IR)

What about the right integral?

j (d(x) et ) 2 dx. X

ra(x) =

So J(r. ) and
�

(�

(�

an a21 0 12 a22 0 0 au 0 a12

22

)

(�

12

)

u 21 0 a12 a22 a12 a22 an (a22aua22 - a12a21a22} - a21 (a12ana22 - a21a12a12) (
.

� u �21

0

- a21
� �

Since J and I are the same and each, evaluated at suitable continuous f's is f(x) dx r (det x) 2 ' }g.c(2 ,IR) 9£(2, JR) is unimodular. Similar computations will result in Q.C(n, JR) also being unimodular with Haar integral of f Q.C(n, JR)-+K that is continuous and has compact support being given by f(x) dx. Riemann mtegral det(x) ln I :

.

.

j

214

7.

Haar Measure on Locally Compact Groups

6.4. Existence of a Complete System of Irreducible Representa­ tions. Let G be a locally compact topological group. Definition 7.33. A unitary representation of G is a homomorphism from G into the group of unitary representations, 9-+U9 , on the same Hilbert space H such that if 9a-+9o in G, then for each x E H, U9a x-+U90 x.

The unitary representation 9-+U9 is irreducible if the only closed linear sub­ spaces S of H (the underlying Hilbert space of representations) that are left invariant (i.e. , U9S � S) by each U9 are S = {O} and S = H. We say that G admits a complete system of irreducible unitary representations if given 90 E G such that 9o i- e (G's identity) , there is an irreducible unitary representation 9-+U9 of G over some Hilbert space H such that U90 i- idH. The Peter-Weyl theorem asserts that every compact topological group ad­ mits a complete system of finite-dimensional irreducible unitary representa­ tions (i.e., unitary representations over a finite-dimensional Hilbert space) . The original proof of the Peter-Weyl theorem, valid for compact Lie groups, relied on the existence of left-invariant differential forms. As soon as Haar discovered his measure, the Peter-Weyl theorem was understood to extend to the proof that every compact metrizable topological group admits a com­ plete system of finite-dimensional irreducible unitary representations. When von Neumann's proof of the existence of a Haar measure over any compact group made its appearance, the extension of the Peter-Weyl theorem for general compact groups was an (almost) immediate consequence. In a parallel development, von Neumann showed that every abelian locally compact group satisfying the second axiom of countability admits a com­ plete system of one-dimensional irreducible unitary representations-such representations are usually called characters. Prophetically, I. M. Gelfand and D. A. Raikov soon presented a proof of this result based on developing machinery of commutative Banach algebras. Soon after Weil's proof in [136] that every locally compact topological group admits a (left-) Haar measure, Gelfand and Raikov showed in [42] that every locally compact topological group admits a complete system of irreducible unitary representations. Their approach is a delightful mix of group theo­ retic ideas, basic notions in Banach algebras, Hilbert space spectral theory, and measure theory; their paper is a model of clarity. We indicate an outline of the Gelfand-Raikov paper and enthusiastically recommend it for those who wish to see truly beautiful mathematics done beautifully. Definition 7.34.

215

6. Notes and Remarks

The Gelfand-Raikov proof proceeds in stages. Stage one establishes the connection between unitary representations of G and positive definite functions ¢ : G---t C . Recall that ¢ : G---t
k=l l=l

for any finite collection g i , . . . , gn E G and A l , . . . , An E
If we identify a and b whenever ( a - b, a - b ) = 0, and complete the resulting inner product space, we get a Hilbert space which we denote by L 2 (¢) . For any g E G, let U9 be the mapping that takes the finitely nonzero a : G---t

for some constant c.

If the unitary representation g--+U9 arises from an ele­ mentary continuous positive definite function ¢, then this representation is irreducible. If ¢ is a continuous positive definite function and ¢ gives rise to an irreducible unitary representation, then ¢ is elementary.

Theorem 7.36.

216

7.

Haar Measure on Locally Compact Groups

Briefly if P is a projection on L 2 ( ) · Thus P is either 0 or idL2 ( 4>) · On the other hand, if the continuous positive definite function

) ' and so x ( g ) = ccp ( g ) . It is immediate from this that we have

In order that G admit a compete system of irreducible unitary transformations, it is both necessary and sufficient of given go E G, go #- e that there is an elementary continuous positive definite function
217

6. Notes and Remarks

functions defined on G. In addition to its Banach space structure (the norm is l l x l l 1 = fc l x(g) I dµ(g) = J l x l dµ), L 1 (G) comes equipped with a multi­ plication (convolution) and a * -Operation: if x, y E £ 1 ( G), then x · y is given by (x · y) (g) = x(h)y(h - 1 g) dµ(h), and the *-operation is x * (g) = lg - i x(g - 1 ) . Here's what's so about the operations and how they treat members of L 1 (G) :

la

For any x, y, z E L 1 (G), a E C, we have 1 • x · y E L (G) and l l x · Y l l 1 :'.S l l x l l1 l lYl l 1 ; • a (x·y) = ( a x) · y = x · ( a y), (x·y) · z = x· (y· z ), x· (y +z ) = x·y + x· z , and (x + y) · z = x z + yz . Fact 7.39. For any x E L 1 (G), x * E L 1 (G) , and l l x * l l 1 = l l x l l i · Moreover for any x, y E L 1 (G), a E C, we have • (ax)* = ax*, • (x + y) * = x* + y *, • (x · y)* = y * · x *, • (x*)* = x. The operator T9 defined by T9x(h) = x(g - 1 h) , acts as a bounded linear operator from L 1 (G) to L 1 (G), and it also acts on L 2 (G) to L2 (G). It is easy to see that T9 is an isometry of L 1 (G) onto itself, and on L 2 (G) , T9 is a unitary operator. In fact g -+T9 is a unitary representation of G over L2 (G) . Definition 7.40. A linear functional L defined on L 1 (G) is called positive if for any x E L 1 (G), L(x * · x) � 0. Now the existence of the left invariant measure µ makes its presence felt. Fact 7.38.

Definition 7.41. A function : G-+C is integrally positive definite if is essentially bounded (with respect to µ) and satisfies (h - 1 g)x(g)x(h) dµ(g) dµ(h) � 0 for all x E L 1 (G) .

jj

7.

218

Haar Measure on Locally Compact Groups

With this definition in hand it is easy to see that the functional L on L 1 ( G) given by L (x) = (g)x(g) dµ (g) is a positive linear functional in the sense described above. We can now state the following

j

Theorem 7.42

( Key Theorem ) . Every positive linear functional L arises

from a continuous positive definite function via the formula L (x) = L¢ (x) =
j

It follows from this that any integrally positive definite function is equal µ-almost everywhere to a continuous positive definite function. We say that the continuous positive definite function is normalized if (e) = 1. Let P denote the collection of all positive definite functions with (e ) � 1 . Then P is plainly a convex set and the identically zero function 0 is an extreme point of P. Critical to the Gelfand-Raikov scheme is the following:

Any nonzero extreme point of P is a normalized elemen­ tary positive definite function, and conversely, any normalized elementary positive definite function on G is an extreme point of P . Stage three utilizes the correspondences derived in stage two to set up an application of the Krein-Milman theorem. To get started, we define the weak topology via telling what a weak neigh­ borhood of 0 and x 1 , . . . , Xn E L 1 ( G) , we define the weak neighborhood W (o; x 1 , . . . , Xn , E ) by Theorem 7.43.

{

I

l/

}·

W = E P : ( o(g) - (g)) x k (g) dµ (g) < E , for 1 � k � n By RealL 1 ( G) = RL 1 , we mean the real linear space which is the closed real linear space spanned by those x E L 1 ( G) for which x = x*. If L is a positive linear functional on L 1 (G) , then L (x*) = L (x); using this, such an L can be interpreted as a bounded real linear functional on RL 1 , and since L (x) = L x � x* + i L x ;ix* , the restriction of L to RL 1 is a one-to-one operation between positive linear functionals on the Banach *-algebra L 1 (G) and positive real linear function­ als on RL 1 .

(

) (

)

219

6. Notes and Remarks

If L = L¢,, ¢ E P, then l l Ll l = 1 1 ¢ 1 1u'° ( µ ) ¢( e) ::=; 1 1
The set P of continuous positive definite functions ¢ sat­ isfying ¢( e) :::; 1 is the smallest weakly closed convex set containing all the normalized elementary continuous positive definite functions and the zero function 0 (9 ) := 0. Because G is locally compact, we have a well-tested manner of constructing nontrivial positive definite functions: L 2 ( G) is a Hilbert space with the usual inner product (x, y) of x, y E L 2 (G ) Theorem 7.44.

J

(x, y) = x(9)y(9) dµ(g) ; the operator U9 : L2 ( G) -+L2 ( G) given by ( U9 x) ( h) = x (9 - 1 h) is a unitary operator and 9 -+Ug is a unitary representation of G over L 2 ( G) . If we take a 9o E G such that 9o -/= e, let W be an open set containing the identity such that g W n W = 0, and choose xw E JC+ ( G) so that ( xw, xw) = 1 yet xw ( h) = 0 outside W, then the function ¢(9 ) = (U9 xw, xw) is a continuous positive definite function such that ¢( e) = ( xw, xw) = 1, so ¢ is normalized. What's more 90W n W = 0, so ¢( go ) = ( U90xw, xw) = xw (90 1 h)xw (h) dµ( h) = 0, since 90 1 h E W precisely when h E 90W which for h E W does not happen. There are plenty of normalized continuous positive definite functions on G! To finish the argument, Gelfand and Raikov now argue by contradiction that if some 9o E G is not e but there is no irreducible unitary representation g -+U9 so that U9 -/= id, it is because every normalized continuous positive definite ¢ has ¢(9 ) = ¢( e) and from this so, too, does each normalized continuous positive definite function ¢.

j

-

220

7.

Haar Measure on Locally Compact Groups

We close this Notes and Remarks section on a high note: a brief description of the still­ wonderful Pontrjagin-van Kampen duality theorem. Let G be a locally compact abelian topological group. We denote by r the collection of all continuous homomorphisms of G into the circle group II = { z E C : l z l = l}. Then r is an abelian group under pointwise multiplication; we can equip r with the Pontrjagin topology: if /'O E I', then a basic open set about /'O is generated by a compact subset K of G and a o > 0, and it looks like P (!'o ; K, o) = { I' E r : 11' (9) - l'o( g ) I < o for all g E K } . With this topology r is a locally compact abelian topological group called the dual group of G. There is a natural pairing of G and r, namely (g, !') = !'(g) . Here I' E r is viewed as a function of g E G but g can also be viewed as a function of I' · It's here that magic happens. Since r is a locally compact abelian topological group we can apply to r the procedure just established to G and and the result will be ostensibly another locally compact abelian topological group, the dual group of r. It is a fundamental phenomenon of mathematics that the result is a topological group that is isomorphic as a topological group to G! Every locally compact abelian topological group is its own second dual group. This was discovered in case the locally compact topological group also sat­ isfies the second axiom of countability by L. S. Pontrjagin in 1934 [103] and in general by E. R. van Kampen in 1935 [131] . The original proof proceeded through the evolving structure of locally compact abelian groups. An excel­ lent modern source for this approach is Sid Morris's Pontrjagin Duality and the Structure of Locally Compact Abelian Groups [86] . Naturally the Pontrjagin-van Kampen duality theorem suggests that each isomorphic invariant of a locally compact abelian topological group has a dual version and the classification and delineation of dual properties has been an active pursuit for many years. We mention a few of the more famous such results: • G is compact if and only if r is discrete (this is due to Pontrjagin and van Kampen). • The weight of G and that of r are always the same, where the weight of the topological space is the least cardinality of a base for its topology. 6.5. The Pontrjagin-van Kampen Duality Theorem.

6. Notes and Remarks

221

• If G is a compact abelian topological group, then G is metrizable if and only if r is countable. • G is connected if and only if r contains no nontrivial compact subgroups. A list of interesting/important duality statements is long and interesting of and by itself. Hewitt and Ross's Abstract Harmonic Analysis [55] serves as an excellent start toward a complete list. The mention of the Hewitt-Ross volume hints at one of the most enduring consequences of the Pontrjagin-van Kampen duality theorem: abstract harmonic analysis. A few words about this direction are worth making. For one thing, the duality theorem allows one to define the Fourier transform. Indeed, if x E L 1 (G) and 'Y E r, then we can define x('Y) as

la

x (g ) = x (g )'Y (g ) dg, where dg indicates integrating against the Haar measure. Many old friends from classical Fourier analysis extend to this abstract setting. To wit: • If x , y E L I (G), then x · y E L 1 (G) with l l x · Y l l I ::; l l x l l i l l Y l l I , where ( x · y ) (g ) = x (g - h )y( h ) dh (we now write the operation in G in an additive manner because of the abelian nature of G) ; of course this is just an extension of what happens in general. • If x E L I ( G) , then the Fourier transform x defined by x('Y) = x ( g )'Y ( g ) dg

la

la

is continuous and vanishes at oo with l l x l l oo ::; l l x l l I · • The map x�x('Y) is a complex-valued multiplicative linear func­ tional on L I ( G) and, in fact, any such functional on L I ( G) comes about in this fashion. • The Fourier transform of x · y, X-y is x f;; that is, for any 'Y E I' and any x , y E L 1 (G), x : Y ( 'Y ) = x('Y) fJ ( '"'f ) . • If x E L I (G) and x('Y) 0, then x = 0. The basic principles of classical harmonic analysis often have interpretations in this general setting, including Plancherel's theorem: the Fourier transform restricted to L 1 (G) n L2 (G) is an isometry (with respect to the L 2-norm) ·

:=

222

7.

Haar Measure on Locally Compact Groups

onto a dense linear subspace of L2 ( r ) and may be uniquely extended to an isometry of L 2 ( G ) onto L 2 ( r ) . In fact, developing the Fourier transform as above allows one to derive the Pontrjagin-van Kampen duality theorem via the Fourier transform. A key feature of this approach is an alternative definition of the topology of the dual group r ' namely if /'O E r ' E > 0 and X 1 ' . . . ' Xn E L 1 ( G) ' then a basic neighborhood of ')'o is given by N (1o ; x 1 , . . . ' Xn, E ) = b E r : I ii(!') - Xi(!'o) < E , i = 1, . . . ' n } . This topology is the same as the Pontrjagin topology introduced earlier, and this fact uses the notions of the Fourier analysis in essential ways. This approach to the duality theorem via the Fourier transform is due to H. Cartan and R. Godement [16] . To see some of the places this takes you, Walter Rudin's Fourier Analysis in Groups [108] is highly recommended. It is natural to wonder how tight is the hypothesis of local compactness to constructing a Haar measure. As we will see in Chapter 10, lack of local compactness, even in very well-behaved groups (such as Polish groups) that are not locally compact leads to pathological conclusions. Nevertheless, local compactness is always nearby in case the underlying measure-theoretics behave, as shown by A. Weil in the appendices of [136] . A measurable group is a group G with a O"-ring M and a O" finite, not identi­ cally zero, nonnegative extended real-valued measure m defined on M where both M and m are left invariant and the map that takes ( x, y ) E G x G to (x, xy) E G x G is measurable. With Weil we suppose that G is separated, so given any x E G other than the identity there is a measurable set A in G such that the m-measure of the symmetric difference of A and xA is positive. Weil showed how to introduce a topology (called the Weil topology) in G making G into a locally bounded topological group with the added features that m(U) > 0 for each open U in G and m(E) is finite for each set E that is bounded in G. (Local boundedness just means that there is a locally compact topological group of which G is a dense subgroup.) The point is that if G is a measurable group, then G is a dense subgroup of a locally compact topological group c' and m is the trace of a left Haar measure m' defined on a' in the sense that m' ( E n S) = m( E) .

Chapter 8

Metric Invariance and Haar Measure

Suppose G is a locally compact metrizable group. Then G admits a metric p which is left invariant and generates G's topology; G also admits a left invariant regular Borel measure (a left Haar measure). Is there any con­ nection between these left invariant objects? In this chapter we show the answer is an emphatic "yes." The highlight of the chapter is the wedding of Haar measure and Hausdorff measure; presiding at the ceremony is C. Bandt. The result is a beautiful offspring: fractional Hausdorff measure . A consequence is that if .A is a left Haar measure on G and p is a left invariant metric generating G's topology, then p-isometric subsets of G have the same .A-outer measure. Now the assumption that G has a left invariant metric that generates its group topology is, of course, not an assumption at all. The Birkhoff­ Kakutani theorem assures us that G's metrizability is enough to ensure such a metric exists. However, since we are looking for relationships between left invariant measures on G and left invariant metrics on G, it is instructive (and fun!) to show how to generate a left invariant metric that generates G's topology directly from its left Haar measure. This proof is due to R. A. Struble [124] and leads off the chapter. (R. A. Struble) . Let G be a locally compact group with left Haar measure .A. Let (Vn) be a decreasing sequence of open sets that form a neighborhood basis of the identity e in G where Vn compact for each n .

Theorem 8.1

-

223

224

8. Metric Invariance and Haar Measure

Then

V V p(x, y) = sup n A(X nL.Y n) defines a left invariant metric on G which is compatible with the topology of G. It's clear that p(x, y) is well defined and that p(x, y ) = p(y, x ) . Moreover p(x, y) 2: 0 and p(x, y) < oo regardless of x, y E G since each Vn is a Borel set with compact closure. Further, p( zx, zy) and p(x, y) coincide because A is left invariant. If x i= y, then there must be an m so that xVm n yVm = 0 since G is Hausdorff; but now p(x, y) 2: A(xVmL. Y Vm) = 2A ( Vm) > 0. On noticing that for any n and any x, y, z E G, xVnL.Y Vn � ( xVnL. zVn) U ( zVnL.Y Vn) , we see that for any x, y, z E G, A(xVnL. Y Vn) < A((xVnL. zVn) U ( zVnL.Y Vn)) < A(XVnL. zVn) + A ( zVnL.Y Vn) < p(x, z ) + p( z , y ) , and with this p(x, y) � p( x, z ) + p( z , y). In sum, p is a left invariant metric on G. If G's topology is discrete, then A( { e} ) > 0 so Vm = { e} for some m ; hence if x i= y, p(x, y) 2: A(xVmL. Y Vm) = A({x, y}) = 2A ( {e} ) > 0, and the topology induced by p is discrete. If G's topology is not discrete, then A(Vn) � A(nn Vn) = A( { e} ) = 0. If V is any open set containing e , then there is an m E N so that Vm V� 1 � V. Claim 1 . x E V whenever p(x, e ) < A(Vm) · To see this, let p(x, e ) < A ( Vm) · Then A(xVmL. Vm) � p( x, e ) < A( Vm) , a positive number. Were xVm n Vm = 0, then A(xVmL. Vm) = 2A(Vm) < A(Vm) , oops! So xVm nVm i= 0 and thus there are v 1 , v 2 E Vm so xv i = v2 E xVm nVm and Proof.

225

8. Metric Invariance and Haar Measure

This is so whenever p (x, e ) < .A(Vm), and our claim is justified. Let's look at all of the points x such that p(x, e ) < r, where r E Q, r > 0. There must be an m E N so that .A(Vn ) < £ , whenever n � m. Each of the functions fk(x) = .A(xVk L:. Vi) is continuous and satisfies fk ( e ) = .A(VkL:.Vk) = .A(0) = 0. But now we know there is an l E N so that if x E Vz , then fi (x), . . . , fm 1 (x) < r. Claim 2. If x E llz, then p(x, e ) < r. Why is this so? Well if x E llz, then by choice of l E N, we have .A(xVi L:. Vi), . . . , .A(xVm- 1 6 Vm- 1 ) < r. What about .A(xVkL:. Vk) for k � m? In this case, .A(xVk L:.Vk) � 2.A(Vk) < 2 4r < r. It follows that p(x, e ) = supn .A(xVmL:.Vn ) < r. Our two claims taken in tandem show that p generates G's topology about e. Since p is left invariant and since G's topology is too, this is enough to D say that p generates G's topology everywhere. -

·

(C. Bandt [8] ) . If p is a left invariant metric on the locally compact metrizable group G defining the topology of G, then any two subsets of G that are p -isometric have the same left Haar measure. To prove Theorem 8.2, we need to develop a fractional Hausdorff measure. Let A be a fixed compact set with nonempty interior. We'd like to construct a Hausdorff gauge function h so that the associated Hausdorff measure µh on G satisfies 0 < µ h (A) < oo . Sadly, finding such a gauge function is elusive. Happily Bandt found a way around this-the fractional Hausdorff measure. Given a Hausdorff gauge function h, we define the fractional Hausdorff measure vh by Theorem 8.2

{

limO inf Z: cjh(diam(Bj)) : Cj > 0, vh (E) = t\, J

.

XE � � Cj X B; , diam(Bj) � t J

}·

Mimicking the proofs encountered in Hausdorff measures, we see that v h is a metric outer measure (ensuring us that Borel sets are vh-measurable) and

226

8. Metric Invariance and Haar Measure

vh is left invariant reflecting p's left invariance. The issue is to judiciously choose h so that 0 < v h (A) < oo. Once this has been achieved, we see that for any compact set K � G, v h (K) < oo (K can be covered by finitely many of A's left translates, each of which has the same vh -measure), and for any nonempty open subset U of G, v h (U) > 0 (we can cover A by finitely many left translates of U, each having the same vh-measure). Before proceeding, we take note that if 0 < s < t, then for any E � G (51)

inf

{� J

Cjh(diam(Bj)) : Cj > 0, XE :::; � Cj XBj ' diam(Bj) :::; t J

is at least as large as inf

{� J

cjh(diam(Bj)) : Cj > O, xE :::; � Cj XBj , diam(Bj) :::; s J

}

}

;

after all, there are at least as many fractional coverings of E when the sets Bj have diameter less than or equal to s as there are with sets Bj having diameter less than or equal to t. So, as a function of t, (51) ascends as t descends to zero. It follows that v h (E) is determined by what happens when the fractional coverings involve sets of small diameter. The gauge function that works is h (t) = sup {A(B) : diam(B) :::; t}, a function that takes finite values on some open interval (0, T0 ) . Indeed, G is locally compact, and so G has a basis for its open sets consisting of sets with compact closure. It follows that at any point of G, we have a basis of open balls with compact closure all with diameter less than To, for some To > 0. The left invariance of p ensures that the same To works throughout G. vh is left invariant. As a matter of fact, the metric p of Theorem 8.1 that generates G's topology is left invariant, and so for any subset B of G, the diameter of B and gB are the same, since they're p-isometric. By the same token, sets in G that are isometric with respect to p are assigned the same vh -values. Note 8.3.

227

8. Metric Invariance and Haar Measure

The real issue with vh is to show that it's nontrivial, that is, 0 < v h ( A ) < oo. Once we know this to be so, then vh is a left Haar measure on G and so is but a multiple of A . Hence sets in G that are isometric ( with respect to p) have the same .A-measure as well, and we will have proved Theorem 8.2. We proceed with several lemmas to get that v h is nontrivial. Lemma 8.4.

Proof.

For any Borel set B � G, .A (B) :::; v h (B) .

Suppose XB :::; "2:,j Cj X Br Then XB :::; "2:,j Cj XIfi and so

l d.A :::; J L Cj X[jj d.A

.A (B)

L CjA(Bj) :::; L Cjh(diam(Bj)) L Cjh(diam(Bj)). It follows that D

Of course a particular consequence of this lemma is Corollary 8.5. 0 < Lemma 8.6.

.A ( A ) < vh ( A ) .

Let E ( t ) be defined by

E ( t ) = inf

{f

If then vh ( A ) < oo.

j=l

cj

:

n

E

}

N, XA :::; f ci X Bj , cj � 0, diam(Bj ) :::; t . j=l

limt inf h ( t ) E ( t ) < oo,

There is a c > 0 and a sequence ( t k ) , t k > 0 with t k � 0 so that h ( t k ) E ( t k) < c for all k. In other words, E ( t k ) < h ct ( k) for all k. For each k, choose a fractional covering of A, n (k) XA :S L c3k \B ( k ) , CJk) � 0 j=l Proof.

J

8. Metric Invariance and Haar Measure

228

with

n (k)

c (k) c :S j � h (tk ) J=l Of course, the definition of the fractional Hausdorff measure v h ( A) ensures that cjh ( diam ( Bj )) : Cj 2 0, XA :S L CjXBi , diam ( Bj ) :S vh ( A) :S inf �

'

{t

tk },

J=l

so

n (k)

vh ( A ) :S L c)k) h ( tj ) :S c < oo.

D

j= l

Lemmas 8.4 and 8.6 show us the way to the end, namely the proof Bandt's theorem, which will follow from the following. 8.7 ( Principal Lemma) . For each E > 0 there is a to > 0 so that if U is an open subset of G with diam(U ) :S to , then for some s 1 , . . . , Sn E G and a 1 , . . . , an > 0 we have

Lemma

n

XA :S L l¥iX si · U i= l

and

n

>.( A ) :S L l¥iA (Si . U ) :S (1 + E)>.(A). i= l

This proof depends on Cartan's approximation theorem ( Theorem 7.27) and is rather delicate. We postpone the proof until after seeing what it buys us­ the completion of the proof of Bandt 's theorem. So with the principal lemma in hand, let's show how to put Lemmas 8.4 and 8.6 into play. Corollary

8.8. The fractional Hausdorff measure is nontrivial; in fact, 0 < vh ( A ) < oo.

Proof. If E > 0, then we can choose t > 0 so that t < min { t0, E} and h is continuous at t. We can do this since h is monotone and so is continuous at all but countably many points of ( O, min { to , E}) . We can find an open set B in G with diam ( B ) :S t so that h ( t ) :S (1 + E) 2 >.(B) .

229

8. Metric Invariance and Haar Measure

How can we do this? Well if we pick t' is continuous at t,

<

t so that h(t' ) � h(t) and since h

h(t) < (1 + t)h(t1 ) , then we choose C so that diam ( C) � t' and h(t1 ) � (1 + t).A(C). Then t t' B = x E G : p (x, C) <

}

�

{

will do. By our Principal Lemma we have a functional covering of A: there is 91 , . . . , 9n E G and a 1 , . . . , an > 0 so that n XA � L ll!iXgi ·B i=l

with

,\( B) It follows that

(t, c>;) t, '.S

a; ,\( a; · B ) '.S (1 + ' ) ,\ ( A ) . n

h(t)E(t) � h(t) L ll!i

h(t)E(t)

i=l

<

n 2 L .A (1 + t ) ( U ) ai i=l

Since E > 0 was arbitrary, limt�inf h(t)E(t) � .A(A) < oo , O and so by Lemma 8.6, vh ( A ) < oo . We already know that 0 < v h ( A ) and so D by Note 8.3, vh is p-invariant with 0 < vh ( A ) < oo . Proof of the Principal Lemma-Lemma 8. 7. Since A is regular and A is compact, we can find an open set V such that V � { x E G : p ( x, A ) < b} that contains A, has V compact, and satisfies .A(V) � (1 + t) 1 13 .A(A). To see this, for each a E A, let Ua be an open set containing a such that Ua � Ua � { x : p ( x, A ) < b} . Let ai , . . . , an E A so that

230

8. Metric Invariance and Haar MeasurE

Then

V = Ua1 n · · · n Uan is open, contains A, V is compact, and V � {x : p(x, A) < b}. Let W = {x E U : p(x, A) < b/2}. Let f : G-+ [O, 1] be a continuous function that is one on A and vanishes outside of W. Choose a > 0 so that a [l + ( 1 + E) 1 13] < ( 1 + E) 1 1 3 - 1, that is, a < 1 - ( 1 + E) - 1 /3. Then l + a < ( l E) l /3 . + 1-a We appeal to Cartan's approximation scheme to get an open set Uo that contains G's identity e for which if

J I (g) - � Ci

9i

'


... , Cn

for all g E G. Let to < min{b/2, d( e, U0)} be a positive number. We'll show that this is the to claimed in the Principal Lemma. Let B be an open set with diam(B) :::; t0. Notice that if g E B, then g - 1 B is an open set that contains e and diam(g - 1 B) = diam(B) :::; t0, since d is left invariant. Each point of B is within to of g and so each point of g - 1 B is within to of e. Can any x E g- 1 B also be in U0? If we try to imagine such an x, then d(e, x) :::; to < d( e, U0), an impossibility. So (replacing B with g - 1 B if necessary) we can assume our B in the opening line of this paragraph contains e and is open with diam(B) :::; to and B � Uo. Now ..\ is inner regular so we can choose a compact C � B so ..\(C) is almost ..\(B) , that is, ..\(B) :::; ( 1 + E ) 1 13 ..\(C) . Suppose

... , Cn

231

8. Metric Invariance and Haar Measure

Alternatively,

n

f(g) a :S I::Ci g;
n

XA(g) a :S L Ci 9; b/2 � diam(B) , so n L di
<

Consequently,

ti=l di J . = J ti=l di. (1 + c) l /3 J xv d>. = (1 + c) l /3 >.(V) .

n

i=l <

n

>.(B) L di :::; (1 + €) 1 1 3 >.(C) L di i=l i=l (1 + c) 1 13 ¢ d>. di :S ( 1 + c) 21 3 >.(V) :::; ( 1 + c)>.(A) , i=l

J t

and that's that.

D

By Corollary 8.8, vh is a nontrivial left invariant Haar measure on G, so it is a multiple of >.. Since sets in G which are p-isometric have the same vh values, it follows that these D sets have the same left Haar measure. Proof of Bandt's Theorem-Theorem 8.2.

8. Metric Invariance and Haar Measure

232

1 . Notes and Remarks

Whenever G is a locally compact metrizable topological group, G has a base for its topology consisting of open sets with compact closure; the collection of open balls with respect to the metric generates G's topology also forms a base for its topology. When can one find a left-invariant metric generating G's topology all of whose balls have compact closure? Of course, for such a thing to be so, G must be separable. After all, G is the union of the n­ balls centered at the identity; if each of these have compact closure then it's easy to see that G has a countable dense subset-the union of the countable dense subsets of each n-ball will do. Our next result, also due to R. A. Struble, tells us that this is the whole story. Though the topic of this result of Struble is not germane to the study of Haar measure, it is simply too satisfying a result not to be included. Here's the theorem of R. A. Struble that led him to consider Theorem 8.1.

locally compact group metrizable topological group has a left invariant metric that generates its topology in which all its open balls have compact closure if and only if G satisfies the second axiom of count­ ability.

Theorem 8 . 9 . A

Let G be a locally compact, second countable {hence metriz­ able, separable) group. Then there exists a family {Ur : r > O} such that ( i ) for each r, each Ur is open and Ur is compact; ( ii ) ur = u-r i . ( iii ) UrUs � Ur+s ( so if r < s, then Ur � UrUr - s � U8), ; ( iv) {Ur : r > O} is a base for the open sets about e ; ( v) Ur >O Ur = G. Once Lemma 8.10 is established, we're ready for business. Indeed, let {Ur : r > O} be the family of open sets about e generated from Lemma 8.10. For x, y E G, set d ( x, y) = inf { r : y - 1 x E Ur } . • Since G = U r >O Ur , for a pair x, y E G, we have y - 1 x E Ur for some r > 0. It follows that d(x, y) � 0. • e E Ur for each r > 0 so d(x, x ) = 0. Lemma 8. 10.

l

1.

Notes and Remarks •

• •

233

If y- 1 x =I- e, then there is an ro > 0 so that y- 1 x 0. Ur = U; 1 so y- 1 x E Ur precisely when x- 1 y E Ur; consequently, d(x, y) = d(y, x) . Suppose x, y, z E G with y- 1 x E Ur, z- 1 y E Us. Then z - 1 x = z - 1 yy - 1 x E Us Ur � Ur + s , so d(x, z) :::; r + s . This is so whenever y- 1 x E Ur so d(x, z) :::; d(x, y) + s ; again this is so whenever z- 1 y E Us so d(x, z) :::;

d(x, y) + d(y, z) . • Finally, if x , y, z E G , then d(zx, zy) = inf{r : (zy) - 1 zx E Ur} = inf{r : y - 1 x E Ur} = d(x, y) .

To summarize: d is a left invariant metric on G. Since d(x, e) < r means x = e- 1 x E Ur, the open d-ball of radius r centered at e is contained in Ur. Also this same d-ball contains Ur' for any 0 < r' < r since if x E Ur' , then e- 1 x = x E Ur' � Ur' Ur - r' � Ur+r' , and so d(x, e ) :::; r�r < r. Therefore if 0 < r < r, then -2-

I

-2-

I

Ur' � {x : d(x, e) < r} � Ur. It follows that the open d-balls of radius r about e are cofinal with the collection {Ur : r > O} so the closure of each open d-ball is compact and d generates G's topology. Proof of Lemma 8. 10. Let p be the left invariant metric resulting from Theorem 8.1. We can assume that each of the open balls Br = {x E G : p(x, e), r} has compact closure for 0 < r :::; 2; after all, there is an ro so that for r < ro, Bro is compact by G's locally compact nature, so recalibrate p to make ro = 2 if necessary. For 0 < r < 2 we let Ur = Br. This assures us clearly of (iv) and since we'll keep these Ur's, (iv) is assumed henceforth. Also (i), (ii), and (iii) hold when r + s < 2 by p's left invariant metric nature.

8. Metric Invariance and Haar Measure

234

G is locally compact and satisfies the second countability axiom so G admits

a countable open base

E N} for its topology, where we can ( and do ) assume that W 2n is compact for each n. We define U2 = B2 n W2 . It's easy to verify that ( i ) and ( ii ) hold for 0 < r < 2 and if r + s < 2, then ( iii ) holds as well. We'll now inch our way from from (i ) , ( ii ) , and ( iii ) , (r + s � 2) holding for 0 < r � 2 to 0 � r � 4. First we have to define Ur for 2 < r < 2 2 . Let 0 < r < 2 2 . Set Ur = u Ut1 . . . Utm ' where the union extends over all t i , . . . , t m so that each t i satisfies 0 < t i � 2 and t i + · · · + t m = r. If 2 < r < 2 2 and t i + · · · + t m = r where each ti > 0, then there must be k, l E N so that 1 � k < l < m and t i + · · · + t k � 2 , t k+l + · · · + t1 � 2, and t1 + i + · · · + tm � 2. Why is this so? Well let k be the least j i so that t i + · · · + tj1 � 2, and let l be the least j2 so that tii + 1 + · · · + t32 � 2. Then l:h +l tj � 2 because otherwise, tjk+l + · · · + tm > 2 and t i + · · · + tj1 + 1 2'. 2 too, where J i + 1 < h + 1 . It follows that Ut1 . Ut2 . . . Utm c ( Ut1 . . . Ut k ) ( Ut k + l . . . Ut1 ) ( Ut1 +1 . . . Utm ) C Ut i+ .. +tk Utk +i + .. +t1 Ut1 +i+ ·+tm ( by ( iv )) c U2 U2 · U2 , so Ur � U2 · U2 · U2 whenever 0 < r < 2 2 . Now U2 is compact so U2 · U2 · U2 � U2 · U2 · U2 is also, and Ur is compact for 0 < r < 2 2 . Since { W2n :

n

·

we see that

u; i

i . . utm ) · LJ u i ( LJ ( Ui · · · Utm ) - i = LJ u;,;,i . . . u;- i LJ Utm · · · Ui = Ur

1.

Notes and Remarks

235

Finally if r > 0, s > 0 and r + s < 2 2 , then upon supposing t i , . . . , t m , T1 , . . . , Tj are positive and satisfying ti + · · · + t m = r, T1 + · · · + Tj = s , then (Ut . . . ut ) (U . . . u . ) = Ut . . . ut u . . . u . so l

m

Tl

T3

l

m

Tl

T3

Ur · Us � Ur + s ·

What if r, s > 0 and r + s = 2 2 ? Suppose r = s = 2 . Then Ur Us = U2 U2 . If 2 < r, then s < 2. But r < 2 2 so Ur � U2 U2 U2

and so Either way define

= (U2 U2 · U2 U2 ) U W22 . Here's what is so for 0 < r < 2 2 : • each Ur is open and U 2 is compact; • ur 1 = ur ,· 2 • if r, s > 0 and r + s :S 2 , then Ur Us � Ur s · + 2 We still have { Ur : 0 :::; r :::; 2 } as a basis for the topology of G about e and of course, W22 � U2 2 . We continue from U22 to U23 in a similar fashion and inch our way forward in a straightforward modification of the above procedure. The fact that at each stage we ensure W2n � U2n allows us to conclude that U22

-

n Ef\l

which is Lemma 8.lO(v) .

n Ef\l

D

(Braconnier [12] ) . If G is a locally compact topological group and G admits a bi-invariant metric d that determines its topology, th en G is unimodular.

Theorem 8.11

Let A be left Haar measure on G and assume that G is not unimod­ ular. Let U be an open set containing e such that A(U) < oo. Let B be an open ball centered at e (of radius p) so that B is compact, and B � U . Then for any x E G, the set xBx- 1 is just B, thanks to d's bi-invariance, so xBx - 1 � U. Proof.

236

8. Metric Invariance and Haar Measure

Let xo E G satisfy

�(x0 1 )>.(B) > >.(U) , where � is the modular function of G. Then xoBx0 1 � U and >.(xoBx0 1 ) = �(x0 1 )>.(xoB) = �(x0 1 )>.(B) > >.(U) . Oops!

D

8.12. The general linear group Q.C(2; JR) is unimodular (see §6.3) . For each m, let Example

Xm =

Then

( � � ) and

( � � ) E Q.C(2; JR) . l )(m � ) ( 1 0) ) ( ;;;, 0 0 i -+ 0 1 . ( � J'.: ) ( � � ) ( � � ) . Ym = _!_

T: m

But Ym Xm =

_!_

m

=

1

=

Therefore no bi-invariant metric can be found so that it generates the topology of Q.C(2; JR). An easy modification of this example shows the same for Q.C(n; JR), when n 2: 2 as well. Theorem 8.13. If G is a locally compact metrizable topological group and p is a left invariant metric that generates G 's topology, then (G, p) is a complete metric. Indeed if U is an open set with compact closure and if e E U, then there is an open ball B centered at e with B both compact and contained in U. Suppose R is the radius of B, and let (gn) be a p-Cauchy sequence in ( G, p) . Then there is an N E N so for m, n :S N, R p(gn , gm ) < 3 · It soon follows that for n 2: N, R p(gn , gN ) < 3 · Therefore for n 2: N, 9n lies in the compact, closed ball of radius � and so (gn) must converge. Observation 8.14. Let U be an open set in the topological group G with e E G, and suppose that K is a compact subset of G. Then there is an open set V in G with e E V so that

for every x E K.

1.

Notes and Remarks

237

To see this, let W denote the collection of all open sets W in G such = w- I . We claim that for any y E G there is a V E W so that if x E Vy, then xvx - I � u. In fact, we can pick Vi E W so that Vi · Vi · Vi � U, and we can pick Vi E W so that yViy - I � Vi . (This is thanks to the continuity of x--+ax --+axa- I for any a E G.) Let v = Vi n Vi. Then if x E Vy, Proof. that W

and and so

(

)

xvx - I � xVix I = (xy - I ) y(Viy - I ) (yx - I ) � VI · Vi · Vi � U. So for each y E K there is a Vy E W so that x E Vx Y implies xVyx- I � U. But K � LJ VyY, yEK and each Vyy is open; hence we can find YI , . . . , Yn E K so that K � Vy1 YI u . . . u Vyn Yn . Let

( )

( )

V = Vy1 n · · · n Vyn · Then if x E K, it must be that x E VykYk for some k, 1 :S k :S n, and so D xVx - I � xVyk Yk � U. Theorem 8.15. A compact metrizable topological group G admits a bi­ invariant metric that generates its topology. Proof. Let p be a left invariant metric on G that generates G's topology. For x, y E G define d(x, y) = sup { p (x z , y z ) : z E G}. Then d is finite for all x, y E G and is easily seen to be bi-invariant.

8. Metric Invariance and Haar Measure

238

Suppose E > 0. By our observation above, there is a 8 > 0 so that z- 1 {x E G : p(x, e ) < 8 } z � {x E G : p(x, e ) < E } , for all z. It follows that p(x, e ) < 8 ensures that d ( z- 1 xz, e ) = d ( xz, z ) < E for all z and so d (x, e ) :::; E. The open p-ball of radius 8 centered at e is contained in the closed d-ball of radius E centered at e. It is plain that the open d-ball of radius E centered at e is contained in the open p-ball of radius E centered at e. Therefore p and d generate the same D topology. ·

·

Moreover, we state the following: • If G is a topological group of the second category and H is a subgroup of G, then G \ H is either empty or of the second category in G. Let y E G \ H. Then yH E G \ H (distinct cosets are disjoint) . Therefore, should G \ H be of the first category, then so is yH, and from this we conclude that G = yH U ( G\ H ) is of the first category. • If G is a topological group of the second category and H is a dense 90 subgroup of G, then H = G. After all, H = nn Hn where each Hn is a dense open subset of G and so G \ Hn is closed and nowhere dense for each n. It follows that G \ H = U n (G\Hn ) is of the first category, and so by the previous remark, G \ H = 0 . (V. Klee [67] ) . Let G be a topological group with a bi­ invariant metric p which gen erates G 's topology. Suppose ( G, p) admits a complete metric d that generates G 's topology. Then G is actually complete under p.

Theorem 8.16

Let ( G* , p*) be the completion of ( G, p) . Then ( G*, p*) is a topo­ logical group into which ( G, p) is naturally isomorphically and isometrically embedded as a dense subgroup. But topological complete metric spaces are always 90 -sets in any super metric space, thanks to an oldie but goodie of D Sierpinski. Hence G = G* and so G is p-complete. Proof.

Chapter 9

St einlage on Haar Measure

1. Uniform Spaces: The Basics Definition 9.1. A uniformity for a set X is a nonvoid family U of subsets of X x X such that the following hold. ( i ) If U E U, then D. := {(x, x) : x E X} � U.

( ii ) If U E U, then u - 1 := {(y, x) : (x, y) E U} E U.

( iii ) If U E U, then there exists V E U so that V o V := {(x, z ) : (x, y), (y, z ) E V, for some y E X} � U. ( iv ) If U, V E U, then

u n v E U.

(v ) If U E U and U � V � X x X, then V E U. We then say that (X, U) is a uniform space. A subfamily B of a uniformity U is a base for U if each member of U contains a member of B. A base B for U determines membership in U: U E U precisely when there is a B E B so that B � U. -

239

9. Steinlage on Haar Measure

240

nonvoid family B of subsets of X x X is a base for some uniformity for X if and only if B satisfies the following: ( i ) Each member of B contains � . ( ii ) If U E B , then u- 1 contains a member of B . ( iii ) If U E B , then V o V � U for some V E B . ( iv ) The intersection of two members of B contains a member of B .

Theorem 9.2. A

Let (X, d) be a metric space, and let B consist of all subsets uf. of x x x where f. > 0 and UE := { (x, y) E X x X : d(x, y) < E } . B is a base for some uniformity for X, namely the metric uniformity. Example 9.3.

Example 9.4.

of the form

Let G be a topological group, and let BL consist of all sets { (x, y) E G x G : y E x U } ,

where U is an arbitrary neighborhood of the identity of G. BL is called the left uniformity of G. We can naturally consider the right uniformity and the two-sided uniformity of G. These examples are, in fact, the raison d'etre for uniform spaces; indeed, Andre Weil created theory of uniform spaces with an eye to providing a common generalization of metric spaces and topological groups. In so do­ ing, he provided a viewpoint from which to discuss Cauchy sequences/ nets, completeness, total boundedness, and uniform continuity. Definition 9.5. A family S is a subbase for the uniformity U when the family B of all finite intersections of members of S is a base for U. The next result provides us with enough firepower to recognize subbases when we see them.

Suppose S is a family of subsets of X x X that satisfies the ollowing conditions: f ( i ) each member of S contains �; ( ii ) if U E S, then u- 1 contains a member of S; ( iii ) if U E S, then there is a V E S so that V o V � U. Then S is a sub base for some uniformity for S. Theorem 9.6.

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Uniform Spaces: The Basics

241

The union of any nonvoid collection of uniformities for X is a sub base for a uniformity for X . Suppose ( X, U ) is a uniform space. The uniform topology is the collection of all T � X such that for each x E T there is a U E U so that U[x] = {y E X : ( x, y) E U} � T. The uniform topology is a topology. Corollary 9.7.

Let A � X. Then the interior of A relative to the uniform topology is the set {x E X : U[x] � A for some U E U}.

Theorem 9.8.

If 0 is an open set that is contained in A, it's because for each x E 0 there is a U E U so that U[x] � 0 � A. It follows that the set B = {x : U[x] � A for some U E U} contains each and every open set 0 for which 0 � A. To see that the set B is open, note that if x E B, then by definition there is a U E U so that U[x] � A. With U we an find a V E U so that V o V � U. Notice that if y E V[x] , then (x, y ) E V, and it follows that V[y] = { z : (y, z ) E V} � { z : (x, y) , (y, z ) E V, for some y E X} = ( V o V ) [x] � U[x] � A. Therefore if y E V[x] , then V[y] � A, and so V[x] � B. In other words, if x E B, then there is a V E U so that V[x] � B. Thus B is open in the uniform topology. Since B is open and contains every open set that is contained in A, B is A's 0 interior. Proof.

So given a uniform space ( X, U ) , U[x] is a neighborhood of x for each U E U, and the family {U[x] : U E U } is a base for the neighborhood system of x. It's an easy consequence of this and the definition of a base for U to see that if B is a base for U, then for any x E X, {U[x] : U E B} is a basis for the neighborhood system of x in the uniform topology. Theorem 9.9. If U E U, then the interior of U (in the product topology) is also in U. Consequently, the family of all open symmetric members of U is a base for U.

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242

We will rely on the following simple features of symmetric sets. Lemma 9.10.

u � x x x,

If V � X x X is symmetric (i . e . , V = v - 1 ) , then for any V o U o V = LJ V[x] x V[y] . (x,y) E U

Proof.

VoUoV

{(u, v) : (u, x) E V, (x, y) E U, (y, v) E V, for some x, y E X} = LJ {(u, v) : (u, x), (y, v) E V} (x,y) E U

LJ

{(u, v) : (x, u), (y, v) E V} (by V's symmetry)

(x,y) E U

LJ V[x] x V[y] , (x,y) E U

and we have proved the lemma.

0

Now to prove the theorem, note that the interior of a set M � X x X is the set (x, y) : U[x] x V[y] � M for some U, V E U But once U, V E U, we have W = U n V E U also, so M's interior in X x X is {(x, y ) : W[x] x W[y] � M for some W E U}. But W E U so there is a W E U so W o W o W � W, where W is symmetric (this takes a moment or two of reflection on the consequences of the axioms of uniformities). Now our lemma steps in to tell us that W o W o W = LJ W[x] x W[y] .

}·

{

(x,y) E W

Each point of W is an interior point of W, so W's interior contains W, which belongs to U. It follows that the interior of W is in U as well. To see that symmetric members of U form a base for U is an easy conse­ quence of the above. After all, we showed that if U E U, then U's interior is the set {(x, y) : W[x] x W[y] � U for some symmetric W E U}.

1.

Uniform Spaces: The Basics

243

But if U is itself symmetric then this set is itself symmetric and open and in U. Of course being U's interior put this set squarely inside of U, too. D

Let A � X . Then the uniform closure of A is precisely nuEU U[A] ' where U[A] = {y E X : ( x, y) E U for some x E A}. Theorem 9.11.

Note that x E A if and only if U[x] n A f= 0 for each U E U. But U[x] n A f= 0 if and only if x E u- 1 [A] . Since each member of U contains a symmetric member of U, x E A if and only if x E U[A] for each member

Proof.

D

U E U.

Definition 9.12. A

for each V E V,

function f :

(X, U) -+ (Y, V)

is uniformly continuous if

{ (x, y ) : (f(x) , f(y)) E V} E U. Note that in testing for uniform continuity, one need only test for V's in a base for V. So f : (X, U)-+� is uniformly continuous if and only if for each E > 0 there is a U E U so l f(x) - f(y) I < E if (x, y) E U. The composition of two uniformly continuous functions is uniformly contin­ uous as usual. Each uniformly continuous function is continuous with respect to the uniform topologies.

Theorem 9. 13.

Suppose f : (X, U) -+(Y, V) is uniformly continuous, and let 0 be a neighborhood of f(x) . Then there is a V E V so that V[f ( x ) ] � 0 and f +--- ( V[ f(x) ] ) = {y : f(y) E V[f(x) ] } = {y : (f(x), f(y)) E V}. If we consider the induced map F : X x X-+Y x Y, F(x 1 , x 2) = (f(x 1 ) , f(x 2)) , then it's plain that f 's uniform continuity says that for each V E V there is a U E U so F ( U ) � V. But {y : (f(x) , f(y)) E V } is naught else but F+--- ( V ) [x] , and this is a neighborhood of x. So r- (o) is a neighborhood of x for any neighborhood 0 of f(x) and f is continuous. D Proof.

If Y � X and (X, U) is a uniform space, then the inclusion of Y into X induces a smallest uniformity making this map uniformly continuous; this is the relative uniformity. It consists of subsets of Y x Y of the form (Y x Y) n U,

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9. Steinlage on Haar Measure

where U E U. Naturally the relative uniform topology on Y is the uniform relative topology as you would want it to be. If { (Xa , Ua ) : a E A} is a family of uniform spaces, then we introduce a uniformity on IIXa by looking at the smallest uniformity such that Pao : IIXa-tXa0 is uniformly continuous for each ao E A; it has as a subbase the sets of the form p1; ( Ua ) where Ua E Ua . The resulting uniformity if the product uniformity. It's plain and easy-to-see that the uniform product topology is the product uniform topology-again, things behave. More is so, and we record this thusly:

The topology of the product uniformity is the product topol­ ogy. A function f on a uniform space (into a product of uniform spaces) is uniformly continuous if and only if the composition of f with each coordinate projection is uniformly continuous. Theorem 9.14.

Theorem 9.15. Let (X, U) be a uniform space, and let d be a pseudo-metric for X . Then d is uniformly continuous on X x X relative to the product uniformity if and only if the set { (x, y) : d(x, y) < r} is a member of U for each r > 0 .

We need to show that d's uniform continuity is equivalent to Vd,r = { (x, y) : d(x, y) < r} E U for each r > 0. Preface. Let U E U. Then the sets {(x, y ) , (u , v ) : ( x, u) E U} , { (x, y) , (u, v) : (y, v) E U} are preimages under the coordinate projections, and so each belongs to the product uniformity. In fact, the family of sets of the form { ( (x, y) , (u, v)) : (x, u) E U, (y, v ) E U}u EU is a base for the product uniformity. Since d is uniformly continuous, for each r > 0 there is a U E U so that if (x, u) , ( y, v) E U, then j d(x, y ) - d(u, v) I < r. But (u, v) = (y, y) is always in U so this ensures us that if (x, y) E U, then d(x, y) = j d(x, y) - d(y, y) i < r, and U � Vd,r, and so Vd,r E U. Conversely, notice that if (x, u) , (y, v) E Vd,r i then j d(x, y) - d(u, v) I < 2r. Proof.

1.

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Uniform Spaces: The Basics

To see this, note that d ( x , y ) :S d ( x , u) + d(u, v) + d (v, y) and d(u, v) :S d(u, x ) + d( x , y) + d(y , v) , which after an inspection tells us d( x , y) - d(u, v ) :S d ( x , u) + d(v, y) and d(u, v) - d (x, y) :S d(u, x ) + d(y , v) , which thanks to the symmetry of d says l d( x , y) - d(u, v) I :S d ( x , u) + d(v, y ) < r + r = 2r. D So if Vd,r E U for each r > 0, then d is uniformly continuous. Lemma 9.16 ( Metrization Lemma) . Let ( Un) be a sequence of subsets of X x X such that Uo = X x X , each Un contains the diagonal !::l. , and

Un+l Un+l Un+l � Un for each n . Then there is a nonnegative real-valued function d on X x X such that ( i ) d ( x , z ) :S d (x, y) + d(y , z ) for all x, y, z, and ( ii ) Un � { ( x, y) : d( x , y) < 2 -n } � Un - 1 for all n 2:: 1 . If each Un is symmetric, then d can be chosen to be a pseudo-metric satis­ fying (ii). o

Proof

and

o

( Outline ) . Define f : X x X---+ JR by n precisely when ( x, y) E Un 1 \ Un f( x , y ) = 02- when ( x , y) E nn un

{

d ( x, y) = inf

{ t. I / (x; , Xi+ i) I

'

}

x = xo, y = x..,_, .

It is easy to see that d satisfies the triangle inequality so we have ( i ) . Since d(x, y) :S f(x, y), we see that Un � { ( x, y) : d ( x , y) < r n }. Note here that if each Un is symmetric ( i.e., Un = Un - 1 ) , then f(x, y) = f(y , x ) and so d ( x , y) = d(y, x ) so d is a pseudometric. What remains? Well, we'll show that n f( xo , Xn+ i) :S 2 L f(xi, XH 1 ) . ( 52 ) i=O From this, it follows that if d(x, y) < 2 -n , then f(x, y) < 2 · 2 -n , and so

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( x, y ) E Un - l thereby ensuring that {(x, y ) : d ( x, y) < r n } � Un - l i and ( ii ) holds. If n = 0, then (52) is plain. Let n > 0, and let k be the largest integer such that n k- l L f(xi, Xi+l) ::; 2l L f(xi, Xi+ i) · i=O i=O Of course, and so Since we see that

n lL 2 f( xi , XH 1 ). i =O n f(x k , Xk + 1 ) ::; L f(xi, Xi+ i), i=O

n L f(xi, XH 1 ) ::; i=k+l

k- l n f (xo, Xn + i) < L f (xi, XH 1 ) + f (x k , Xk+ i) + L f ( xi , Xi+ i) i =k+l i=O n l n n < 2 L f(xi, XH 1 ) + L f( xi , XH 1 ) + 2l L f( xi , Xi + i) i =O i =O i =O n = 2 L f( xi , XH 1 ) , i=O and (52) holds.

D

Of course, if a uniformity U for X has a countable base ( Vn)n ;o:: 1 , then the axioms of a uniformity allow us to construct symmetric members ( Un)n > O of U such that Un � Vn and ( Un)n ;o:: o satisfy the hypotheses of Lemma 9.16, the metrization lemma. So we get that a uniformity with a countable base is a pseudo-metrizable uniformity. Of course the converse is also so: if d is a pseudo-metric, then { (x, y) : d ( x, y) < l/n} is a countable base for the pseudo-metric uniformity. Thus we have

uniform space is pseudo-metrizable if and only if its uniformity has a countable base.

Theorem 9 . 1 7. A

2. Some Miscellaneous Facts and Features about Uniform Spaces

247

Keep in mind our characterization of those pseudo-metrics that are uni­ formly continuous with respect to the product uniformity: if (X, U) is a uniform space and d is a pseudo-metric on X x X, then d is uniformly con­ tinuous of X x X relative to the product uniformity if and only if for each r > 0 the set { (x, y) : d (x, y ) < r} E U. So suppose ( X, U) is a uniform space, and let P be the family of all pseudo­ metrics on X that are uniformly continuous on X x X (with respect to the product uniformity, of course ) . The uniformity generated by P is no bigger than U by our characterization of uniformly continuous pseudo-metrics. On the other hand, our metrization lemma assures us that given a member U E U there is a member d E P such that { (x, y) : d(x, y) < 1/4 } � U, and so U is no larger than the uniformity generated by P. In other words, we have the following. Theorem 9. 18. Each uniformity for X is generated by the family of all pseudo-metrics that are uniformly continuous on X x X . A bit of standard argumentation soon reveals the following.

Weil ) . Each uniform space is uniformly isomorphic to a subspace of the product of pseudo-metric spaces. Consequently, a topology r for a set X is the uniform topology for some uniformity for X if and only if r is a completely regular topology. (Warning: "Completely regular" is sans Hausdorff! ) Corollary 9.19 (A.

2 . Some Miscellaneous Facts and Features about Uniform Spaces

We don't propose to study uniform spaces in detail, but since we mentioned "Cauchy sequences /nets" and "completeness" we feel obligated to at least define these terms and state a few of the results of immediate interest to analysts. Definition 9.20. A net (xa) v in a uniform space (X, U) is a Cauchy net if given U E U there is a du E D so that if d, d1 2". du, then (xd , xd' ) E U. Convergent nets relative to the uniform topology are Cauchy nets. (X, U) is complete if Cauchy nets are convergent.

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248

It is important to understand that completeness entails nets and not just sequences. This distinction is alive and dangerous to the ill informed, es­ pecially in analysis. For instance, consider an infinite-dimensional Banach space X in its weak topology. The associated uniformity is generated by the collection of sets of the form Uxi 1 ,xii;€ = { (x, y ) : l x i ( x - y) I , . . · , l x� (x - y) I < E } , where x i , . . . , x� E X * , E > 0. It's a fact that an infinite-dimensional Banach space in its weak topology is never complete. All is not lost though because it is a phenomenon of­ ten (but not always) encountered that a Banach space in its weak unifor­ mity be sequentially complete; indeed, in addition to the reflexive spaces, L 1 (µ) , L 1 (1f)/ H 1 , H 1 ('JDJ) *, C(S)*, Orlicz spaces generated by � 2 -formation, preduals of von Neumann algebras, find themselves enjoying the weak com­ pleteness of weakly Cauchy sequences. Again if X is a Banach space and X* is its dual then the weak* -uniformity is generated by sets of the form •••

where x 1 , . . . , Xn E X, E > 0. Remarkably enough, we again have that if X is an infinite-dimensional Banach space, then its dual space X* with the weak* -uniformity is never complete but always sequentially complete. 3.

Compactness in Uniform Spaces

If (X, U) is a compact uniform space, then every neighborhood of the diag­ onal � in X x X is a member of U. Let B be the family of closed members of U, and let V be an open set in X x X that contains �. Suppose ( x, y) E nuE B U. Since B is a base for U, y E U[x] , we see that y belongs to every open set in X that contains x. Hence (x, y) lies in any open set in X x X that contains �. Therefore

n u � v.

UEB

Claim. There exists U1 , . . . , Un E B so U1 n · · · n Un � V. Suppose not. Then for any U1 , . . . , Un E B , we can find a point

3. Compactness in Uniform Spaces

With this natural ordering

( U1 , . . . , Un) 2: ( U� , . . . , u:n) when we get a net in X : ( xu1 ,

•

•

•

, xun )

XE

249

n

n Ui �

l=l

m

n u; ,

j=l

which necessarily has a limit point

n U � V.

UEB

But x E V says for ( Ui , . . . , Un ) big enough, and so there are so U1 n · · · n Un � V, and as a result V E U. It follows that

n u�v

UEB

But each member U of B is closed in the compact space X, and so each is compact and V is open. Thus the intersection of some finite subfamily of B is also a subset of V, and from this we see that V itself belongs to U. If d is a pseudo-metric on X that is continuous on X x X, then for any E > 0 { (x, y) : d (x, y) < c} is an open set that contains the diagonal. This is tantamount to d being uniformly continuous. Therefore if (X, U) is a compact uniform space, then every continuous f : (X, U) --+R is uniformly continuous. Look at ( !, f) where (!, f) : X x X--+lR x JR; if we look at d, d(x, y) = l f( x ) - f(y) I , then d is a continuous pseudo-metric on X x X , and so is uniformly contin­ uous. A quick check shows that this is just f's uniform continuity.

If (X, U) is a compact uniform space and f : X--+lR is continuous, then f is uniformly continuous. Indeed d(x, y ) = l f( x ) - f(y) I is a continuous pseudo-metric on X x X , and hence uniformly continuous.

Corollary 9.21.

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4. From Contents to Outer Measures

Let S be a locally compact Hausdorff space. Denote by C(S) the collection of all compact subsets of S. A content is a set function .A on C (S) which satisfies (i) O s .A(C) < oo; (ii) if C1 , C2 E C (S) with C1 � C2 , then .A(C1 ) s .A(C2 ); (iii) if C1 , C2 E C (S) , then .A(C1 U C2 ) S .A(C1 ) + .A(C2 ) ; (iv) if C1 , C2 are disjoint members of C(S) , then .A(C1 U C2 ) = .A(C1 ) + .A(C2 ) . Since .A(0) + .A(0) = .A(0 U 0) = .A(0) < oo , .A(0) = 0. A content .A induces .A * on the topology of S: .A * (U) = sup{.A(C) : U :J C E C(S) } . Definition 9.22.

The function .A* vanishes at 0, is monotone nondecreas­ ing, countably subadditive, and countably additive {on disjoint open sets, of course).

Theorem 9.23.

Proof. The fact that .X * (0) = 0 and the monotone nature of .A * are both plain. Suppose U, V are open and C is a compact subset of U U V. Then C \ U an� C \ V are disjoint compact sets in S, and so there exist disjoint open sets U and V such that C \ U � fJ and C \ V � V. Let D = C \ fJ and E = C \ V. Then D � U, E � V, and both D and E are compact. Since fJ n V = 0, D u . E = ( c \ fl) u ( c \ V ) = c \ ( fJ n V ) = c. Now

and so .A * (U U V) = sup{.A(C) : C � U U V } S .A * (U) + .X * (V) , and .A * is (finitely) subadditive. If (Un ) is a sequence of open sets in S and C is a compact set such that C � U n Un , then there is an N so that N

C � LJ Un . n= l

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4. From Contents to Outer Measures

It follows that

Therefore, since C E C is an arbitrary compact set contained in Un Un ,

and >. * is countably subadditive. Next suppose that U and V are disjoint open sets, and let C, D be compact sets such that C � U and D � V. Then C U D � U U V, and so It follows that and >. * is additive on disjoint open sets. Of course if U1 , . . . , Un , . . . is a sequence of pairwise disjoint open sets, then for any N >. * It follows that

N

N

( y Un) � >.* ( � Un) = � >.* (Un ) · n l

( )

>. * LJ Un � L >. * (Un ), n n and so the proof is complete.

D

From >. we pass to >.* and now we define µ * µ* (E) = inf {>. * (U) : E � U, U is open }. Theorem 9.24.

µ* is an outer measure.

Since 0 is open and >. * (0) = 0, µ* (0) = 0. Clearly µ* is monotone nondecreasing. If (En ) is a sequence of subsets of S with each having µ* (En ) < oo, then given an € > 0 there is an open set Un containing En and such that µ* (En ) '.S >. * (Un ) '.S µ * (En ) + : · Proof.

2

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9. Steinlage on Haar Measure

It follows that

( ) (

A * LJ Un :::; L A * (Un ) n n < L µ * (En ) + 2: n and µ* is countably subadditive.

)

n D

A set E is µ* -measurable if and only if µ * (U) 2: µ * (U n E) + µ* (U n Ee) (5 3 ) for each open set U � S. In fact, take any A � S, and suppose A � U, U open. Then assuming (53) , .\* (U) = µ * (U) 2: µ* (U n E) + µ * (U n Ee) 2: µ * (A n E) + µ* (A n Ee). It follows that µ * (A) = inf{.\* (U) : A � U, U is open } 2: µ * (A n E) + µ * (A n Ee). Naturally the reverse inequality follows from µ * subadditivity and the very D definition of µ* -measurability. Fact 9.25.

Fact 9.26.

Proof.

If C E C (S) , then µ * (int(C)) :::; ,\(C) :::; µ* (C).

We first show that

µ * (U) = ,\ * (U) . Since U � U it follows that µ* (U) :::; A * (U) . On the other hand, if V is open and U � V, then .\ * (U) :::; A * (V), and so .\ * (U) :::; inf{A * (V) : U 2 V, V is open} = µ * (U) . There is only one conclusion: If C E C (S) and U is open with C � U, then .\(C) :::; .\ * (U) so .\(C) :::; inf{A * (U) : C � U, U is open} = µ* (C).

4. From Contents to Outer Measures

If C, D E C ( S ) and D � int(C), then >.(D) :::; >.(C), and so µ* (int(C)) = >. * (int(C)) = sup{A(D) : D � int(C)} :::; >.(C). In summary, µ* (int(C)) :::; >.(C) :::; µ * (C).

253

D

Fact 9.27. Compact sets are µ*-measurable.

Let C E C ( S ). We need to show that for any open set U, µ* (U) 2:: µ * (U n C) + µ * (U n cc). Let D be a compact subset of U n cc , and let E be a compact subset of U n D c . Then U n Cc and U n D c are open and D n E = 0 with D U E � U. so µ * (U) = >. * (U) 2:: >.(D U E) = >.(D) + >.(E) . It follows that µ* (U) > >.(D) + sup{A(E) : E � U n Dc} >.(D) + >. * (U n Dc) >.(D) + µ * (U n Dc) > >.(D) + µ * (U n C). In turn, this says µ * (U) > µ * (U n C) + sup{A(D) : D � U n cc} µ * (U n C) + >. * (U n cc) µ * (U n C) + µ* (U n cc) , D and D is µ*-measurable. Proof.

Here's an important feature of µ*: µ* (C) < oo for each C E C ( S ) . In fact, we know that there is a compact set F � S so that C � int(F) � F. It follows that µ* (C) :::; µ * (int(F)) :::; >.(F) < oo . Finally we take note of the following. Theorem 9.28.

content, then

If

. is a

>.( C) = >.( ¢( C)) is a content too. If µ; is the outer measure generated by A, then µ; = µ * (¢(E)). So if >. is invariant under ¢ , then so is µ *.

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254

5.

Existence of G-invariant Contents

Let X be a locally compact Hausdorff space, and let G be a group of homeomorphisms of X onto itself. We say that G is transitive if given x, y E X there is a g E G so that gx = y. We say that G is weakly transitive if for any nonempty open set U � X, {gU : g E G} covers X. We say that G separates compact sets if given a pair A, B of disjoint compact sets in X and g E G, there is a nonempty open set U � X such that gU never meets both A and B. Definition 9.29.

Let X = �. and let .G be translation by rational numbers. Then G is a weakly transitive group of homeomorphisms from X onto itself that is not transitive. Here's the first central result of Steinlage; it pertains to the existence of a G-invariant measure. The attentive reader will soon realize that this proof is modeled on Banach's proof. Example 9.30.

Let G be a weakly transitive group of homeomorphisms of X onto X, and suppose that G separates compact sets. Then there is a content A on X such that A is G-invariant and A(C) > 0 whenever C is a compact subset of X with nonempty interior. Recall that a content is a nonnegative real-valued map defined on C(X) , the compact subsets of X, which is monotone nondecreasing, subadditive, and additive on disjoint sets. Lemma 9.31.

To build A we proceed by stages. Stage 1 . Fix xo E X, and let n (±o) = { U � x : Xo E u, u open}. Let A and B be disjoint compact sets in X. There exists a nonempty set V such that for any g E G, gV can intersect A or B-but not both. Since G is weakly transitive, there is a g E G so that gV contains x0 . Therefore gV E O(xo) and gV cannot intersect both A and B. In other words, we can separate compact sets by members of O (xo) . Let E be a compact subset of X and B a subset of X with nonempty interior. Define h(E, B) = the least number of images g B needed to cover E with g E G. Plainly, h(E, B) is well defined and 0 ::::; h(E, B) < oo . If E =I= 0, then h ( E, B) � 1 . Proof.

·

5. Existence of G-invariant Contents

255

More is so: if E, E1 , E2 are compact subsets of X, then h ( E1 , B) .S h ( E2 , B) whenever E1 � E2 ; h (gE, B) = h ( E, B) for every g E G; h ( E1 U E2 , B ) .S h ( Ei , B ) + h ( E2 , B) ; if D is compact with nonempty interior, then h ( E, B ) .S h ( E, D ) · h (D , B) ; if E1 n E2 = 0, then there is a Bo E O ( x o ) such that for any B E O(x0) with B � Bo, h ( E1 U E2 , B ) = h ( E1 , B ) + h ( E2 , B ) .

Fact 9.32.

(i) (ii) (iii) (iv) (v)

As one might expect, all but Fact 9 . 3 2 (v) are self-explanatory and easily established; Fact 9 . 3 2 (v) needs some work. We know by the fact that G separates compact sets that there is a Bo E O ( xo ) so that gBo never meets both E1 and E2 . Let n = h ( E1 U E2 , Bo) . Then there are points 91 , . . . , 9n E G so that

i=l l) {giBo : 9iBo n E1 =I= 0} u l){giBo : gi Bo n E2 =I= 0} .

(

i

) (

i

)

If 1 .S i .S n, then either gi Bo n E1 f= 0 or gi Bo n E2 =I= 0, but not both. Now {giBo : gi Bo n E1 =I= 0} must cover E1 , and {giBo : 9iBo n E2 f= 0} must cover E2 . So n � h ( E1 , Bo) + h ( E2 , Bo) , and Fact 9.3 2 (v) is established. Stage 2. Fix a compact set Ao � X witP. a nonempty interior. For each compact E and for each U E O ( xo ) define E, U) . lu ( E ) = hh (Ao, ( U) The attentive reader will realize that a similar normalization occurred in Banach's presentation and in the Cartan-Weil construction. It goes back, in fact, to Haar's original ground-breaking work. Here's the truth: If E, E1 and E2 are compact sets in X, then (i) if E1 � E2 , then lu ( E1 ) .S lu ( E2);

Fact 9.33.

9. Steinlage on Haar Measure

256

(ii) lu(gE) = lu(E) for g E G; (iii) lu(E1 U E2 ) ::; lu(E1 ) + lu(E2 ); (iv) 0 ::; lu(E) ::; h(E, Ao) and if E has nonempty interior, then 1 h(Ao, E) ::; lu(E) ::; h(E, Ao) ; (v)

if Ei and E2 are disjoint compact sets, then there is a Uo E n(xo) such that lu(E1 U E2 ) = lu(E1 ) + lu(E2 ) for any U E n(x o ) with U � Uo; (vi) lu(0) = 0 for all U E n(xo) . Stage 3. Let E be a compact set in X. Set 0 if E has empty interior 1 h ( Ao ,E) if E has nonempty interior. Then K = II [8E , h(E, Ao )] E EC ( X )

is a compact Hausdorff space. For U E n(xo ), we let lu be the point lu := (lu(E) : E E C(X)) E K. For each V E n(xo ) , let A(V) = {lu : U E n(xo), U � V } � K. The family {A(V) : V E n(xo)} has the finite intersection property since the finite intersection of members of n(xo ) is also in n(xo ). So by K's compactness there is a >. >. E n A(V) . VE!1(:z:o)

This >. does the trick.

D

What does >. do? To start, >. E K so for any E E C(X), 8 E ::; >.(E) ::; h(E, Ao) . It follows that 0 ::; >.(E) < oo for any E E C(X). Since >. lies in each A(V) , V E n(xo) , and A(V) consists of all the points lu where U E n(xo ) satisfies

257

5. Existence of G-invariant Contents

U � V. Hence the facts that each lu is G-invariant and monotone nonde­ creasing soon reveal that the same is so for >.. Again each lu is subadditive, and so the same holds for all >.. Finally, if E1 and E2 are disjoint compact sets, then there is a Uo E O ( xo) such that for any U E O ( xo) with U � Uo , >.u(E1 U E2 ) = >.u(E1 ) + >.u(E2 ). Since >. E n A(V) , VE n (xo)

this additivity passes to >.. An aside:

Proposition 9.34. Let G be a group of homeomorphisms from the topolog­ ical space X onto itself. Then G is weakly transitive if and only if for each x E X, the orbit Gx := { g x : g E G} of x is dense in X .

Suppose G is weakly transitive but there is an xo E X such that Gxo is not all of X. Look at the open (nonempty) set X \ Gxo . By weak transitivity there is a g E G so that Proof.

xo E

It follows that

g

(

X \ Gx o

).

-g - 1 xo E X \ Gxo.

Oops! Now suppose that G is not weakly transitive. Then there is a nonempty open set U so GU := LJ gU =/= X. gEG

Plainly, gGU = GU for each g E G so g ( X \ GU) = X \ GU for each g E G. If x E X \ GU, then Gx � X \ GU, and Gx is not dense.

D

Recall that if ( X , U ) is a uniform space, then the uniform topology is the collection of all T � X such that for each x E T there is a U E U such that U[x] := {y E X : (x, y) E U} � T.

9. Steinlage on Haar Measure

258

Definition 9.35. Let F be a family of functions from a topological space X into a uniform space (Y, V) . Suppose x E X. We say that F is equicontinuous at x E X provided, given V E V, there is an open set 0 � X containing x such that f(O) � V[f(x)] for each and every f E F. We say that F is equicontinuous if F is equicon­ tinuous at each point x E X; i.e., regardless of x E X, given V E V there is an open set 0 � X that contains x and satisfies f(O) � V[f(x ) ] for each f E F. Equicontinuity provides new insights into weak transitivity.

Suppose ( X, U ) is a uniform space and G is an equicontinu­ ous group of homeomorphisms from X (in its uniform topology) onto itself. If Gxo is dense for some xo E X, then G is weakly transitive.

Lemma 9.36.

Suppose that G is not weakly transitive. Then there is an x E X and a nonempty open set U � X so that x
gEG

Since Gxo is dense in X, there is a 9 1 E G so that Gxo n 91 U f= 0. Hence ( again by Gxo ' s by density ) there is a 92 E G so that 92 xo E 9 1 U, and so Xo E 92- 1 91 u·

Xo E 90U. There is a symmetric V E U so that V[xo] � 90U. By equicontinuity there is an open set 0 � X that contains x such that 9 0 � V[9x] for all 9 E G. But since x
it follows that

Gx n 90U = 0.

5. Existence of G-invariant Contents

259

Since V[xo] � goU,

Gx n V[xo ] = 0. It follows that for any g E G, xo � V[gx] ; otherwise, if xo E V[gx] , then (gx, xo) E V, which by symmetry tells us (xo, gx ) E V or gx E V[xo] yielding, Gx n V[xo] =/:- 0, which is impossible. By a similar argument it follows that since xo � V[gx] for all g E G, Gxo n V[gx] = 0 for each g E G, and since g O � V[gx] for all g E G,

Gxo n

( LJ gO) = 0, gEG

contradicting the density of the orbit Gxo of xo. Our conclusion is clear: G must be weakly transitive.

D

We can lighten the load of equicontinuity on G's shoulders. Here's how.

Let ( X, U ) be a uniform space, and let G be a group of homeomorphisms of X (in its uniform topology) onto itself. Suppose that there is an xo E X such that G is equicontinuous at xo and for any open set U containing xo, GU = X. Then G is equicontinuous.

Lemma 9.37.

Fix U E U. There is a symmetric V E U such that V o V � U. Since G is equicontinuous at x 0 there open set 0 � X, containing xo such that gO � V[gxo] for all g E G. Fix x E X. We'll show that G is equicontinuous at x. Since GO = X, there is a g0 E G so that x E goO. Fix g E G. Therefore ggoO � V[ggoxo] , so that gx E ggoO � V[ggoxo] . Since (54) (ggoxo, gx ) E V, Proof.

9. Steinlage on Haar Measure

260

by symmetry If y E goO, then as

above ( by ( 54))

Since

(gx, ggoxo) E V. gy E 99o 0 � V[ggoxo] ; (ggoxo, gy) E V. (gx, ggoxo), (ggoxo, gy) E V,

we have

(gx, gy) E V V � U. All of this is true for any arbitrary g E G. Therefore, for any g E G, if y E goO, then (gx, gy) E U; i.e., if y E goO, then gy E U[gx] for all g E G. Therefore for any g E G, ggoU � U[gx] , and G is equicontinuous at x. o

D

From uniform spaces, we borrow the following Urysohn-like result. Lemma 9.38

( Uniform Space Lemma) . Let (X, U) be a uniform space, and

let A, B be disjoint compact subsets of X (in its uniform topology). Then there is a symmetric U E U such that for any x E X either U[x] n A = 0 or U[x] n B = 0. Of course the key use of this lemma is to lighten the load of establishing that a group of homeomorphisms on X separates the compact subsets of X. Suppose (X, U) is a uniform space whose uniform topol­ ogy is locally compact and Hausdorff. Let G be an equicontinuous group of homeomorphisms of X onto itself. Then there is a nonzero G-invariant content on X .

Theorem 9.39.

Proof.

Let xo E X, and set

Xo = Gxo. Then Xo is a locally compact Hausdorff space in its relative topology which is the uniform topology generated by the relative uniformity Uo on Xo x Xo inherited from U, and G is an equicontinuous group of homeomorphisms

6. Steinlage: Uniqueness and Weak Transitivity

261

taking Xo into Xo. Of course the orbit of xo is dense in Xo, and so by Lemma 9.36, G is weakly transitive in Xo . Thus in order to use Lemma 9.31 , we just need to show that G separates compact sets of Xo. If C and D are disjoint compact subsets of Xo, then by the Uniform Space Lemma, we can find a symmetric V E Ux0 such that for any x E Xo, either V[x] n C = 0 or V[x] n D = 0. By equicontinuity, there is an open set 0 containing xo such that gO � V[gxo] for any g E G. It follows that either gO n C = 0 or gO n D = 0 for each g E G. Thus G separates compact sets in Xo. We can appeal to Lemma 9.31 to conclude that there is a nonze!o G-invariant content A on Xo. We can extend A to a G-invariant content A on X simply by defining .>:(K) = A (K n Xo) D for any compact set K � X. 6. Steinlage: Uniqueness and Weak Transitivity

(Steinlage). Let G be an equicontinuous group of homeo­ morphisms of the nonempty locally compact space X onto itself, where X 's topology is generated by the uniformity U . Then the G-invariant Haar mea­ sure on X is unique if and only if G is weakly transitive. On the other hand we have Theorem 9.40

Let ( X , U ) be a uniform space whose uniform topology is locally compact and Hausdorff, and assume that G is an equicontinuous group of homeomorphisms of X onto itself. If G does not act in a weakly transitive manner on X, then the G-invariant content of Theorem 9.39 is not unique. Note 9.41.

Proof. If G is not weakly transitive on X, then by Lemma 9.36 no point of X has dense orbit. Let xo E X, and set Xo = Gxo. As in the proof of Theorem 9.39, Xo is a locally compact Hausdorff space in its relative topology which is the uniform topology generated by the relative uniformity Uo on Xo x Xo inherited from U and G acts as a weakly transitive, equicontinuous group of homeomorphisms of X0 onto itself (again G's weak transitivity holds on Xo through the good graces of Lemma 9.36) . Hence by

262

9. Steinlage on Haar Measure

Lemma 9.31 there is a nonzero G-invariant content Ao on Xo which extends to all of X to a content .:X0 where values at a C E C(X) are given by .:X o (C) = A(C n Xo). Now Gxo =/= X, so pick a point x 1 E X \ Gx o , and look at X1 = Gx 1 . Again as in the proof of Theorem 9.39, we get a locally compact Hausdorff space in the relative topology which is the uniform topology of X1 generated by the relative uniformity U1 on X1 x X1 inherited from U, and G acts as a weakly transitive, equicontinuous group of homeomorphisms of X1 onto itself ( thanks again to Lemma 9.36 for this ) . As before, apply Lemma 9.31 to get a nonzero G-invariant content A l on X1 which extends to a content X 1 on all of X by letting for any C E C(X). Neither of these contents it a scalar multiple of the other; indeed Xo is supported on Xo and X 1 is supported on X1 with X1 � X \ Xo. The lack of weak transitivity costs the uniqueness of nontrivial G-invariant D contents on X. Now we're ready to prove the uniqueness aspect of Steinlage's work: Theorem 9.42. If (X, U) is a uniform space whose topology is locally com­ pact and Hausdorff and if G is an equicontinuous weakly transitive group of homeomorphisms of the space X (in its uniform topology) onto itself, then there is but one G-invariant integral (up to multiplicative constants) on X . The fact is that uniqueness is evidenced, if at all, on the collection of continu­ ous nonnegative real-valued, nonzero functions defined on X having compact support. We'll refer to members of this collection as test functions. We start with a few basic facts, features regarding the notions of equiconti­ nuity and weak transitivity.

Let (X, U) be a uniform space, and let G be an equicontinuous group of homeomorphisms of X (in its uniform topology) onto itself. For each U E U and xo E X, there is a nonnegative real-valued function f defined on X such that Lemma 9.43.

6. Steinlage: Uniqueness and Weak Transitivity

263

( i ) f (xo) # O; ( ii ) f(x) = 0 for each x E X \ U[xo] ; ( iii ) f g ( xo) = f g- 1 ( xo) for each g E G. o

o

Proof. Fix U E U, and x0 E X. There is a uniformly continuous pseudo­ metric d on X such that for each r > 0 , Ud,r = {(x, y) : d ( x, y) < r} E U, and there is an ro > 0 so that Ud,ro � U. For x, y E X , define h* (x, y) = sup{d ( g(x) , g(y)) : g E G} and h ( x, y) = min { h * ( x, y) , ro}. Clearly, h is a pseudo-metric on X x X and h satisfies h( x, y) = h (g(x) , g(y)) for each g E G. Moreover, h (x, x0 ) is actually a continuous function of x. Indeed, if we fix x and let 0 < E < ro, then the equicontinuity of G at x assures us that there is an open set 0 containing x such that for all g E G, gO � Ud, E [g (x) ] . It follows that for any y E 0 , l h (y, xo) - h (x, xo) I :::; h (x, y) :::; E. Define f : X-+[0 , oo ) by f(x) = ro - h ( x, xo), so f is continuous. Also f (xo) = ro - h ( xo, xo) = ro. Further, if x t/. U[xo] , then x t/. Ud,ro [xo] ; hence, d ( x, xo) :2: ro and h ( x, xo) = ro, forcing f(x) = 0. Finally, regardless of g E G, we have ro - h (g( xo), xo) f(g(xo)) ro - h( g - 1 g(xo) , g - 1 ( xo)) ro - h( xo, g - 1 ( xo)) = f(g - 1 (xo)), 0 and we have proved ( iii ) .

9.

264

Steinlage on Haar Measure

Suppose G is a weakly transitive group of homeomorphisms of the locally compact space X onto itself. Suppose f, h are test functions defined on X. Suppose that h is not identically zero. Then there exist 91 , . . . , 9n E G and c > 0 so that n f(x) :::; c :2: h (9i( x)) i=l for each x E X. Lemma 9.44.

Proof. Let K = suppf. There is an 'f/ > 0 and a nonempty open set 0 � X so that h ( x ) > 'f/ for all x E 0. Since G is weakly transitive, K � LJ 9 0 gEG

so there are 91 , . . . , 9n E G such that K � 91 0 U · · · U 9n 0. Let M = sup{f(x) : x E K} , and set c = M/TJ. If x E K , then x E 9i 0 for some i , 1 :::; i :::; n . It follows that 9i 1 ( x ) E 0 and so h (9i 1 (x)) 2:: 'f/ · Hence, 9i 0 M h i 1 x 2:: M 2:: f(x). (9 ( )) 'f/

The constant c = M/TJ works. Now we're ready to prove Theorem 9.42. Fix xo E X and U E U. By equicontinuity of G at xo, there is a V E U such that for each 9 E G, 9 V[xo] � U[9( xo)] . We may assume that V[xo ] is compact. Let f* be a test function that vanishes off V[xo] , is positive at x o, and satisfies f*9(xo) = f * 9 - 1 ( xo) for each G E G. Such an f* exists by Lemma 9.43. Let I be any G-invariant Haar integral on X , and let f be a test function. For each 9 E G, let h (9) = I(f · ( ! * 9 - 1 )), and let Ku denote the oscillation of f on U; i.e., Ku = K = sup{ l f(x) - f(y) I : (x, y) E U} < oo. o

6. Steinlage: Uniqueness and Weak Transitivity

265

It's plain that for any x E X, we have f(g(xo)) - K :S f( x ) , and so h (g ) = I(f · (f* g - 1 )) � I( [f (g( xo)) - K] · (f* g - 1 )) (55) = [f(g(xo)) - K] I(f * g - 1 ) = [f(g(xo )) - K] I(f * ) by I's invariance. Let C be any compact set containing f's support, and let

0. Since f* is uniformly continuous, there is a U* E U so that if ( x, y) E U *, then l f*(x) - f* (y) I < € . Equicontinuity steps up to provide a V * E U such that for each g E G, gV* [xo] � U* [gxo] . We can partition unity over C via [O, 1]-valued test functions Ji , . . . , fn , where fk ( x ) vanishes off 9k( V * [xo] ) . If x E 9k V * [xo] , then for every g E G g - 1 ( x) E g - 1 gk ( V* [xo] ) and g - 1 (x) E U* [g - 1 gk ( xo ) ] . Since this is true for each and every g E G, we conclude that o

o

o

for all g E G. So for any g E G and any x E X , we have fk (x) · f(x) · f* (g - 1 ( x )) :S fk (x) · f(x) · (f* (g - 1 gk ( xo)) + t: ) . With (55) in mind, we see that

(56)

:S l: U * (g - 1 gk (xo)) + t:] I( fk f) . k�n

9. Steinlage on Haar Measure

266

But I(! * ) > 0 and f* (g(x0)) = f*(g - 1 (xo)) for each g E G so (55) soon unveils the relationship I(fk · f) [ J *(9"k 1 g(xo )) + E] l: h (g) n k :::; < < . f(g (xo )) K I(f*) I(f*) Gx0 is dense in X by G's weak transitivity, and so we deduce from this inequality I(fk · f) [f*(g - 1 (x)) + E] (57) f(x) K -< h(g ) -< '"" ' � I(f*) I(f*) k:=:;n and this is for each x E X and g E G. Let I(l: k( x ) + k 0 and for such x's, ( x ) = 1. We now keep track of the relative size of nonzero test functions much like we did with the fractional covering functions in Chapter 8. Now if J # is a test function, we define N ( J # , f*) = inf L Ck , k :::;n where this infimum is taken over all c 1 , . . . , Cn such that for some gf, . . . , g'f! E G, _

_

{

}

Lemma 9.43 tells us that N (f, f*) < oo, and our work above soon translates into saying N (f, J * ) � ( K + Eco) · N (, f* ) + c o. But E > 0 was arbitrary, so we can take it away at will-just let E � 0, and we get N (f, f*) � K · N ( <J> , f*) + co. Now I is still G-invariant, and so any inequality such as, f(x) � L ck f * (gf x) , k:::;n

6. Steinlage: Uniqueness and Weak Transitivity

soon becomes

267

I(f) � I: Ckl(f * ) . k�n

It follows that

I(f) � N (f, f * )J(f * ) . Here's what we have: Start with xo E X and U E U. Find V E U so that gV[xo] � U[g(xo)] for each g E G with V[xo] compact. Find a test function f* that is positive at xo, vanishes off V[xo] , and is symmetric at xo. Let I be any G-invariant integral. Let J be any test function with support contained in the compact set C. Let cf> be a test function that is [O, l ] -valued and is identically one on C. Then � N (f, f * ) . N (f, J * ) - ( oscillation of J on U) · N (cp, f*) �

:(�})

Now fix the test function J0. Both J and Jo are uniformly continuous, so for each n there is a Un E U such that 1 1 IJ (x) - J (y) I < -n and I Jo(x) - Jo( Y ) I < -n , so long as ( x , y) E Un . Applying the equicontinuity of G, find Vn E U so that gVn [xo] � Un [gxo] for each g E G, and V[x0] is compact. Find a test function J� that is positive at xo, vanishes off Vn [xo] and is symmetric about xo. Let Co and C be compact sets containing the support of Jo and Ji , respectively, and let cf>o and cf> be [O, 1 ] -valued test function with cf>o identically one on Co and cf> identically one on C. Then by our work summarized above we have (58) I(f� ) N (f, J� ) - 2_ N (cp, J� )I(f� ) � I(f) � I(f� ) N (f, J� ) n

and

I(f� ) N (fo, J� ) - 2_n N ( c/>o, J� )I(f� ) � I(fo) � I(f� ) N (fo, J� ) . It's part of the nature of the counting function N that if Ji , f2 and f3 are test functions, then N (f1 , f3) � N (f1 , f2) N (f2, f3). Hence, N ( c/>o, J� ) � N ( cf>o, Jo) N (fo, J� ) . Now N (fo, J�) > 0 for each n since I(fo) > 0. (59)

9. Steinlage on Haar Measure

268

If N (
� N (fo, J� ) - n N ( 0. If N ( 0 and n > N (
��:�: j�j

2: 0. N (fo, J� ) - � N ( N (fo, J� ) Hence whenever n > N ( o . J:;. ) · n l N (fo, Jri ) I(fo) N (fo, Jri ) 1 - n N((fo. N (fo, Jri ) Jri) Now regardless of n, N (
[

]

·

7.

Notes and Remarks

The paper of R. Steinlage [121] included earlier work of S. Banach (see Chapter 4) and I. E. Segal [118] . The simple condition that the group G separate compact subsets of X was noted by E. Hewitt in his extremely positive review of Steinlage's paper. D. Ross [107] spoke to invariance relative to a collection G of local homeo­ morphisms (closed under inverses) on a compact Hausdorff space X. Using a

7. Notes and Remarks

269

variation of the theme of separating compact sets, Ross showed the existence of a G-invariant Borel probability on X. His arguments are nonstandard. P. Niemiec [95] studied G-invariant regular Borel probabilities on a compact Hausdorff space X, where G is an equicontinuous semigroup of maps from X to itself. Of particular interest is Niemiec's discovery that if the collection of G-invariant regular Borel probabilities is nonempty, then it is a Choquet simplex.

Chapter 1 0

Oxtoby 's View of Haar Measure

1 . Invariant Measures on Polish Groups

In this short chapter we broach the subject of invariant Borel measures in groups which are not locally compact. More precisely, we look at separable metrizable topological groups where the topology is Polish; i.e., the group topology is that of a complete metric space. In addition to the well-developed theory of Polish spaces, this class of topological groups seems regular enough to offer hope of a reasonable theory. The three theorems presented here indicate that without local compactness, invariant measures can be quite pathological. These results are due to J. Oxtoby and are, we believe, beautiful and surprising. In this discussion a Borel measure will mean a countably additive, [O, oo]­ valued set function defined on each Borel set, vanishing on singletons, and having at least one nontrivial value. Our first result indicates that there are some surprises in store for us when discussing invariant measures in nonlocally compact topological groups. (Oxtoby and Ulam). Let G be a complete, separable metriz­ able topological group that is not locally compact, and let m be any left­

Theorem 10. 1

invariant Borel measure on G having at least one nontrivial value. Then every nonempty open set in G contains uncountably many disjoint congru­ ent compact sets of equal finite positive m-measure. -

271

272

1 0.

Oxtoby's View of Haar Measure

Suppose B is a Borel set for which 0 < m(B) < oo. Let U be a nonempty open subset of G. Since G is separable and { xU : x E G} is an open cover of G, at least one of the sets xU intersects B in a set of positive m-measure, say m( a U n B) > 0. Then a - 1 B n U is a Borel subset of U having positive m-measure. It follows that a - 1 B n U contains a compact set C with 0 < m(C) < oo . In fact m(E) = m(E n ( a - 1 B) n U) is a finite Borel measure on G so Ulam's theorem applies to ensure m is regular. Consider the subgroup G 1 = ( C) generated by C. Keeping in mind the compactness of C1 · C2 and C! 1 given that of C1 and C2 , we see that G 1 is the countable union of Un Cn of an ascending sequence of compact sets: C1 = c u c - 1 , C2 = C1 . C1 , C3 = C2 u c2 1 , C4 = C3 . C3 , . . . . Now compact sets in nonlocally compact groups are nowhere dense, so G 1 is of the first category in G, and so too are all of G 1 's left cosets. Let V be an open subset of G that contains the identity and satisfies V · C � U. The left cosets of G 1 of the form vG 1 , where v E V, cover V. But V is of the second category, as is any open subset of a complete separable metric space, thanks to Baire's category theorem. Since each left coset of G 1 is of the first category, the cosets v G 1 (v E V) that are distinct (and hence disjoint) are uncountable in number. Let's stop to admire what we've done: each of the left cosets v G 1 contains a set v C that's congruent to C and satisfies vC � V · C � U, so each of the v C's is contained in U. The vC's are pairwise disjoint, congru­ ent to C, uncountable in number, and have positive finite m-measure. D Proof.

Remarkable. And true. Corollary 10.2. If X is an infinite-dimensional separable Banach space, then there does NOT exist a translation invariant Borel measure m on X for which m(Bx) < oo .

The surprises found in the measure theory of invariant Borel measures on complete separable metric groups are not done yet. Here is a stunning refinement of the Oxtoby-Ulam Theorem. (Oxtoby) . Let G be a complete separable metric topological group that is not locally compact, and suppose m is a translation invariant Borel measure in G. Then every nonempty open set in G contains a compact

Theorem 10.3

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Invariant Measures on Polish Groups

273

set which is the union of uncountably many disjoint congruent compact sets of (equal} finite positive m-measure. Proof. Let U be a nonempty open subset of G . Apply the Oxtoby-Ulam result above to find a compact set C � U with 0 < m( C) < oo, and again let G 1 = ( C) be the subgroup of G generated by C. As before, pick V to be an open subset of G that contains the identity e of G and satisfies V · C � U. Then C · c - 1 is compact, and so is nowhere derise in G . It follows that for each € > 0, there is an open set Ve � V of diameter less than E , containing e with Ve g; C c - 1 ; therefore, we can find a sequence ( an ) in G that converges to e with an
•

•

•

274

1 0.

{

Oxtoby's View of Haar Measure

}

dn = min p(x, y) : x f:. y, x, y E An , and

� {

}

6n = min p (ain ' c . c - 1 ) , p(a;;,1 , c . c - 1 ) , dn · Then ain+ i is the first an with i n+l > in so that for any a E A� 1 · An , 1 aain+ i ) < 6n . 1 a ) , p (a, a.;-•n+l p (a, aain+ 1 ) , p (a, a.;-•n+l Remember A� 1 An is a finite set so we can arrange the choice of ain+ i without much fuss. Now r is the closure of the set al: · al� · · · al: : n E N, (jk) E {O, l} n ·

{

}

Enough already with the subscripts! We have done our pruning so let's just call (ain ) , ( an) , and proceed. It is worth nothing that e E A� 1 An , so p( e , an+l ) = p( e , ean +l ) < 6n , which in turn implies that dn+l = min {p(x, y) : x, y E An+l , x f:. y} :S p ( e , an+l ) :S 6n t t

From this we see that

each in An+l ·

46n+ 1 :S dn+l :S 6n , and the sequence (6n) tends to zero fast-geometrically, in fact. Lets run with this for a bit. Take (jn)n E {O, 1 }N. Look at the sequence ( xn) = (a{1 a�2 · · · a�n ) . Notice that p(xn, Xn+k ) :S p( xn , Xn+ i) + p( xn+l > Xn+ 2) + · · · + p(Xn+k - 1 > Xn+k ) :S p( a{1 a�2 . · · a�n , a{1 a�2 · · · a�n a�";f ) n+ I ai i ah . . . ain+ i ain + 2 ) + P (aj11 ai22 . . . ainn ain+l n+l n+ 2 ' 1 2 n+kn+ l l k+ . . . + p (aj1 . . . ajn+k - 11 > aj1 . . . ajn+k - 1I ainn+ +kk ) · A typical term herein is p( a{1 a�2 . . . a;1 ' a{1 a�2 . . . a;1 a;'.tt ) '

1.

Invariant Measures on Polish Groups

275

where n ::::; l ::::; n + k - l ; this term is of the form p ( a, aali+1+i1 ) for some a E An and J1 + 1 = 0 or 1 . So it is either 0 or p(a, aa1 + i) · Either way, it is less than or equal to i5n . The result is i5n+l + · · · + · · :S 4 i5m . i5n + 7 p ( Xn, Xn+k ) :S i5n + i5n+l + · · · + i5n+k - 1 :S i5n + 4 3 It follows that the sequence ( a{1 a�2 ahn )n is a p-Cauchy sequence regard­ less of (jn)n E {O, l}N. It follows that this sequence converges to some x ((Jn)) E r. Now if (j� )n is a different sequence in {O, l }N and Yn = a{� · · · ah� , then (Yn) converges to some y((j�)) E {O, l }N. If jk =/= j� while J 1 = j � , . . . , Jk - 1 = j� _ 1 , then 4i5k :S dk :S p(xk , Yk ) :S p(x k , x) + p( x, y) + p(y, Yk ) :S 34 i5k + p(x, y ) + 34 i5kr It follows that 4 p(x, y ) � 3 i5k . From this we see that r is homeomorphic to the Cantor set. After all, ¢( (jn)n) = ( x(jn)) is a continuous, one-to-one surjection of {O, 1 }N onto r. To see that ·

•

•

•

r � v,

we take any a E Am. Then for some J 1 , . . . , Jn E {O, 1}, p( e, a ) = p( e, a{1 . . ahn ) ::::; p( e, a{1 ) + p( a{1 , a{1 a�2 ) + · · · + p( a{1 a��11 , a) :S p( e, a l) + '51 + · · · + i5n - 1 p( e 1 a 1 ) p( e 1 a 1 ) < - p ( e, a l ) + 4 + . . . + 4n - l ' by the definition of i5k 's and the presence of e in cc - 1 . We have that 1 4 p( e, a) :S p( e 1 a 1 ) ( l + 4 + · · · ) = 3 P( e 1 a 1 ) . It follows that for any 'Y E r 4 p( e, 'Y) :S 3 P(e, a l) < 2 p( e, a l) and 'Y E V. Therefore r � V. Now we can establish that r- 1 r n c c - 1 = {e}. •

•

•

•

276

1 0.

Oxtoby's View of Haar Measure

Take z E r- 1 r . Then

z = lim n bn , where bn E A;;- 1 An . If z i- e, then sooner or later bk i- e as well. Let's look at how bk differs from e when bk E A;; 1 Ak : bk = (ai1 . . . a�k ) - 1 ( a{1 . . . a{k ) , where ii , . . . , i k , j 1 , . . . , jk E {0 , 1}. So bk = ak- ik · · · a -1 i 1 aii1 · · · ajkk J. , but if ii = j i , . . . , i k = jk , then bn = e. So suppose bk is the first bj that is not e and so k is the first time i k i- jk . At this step, bk = ak_ i k . . . a 1 i 1 a{1 . . . a{k = . . . = ak_ i k a{k . But i k , jk E {O, 1} and i k i- jk , so b k = ak- i k ajkn = ak± 1 . It follows that 4 b . -1 p( bk , z ) ::; <5k + <5k+ I + . . . ::; 3 <5k ::; p( k , c c ) . 3

Hence, if z E C · c- 1 ,

b . -1 p( bk , c · c - 1 ) ::; p( bk , z ) ::; p( k , c3 c ) , so if z E r - 1 · r and z i- e , then z f/_ cc - 1 , and so r- 1 . r n c . c - 1 = { e } .

D

Corollary 10.4. If X is an infinite-dimensional separable Banach space, then there does not exist a translation invariant Borel measure on X that 's finite on all compact subsets of X .

On the positive side of the ledger, we have the following theorem: ( Oxtoby ) . In any complete separable metric group G which is dense in itself, there exists a left invariant Borel measure having a non­ trivial value.

Theorem 10.5

Proof. We start by applying the Birkhoff-Kakutani metrization theorem to find a left invariant metric on the group. Now in so doing we may well lose the completeness, but we gain some flexibility. Let's agree that in this new left invariant metric. we will denote the distance between x , y E G by

(x, y) .

1.

277

Invariant Measures on Polish Groups

Now we will begin the construction in earnest. The knowledgeable and attentive reader will recognize similarities with the construction of Hausdorff measure. Let (Un ) be a sequence of open sets whose diameters (with respect to ( , )) tend to zero, and let ( wn) be a sequence of positive numbers which descends to zero. Let U denote the family of (open) subsets of G that are congruent to one of the Un 's. Let A � G and r > 0. We define Ar(A) = inf L Wni : A � LJ bi Uni , diam(UnJ < r .

}

{

Notice that the smaller the r , the fewer the coverings of A via members of U, and so the bigger is Ar ( A ) . We also define * m ( A ) = lim Ar ( A ) . r �O * It may be that m ( A) = +oo, but we will just have to live with this. It is reasonably straightforward to verify that m* is a metric outer measure that vanishes at singletons and is left invariant. Let m be the countably additive Borel measure on G generated by m * a la Caratheodory. Now we are face-to-face with the crux of the matter: Does there exist an A such that 0 < m(A) < oo? We will show how to find a Cantor set with measure 1. It will take some careful tinkering, but it is possible. So, we will construct a family

{ C(n; ii , . . . , in) : n E N, ii , . . . , ini{O, 1} }

of compact subsets of G so that (i) for any m, C(n; ii , . . . , in) = C ( n + 1; ii , . . . , in, 0 ) U C(n + 1; ii , . . . , i k, 1 ) ; (ii) all of the 2n C's of rank n are congruent; (iii) the diameters dn of each of the rank n sets tend to zero; (iv) any two of the sets of rank n are separated by 3dn and so are, in particular, disjoint. The resulting set C satisfies C = C ( l; 0) u C ( l; 1) = u ii ,

...

,in E { O , i }

and must be homeomorphic to the Cantor set.

1 0.

278

Oxtoby's View of Haar Measure

Since G is dense in itself, there exists a sequence ( xn) in G so that for each

n

0 < (e, xn+ i) < (e, 9Xn) Let i i , i 2 , . . . , in E {O, 1 } and ai. 1 , ... , in xi1i x i22 . . . xnin · We define A ( n; i i , . . . , i n) = { ai , .. . ,in +p : P 2: O}. So A(n; i i , . . . , in) is the set of all ai1 , , ain+p whose first indices are equal to i i , . . . , in . If ai· 1 , ... , in+p · > aJ1· , .. . ,Jn+q E A(n '· i 1 ' . . . ' in ) ' then --

.

.

_ -

•

•

•

·

and

i n+q xi1i xh2 . . xinn xinn+ + I . . . xin+q satisfy i i = J I , h = h , . . . , in = Jn · We define C(n; i i , . . . , i n) by aJ , ... ,Jn+q .

.

_ -

.

C(n; i i , . . . , i n) = A(n; i i , . . . , i n) ·

Notice that A(n; i i , . . . , in) = A(n + 1 ; i i , . . . , i n , 0 ) U A(n + 1 ; i i , . . . , i n , 1)

and

A(n; i i , . . . , i n) = ai i , i2 , .. ,in A ( n; 0 , 0 , . . . , 0 ) . .

Statements ( i) and ( ii) follow from their observations. Next, let's play with estimating dn . Using the left invariance of the metric ( , ) , we see that (e, YiY2) � (e, Y 1 ) + (y1 , YiY2) = ( e, Yi) + (e, Y2) . Further, so for k = - 1 , 0 , 1 we have Now a simple induction argument shows that for any k i, k2 , . . . , kn = - l, 0, 1 , (e, y� 1 y�2 · · · y�n ) � (e, y1 ) + (e, y2) + · · · + (e, yn) ·

1.

279

Invariant Measures on Polish Groups

• ts of A( n; 0 , 0 , . . . , 0) , say Xin+l n+l · · · Xinn+p and Xin+ i · · · Xin+ q , ke two porn n+ q n+l +p where the i's and j's are - 1 , 0 or 1 . Because the exponent 0 is allowed, we can assume p = q. How far apart are these points? Well .1.a l""f'

Therefore dn ::; ( e ,�n ) .

But both e and Xn+ I are in A(n; 0, 0, . . . , 0) , so

and

and (iii) follows. How far apart are distinct sets A(n; ii , . . . , in ) and A(n; j 1 , . . . , Jn ) of the same rank? Take the typical member of each of these sets, say ai 1 , . . . ,ip E A(n; i 1 , . . . , in ) and aji , ... ,jq E A(n; j 1 , . . . , jn ) , where once again, because all the i's and j's are - 1 , 0, or, 1, we can assume p = q. Since the sets A( n; ii , . . . , i n ) and A( n; J I , . . . , Jn ) are distinct, the indices of ai1 , ... ,ip and aj1 , ... ,jp differ at some index k ::; n. We may as well assume that i k = 0 and Jk = 1 . Look at p +1 x k xik V k+ l . . . xjp . -

Of course ( u ' v ) - ( ai i , ... , ip l aJ 1 , ... ,Jp ) -

·

·

·

·

280 and

1 0.

Oxtoby's View of Haar Measure

(e, xk) ::; (e, ) + ( v ) + (v, xk) = ( e, Xikk++Ii · · · Xpip ) + ( ) + ( XkXjkk++li · · · Xpjp , Xk ) = ( e, xikk++Ii · · · Xpip ) + (u, v ) + ( xikk++Ii · · · Xpjp , e ) = ( e, xikk++Ii · · · Xpip ) + (u, v ) + ( e, xik+I k+ i · · · XpJp ) ::; ( v ) + 2 (e, Xk) [ (e, Xk+ i) + · · · + (e, xp) ] . . + p1 k ] < · + 2 - ( ' v ) + 2 (e ' xk) [ �9 + __!__ g9 n ::; ( v ) + 2 (e, Xk) L g n k) (e, ::; ( v ) + : . u

u,

U, V

u,

u

--

u,

u,

So ( u , v ) � i(e, xk) � i( e, Xn) � 3dn , and ( iv ) follows. ( Semi- ) finally, each C (n; ii, . . . , in ) is the union of 2p - n sets C(p; ii, . . . , ip), each having diameter dp , where dp ---t 0 . It follows that each C(n; ii, . . . , in ) is totally bounded. Therefore C(n; ii, . . . , in ) is closed and totally bounded. If G were ( ) complete, then we would be done. But we mentioned earlier that in this new metric ( , ), we had no assurances that G is complete. Did we throw out the baby with the bath water? Not if we follow Oxtoby's able leadership; we didn't. Embed ( G, ( , )) into a complete metric space Q. Let C (n; ii, . . . , i n) denote the closure in g of C(n; ii, . . . , in) · The point is j ust this: If we work a bit harder, we can rig things so that C(n; ii, . . . , in ) is already closed ( hence, complete) in Q! To be sure, denote by p the original complete metric that generates G's topology. With revisionist behavior so acceptable, take a careful look at our choice of (xn ) · We used the dense-in-itself character of G to pick (xn ) so that for each n, 0 < (e, Xn+ i) < ( e ,�n ) . But the dense-in-itself character of G is topological, and so we could also have insisted, regardless of ii, . . . , in = 0, 1 , that ,

­

(60)

also. But (60) has added benefits; indeed, regardless of the sequence (in ) of indices (in = 0, 1 ) , the sequence ( aii , ... ,in ) is p-Cauchy! After all,

1.

281

Invariant Measures on Polish Groups

so But

(

)

P xi1i xi22 · · · xnin xi1i xi22 · · · xnin xnin++li if in+ l = 0, p ( xi1 x�2 · · · X� , xi1 x�2 · · · x�n ), = or p ( xi1 x�2 · · · x� , xi1 x� · · · X� Xn + i ) , if i n + l = l . Whichever the case, the result is less than 2- n by our judicious revisionist '

{

choices. It follows that if m > n, then

p ( ai1, . . . , in > ai i , .. . ,im ) < 2n - m ,

and ( ai1 , . . ,in ) � i is p-Cauchy regardless of the choice of indices ( in ) · Hence ( ai i , . . ,in ) converges to a member of G modulo the adjustments made above in selecting ( xn ) · But any point of accumulation of C in G is the limit of just such a sequence and so must be in G. So, too, is all of C. So we may suppose that C is complete (in G) and totally bounded, and C is, indeed, a Cantor set. As yet we have not specified the weights ( wn ) ; we will do so in such a way that m(C) = 1 . By (iv) we can find a sequence ( en ) of positive numbers descending to zero so that for any two distinct components C(n; i i , . . . , in ) and C(n; ji , . . . , Jn) of C of the same rank n we have dist ( , ) C(n, ii , . . . , im ) , C(n, j i , . . . , jn ) > dn + 2En . =

)

(

Let

{

}

Un = x E G : d ( , ) ( x , C(n; O , O , . . . , 0) ) < En , and let Wn = 2- n . We know the ( , )-diameter of Un is less than or equal to diam( , ) C(n; 0, 0, . . . , 0) + 2En :S dn + 2En ---+ 0. It follows that any set congruent to Un can overlap at most one set C(n; i i , . . . , in ) of rank n. Let r > 0. Choose n so that diam Un < r . From (iii), we know that C can be covered by 2n sets congruent to Un ; hence, Ar(C) :S 2n · Wn = l .

1 0.

282

Oxtoby's View of Haar Measure

On the other hand, if we cover C by a sequence of sets each of the form bi Un; , then we can always assume the covering is finite: Set

C � bi Un 1 U · · · U bk Unk ·

p = max { ni , . . . , nk } , and let qi be the number of components of C(p; ii , . . . , ip ) that are overlapped by bi Un; . Since C � bi Un 1 U · · · U bk Unk ' qi + qz + · · · + qk 2: 2P · After all, to cover C, you have to cover each of the 2P components of C(p; ii , . . . , ip ) · But bi Un; can overlap at most one component of rank ni, and so at most 2p - n; components of rank p . It follows that 2p- n; 2: qi .

So

and Hence, Therefore, and

Ar (C) 2: 1 . Ar ( C) = 1 for all r > 0 m(C) = 1 .

D

2. Notes and Remarks

We have presented only a few of the results found in Oxtoby's wonderful paper [97] . Here are a couple of others worthy of close study. Theorem 10.6 ( Oxtoby ) . Let H be a Borel subset of the metric group G, and suppose m is a Borel measure on H such that m(xA) = m(A) for any Borel set A � H and any x E G with xA � H. Then there exist two left invariant Borel measures ,\ and v on G, both of which are extensions of m such that if µ is any other such extension of m, then -\ (A) � µ(A) � v(A) for each Borel subset A of G.

2. Notes and Remarks

283

Theorem 10. 7. If a metric group G contains a locally compact subgroup H

that is dense in itself, then there exists a left invariant Borel measure on G which is an extension of the Haar measure on H.

This classic paper of Oxtoby is chock-full of delicate examples, delicious ob­ servations, and curious results. We highly recommend it to youthful readers. Corollary 10.4 might lead one to believe that the study of invariance of Borel sets in (separable) infinite-dimensional Banach spaces leads to a dead end. Such is not the case. Motivated by classical results related to differentiability properties of convex functions and Lipschitz maps on !Rn , a number of mathematicians have iso­ lated various notions of smallness that have been useful in the classification of Banach spaces-both linear and nonlinear. Let B denote a separable infinite-dimensional Banach space. P. Mankiewicz [74] introduced cube measures. A cube measure on X is the distribution of B-valued random variables of the form where x, ei , . . . , en E B, n

and the sequence (Xn) is a sequence of uniformly distributed independent [O, l]-valued random variables (on some probability space). Cube null sets are those Borel subsets of B that are null sets for every nondegenerate cube measure on B. Next, N. Aronszajn [2] , in his study of Gateaux differentiability of Lip­ schitz mappings, considered the following: given ei , . . . , en , . . . E B, let A(ei , . . . , e n , . . ) consist of those Borel sets E which are the union U n Fn of Borel sets En such that En is "null" on every line in the direction en . So for each x E B, the Lebesgue measure of the set { t E JR a + ten E En } is zero. A Borel set is Aronszajn null if E belongs to A(e 1 , . . . e n , . . . ) whenever the linear span of the e n 's is dense in B. Again, R. R. Phelps [99] isolated the Gaussian null sets as those Borel sets in B that are null for every Gaussian measure. Each notion of nullity comes with accompanying results about the differ­ entiability of Lipschitz maps defined on B. There are similarities, to be sure. .

:

,

284

1 0.

Oxtoby's View of Haar Measure

In M. Csornyei [22] these similarities were explained when she proved that in every B, the o--ideals of Aronszajn null sets, Gaussian null sets, and cube null sets coincide. Another notion of smallness has gained considerable interest-Haar null sets, a notion from the fertile mind of J. P. R. Christensen [19] . A Borel set A in an abelian Polish group G is said be Haar null if there is a Borel probability µ on G such that µ ( x + A) = 0 for each x E G. Christensen establishes some basic (expected) properties of the Haar null sets, but each is shown only with considerable difficulty. The abelian nature of the group simplifies things. Haar null sets have been studied in more general Polish groups and related to classical results. Almost always the arguments are nontrivial and intri­ cate. We refer those interested to the papers of P. Dodos [30] and [29] , E. Matheron and M. Zeleny [79] , and S. Solecki [120] for a good start in this fascinating subject. Of related interest to the work of this chapter (indeed, to all we've been discussing in these deliberations) is work initiated by G. W. Mackey [73] in the late 1950s. Mackey was looking for a proper home for "duality" , similar to what is encountered in the case of commutative locally compact groups with the Pontrjagin-van Kampen duality. He came to believe the proper setting for such a duality might be found in classical descriptive set theoretical ideas. Mackey reformulates ideas from that theory to fit his approach to general representation theory. To be precise, a Borel space is a set S along with a o--field B of subsets of S. Members of B are called Borel sets. The Borel space is called standard if the o--field B is isomorphic (as a o--field) to the Borel o--field of an uncountable Borel subset of a Polish space (a homeomorph of a separable complete metric space). Along with Borel spaces comes Borel functions: if S and T are Borel spaces and f is a function from S to T such that the preimage of any Borel subset of T is a Borel subset of S, the f is called a Borel function. Cementing ties with the classical theory, he establishes that a a Borel space has a countable family of sets that generates B precisely when (S, B ) is isomorphic to a subspace of the Cantor set with its natural collection of Borel sets. Again, following the path of the classical theory, Mackey isolates the analytic Borel spaces as those countably generated Borel spaces that are the image

2. Notes and Remarks

285

of a standard Borel space under a Borel map. Another link to the classical theory: a necessary and sufficient condition that a Borel space be analytic is that it be Borel isomorphic to the Borel space generated by the topology of an analytic subset of the reals. Groups? Suppose G is a Borel space which is also a group; it is called a Borel group if multiplication and taking inverses are Borel functions. Standard Borel groups and analytic Borel groups have the natural meanings. Measures? If S is a Borel space, a Borel measure on S is a O"-finite, countably additive, nonnegative, extended real-valued map on the Borel O"-field of S. A Borel measure is standard if it is supported by a standard Borel set. A measure class is the collection C of all Borel measures on a Borel space having the same null sets as a fixed, nontrivial Borel measure. If G is a Borel group and C is a measure class, C said to be right (left) invariant if whenever µ is in C so is each measure µ9 given by µ9 ( E) = µ(Eg) (gµ ( E) = µ(gE)) . To whet the appetite of those who've enjoyed our discussions of Haar mea­ sures, here's a gorgeous theorem of Mackey that generalizes the earlier the­ orem of Weil. 10.8. Suppose G is an analytic group, and assume G admits a left invariant measure class. Then G has a unique locally compact topology whose Borel sets are the given ones and under which G is a topological group.

Theorem

This was but the start of Mackey's work. He went on to apply these de­ scriptive set theoretic ideas to the representation theory of locally compact groups satisfying the second axiom (thereby encompassing all Lie groups) and to develop a duality theory for such groups that was to bring to bear measure-theoretic ideas to supplant the topological basis of duality in the commutative case. To give a synopsis of Mackey's work is beyond the capabilities of the present authors. A clear overview can be found in the special volume of Contempo­ rary Mathematics [32] , paying tribute to his work.

Appendix A

In 1933 Alfred Haar gave the first abstract construction of a countably ad­ ditive translation invariant measure on a locally compact separable metriz­ able topological group. This appendix is devoted to an exposition of Haar's construction in the compact case. In this setting Haar's construction is particularly elegant and transparent. Haar's construction starts with the introduction of his covering function, which gauges the relative size of a given subset of the compact group vis­ a-vis a nonempty open subset of the group. Next, to put all sets on a level playing ground, he averages against shrinking sequences of open balls. This averaging procedure is shown to be especially effective when applied to open sets whose boundary is "null" with respect to the notion of content ( a natural generalization of the Jordan content ) . These open sets are called intervals, and they are used much as intervals in Euclidean spaces are in the generation of an outer measure there. Let G be a compact metrizable topological group. Fix a metric on G that generates its compact metrizable topology. If A and B are subsets of G, we say that A and B are congruent if B = Ax for some x E G. Of course, if A and B are congruent, then they are open together or closed together being homeomorphic. For B � G and U an open set in G we define h(B, tJ) to be the smallest number of sets congruent to tJ needed to cover B . Since B is totally bounded, h(B, tJ) < oo .

Theorem A.1. Some pertinent features of h(·, ·) follow:

287

Appendix A

288

(i) ( ii ) ( iii ) ( iv ) (v) ( vi )

h ( U, U ) = l . If C � E, then h(C, U) :::; h ( B , U) . If Ei , E2 � C, then h(B1 u B2 , U) :::; h(B1 , U) + h(B2 , U) . If Ei and E2 are congruent, then h(B1 , U) = h ( B2 , U) . If U and V are open, then h ( B , U) :::; h ( B , V)h(V, U) . If ( Un) is a descending sequence of open sets with exactly one point in common, then whenever E and C are open sets with B n C = 0, there is an no so that for n 2'. no,

h ( B u c, Un) = h(B, Un) + h(C, Un) ·

The proofs are straightforward with the proof of the last statement following Banach's variation of Haar's initial approach, so we omit them. Fix a sequence ( Un) of open balls with common center and with radius � · Let E be an open set in G. Define ln( B ) = h ( B , Un) . h ( G , Un) Notice that 1 l ( E ) :::; h ( E , G ) . :::; h( G, E) So if E is a nonempty open set in G , then ( ln( B )) is a bounded sequence of real numbers, each greater than h (J, B ) . Thus there is an increasing sequence ( n k ) of positive integers so that ( lnk ( B )) converges ( to a positive number between h ( J, B ) and h ( B , G )) . What's more, if C is congruent to E, then lnk ( C) . lim lnk ( B ) = lim k k -

Now we choose a countable collection {Em } of nonempty open sets as fol­ lows. Choose a countably dense sequence ( an) in G , and look at any open ball centered at am with rational radius. The collection of all such sets is countable. If we look at the collection formed by taking all finite unions of this collection, then we arrive at our desired collection {Em : m E N} . A straightforward diagonal argument allows us to find a sequence ( nk ) so that lim lnk ( En) k exists for each n. For simplicity, we assume nk = k.

289

Appendix A

A subset N of G is called a null set if given E > 0 there is an open set Uc containing N so that limk lk(Uc ) ::; E . Subsets of a null set are null sets, finite unions of null sets are null sets, and sets congruent to a null sets are null sets. We plan to show that lim k lk(C) exists whenever the boundary {)C = C \ C of the open set C is a null set. First though we show that such open sets C with null set boundaries exist at all. Take a typical ak from our countable dense subset of G, and let B( r) be the open ball centered at ak with radius r. If r is rational, then B (r) is among the Em's for which limk lk(Bm) exists. In any case, is nondecreasing in r. As a consequence, the monotone functions An (r) con­ verge at all but countably many r's to a function A(r) . A is also nondecreas­ ing and so, except for countably many r's, is continuous. If r is such that limn An (r) = A(r) exists and r is a point of continuity of A, then the open ball B (r) has a boundary that is a null set. In fact, by the continuity of A at such r's, if E > 0, then there are r' , r" , and r"' such that r' < r" < r < r"' and 0 ::; A(r"' ) - A(r' ) < E . On the one hand, the closed sets B (r"' ) \ B (r" ) and B (r' ) are disjoint, so for n large enough h([B (r"' ) \ B (r" )J U B (r' ), Un ) = h( B (r"' ) \ B (r" ), Un ) + h( B (r' ), Un ) · But now h([B (r"' ) \ B (r" ) J U B(r' ), Un ) ::; h( B (r"' ), Un ) so that h( B (r"' ) \ B (r" ), Un ) ::; h( B (r"' ), Un ) - h( B (r' ), Un ) · It follows that " U B "' lim sup h( (r ) \ B (r ) ' n ) -< A(r"' ) - A(r' ) < € . h(G, Un ) n Since the compact set B (r"' ) \ B (r" ) contains the boundary of B (r), this boundary is indeed a null set. A few moments of reflection reveals that we've shown not only that open sets whose boundaries are null sets exist, but each open ball B with center one of the points of the dense set {an n E N}-with but countably many exceptions-has limn An ( B) existing as well. :

Appendix A

290

Let U be any open set, and let F be a closed subset of U. Then the boundary of U, a u = U \ U, and F are disjoint, and the distance between a u and F is positive, say dist( a u, F) = rJ > 0. Look at points of {an : n E N} . The open balls centered at points of {an : n E N} that lie inside F and have radii between ¥ and ¥ cover F. So then F is covered by a finite union of such open balls. We can even assume the balls have rational radii and have boundary that is a null set, and so we can assume that the finite union that covers F is one of the Bm 's. What we have (because of the judicious choice of possible radii being between ¥ and ¥) is the existence of m with F � Bm � Em � u. Along with this, we can assume exists. Here's what we have at this stage: If U is open and F is a closed set, F � U, then there exists an open set C, F � C � C � U with ac a null set and there exists an open set B so that F � B � B � U where limn ln (B) exists. We plan to show that when the boundary of an open set U is a null set, then the sequence (ln (U)) is convergent, and we will define limn ln (U) to be the content, I(U), of U. Let C be an open set whose boundary is a null set. By definition for each € > 0 there is an open set UE containing ac such that . :up h(UE , Un ) ::; E . hm h(G, Un ) Let C" = C n UE and C' = C \ C". Then C" is open and C' is closed. Since C" is a subset of UE . sup h( C" , Un ) ::; E . hm n h(G, Un ) We know there is an open set B containing C' so B � C and lim n ln (B) exists. Now and so it follows that

� �

lim sup h C, Un 2: lim n ln (B). n h G, Un Since C � B U C", we have h(C, Un ) ::; h( B , Un ) + h( C" , Un ),

Appendix A

and so

291

h(C, U:---'n) h(G, Un)

---..,

Therefore

< -

h(B, Un) h(G, Un)

+

h( C", Un) . h(G, Un)

1.

. :up h( C" , Un) im:up h ( C , Un) :::::; h. �sup h(B, Un) + hm h(G, Un) h(G, Un) h(G, Un) :S lim n ln( B ) + E . On applying the theory of subtraction, we see h ( C , Un) 1 . . h ( C , Un) < E rim: up imnmf . h(G, Un) h(G, Un) Therefore h(G, Un) lim n ln( C) n h(G, Un) = lim exists and is I( C ) . The content I( C ) exists for any nonempty open set C whose boundary is a null set. The following are more or less clear features of the content. -

-

A.2. (i) I( C ) > 0. (ii) If C1 and C2 are congruent sets with one of them open and having boundary that is a null set, then so too is the other, and J( C1 ) = J( C2) . (iii) If Ci , C2, . . . , Cn are open sets with boundaries that are null sets, then so too are C1 n C2 n n Cn and C1 u C2 u u Cn . A bit more complicated: Fact

·

·

·

·

·

·

Fact A.3. If C1 and C2 are disjoint open sets with boundaries that are null sets, then C1 n C2 is an open set with boundary that is a null set and

More generally, if C1 and C2 are simply open sets with null set boundaries, then Proposition A.4. If C1 and C2 are disjoint open sets, each with a null set for boundary, then C1 U C2 is likewise an open set with null set for boundary

and

Appendix A

292

Let E > 0 be given. We've seen that there exists an open set U with tJ � C for which limn l(tJn ) exists and for which l ( C1 ) :::; lim n ln (tJ) + € .

Proof.

Now tJ and C2 are disjoint closed sets, and so for n large enough, (61)

Since tJ U C2 � C1 U C2 , we have h(tJ u C2 , Vn ) :::; h(C1 u C2 , Vn ) :::; h(C1Vn ) + h(C2Vn ), and for n big enough for (61) to hold, h(tJ, Vn) + h(C2 , Vn ) :::; h(C1 , Vn ) + h(C2 , Vn ) · From this and our placement of U within C1 , we have The usual epsilonics finishes the proof.

D

Fact A.5. Suppose ( Cn) is a descending sequence of open sets, each of which has a null set for a boundary. If nn Cn = {c} , then lim n l(Cn ) = 0.

Let H be an open set with boundary a null set. Take distinct points hi , . . , h k E H. Look at the maps x t--7 x c - 1 hp for p = 1 , 2 , . . . , k from G onto itself. Suppose CA1) , . . . , CAk) are the images of Cn under these maps so CA1) , . . . , CAk) are mutually congruent. For n sufficiently large, CA1) , . . . , CAk) are pairwise disjoint, and so (since I( Cn) = I( CA1) ) = = I( CAk) )) k kl( Cn) = L l( C�) ) = I( CA1) u . . . u c�l ) :::; I( H ) . .

.

.

.

p=l

It follows that for n large enough,

H I( Cn ) -< I(k )

for each k. We now introduce an analogue of Lebesgue measure in G in which congruent sets have the same measure. To emphasize the analogy, we will call any nonempty open set whose boundary is a null set an interval.

Appendix A

293

Let M � G. Define the outer measure m * ( M) by

{ I C },

m * (M) = inf L ( n ) n

{C1 , C2,

where the infimum is taken over all coverings . . . , Cn , . . . } of M by intervals. Since M is totally bounded, it can be covered by finite collections of intervals, so it is plain that 0 � m * (M) < oo . Further, if M � M' � G, then m * (M) � m *(M' ). Also, m * is countably subadditive (something that's easy to see from its def­ inition) . Since sets congruent to intervals are intervals, and a set congruent to M can be covered by intervals that are congruent to intervals that cover M, it is plain that m* (M) = m * (M' ) whenever M and M' are congruent.

C

I(C). C'

The outer measure m* (C) of an interval is its content Indeed, let E > 0 be given. Given the interval there is an interval such that C' � and > - E . Cover and so C', with intervals . . . , Cn , so that - E < � + + · · · + ( n ), and conclude that � m * (C) follows. Since - E � m* (C). Clearly, * is plain, our claim is established. m (C) � If is an interval, then m* (C) = In fact, if E > 0 is given, then there is an interval containing C so that < + E . Now m* (C) � m* (C" ) = < + E. Fact A.6.

C, C I(C' ) I(C) C, C1 , C2, I(C) I(C') J(C1 ) J(C2) I C I(C) J(C) I(C) C C" I(C).J(C") I(C) I(C") I(C)

Epsilonics ilber allesf

Next, we note that Fact A.7.

m' (M) � inf

CL

{ � I(C�) } ,

where . . ' c� , . . . are pairwise disjoint intervals with M � Un c� . Suppose M � U U · · · u n u · · · , with all the 's intervals. We "disjointify" : look at \ Take a countable dense family {xn : n E N} of points of \ Around x 1 there is an interval V that is contained in \ Look to the next that is in \ U V) . Around that there is an interval that is contained in \ U V) . Continue. We exhaust .

C1 C2C2 C . C 1 C2 C . 1 C2 C1 . Xk C2 C2(C (C1 1

Ck

Xk

294

Appendix A

with disjoint intervals. We now turn our attention to C3 \ ( C1 U C2 ) and continue, ending up with a sequence ( C�) of disjoint intervals such that C2 \ C1

M � LJn C� .

Our next task is to isolate what "measurability" means. For any subset M of G, define the inner measure of M to be m * (M) = I(G) - m* (G \ M) . A set M � G is measurable if and only if m* (M) > m* (M) . It is classical (and an excellent exercise) to show that the collection of measurable subsets of G is a o--field and that on this o--field m* (or m* ) is a countably additive measure that assigns to any set M' congruent to M the same value of m* .

Appendix B

In this final appendix we broach the possibility of extending Haar measure to a larger domain while preserving its countable additivity and its translation invariance. We restrict our attention to the situation where G is an infinite compact metrizable topological group. In the main, we follow the path blazed by S. Kakutani and J. Oxtoby for the reals and generalized by E. Hewitt and K. Ross to the current setting. In truth, out exposition is but a slight embellishment of that found in the text Abstract Harmonic Analysis, volume I [55] of Hewitt and Ross. Note B . l . If G is a nondiscrete locally compact topological group, then G's cardinality is at least c , the cardinality of the continuum. To see this, we follow a Cantor-inspired path. Let U be a nonempty open set in G with U compact. Then U must be infinite since G's topology is nondiscrete. Let xo , x 1 E U, xo # x 1 . Envelop xo and x 1 in open sets Uo and U1 , where Uo n U1 = 0 and Uo U U1 � U. Each of Uo , U1 is infinite. Let xoo, xo 1 be distinct points of Uo, and let x 1 0, xn be distinct points of U1 . Envelop xoo and xo 1 in open sets Uoo and Uo 1 , where U oo n U0 1 = 0 and Uoo U Uo 1 � Uo. Similarly envelop x 1 0 and xn in open sets U1 0 and Un , where U 1 0 n Un = 0 and U 1 0 U Un � U 1 . Continue in this Cantor-like fashion. For (En) E {O, 1 }1'\ let Do m = LJ {Uq ,. : q, . . . , E m E {O, 1 } } . Proof.

. . , €m

-

295

Appendix B

296

Then each �m is compact and �m contains �m+ l Let ·

m Since the �m's are descending nonempty compact sets, � f:. 0. Moreover, if (En ) E {O, l } N , then n Uc i , i- 0. ...

n

,
Further if (En ) f:. ( E�) , then and so � contains a set of cardinality at least c .

D

Note B.2. Now Suppose G is an infinite compact metrizable topological group. We know that G has at least c many members. In fact, the car­ dinality of G is exactly c. After all, G is separable, and so G contains a countably infinite dense subset D. Each member of G is the limit of a sequence of points from D, and so G has no more than c many elements. Not only does G have exactly c many elements, G has exactly c many distinct compact sets having cardinality c . Indeed, any open subset of G is the union of (countably many) sets in a countable open base for its topology, so the totality of compact subsets of G (which is the same as the totality of closed subsets of G) is at most c. That there are exactly c distinct compact subsets of G having cardinality of the continuum follows from the nature of the Cantor set. As noted in the Notes and Remarks section of Chapter 2, a classical theorem of Alexandrov and Hausdorff assures us that G contains a homeomorphic copy of the Cantor set � = {O, l } N which is itself homeomorphic to � x � = {0, l } N x {0, l } N . It is this last version of � that tells us the truth if we but wish to know it: if E E � = {O, l } N , then {E} x {O, l } N is homeomorphic to � ' and so {E} X {0, l } N � � = {0, l } N x {0, l } N . We can find c many distinct compact sets {{E} x {O, l } N : E E �} , each of cardinality c . Next, recall the Frechet-Nikodym metric space: n is a set, E a o--field of subsets of n, and µ is a probability measure on E. For E, F E E we define the pseudo-metric dµ as follows dµ (E , F) = µ(E�F) ,

297

Appendix B

where EtlF = (E\F) U (F\E) is the symmetric difference of E and F. If we identify sets E, F with dµ (E, F) = 0, then we get a complete metric space (E/µ+- ( {O} ) , dµ ) · The density character of this metric space is called the character of (n, E, µ) ; it is the smallest cardinal m such that there is a subset S of E with the cardinality of S equal to m so that given E > 0 and E E E there is S E S with dµ (S, E) = µ(StlE) < E .

Let G be a fixed infinite compact metrizable topological group, and let E be the CT-field of all µ-measurable subsets of G, where µ is the normalized Haar measure on G. It is plain that the character of ( G, E, µ) is �O · We now state the stunning theorem of Kakutani and Oxtoby a la Hewitt and Ross. Theorem B.3 (Kakutani and Oxtoby). There is an extension ( G, E* , µ*) of (G, E, µ) such that the character of (G, E* , µ*) is 2c and µ* is a left-, right-, and inverse-invariant countably additive probability measure on E* which extends µ.

The proof will entail careful bookkeeping and devilishly clever-yet­ simple constructs. We will denote by F the collection of all functions f G --+ G of the form f(x) = ax, f(x) = xa, or f(x) = x - 1 . Proof.

:

Step 1 . We extend (G, E, µ) to ( G, E', µ') , where if A � G has fewer than many members, then A E E' and µ'(A') = 0. This extension will be a suitably invariant and complete probability on E 1 •

c

How? Let P denote the family of all P � G such that P has cardinality less than c . If P E P n E, then µ( P ) = 0. After all, if µ( P) > 0, then P contains a Borel set B with µ(B) > 0. In turn then B contains a compact set K with µ(K) > 0. Such a compact set K with µ(K) > 0 must be uncountable, so it contains a homeomorphic copy of the Cantor set, by a classical result of Alexandrov and Hausdorff. Next note that if P1 , P2 , . . . , Pn · · · E P, then n

This is a consequence of Konig's theorem in the theory of sets.

Appendix B

298

Let :E' consist of all those sets M' � G such that M' D.. M E P for some M E :E. So M' E :E' precisely when M' = (M n Ac ) U B for some M E :E and A, B E P. Note that :E' is a a-field: If M E :E, then for any M' E :E' we have (M' ) cb.. Mc = M' b.. M, and so M' E :E' is equivalent to (M') c E :E'. Suppose (M�) is a sequence of members of :E'. Let (Mn ) be a corresponding sequence in :E so that Then SO

( LJn M� ) D.. ( LJn Mn) � LJn (M�D..Mn ) E P ,

Un M� E :E'. Let M' E :E'. Then M' D.. M E P for some M E :E. Naturally, we would like to define µ'(M') by µ'(M') = µ(M) , if M' D.. M E P. Is this legal? Well suppose Mi , M2 E :E, and suppose M' D.. Mi , M' D.. M2 E P. Then Mib.. M2 � (Mib.. M' ) U (M2 D.. M' ) E P. So Mib.. M2 E P n :E, and, as noted earlier, µ(Mib.. M2 ) = 0. In particular, µ(Mi ) = µ(M2 ) , and so µ' (M') = µ(M) for M' D.. M E P is not only legal but ethical as well. Now we have M defined on :E', let's show µ' is also countably additive thereupon. Suppose (M�) is a sequence of pairwise disjoint members of :E'. Let (Mn ) be a corresponding sequence of members of :E with M�D.. Mn E P for each n. We disjointify the Mn 's: Let Q i = Mi , Q2 = M2 \ Mi , . . . , Qn = Mn \ (Mi U U Mn-i) . Then ( Qn ) is a sequence of pairwise disjoint members of :E, and since n M� D.. Qn � LJ (M� D.. Mk ) E P , k= i ·

·

·

Appendix B

299

M�tl. Qn E P. What's more,

Hence

(

) (

)

µ' LJ M� = µ LJ Qn = L µ( Qn ) = L µ'( M� ) , n

n

n

n

so µ' 's countable additivity on E' is secure. To see that ( G, E', µ') is a complete probability space, suppose µ' ( N) = 0 and Ni � N. Since N E E', there is an M E E such that Nf:l.M E P. That µ'(N) = 0 says we have µ(M) = 0. Now Ni n M � M, and (G, E, µ) is a complete probability space, so µ(Ni n M) = 0 with Ni n M E E. Since Nitl.(Nitl.M) � N f:l.M, and N tl.M E P, we have Nitl.(Nitl.M) E P also. But this ensures us that Ni E E' with µ'(Ni) = µ(Ni n M) = 0. The completeness of ( G, E', µ') is established. Finally, µ' is invariant on E', left-, right-, and inversion-invariant. After all, if P E P and f E F, then f(P) E P. If M' E E' and M E E is chosen so that M' tl.M E P, then, thanks to the bijective nature of members of F, f (M' )f:l.f (M) = f (M' f:l.M) E P. It follows that f(M') E E' (each f(M) E E) and µ' (f(M' )) = µ(f(M)) = µ(M) = µ1 (M' ) , thanks to the invariance properties of µ. Our first step has been successfully taken. An observation regarding members of E' of positive µ' measure: if M' E E'

and µ'(M') > 0, then M' contains c many distinct compact sets each having cardinality c. Indeed, if M' E E and µ'(M') > 0, then there is M E E so that M' tl.M E P and µ(M) = µ'(M') > 0. Now M' f:l.M E P just says M' tl.M has fewer than c members. Since M E E, M contains a Borel set

of positive µ-measure, which is necessarily uncountable. Hence, again by the Alexandrov-Hausdorff result, this set contains a homeomorphic copy of the Cantor set. But M must itself contain a continuum of pairwise disjoint compact sets each of cardinality of the continuum (look at our preliminary remarks). Now each of these lies in M, and so they are each disjoint from M' tl.M; that is, each is in M' n M. How about that?

Appendix B

300

Step 2. Let We denote the first ordinal having cardinality c, and let (Fa : 1 ::; a ::; we) be a corresponding well-ordering of order type We of all the distinct compact subsets of G having the cardinality of the continuum. Then for any M' E 'E' with µ' (M' ) > 0 and any ordinal a < We, there is an ordinal f3 with a < {3 < We so that F13 � M' .

After all, we just saw that each M' E 'E' with µ' (M' ) > 0 contains c many (distinct) compact subsets of G each having cardinality of the continuum. Since We is just the first ordinal having cardinality c, for any a < We, there must be an F13 following Fa with F13 � M' . Let A � G and define µ(A) by µ(A) = inf{µ' (M' ) A � M' E 'E' }. We say that A is µ-F invariant if for any f E F we have f(A)�A E 'E' and µ'(f(A)�A) = 0. We are obviously going to try to mimic our first step to continue our con­ struction. Step 3. There exists a family {Xv : v E JR.} of pairwise disjoint subsets of G :

with

µ(Xv) = 1

for each v E JR., and so that for any lR.o � JR.,

is µ-F invariant.

Let (Fa 1 ::; a ::; We) be as in Step 2: a well-ordered set of all the c many compact subsets of G each having cardinality of the continuum, so that if M' E 'E' has positive µ1 measure and a is an ordinal less than We, there is a {3, a ::; f3 < We so that F13 � M' . Well-order F with order type We . For each x E G and ordinal a < We, let Ca (x) be the collection of all members of G of the form :

0"2 f/3O"l1 f/32 0

0 . . . 0

Un (x) ' ff3n

where 1 ::; f31 , . . . , f3n ::; a (the {3's need not be distinct) and 0-1 , . . . , O"n E {±1}, f131 1 , ff3n E F, n E N. It is plain to see that x E Ca (x) , and that if 1 ::; f3 ::; a < We, then f13(Ca (x)) = Ca (x) . Moreover, the cardinality of Ca (x) is no more than the maximum of the cardinality of a and �o. and so Ca (x) has fewer than c members, and Ca (x) E 'E' . •

•

•

Appendix B

301

We will now construct a transfinite doubly infinite sequence (x3 : 1 � f3 � a < we) satisfying (i) x3 E Fa and (ii) {Ca (x3 ) 1 � f3 � a < we} is a disjoint collection of subsets of :

G.

We will call on the lexicographic ordering -< of pairs (a , {3) of ordinals where 1 � f3 � a < We and define (x3 ) by transfinite induction. Pick x� E Fi . Suppose 1 � f3 � a < We, and suppose that xJ has been suitably chosen for all (1, 0) -< (a, {3) with (i) and (ii) in mind. Consider the set D(a, {3) = LJ Ca (xJ ) . ("Y 6 )-< ( a ,.B) ,

The cardinality of D( a, {3) is no more than

(#(a, {3) ) (#C0 (xJ) ) , ·

which is strictly less than or equal to (#a) 2 · max{#a, No } < c . Since Fa has a continuum of members it is not a subset of D(a, {3) . There must be a point x3 E Fa\D(a , {3) . Suppose (1 , o ) -< (a, {3) . Then Ca (x3 ) U C-y (xJ) = 0. In fact, if this intersection is nonempty it is because there are f31 , . . . , f3n � a, Oi , . . . , Orn � 1, O'i , . . . , O'n , O' � , . . . , O':n E { ± 1 } and so that It follows that x3

=

Ji::n

o

·

·

·

o

fi1u1 J;f o

o

·

·

·

o

J%f;: (xJ) E Ca (xJ) � D (a, {3) .

But x3 was chosen from Fa \D(a, {3) . It follows that Ca (x3 ) U C-y (xJ) = 0, whenever (T, o) -< (a , {3) .

Appendix B

302

Now the collection of ordinals v < We is equinumerous with �. and so we will use this (well-ordered) collection to do some indexing. Let 1 ::; v < We, and define Xv as Xv = LJ{Ca ( x�) : V ::; 0: < We} · By our earlier observation about the Ca (x� ) 's, the Xv's are pairwise dis­ joint. Further, P,Xv = 1 for each v. In fact suppose P,Xv < 1 for some v. Then by definition of µ, there must be a M' E � so that (Xv) � M' and µ' (M') < 1 . This means, in particular, that µ' (M'c) > 0. So there is an ordinal o: , V < o: < We with Fa � (M')c. But X� E Fa SO x� E Ca(x�) � Xv � M' , an impossible situation for a member of Fa which is a disjoint from M' . Finally, if �o � {v : 1 � v < we} , then LJ Xv v EIRo is P,-F invariant. Indeed, pick any f-r E F. Then LJ Xv = LJ{Ca(x�) ; v E �o , v � o: < we} v EIRo is a disjoint union of its component sets Ca(x�), each of which is fixed by the action of f-r whenever "Y ::; o:. Hence But the cardinality of this latter set is less than or equal to #{ ( o:, v ) : v < o: < "Y} · max{#o: , �o}# (f-y ( Ca ( x� ))), which is less than ( #o: ) 2 . c . c = c . So the set Xv Xv �

(1{� )) � "

also has cardinality less than c, and thus this set belongs to P. The p,-F invariance of U v EIRo Xv is established, and we have taken Step 3 without serious mishap. Step 4. We will prove a striking lemma of Tarski, tailored to our current purposes.

Appendix B

303

Lemma B.4. There exists a family {Ne :

() E

8} of distinct subsets of G

each having cardinality c such that (i) = and (ii) given any sequence (en of distinct members of the family any sequence ( O'n of + l 's or 's we have

#8 2c

) ) -1

8 and

n

where A + 1 = A and A- 1 = G\A = Ac for A � G.

1),

Proof. Some basic notational conveniences will be observed. For an under­ lying set we will choose [O, the half-closed, half-open interval from 0 to l; will be the collection of all subsets of [O, so (i) is thereby secure. For () E let B(()) = (J U n {x + x E () c , so B(()) � is just () plus the translates of (JC in [l, Notice that if e1 , ()2 are distinct members of then B(e 1 ) n B(e2 r =/= 0. After all, if x E () 1 n ()� , then x E B(()1 ) , since () 1 � B(()1 ) and x tf. ()2 so x + is a member of [l, that is a translation of a member of ()�. The case x E ()� n ()2 is dealt with similarly. Now let C be the family of all countable subsets of [O , Then the cardinality of C is c. Let N be the family of all countable subsets of C so the cardinality of N is c also. For any () E denote by the Ce the family of all countable subsets of B(e), and denote by Ne the family of all countable subsets of C n q. Plainly, for any () E Ce � C, and Ne � N. Let (()n )be a sequence of distinct members of and let ( O'n be a sequence of ±l's. Choose () E so that () is not one of the ()n 's, and let O"o =/= 0" 1 . Consider (()fi : i � 0) , and let I = {i � 0 : O'i = and J = {i � 0 : O'i = Since () was chosen not to be in the list e 1 , e2 , . . . , each of I and J are nonempty. Moreover I n J = 0 and I u J = {O, 1 ,

8

8,

[O , 2]

1

8'

([1 ,2)

1) 1: }) 2).

2)

8

2) .

8

8

8,

)

+l}

-1}.

2, . . . }.

Appendix B

304

For each pair ( i , j ) E I x J, pick a point x(i ,j ) E B(Oj) \ B(Oi ) · Let Cj = { x(i ,j ) : i E I }, and let v = { Cj : j E J}. It is plain that each Cj E C and v E N. Moreover if i E I, j E J, then X (i ,j ) E B(Oj), and so Therefore It follows that

I/

(62)

Also if i E I and j E J, then

E n N%;i = n Njj . jEJ

jEJ

So and It follows that and so (63)

Taking (62) and we see that

(63)

iE/ into account along with the relationship of I and J, n

We have successfully navigated Step 4.

0

Our critical Step 5 will entail an upgrading of Tarski's lemma to incorporate

jl-F invariance. Here is Step 5 . Step 5. There exists a family {Ee ; e E 8} of distinct subsets of G such that ( i ) #8 = 2c , ( ii ) n�=l Ei;: is jl-F invariant for any sequence (On ) from 8 and every sequence ( an ) of ± 1 's, and

305

Appendix B

(iii) p,(n�=l E;;: ) = 1 for each sequence (On ) of distinct members of E> and every sequence (un ) of ±1 's. Let {Xv : v E JR} be as in Step 3, so {Xv : v E JR} is a family of pairwise disjoint subsets of G with µ(Xv ) = 1 for each v, and regardless of Ro � JR, is P,-F invariant. In tandem with this we call in Step 4 to find the family {Ne : 0 E E>} . For any 0 E 8, we let Ee = LJ Xv . vENo

The sets Ee are all distinct since the Xv 's are pairwise disjoint. Item (i) follows from the first claim in Tarski's lemma. By construction (Step 3) each Ee is p,-F invariant, and so (ii) follow from the fact that the collection of P,-F invariant sets is a u-field. To see (iii), let vo E n�=l N%n , a nonempty set thanks to Tarski's smarts. n By definition Ee = LJ {Xv : v E Ne} , and the Xv 's are pairwise disjoint, so

(

Ee = LJ {xv : I/ E Ne}

Regardless of u = ±1 , Hence

00

n ECTn e n l n

:::::>

=

:::::> :::::>

) c 2 LJ {Xv :

01 ( u { 01 u{

I/

Xv : I/ E NB:n

Xv : I/ E

Ntnn

Xvo ·

}

Step 3 assures us that 1 = il (X.,,, )

:".

µ,

(01 EB:)

:".

1.

E NB} .

})

Appendix B

306

An aside: The crucial Step 5 hints at the enormity of a set having a contin­ uum of elements. It gives proper respect to this event. We are now headed for the final turn. Let £ denote the family of all subsets E of G of the form

E = LJ

(( 6 )

)

E;: n M�i , ... ,an , {a1 , ... ,an} k - 1 where ()i , . . . , On E 8 are distinct, ai , . . . , an are ±l's, and M�1 , ... ,an E :E'. For each such E define >.(E) = rn µ'( M0-' 1 , ... , 0"n ) a1, ... ,anE{ +1 ,- 1 } Then £ is a field of sets that contains :E', ).. is well defined on £ and finitely additive thereupon with >.!�' = µ', and for any E E £ we have f(E) E £ with >.(f(E)) = >.(E) wherever f E F. This in hand we stand ready to finish the proof of the Kakutani-Oxtoby theorem. Let :E* be the a-field of subsets of the G generated by £. To show ).. can be extended to a countably additive measure µ* on :E* , it is enough to show that if (Ep) is a sequence from £ that is descending and >.(Ep) � a > 0 for all p, then •

n Ep � 0. p

We can (and do) suppose (On) is a sequence of distinct members of 8 , (np) is a strictly increasing sequence in N, and for each choice a1 , . . , O'np of ±1 's, .

Ep =

U +l ,-l} a1 , ... ,anpE{

Replacing M�1 , ... ,anp by

((n E;: ) k=l

p n M�1 .... , Unq

)

n M�1 .... ,anp .

q=l does not affect the above equality since the Ep 's are descending. So we suppose that Ma1' , ... , CTnp , unp+ l C- Mcr' 1 , ... , crnp regardless of ai , . . . , O'np • O'np+ i and p. By definition 1 >.(Ep) = 2np µ'( Ma1, ... ,anp ) � a , L a1 , ... ,anpE{ +l ,-l}

307

Appendix B

so for some fixed O"iv) , . . . , O"Y1j E {+ 1 , - 1 } we have µ'( M' 1(pl , ... , O'n(ppJ ) � a , and this we can do for each p. It is time to apply the pigeon-hole principle. There is a choice '51 , . . . , r5n 1 of ±l's so that Z1 - { q · ,.,. v l( q) - u1 , . . . , ,.,.v n( q1) - Un 1 } is infinite. After all, regardless of n 1 's size, there are but 2n 1 possible choices . the sequence £T (p) , . . . , O"n(p1) , O" (p) ' . . . £or each p; at of the first ni p1aces m n i+l 1 least one of those choices must be made infinitely often. By the same token, we can choose r5n 1 +1 ' . . . , '5n2 ( each + 1 or - 1 ) so Z2 = { q E Z1 : q � 2 , O"��)+l = r5n 1 + i , . . . , O"�� = r5n2 } is infinite. Etc., etc., etc. So regardless of p, there is a q � p so that µ'(M' >.. ( M��q J �� ) ' CT

- �

- �

•

-

,

. . ,a

, . . . ,a

> a > O.

But (M81 ,. . , on) is a descending sequence in :E' so .

)

(p

µ n M81' , ... ,0p -> a > 0. I

Since

µ( n Ez: ) ( n M81 , . .. ,0p ) ( n EZ: ) i= 0. =

k

n

It follows that

p

1,

k

n Ev , p

which contains this set, is also nonvoid. Conclusion. >.. can be extended to a countably additive measure µ* on :E* . Finally we show that ( G, :E* , µ *) has character 2 c .

Appendix B

308

We know that if 8 1 , 82 are distinct members of 8, then >. ( Eo1 n E02 ) = l = µ* ( Eo1 n E02 ) . It follows from this that if 8 1 , 82 are distinct members of 8 , then µ * ( Eo1 D. Eo2 )

=

�-

Now should S E L: * be a basis for the Frechet-Nikodym metric space (L:* , dµ* ), then for any 8 E 8 , there is an So E S so that µ* ( Eob. So ) <

If 81 =I 82 and So1 1 2

=

So2 , then

l·

dµ• ( Eo1 , Eo2 ) < dµ• ( Eol ' So1 ) + dµ• (Sol ' So2 ) + dµ• (So2 , Eo2 ) dµ• ( Eo1 , So1 ) + 0 + dµ* (So2 , Eo2 )

< 41 + 41 1

2·

Hence if 8 1 # 82 , then So1 # So2 . Now {So : 8 E 8 } has 2c members, and the character of ( G, L:* , µ*) is at least 2c. It cannot be any bigger than 2c so we can conclude the character is exactly 2c. Alas, the p,-F invariance of members of L:* is straightforward. As a matter of fact, if we look at the collection B of all those members B of L:* such that f (B) E �* for each f E F with µ* (B) = µ* ( f (B) ) , then B contains £. Moreover B is a monotone class of sets: if (Bn ) is an ascending sequence of members of B and f E F, then ( LJ sn ) = LJ J (Bn ) , and

n

n

lim n µ* ( f (Bn )) = l im µ * (Bn ) n =

This ensures B = L:* . This proof is complete.

D

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[109] C. Ryll-Nardzewski, Generalized random ergodic theorems and weakly almost peri­ odic functions, Bull. Acad. Polon. Sci. Ser. Sci. Math. Astronom. Phys. 10 (1962) , 271-275. [110] , On fixed points of semigroups of endomorphisms of linear spaces, Proc. Fifth Berkeley Sympos. Math. Statist. and Probability, Vol. II (Berkeley, CA) , Con­ tributions to Probability Theory, Part I, Univ. Calif. Press, 1967, pp. 55-61 . [1 1 1] S. Saks, Theory of the integral, vol. VII, Monografie Matematyczne, WarazawaHw6w, 1937. [1 12] , Integration in abstract metric spaces, Duke Math. J. 4 (1938) , 408-4 1 1 . [113] , Theory of the integral, second edition, Dover, 2005 . [1 14] I. Schur, Neue Begrundung der Theorie der Gruppencharaktere, Sitz. Pr. Akad. Wiss Berlin (1905) , 406-432. [115] Issai Schur, Neue anwendungen der intergralrechnung auf probleme der invarianten theorie, I, Sitz. Preuss. Acad. Wiss. phys-math-kl (1924) , 189-208. [1 16] ___ , Neue anwendungen der intergralrechnung auf probleme der invarianten the­ orie, II, Sitz. Preuss. Acad. Wiss. phys-math-kl (1924) , 297-321. [117] , Neue anwendungen der intergralrechnung auf probleme der invarianten the­ orie, III, Sitz. Preuss. Acad. Wiss. phys-math-kl (1924) , 346-353. [1 18] I. E. Segal, Invariant measures on locally compact spaces, J. Indian Math. Soc. (N.S.) 13 (1949) , 105-130. [119] K. Shiga, Representations of a compact group on a Banach space, J. Math. Soc. Japan 7 (1955), 224-248. [120] S. Solecki, Size of subsets of groups and Haar null sets, Geom. Funct. Anal. 15 (2005 ) , 246-275. (121] R. C. Steinlage, On Haar measure in locally compact T2 spaces, Amer. J. Math. 97 ( 1975) , 291-307. [122] A. H. Stone, Paracompactness and product spaces, Bull. Amer. Math. Soc. 54 (1948) , 977-982. [123] Karl Stromberg, The Banach- Tarski paradox, American Math Monthly 86 ( 1979) , no. 3, 151-161 . [124] R . Struble, Metrics in locally compact groups, Compositio Math. 2 8 (1974) , no. 3 , 217-222. [125] Dennis Sullivan, For n > 3 there is only one finitely additive rotationally invariant measure on the n-sphere defined on all Lebesgue measurable subsets, Bull. Amer. Math. Soc. (N.S. ) 4 (1981) , no. 1 , 121-123. [126] E. Szpilrajn, La dimension et la mesure, Fund. Math. 25 (1937) , 81-89. [127] N. Tomczak-Jaegermann, Banach-Mazur distances and finite dimensional operator ideals, Pitman Monographs and Surveys in Pure and Applied Mathematics, Harlow and Wiley, New York, 1989. [128] S. Ulam, Zur masstheorie in der allgemeinen Mengen lehre, Fund. Math. 16 (1930) , 141-150. [129] D. van Dantzig, Uber topologisch homogene Kontinua, Fund. Math. 15 (1930) , no. 1 , 102-105. [130] , Zur topologisch algebra i. komplettierungstheorie, Ann. of Math. 107 ( 1932), 587-626. [131] E. R. van Kampen, Locally bicompact abelian groups and their character groups, Ann. of Math. (2) 36 (1935) , 448-463. ___

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___

___

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[132) Guiseppe Vitali, Sul probleme della misura dei gruppi di punti di una retta, Gam­ berani e Parmeggiani, Bologna, 1905. [133) J. von Neumann, Die Einfiihrung analytisches Parameter in topologischen Gruppen, Ann. of Math. 34 (1933) , 170-190. [134) John von Neumann, Invariant measures, American Mathematical Society, Provi­ dence, RI, 1999. [135) Andre Weil, Sur les espaces a structure uniforme et sur la topologie generale, Actual. Sci. Ind., no. 55 1 , Hermann et Cie., Paris, 1938. [136] , L 'integration dans les groupes topologiques et ses applications, Actual. Sci. Ind., no. 869, Hermann et Cie., Paris, 1 940. [137] H. Weyl, The classical groups, their invariants and representations, Princeton Uni­ versity Press, Princeton, 1939. ___

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Aut hor Index

Alexandroff, A. D . , 133 Alexandrov, P. , 43 Alfsen, E. M., 207 Aronszajn, N., 283 Arzela, C., 1 1 5 Ascoli, G . , 1 1 5 Ball, K . , 20 Bandt, C., 225 Bernstein, F., 16, 17 Birkhoff, G., 5 1 , 62 Braconnier, J . , 235 Brunn, H., 10

Godement, R. , 222 Gordon, Y. , 173 Grothendieck, A., 168, 1 73 Gurevic, A., 1 73 Halmos, P. , 137 Hausdorff, F., 43, 46 Hewitt, E., 62, 221, 295 Higgins, P. J., 62 Hoffman, K. H., 62 Hurewicz, W., 46 Jarchow, J . , 173

Caratheodory, C., 22, 134 Cartan, H., 198, 207, 222 Chevalley, C., 60 Christensen, J. P. R. , 284 Cohn, D . , 207 Csi:irnyei, M . , 284

Kakutani, S . , 5 1 , 62, 73, 1 1 2 , 144, 209, 295 Kantorovich, L., 127 Kazdan, D., 1 1 2 Klee, V . , 238 Koosis, P. , 173 Kwapien, S . , 173

Dodos, P. , 284 Drinfeld, V. G . , 1 12 Dvoretzky, A., 172

Leja, F . , 61 Lindenstrauss, J., 168, 1 72

Erdos, P. , 209 Federer, H. , 20, 46 Figiel, T., 172 Freudenthal, H., 61 Frink, 0 . , 60 Garling, D. H . , 145 Gelfand, I. M . , 214

Maak, W., 136 Mankiewicz, P. , 174, 283 Marczewski, E., 44 Margulis, G., 1 1 2 Markov, A. , 144 Matheron, E , 284 Mattila, P. , 20 Maudlin, D . , 209 Michael, E. A., 62 -

317

Author Index

318

Milman, V., 136, 172 Minkowski, H., 10 Montgomery, D., 62 Morris, S., 62, 220

Wagon, S., 112 Wallman, H. , 46 Weil, A . , 62, 147-149, 1 72, 190, 207, 209 Weyl, H . , 62, 154

Nachbin, L . , 172, 173 Nadkarni, M. G . , 19 Naimark, M. A . , 207 Nakamura, M., 73 Natanson, I. P. , 1 1 2 Niemiec, P. , 269

Zeleny, M . , 284

Oxtoby, J . , 1 12, 271 , 282, 295 Pekzynski, A . , 168, 173 Peter, F., 154 Phelps, R., 283 Pietsch, A., 168, 169, 173 Pontrjagin, L., 6 1 , 136, 173, 220 Raikov, D. A . , 214 Riesz, F., 45 Rogers, C. A . , 42 Rosenblatt, J . , 112 Ross, D . , 268 Ross, K . , 62, 221, 295 Rudin, W., 222 Ryll-Nardzewski, C . , 144, 145 Saks, S., 63, 73 Schechtman, G . , 136, 173 Segal, I., 268 Solecki, S . , 284 Steinlage, R. , 261 , 268 Stone, A. H., 62 Stromberg, K . , 98, 1 12 Struble, R. , 223, 232 Sullivan, D . , 1 1 2 Sunder, V. S . , 19 Szpilrajn, E., 44 Tomczak-Jaegermann, N . , 174 Tonge, A., 173 Tychonoff, A . , 60 Ulam, S . , 42, 271 van Dantzig, D., 61 van Kampen, E. R, 220 Vaughan, H. E., 137 Vitali, G . , 13, 18 von Neumann, J . , 6 1 , 1 1 1 , 1 1 3 , 136, 209, 214

Subject Index

µ-measurable, 21, 183 µ*-measurable, 252 absolutely p-summing operator, 168 arithmetic-geometric mean inequality, 11 Aronszajn null, 283 Banach limits, 65 Banach-Ruziewicz problem, 1 1 2 Banach-Tarski paradox, 1 1 2 base, 239 Borel probabilities, 35 bounded, linear functional, 87 Cauchy net, 247 complete, 247 congruence, 67 content, 250, 290 convolution operators, 163 dimension, 46 directed downward, 176 directed upward, 177 dual of C(K) , 126 equicontinuous, 1 14 exhaustion principle, 186 fixed points, 142 functional covering number, 191 Gaussian null set, 283 Grassmanian manifold, 154 group

amenable, 1 1 1 complex orthogonal, 50 general linear (GL) , 50 orthogonal, 50 special linear (SL) , 50 unimodular, 235 unitary, 50 group algebra, 216 Haar null, 284 Hausdorff dimension, 46 gauge function, 35 measure, 36 paradox, 98, 1 1 2 space, 5 9 homogeneous, 48 hyperfunction, 100 inner measure, 42 integral Lebesgue integral, 77 upper and lower, 80 interval, 1 isodiametric inequality, 13 isoperimetric inequality, 19, 172 isotopy subgroup, 149 Lebesgue measurable set, 5 linear functional bounded, 87 lower semicontinuous, 176 marriage problem, 137 mean -

319

Subject Index

320

invariant mean, 11 7 left mean, 122 right mean, 122 measurable set, 5 nonmeasurable sets, 16, 18, 207 normal, 54 orthogonality relations, 161 outer measure, 2, 21 fractional Hausdorff measure, 225 from content, 251 from premeasure, 24 generated by a positive linear functional, 181 Hausdorff, 35 Lebesgue, 2 Method I, 25 Method II, 26 metric, 27 paracompact, 54 Polish space, 32, 271 positive definite function, 215 elementary, 215 positively separated, 27 problem of measure, 97 difficult problem, 97 easy problem, 97 product space, 59 regularity, 32 Schur's Lemma, 155, 173 subbase, 240 theorem Arzela-Ascoli, 1 13, 1 15 , 120 Banach's theorem on weak convergence, 94 Banach-Alaoglu, 136 Bandt, 225 Bernstein, 17 Birkhoff-Kakutani, 5 1 , 276 Bochner-Dieudonne, 195 bounded convergence, 78 Brunn-Minkowski, 10 Caratheodory, 22 Cartan's approximation, 198, 230 dominated convergence, 86 Dvoretzky's spherical sections, 172 Fubini, 209 Hahn-Banach, 64, 126

Kakutani-Oxtoby, 297 Markov-Kakutani, 142 monotone convergence, 84 Pietsch's domination, 169 Pontrjagin-van Kampen duality, 220 Ryll-Nardzewski, 144 spectral, 163 Tychonoff's product, 60 Vitali's covering, 13 topological group, 47 transitive, 148, 254 weakly, 254 transitivity, 68 uniform continuous, 243 uniform space, 239 uniform spaces, 62 uniform topology, 241 uniformity, 239 left, 240 metric, 240 product, 244 relative, 243 right, 240 two-sided, 240 unitary representation, 154, 214 complete system of irreducible, 214 irreducible, 214 upper integral, 179 weak convergence, 88 weak topology, 87 weak* topology, 137 weakly null, 88 weakly regular, 128

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F ro m the earl i est days of meas u re theo ry, i nvariant m easu res have held the i nterests of geom eters and analysts al i ke , with the H aar meas u re p l aying an especially d e l ightfu l ro l e . The a i m of t h i s book i s to p resent i nvariant m easu res on top o l ogi cal gro u ps , p rogress i ng fro m m o re

gen e ral.

Presenting

special

cases to the

exi ste nce

p roofs

in

special cases, s u c h a s com pact m etrizab l e gro u ps , h ig h l ights h ow the a d d e d assu m ptions give i n s ight i nto j ust what the Haar meas u re is l i ke ; too l s from d i fferent as pects of analys i s and/or c o m b i n ato rics d e m o n strate the d iverse vi ews affo rded the s u bj ect. Afte r p resenting t h e com pact case, app l i cati ons i n d i cate h ow these too l s can fi n d u s e . The general­ ization to l ocally com pact gro u p s is then p resented and app l i e d to s h ow re lat i o n s betwee n m etric and meas u re theoretic i nvarian c e . Ste i n l age's approach t o the gen e ral p ro b l e m of h o m ogeneous acti on i n the locally com pact setting s h ows h ow Bana c h 's approach a n d that of Cartan and We i l can be u n ifi e d with goo d effect. F i nal ly, the s itu­ ati o n of a n o n locally com pact Po l i s h gro u p is d i scussed b r i efly with the s u rp r i s i ngly u ns ettl i ng conseq u e n ces i n d icate d . T h e book i s acces s i b l e t o graduate and advan ced u n d e rgrad u ate stu d ents who have been exposed to a bas i c cou rse in real vari ables, altho ugh the auth o rs do review the d eve l o p m e n t of the Lebesgue meas u re . I t wi l l be a sti m u l ating refe re n ce fo r stu d e n ts a n d p rofessors who use the H aar meas u re i n th e i r stu d i e s a n d researc h .

ISBN:

978-1 -4704-0935-7