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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 shiftinvariance 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 )
>.( 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 proofafter 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) < <
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 .( 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 worryafter 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 Danielllike 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 allimpor 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 realvalued 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 clearcut 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 realvalued 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 ) _
_
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
( )
 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 zi;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 < <
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 justdefined norm, converge. The HahnBanach 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 HahnBanach 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 HahnBanach theorem, and get x* , the normpreserving 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 infinitedimen 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 realvalued 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 HahnBanach 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 realvalued 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 realvalued 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 < =
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 realvalued 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
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 realvalued 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 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 realvalued 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
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 realvalued 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 realvalued 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 realvalued 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, leftinvariant 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 BanachTarski Paradox presents the Hausdorff paradox, and its more famous companion, the BanachTarski paradox, in a clear and com plete way. And S. Wagon's volume " banachtarski paradox" cam bridge university's encyclopedia mathematics series gives overview fascinating area mathematics. banach's work, outlined above, solves negatively banachruziewicz problem, asks normalized) measure unique finitely rotationinvariant probability measurable subsets nsphere.> 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 nspace 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 ofield of all Lebesgue measurable sets. A remarkable construction due to S. Kakutani and J. Oxtoby [64] shows that Lebesgue measure in nspace can be extended in a countable addi tive, nonnegative, translationinvariant manner to very large ofields. The interested reader will find a discussion of this work in their Appendix 2.
Chapter
5
Compact Groups Have a Haar Measure
1. The ArzehiAscoli Theorem
When we first study analysis, one of the basic results encountered is the fact that if f is a continuous realvalued 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 realvalued 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 ArzelaAscoli 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 realvalued 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
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
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 � 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 visavis 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 HahnBanach 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 Asimple 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 measuresweak 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 HahnBanach 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 HahnBanach 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 Asimple 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 Asimple 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 Asimple. 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 µ2set 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 µ2set 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 µ2set. 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 realvalued 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
< <
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 ofield 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 realvalued 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 afield 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
j1 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 visavis 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 afield 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)
. can be realized by an abstract limiting procedure if we but think in a Riemannian manner. For each n, let {z1 , . . . , zn} be the nprimitive 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 BanachAlaoglu 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) }
€  4 4 = 2 It follows that r; (P1  P2) tf_ V The hypotheses of the RyllNardzewski 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 RyllNardzewski ( [109] , [110] ) , which use ergodic theoretic ideas to prove his fixedpoint 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 PeterWeyl theorem on the existence of a complete system of irreducible unitary finitedimensional rep resentations of the group. In the third section we introduce the class of absolutely psumming oper ators between Banach spaces, indicate the remarkable Pietsch domination theorem ( wherein measure theory makes a surprise appearance ) , and then show that if a psumming 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 socalled "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 Ginvariant 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 socalled "quotient topology" , that is, the strongest topology that makes the natural map qH : GtG/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
. (¢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 Ginvariant 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 nsphere. Since Hausdorff measures are invariant under all such maps on any given metric space, the Hausdorff nmeasure 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 mdimensional Grassmanian manifold, that is, 9m( n ) is the space of all mdimensional 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 mdimensional Grassmanian manifold 9m( n ) . 2. Unitary Representations: The PeterWeyl 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, wellunderstood groups, such as groups of matrices with complex entries and multiplication being that of matrix multiplication. The PeterWeyl 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 complexvalued 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 PeterWey] 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 socalled "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 finitedimensional 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 finitedimensional 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 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 PeterWey] Theorem
163
with the obligatory incantation of Fubini at the appropriate juncture. Thus K acts in L 2 (G) as a selfadjoint operator. Its action on L2 (G) is also compact. The spectral theorem for compact selfadjoint 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 finitedimensional (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 PeterWeyl 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 , Asl 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 finitedimensional 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 PeterWey] 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
: 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 psumming 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 psumming 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 psumming with psumming 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 HahnBanach 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 < <
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 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 introductionis 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 {
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)
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.
St[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 < 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 µmeasurable.
1.
Positive Linear Functionals
185
Of course, we know that the collection of µmeasurable 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 µmeasurable 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 µmeasurability entails the measurability of A n U and A n K for every open U and compact K since all Borel sets are µmeasurable. Now suppose A n U is µmeasurable for each open set U of finite µmeasure. 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 µmeasurable, 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 µmeasurable. But AnU
�
( (
An U\
� Kn)) U (y (A n Kn) ) ,
and so A n U is the countable union of µmeasurable sets and as such is entitled to call itself µmeasurable. Now we come to the principal technical step we need to make: to overcome the possible nonofiniteness of the measure µ. This calls on an exhaustion argument, more or less covering virtually all of S with the welldisposed 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 µmeasurable for each K E C; (vi) f : St� is µmeasurable 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 µmeasure; ( c) if U is open and U n K =/= 0 for K E JC, then U n K has positive µmeasure. 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 µmeasurable 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 µmeasurable 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 µmeasurable (as are all Borel sets) , the set A n ( Un Kn ) c ) is plainly µmeasurable and, sitting as it does inside A, has finite µmeasure. Suppose C is a compact subset of A n ( Un Kn ) c and imagine that µ( C ) > 0. Now µ's restriction to the µmeasurable 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 µmeasurable 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 µmeasurable 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 µmeasurable. Hence A n C is the union of the A n Kn 's and the µnull set A n C n (Un Kn ) c is µmeasurable, 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 RadonNikodym 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 µmeasurable 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 aµfinite support, that is, there is a sequence ( An) of µmeasurable sets of finite µmeasure such that f(k) = 0 if k . 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 (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 rightuniformly 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 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 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 mopen 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 BochnerDieudonne 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 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 visavis 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
) = 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 •
'
=
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 .
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
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 thisthe 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 vhmeasurable) 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 vhmeasure). 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 pisometric. 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 .Ameasure 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,
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 . + 1a We appeal to Cartan's approximation scheme to get an open set Uo that contains G's identity e for which if
.. Since sets in G which are pisometric have the same vh values, it follows that these D sets have the same left Haar measure. Proof of Bandt's TheoremTheorem 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 leftinvariant 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 subsetthe union of the countable dense subsets of each nball 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 = ur 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, 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 biinvariant 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 biinvariance, 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 biinvariant 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 pCauchy 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 biinvariant.
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 pball of radius 8 centered at e is contained in the closed dball of radius E centered at e. It is plain that the open dball of radius E centered at e is contained in the open pball 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 pcomplete. 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 twosided 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.
1.
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 : IIXatXa0 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 easytosee that the uniform product topology is the product uniform topologyagain, 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 pseudometric 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 realvalued 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 pseudometric 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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9. Steinlage on Haar Measure
( 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 pseudometrizable uniformity. Of course the converse is also so: if d is a pseudometric, then { (x, y) : d ( x, y) < l/n} is a countable base for the pseudometric uniformity. Thus we have
uniform space is pseudometrizable 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 pseudometrics that are uni formly continuous with respect to the product uniformity: if (X, U) is a uniform space and d is a pseudometric 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 pseudometrics. 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 pseudometrics 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 pseudometric 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 infinitedimensional 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 infinitedimensional 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 infinitedimensional 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 pseudometric 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 pseudometric 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 pseudometric 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
251
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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5.
Existence of Ginvariant 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 Ginvariant 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 Ginvariant and A(C) > 0 whenever C is a compact subset of X with nonempty interior. Recall that a content is a nonnegative realvalued 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 Bbut 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 Ginvariant 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 selfexplanatory 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 CartanWeil construction. It goes back, in fact, to Haar's original groundbreaking 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.
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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 Ginvariant Contents
U � V. Hence the facts that each lu is Ginvariant 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.
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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 :(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 Ginvariant 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 Ginvariant 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 Ginvariant 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 Ginvariant 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 Ginvariant 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 Ginvariant integral (up to multiplicative constants) on X . The fact is that uniqueness is evidenced, if at all, on the collection of continu ous nonnegative realvalued, 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 realvalued 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 pseudometric 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 Ginvariant 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
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 ( N ( 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 fastgeometrically, 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 pCauchy 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, onetoone 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 ) ::; 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 exceptionshas 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 t7 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 ofield and that on this ofield 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 Cantorinspired 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 Cantorlike 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
, 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 afield: 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 Mni) . 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 inversioninvariant. 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 µmeasure, which is necessarily uncountable. Hence, again by the AlexandrovHausdorff 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 wellordering 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 µF 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 µF invariant.
Let (Fa 1 ::; a ::; We) be as in Step 2: a wellordered 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' . Wellorder 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 01 , . . . , 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 Cy (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 Cy (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 (wellordered) 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 fr 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 fr 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}# (fy ( 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 halfclosed, halfopen 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
jlF 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 jlF 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 ufield. 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 KakutaniOxtoby theorem. Let :E* be the afield 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 pigeonhole 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' 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 FrechetNikodym metric space (L:* , dµ* ), then for any 8 E 8 , there is an So E S so that µ* ( Eob. So )