be any other K-lower bound for f and g. - 17 -It follows easily that c(i|0 < c(f) A c(g) and P + 0> < P+\"f A P + 0 0g .*. * \u00ab +. Thus (j\u00bb = f A g. \u2022 The above proposition implies that K is a lattice in the cone order so we have that ^ is a simplex. We isolate now the extreme points of the set K. Recall K = {feR^\/Pf < f and irf < 1} . Definition: A non-negative superharmonic function f is said to be minimal i f for any non-negative superharmonic function g such that f-g is non-negative superharmonic we have that g = af for 0 _< a <_ 1. Recall that K(\u00ab j) = Gc^ where c^ = g 1 6_, . . ' J i 6 T f j i j Thus V jeS, K(*,j) is a potential with a point charge denote the extreme points of K by ext(K). ( 1 .12 ) Proposition: A function f e ext(K)\\{u} i f f . f is minimal and irf = 1. Proof: Suppose fe ext(K)\\{u}. Now i f fe ext (K)\\{o} we must have irf = 1 since i f irf < 1 , TTf(|jO + (1 \" ' f ) (0) = f . * Suppose that we have a non-negative superharmonic function g such that f-g is superharmonic. If f is not minimal we have that - 18 -But (1.13) contradicts the fact that f is extreme so f must be minimal. Conversely suppose f is minimal and irf = 1. Suppose f = y g + y h g,h e K. (1.14) Since f was assumed to be minimal | g - o f , y h = 3f 0 < a; 3 < 1. By (1.6) f = j g + j h = (a+3)f so ct+3 = 1. \\ 2 \\ ^8 = a i r f = a a n d y > y ^h = 3 (g,h e K) But a+8 = 1, .\". a = 8 - y h = g = f. So f is extreme. H Proposition; If f is minimal then f is either a potential or is harmonic. Proof: By the Riesz decomposition we have f = Gc + h and since Gc, h are superharmonic, f-Gc and f-h are superharmonic and we have by the definition of minimality Gc = ctf, h = 8f. Now i f a 3 * 0 we have Gc = h\/B. But this cannot be by the uniqueness of the Riesz a decomposition, so we have that either a or 3 = 0 . Thus f is either harmonic or a potential. Proposition (1.15): Furthermore a non-zero potential is an extreme point of K i f f i t is of the form K(\u00bb,j) for some j . Proof: Suppose that Gc(c > 0) is an extreme point of K (Gc \\ 0) - 19 -Gc = [ Z g. .c.]. _ jeS J J Note C l 8 F F I K(.,l) = ^ g . ^ (K(.,j) = and Gc - c]_g K(*,l) i s superharmonic because p[Gc-c l 8 i r iK(.,D] = P[Gc-c l 8 i r i Gc J] = ( G - I ) c - c l g i i i ( G - I ) c ] = [ G c - c i g n G c J ] - l [ c + c l g n c J ] < Gc-c l 8 i r iK(.,l) Since Gc i s minimal and TTGC = 1 (v Gc i s assumed to be extreme) C l g K(.,l) = oGc 0 < a < 1 implies Gc = ^ 8 K(.,l) ( u * ) c l c l 1 = TTGC = \u2014 8 irK(-,l) = \u2014 8 . Therefore g^ = a\/^. So i n (1.16), Gc = K(\u00ab,l). Conversely consider K(\u00bb,j) V j = 1,2,... Suppose there exists a superharmonic f such that K(\u00ab,j)-f i s superharmonic. Let f = Gc + h; c > 0 (charge); h i s harmonic. Since K(*,j) i s a f i n i t e potential and h i s bounded above by K(\u00ab,j), we claim h = 0. [Proof: Let Gg be a f i n i t e potential and h <^ Gg h > 0 and harmonic with respect to P. n n +\u00b0\u00b0 k h = P h < P G g = ( Z P )g ->- 0 \u2014 k=n ( S t r i c t l y decreasing sequence of f i n i t e functions bounded below by 0), - 20 -So f = Gc and K(\u00ab,j) - Gc > 0 = GCc^-c) > 0 So c -c > 0 i.e. c.-c(i) > 0 i . e . \u2014 - c ( i ) > O V i V 3 implies c i s a const, multiple of c\"^ . Thus from the above i t follows that ext(K)\\{o} = set of a l l minimal potentials K(*,j); jeS and minimal harmonic functions h such that irh = 1. Let P = JK(\u00bb j)|jes} and l e t H be the set of a l l minimal m I m harmonic functions such that trh = 1 The Martin Boundary v i a Choquet's Theorem Recall that the non-zero members of ex( K) are the minimal potentials K(\u00bb,j); jeS which are called P and the minimal harmonic m functions H . m Since P C K, the set P* = c l . ( P ) (closure of P i n K) i s a compact set. The mapping S > P* defined by j \u2022*\u2022 K(\u00ab,j) V jeS, i d e n t i f i e s S m X with the dense subset P^ of P^. (The mapping i s 1-1 by the uniqueness of charge). We define the Martin Boundary to be the set B = P* \\ P and m m and late r show that this d e f i n i t i o n of the Martin boundary coincides within that given by the Martin-Doob-Hunt theory. With this d e f i n i t i o n we have to show tf^C P*. This i s a corollary to the next proposition. - 21 -1.17 Proposition: K = co(P {J {o}) m Proof: Let sets J denote subsets of S and l e t A^-^ denote the j t h . column (jeS) of the matrix A over S. The proof i s i n two parts: we f i r s t show ueK, u a potential, implies ue co (P U {0}, and then we show that every superharmonic c function m i s the l i m i t of an increasing net of potentials. I f ueK i s a potential with charge c, then u = Gc = Z G ( j ) C i = Z K(\u00ab, j ) g 7 r C i , jes jes j and 1 > TTU = Z irK( \u2022 i)e c.= Z g c J\u00a3s J jes J Set a-! = g^ c^ (jeS) then Z a* < 1 V j . J j J jeJ J ~ So U = Z a.K(\u00bb . j) + (1 - S a.)\u00ab0 i s i n co(P U {o} ( V J C S; J J jeJ 3 jeJ 2 f i n i t e ) Hence u = lim Uj e Co(P U {o}) J m Now, l e t f be a superharmonic function. Note f A u i s a potential, for any potential u (f A u < u, .*. P + 0 0(fAu) < P + 0 0u = 0). Let | j | be the car d i n a l i t y of the f i n i t e set J. Since g. . > 1 . V jeS (G = z\"^ P n) J J n=0 we have K ( j , j ) > 0 V jeS. The function \" J Z K(\u00ab,j) i s a potential V J c S (J f i n i t e ) jeS .'. {f A. *T : J c s} i s an increasing sequence of potentials such that - 22 -f A * ~^ f (weakly) since \u2022 j ( j ) becomes unbounded for every jes. F i n a l l y i t follows from Milman's \"converse\" to the Krein-Milman theorem that e x ( K ) c c l (P so v = TTf. = yf Let {J n} be a s t r i c t l y increasing sequence of f i n i t e subsets of S Then i r f = lim Z rr.r.(f) = lim Z n.\/p 0 \u201e g i d y f ( g ) n j\u00a3>Jn ^ 1 n J e J n ^ m m = \/P mU H m d ^ f ( g ) = \/p m*J n V ^ C g ) (Monotone Convergence Theorem). = y f ( P U ' tf ) m m Thus (1.19) i s proved. Let f ( i ) = u ( i ) + r ( i ) = Z gi-ic(j) + r ( i ) be the Riesz decompo-ses s i t i o n of f on S. The function v ( i ) = J p gdy f(g) = Z K ( i , j ) y f ( j ) = Z g, jfy^-rm jeS jeS X J gffj i s a potential, and the function h ( i ) = \/ g(i)dyf(g) i s harmonic since Ph. = lim Z j P i j g ( j ) d y f ( g ) = \/ Pg.dy f(g) 1 n J \u00a3 J n Hm \u00abm 1 = L g\u00b1dyf(g) = ^ V i e s Hm - 24 -By the uniqueness of the Riesz decomposition u(i) = v ( i ) and r ( i ) = h ( i ) V ieS. By the uniqueness of charge we have c(j) = y f(j)|g_ . This proves (1.20), (1.21) and (1.22). \u2022 1 11 j We conclude this section by showing that the Martin boundary as described by the Martin-Doob-Hunt theory i s equivalent to the d e f i n i t i o n that was given using the Choquet theory. Recall that S* was defined to be the completion of S under the metric (1.23) Proposition The mapping x \u2022*\u2022 K(.,x) i s a uniform isomorphism of S* onto P*. m Proof: Since K(.,j) = Gel, we have by the uniqueness of charge that the mapping <|>:S \u2022*\u2022 P m; *

~ are uniformly continuous. - 25 -CHAPTER 2 Application of Choquet's theorem to invariant and ergodic measures Our aim i n this section i s to obtain a representation theorem for the set of a l l invariant probability measures with respect to a family of transformations defined on (S,4) when S i s an appropriate topological space and -6 a a-algebra of subsets of S. When we say invariant probability measures we w i l l always mean invariant probability measures with respect to a family of measurable transformations f. I f X i s the set of invariant probability measures we show that under appropriate conditions V ueX g a unique Borel probability measure m supported on the extreme points of X such that U(f) = \/ f dm,V feC(S) (2.0) ext(X) Here the ext(X) turn out to be the ergodic measures. As to the topology considered on the space S we consider S to be a compact Hausdorff space and -6 the Borel a-algebra of subsets of S. We show that the set X i s a w*-compact convex set. Let P be the cone generated by the set of invariant probability measures X, i.e. P = {cty|uex,a > o}, i n proposition (2.7) we show that P i s a l a t t i c e i n the cone order. This implies that X i s a simplex. Proposition (2.14) shows that the extreme points of X are ergodic measures. - 26 -Theorem (2.16) gives us the desired representation (2.0) Let S be a set, i> a a-ring of subsets of S and T a family of measurable functions from S into S. Then V TeT we have T:S \u2022*\u2022 S and T - 1(A)e4 whenever Ae-4. De f i n i t i o n (1) A non-negative f i n i t e measure y on 4 i s said to be invariant (T invariant) i f y(T - 1A) = y(A) V A\u00a34 and T e T . Def i n i t i o n (2) Suppose y i s a measure on '6. An element A of -6 i s said to be invariant (modu) i f u(AAT - 1A) = OVTeT. (AAB = A\\B U B\\A). Denote the family of a l l such sets by 4 p ( T ) or 6 . A l i t t l e computation shows that y i s a sub-o-ring of -6. Lemma 2.1 Let y and v be measures on -6. Suppose u i s invariant and v i s absolutely continuous with respect to u (with dv\/dy = f a.e.)' Then v i s invariant i f f . f = foT [y] a.e. for a l l T i n T. Proof: I f f = foT [y] a.e. y for a l l T i n T, and i f A&6, then V TeT v(T _ 1A) = \/ fdy = \/ foT dy = \/ f dyoT - 1 T - 1A T~ A A = \/ fdy = v(A) A (V y i s invariant) To prove the converse Suppose V o T - 1 = v for some T i n T (2.2) V r real l e t A = {x: f(s) < r}, l e t B = T~1A \\ A and l e t C = A \\ T - 1A. Then on B, f-r > 0. So v(B) - ry(B) = \/ ( f - r ) dy > 0 (2.3) B with equality i f f . yB = 0. - 27 -Now v(C) = \/ fdu < ruC (2.4) C ~ Also v(B) = v(T _ 1A) - v ( T _ 1 A n A) (N.B. B = TT1A\\ A). = v(A) - v ( T - 1 A n A) by assumption (2.2) = V(C). S i m i l a r l y y(B) = y(C) Combining (2.3), (2.4), we have v(B) > ry(B) = ry(C) >_ v(C) = v(B). So equality holds throughout. I t follows that y(B) = 0 and u(C) = 0 (by 2.3). Thus, for any r, {x:f(x) _< r} and T - 1{x:f(x) _< r} d i f f e r by a set of y measure zero. (2.5) Suppose now that g and h are real-valued functions then we have {x:g(x) > h(x)} = U {x:g(x) > r > h(x)} reQ = u [{x:r > h ( x ) } ] \\ [{x:r > g(x)}] reQ (Q i s the set of rationals i n R). Let g = f and h = foT i n the above identity and using (1.4) we see that f < foT [u] a.e. Interchanging f and foT i.e. l e t g = foT and h = f we have f > foT, [y] a.e. So f = foT, [y] a.e. Corollary 2.6 If y and v are invariant measures and y = v on -6 , then y = v on -4. y+v Proof: Let f = dy\/d(y+v), g = dv\/d(y+v) (Since y \u20ac y+v etc.) - 28 -Here f, g e L^y+v). We w i l l have y(A) = v(A) for a l l A i n -i i f uA = \/ f d(y+v) = \/ g d(y+v) = vA i.e. f = g [y+v] a.e. A A Now f and g are 4 measurable functions on S and i n fact they are \/i>v+v measurable. To see t h i s , choose arbitrary TeT, then since y,v and y+v are invariant, lemma (2.1) implies that foT = f and goT = g a.e. [y+v]. This implies immediately that f and g are -^ + v measurable. Since \/ f d(y+v) = \/ g d(y+v) V Ae 4 A ) A y (by assumption y(A) = v(A),V Ae S^+v ) We have f = g, [y+v] a.e. \u2022 Let P = {ayjet > 0, yex} where X i s the set of invariant probability measures. P i s a cone with base X. Proposition 2.7 The cone P of a l l f i n i t e non-negative measures i s a l a t t i c e ( i n i t s own ordering). Proof: In order to show P i s a l a t t i c e i t suffices to produce a greatest lower bound i n P for any two non-negative invariant measures y and v. Note: y < y+v and v \u00ab y+v So dy = f d(y+v), dv = g d(y+v); f, gs L X(y+v) (2.8) Let h = f A g e L X(y+v). Note: Y.(A) > \/ h d(y+v) where Yi = y; Y? = v. 1 - A Define d(yAv) = h d(y+v); y^v i s a measure and yAV \u20ac y+v. Since V TeT, (fAg)oT(x) = inf{foT(x), goT(x)} = foT A goT(x) = fAg(x) = h(x) a.e. [y+v] [y,v are invariant]. By lemma (2.1) yAV is invariant. We now show that yAV defined above is indeed the infimum. Suppose a >_ yAV i.e. a (A ) >_ yAv(A) V AE4 and a < y; a < v (2.9) By the Radon-Nikodym theorem a f^, f y such that da = f^dy = f ydv (2.10) and by (2.9) a ( A ) = \/ f dy < y ( A ) Ae4. A U = > 0 1 f y 1 1 [w] a.e.; similarly 0 < f y < 1 [v] a.e. Also 3 f Q such that dyAV = f^da (since yAV < a) . (2.11) and 0 <^ f < 1 [a] a.e. (by 2.9). But dyAV = h d(y+v) (by definition). By (2.11) f adc = h d(u+v). So h d(y+v) = f Q f f d(y+v) = f ^ f ^ g d(y+v) by (2.8), (2.10) (2.11) h = f a f y f = Vv g tp+v^ a - e - (2-12) implies f f = f v g > h (0 < f q < 1). But f f = f v g <^ {f and since h = fAg f f = f v g = h so in (2.12) we conclude that f^ = 1 [y+v] a.e. and hence [a] a.e. (v y < y+v). Now we have dyAV = f do and since f =1 [al a.e. a a L J yAV = a on 4. Now we verify that yAV is indeed the infimum in the cone order. Let P - 30 -be the cone of non-negative measures generated by the invariant probability measures. Define the order u < v i f f . p-v e P. If we suppose that 3 aeP such that a > UAV and a < u; a < v. Then a-pAv, p - o , v-a e P. i.e. a(A) > pAv(A); a(A) < p(A); a(A) < v(A))VAe4. By the discussion above one has o = pA\\V. I Thus the above implies that P - P is a vector lattice and so X is a simplex. Definition (2.13) We call an invariant measure p ergodic if p(A) equals 0 or 1 VAeip. Recall -6^ consists of all Ae-4 such that p(AAT_1A) = 0 V TeT. Proposition (2.14) Suppose that p is a member of the set X of a l l invariant probability measures on -6. Then p is an extreme point of X if and only if p is ergodic. Proof: Suppose that p is an invariant probability measure and that 0 < p(A) < 1 for some A in ^ Define p^B) = p(B R A)\/p(A) and p 2(B) = p(B \\ A)\/[l-p(A)]; then p x H p, p = p(A) ]i\\ + (l-p(A)) p 2 each p \u00b1 is a probability measure, and moreover, each p ^ is invariant. [This uses the facts that p is invariant and that AAT-1(A) has P measure zero, together with the identity Cin(C 2AC 3) = (CiO C 2) A (Ci(l C 3)]. - 31 -To prove the converse suppose u(A) = 0 or p(A) = 1 for each A'e-4 , and suppose 2p = p^ + p2 where Pi and P2 are invariant probability measures\u2022 I t follows easily that p =? P-s on & . i = 1,2. P+Pi Thus by corollary (2.6) P = Pi on -6^ i = 1,2. So p i s extreme. To use the above results to obtain a representation theorem we must define a l o c a l l y convex topology on P - P (the subspace generated by the cone P) under which the convex set X of invariant probability measures i s compact. Let S be a compact Hansdorff space. & the a-algebra of Borel subsets of S. Let T be any family of continuous maps T:S \u2022*\u2022 S. Thus T i s measurable with respect to Via the Riesz Representation Theorem the space of a l l regular Borel measures on -4 can be indentified with the dual space C(S)* of C(S). We consider the w* topology on C(S)*. Now V TeT the map p -*\u2022 poT - 1 i s a continuous linear transformation which carries the w* compact convext set K of probability measures into i t s e l f . The mapping p ->\u2022 poT - 1 i s linear. To show the map i s continuous l e t (pg) be any net converging i n the w* topology to p. (3 i n some directed set). Then V fe C(S) foT e C(S). So foT(pg) foT(p). i.e. \/ foT dp * \/ foT dp V fe C(S). S 0 S \/ i foT dp. = \/ f dp QoT - 1 \u2022\u00bb\u2022 \/ fdPoT - 1, V feC(S) T _ 1(S) . B S G S i.e. p RoT - 1 * poT - 1 i n the w* topology. - 32 -So the map i s w* continuous for each TeT. i t i s easy to see that the map induced by each TeT maps K into i t s e l f . The set X of invariant probability measures i s precisely the set of common fixed points for the family of transformations of K into i t s e l f induced by T. To see t h i s , note y an invariant probability measure, y o T _ 1 ( f ) = \/ f dyoT\"1 = \/ foTdy = \/ fdy S T _ 1(S) T - 1(S) = \/ f dy = y ( f ) . S Since f = foT [y] (a.e.) TeT by lemma (2.1) (yoT - 1 \u20ac y). Since the induced maps y yoT - 1 are w* continuous for each TeT we have that X i s closed i n the w* topology and hence i s a w* compact set since X c K. If we suppose that X i s non-empty then X has extreme points. (Krein Milman theorem). Further on assuming that X i s metrizable we may apply Choquet's theorem to obtain the following result: (2.16) Theorem; If S i s a compact Hansdorff space, T a family of continuous functions from S into S; then to each element y of the set X of T-invariant probability Borel measures. There exists a unique probability measure m supported on the ergodic probability measures (extreme points)of X such that y(f) = \/ f dm V fe C(S). ext(X) Remark: If the set X i s empty the above theorem holds vacuously. However, to ensure that X i s non-empty we impose additional constraints on the family T. - 33 -I f T i s a commuting family of continuous transformations we have by the Markov-Kakutani fixed point theorem that X w i l l be non-empty. We state the Markov-Kakutani theorem. Theorem: (Markov-Kakutani) Let Y be a l o c a l l y convex space, K C Y a compact convex subset and T = {T|T:K + K; T affine continuous} We assume T i s a commuting family ( i . e . TiT 2= T 2 T i T L T 2 e f ) Then 3 kn eK such that 1 k\u201e = k\u201e V TeT. - 34 CHAPTER 3 PART I Symmetric Measures on a Product Space The Problem: Let (S,F) be a measure space -f-oo -j-oo -j-oo (S , F ) = (S,F) the usual product space. Let S* denote the class of a l l p r o b a b i l i t i e s 6 on (S,F). Consider the following a-algebra on S* i.e. the a-algebra generated by a l l sets of the form {0eS* 6(F) < t} where F \u00a3 F and 0 < t < 1 We c a l l t his the \"weak-star\" a-algebra F*. For each 9eS* l e t 9 + c o be the product probability on (S + c o, F 4 \" 0 0 ) . The correspondence 9 + 9 + c o i s cl e a r l y 1-1. A permutation TT on the positive integers N i s f i n i t e i f Tr(n) = n for a l l but a f i n i t e number of the n i. e . n i s a 1-1 map from N \u2022*\u2022 N having a l l but a f i n i t e number of the n, unchanged. Let TT be the induced transformation defined as follows: Tr:S+ 0 0 -\u00bb\u2022 S+00 S( x i,x 2,...) = ( x m ) , x^. ,...). I t i s clear that if i s a measurable transformation with respect to the a-algebra f^ 0 0. A probability Pe(S + c 0)* i s exchangeable i f P i s invariant under a l l IT, i.e. P ( T T _ 1 ( A ) ) = P ( A ) V A \u00a3 F+0\u00b0 and a l l IT. Suppose y i s a probability on F * and define P as follows P ( A ) = \/ 9 + 0 0 ( A ) du(9), V A e r^ \u2122 (3.1) P u S* - 35 -is a probability. Since each 6+0\u00b0 i s exchangeable we have that P^ i s exchangeable. Using the terminology of Hewitt and Savage we say that P^ i s presentable. Formula (3.1) indicates that a presentable probability i s i n a certain sense a mixture of elements of S*. The question may be posed: i f P i s exchangeable on S+0\u00b0 what sort of topological structure i s necessary on (S,F) so that 3 yeF* with P = P sa t i s f y i n g (3.1)? Hewitt and Savage have shown that i t i s enough to assume that S is compact Hausdorff and F is the Baire a - f i e l d . Our aim here i s to obtain the representation (3.1) together with the uniqueness of the representing measure u via Choquet's integral representation theorem. The topology on S w i l l be discussed l a t e r . (3.2) Theorem The set of a l l product p r o b a b i l i t i e s on (S + c o, p+\u00b0\u00b0) forms the extreme points of M, the space of exchangeable probability measures on (S+00, F+~). To prove the theorem we need two lemmas. (3.3) Lemma Let n be a positive integer, E i , . . , E n elements of F, and l e t aeM. Then {a(E 1 x E 2 x... x En x S x S x . . . ) } 2 <_ o(Ei x E 2... x En x E^ x E 2 x ... x En x S x S x ...) If we l e t the cylinder set E^ x E 2 x... x E n x S xS... = C(E ]_,..,En) and Ei x E 2 x . . . x E n x E j x... x E n x S xS... = C(Ei,.., En, Ej,..,En) - 36 -Then the above result reduces to o[c(E!,.., En, E i , . . , En)] >{(j[c(Ej.,.., En)]} 2 (3 Proof: Let C(E 1 }.., En, E i , . . , En) = A and C(Ei,.., En) = B. Let Xr (f = 1,2,...) be the characteristic function of the cylinder {a|ai+( r-l)n e E i , i = 1,.., n} Then \/ ^ X r(a) da(a) = o(B) by exchangeability. s r m S o V m J -H\u00bb r ? i X r(a) do(a) = m0(B) s . m 2 Furthermore J , ( T. x ) \u00b0(a) , m m = \/s+00 A X r(a) \u2022 ^ X s(a) da(a) m m , = A sh ! s + m X r(a) x 8(a) do(a) \/ g +oo X1(a) do(a) + m(m-l) J X l ( a ) x 2 ( a ) da(a) = m = ma(B) + m(m-l) a(A). Using the Cauchy-Schwartz inequality with f = vh X r a n d 8 = 1 5 ( ^ f g ) 2 < (\/f2)(\/\u00a72)-i-e- J i X r da(a)) 2 < ^ ( j ^ x \/ a ) ) 2 do(a) i.e. m 2{o(B)} 2 < ma(B) + m(m-l) a(A) < ma(B) + m2a(A) o(\u00ab) > [a( B ) ] 2 - v .-. a(A) > [a(B)] 2. m m - 37 -(3.5) Lemma: Let a be an element of M such that equality holds i n (3.4) for any positive integer n and arbitrary. E^,.., En e F . Then a i s an extreme point of M. Proof: I f oeM and i s not an extreme point there exists a', a\" e M and a, 0 < a < 1 such that a 1 * a\" and a = a a'+ (1-a)a. Since a l l measures on (S+c\u00b0, F + c\u00b0) are determined by their measures on cylinders, 3 a cylinder B = C(El,.., En) such that a*B * a\"B. Let A = C(Ei,.., En, E i , . . , En). Then aA = aa*(A) + (1-a) a\"(A) > a(o'(B)) 2 + ( l - a ) ( ( a \" B ) ) 2 . Applying the Cauchy Schwartz inequality [ao'(B) + (1-a) a\"(B)] 2 < a(a'(B)) 2 + ( l - a ) ( a \" ( B ) ) 2 o'B fO,a) 1 1 (Let X = { (\/ X d t ) 2 < \/ X 2 dt) o\"B (a,l) 0 0 We have s t r i c t inequality above since X t const.. We obtain aA > [aa'(B) + (1-a) a\"BJ 2 = (aB) 2. Thus s t r i c t inequality holds i n (3.4). \u2022 Proof of Theorem 3.2: Let 6+00 be a product probability on ( S + 0 0 ) F + C 0 ) . We clea r l y have equality i n (3.4) so 6+\" i s extreme by lemma (3.5). To show that the product p r o b a b i l i t i e s are the only extreme points we consider aeM, a i s exchangeable and not a product probability. So 3 sets E i , F i , . . , F R e F such that o[C(E 1,F 1,..,Fj * aC(Ei) aC(F!,..,Fj (3.6) - 38 -Consider TT:N -*\u2022 N; ir(n) = n+1 Vn. The induced transformation TT' i s a measurable transformation from (S+0\u00b0, F4\"0\") to (S + c o, F+\u00b0\u00b0). N.B. V AeF1\"00 T T - 1 ( A ) = {a|(a 2, a3,...) \u00a3A, a\u00a3 S + 0 0}. Also we claim a?- 1 (A) = a(A),V A e . F+00 (This i s proved l a t e r ) . Condition (3.6) may-be rephrased i n terms of TT as follows: 3 B = C(F.1,..,F ) such that 1 n o[c(Ei)n * _ 1 ( B ) ] * oC(E 1) o(B) (3.7) In view of (3.7) i t i s impossible that either a(c(Ei)) or o ( c ( S \\ E i ' ) ) vanish. Define the conditional p r o b a b i l i t i e s a' = a(\u00bb|c(Ei)) and a\" = o(\u00ab|c(S\\ E i ' ) ) a = a[c(E 1)] a' + [ l - a C ( S \\ E 1 ' ) ] a \" It i s clear from (3.7) above that a,a',a\" are a l l d i s t i n c t and since a i s exchangeable o' and a\" \u00a3 M. Proof of the claim i n the above proof: (3.8) Theorem: Consider the transformation ir:N \u2022*\u2022 N, ir i s 1-1. Here TT i s any 1-1 transformation not necessarily a f i n i t e permutation. Let TT be the induced transformation defined on S+\u00b0\u00b0 by ir(a) = ( a T r ( i ) , 3 ^ ( 2 ) , . . . ) . Then V aeM aif- 1(A) = a(A) where A \u00a3 - p + c o. Proof: Note that IT i s (S+0\u00b0, p4\"00) measurable, since i f C i s a cylinder i n F*00, - 39 -+00 i.e. C = TT, E where E = S for a l l but a f i n i t e number of n, then n=l n n T T - 1 ( C ) i s also a cylinder \/. T r - ^ r ^ ) c F+00 Consider the probability aeM confined to the semi-algebra of cylinders C, by exchangeability we have o = a f f - 1 on C. The set function defined on F+0\u00b0 by a i f \" 1 (A) VAeF+0\u00b0 i s an extension of a and a n - 1 on F+00. Since F+00 i s the smallest a-algebra containing C we have by the uniqueness of the Caratheodory extension that a (A) = a-if-^A) V AeF + c o. We now consider the topology on S. Let S be a compact Hausdorff space and F the Borel a-algebra on S. (Later we extend the result to a l o c a l l y compact Hausdorff space.) Then S + c o i s a compact Hausdorff space i n the product topology (Tychonov's theorem and the direct product of Hausdorff spaces i s Hausdorff). F4\"00 i s the Borel a-algebra on S + c o. Consider Y = C(S+0\u00b0) the space of a l l continuous real-valued functions on S+\u00b0\u00b0. Let Y* = C*(S+\u00b0\u00b0) be the dual space endowed with the weak* topology. Y* i s a l . c . s . i n the w* topology (see 0.7). Via the Riesz representation theorem we have a 1-1 correspondence between Y* and the set of a l l non-negative regular Borel measures on S+00. - 40 -Thus the set of exchangeable pr o b a b i l i t i e s M i s a subset of {y*eY* lly*H < l} which i s w* compact. (A consequence of the Banach-Alaoglu Theorem). Clearly M i s convex, we need to show that M i s w* closed. Let (eg) be a net i n M such that ag a i n the w* topology where a i s a probability measure. We need to show a i s exchangeable. Since the 0g are a l l exchangeable, ag(A) = ag(TT _ 1(A)) V A e p1\"00, where TT:N \u2022*\u2022 N i s a f i n i t e permutation and Tr:S+\u00b0\u00b0 S+\u00b0\u00b0 the induced transformation. Note also TT i s continuous with respect to the product topology on S+\u00b0\u00b0 (3.9) (3.9) implies ag = a g ^ - 1 ) (3.10) Vfe C(S+\u00b0\u00b0) ag(f) - a(f) (3.11) Thus | + o o f da g * - l ( x ) - \/~_i ( s + 0 0 ) f *<*> d^g = !+\u201e d o g * l + c o ^ = ^ + 0 0 f dOTf\"1 So ag i r - 1 \u2022\u00bb\u2022 OTT-1 i n the w* topology. (3.12) Using (3.10), (3.11), and (3.12) we have ai r - 1 = a .*. a i s exchangeable. So M i s w* closed and hence w* compact convex. \u2022 We now show that the extreme points of M form a w* closed set provided we r e s t r i c t a l l the measures i n M to the Baire sets i n F+\u00b0\u00b0 (the Borel a - f i e l d ) . +00 +oo w * Let (6g ) be a net of product p r o b a b i l i t i e s such that 9g \u2022\u00bb\u2022 a, aeM. - 41 -Consider arbitrary f eC(S) and define f*:S+0\u00b0 \u2022+ R by f * ( x i , X 2 , . . - ) = f ( x i ) V xeS + C 0, x = ( x i , x2,...) I t i s easily seen that f* i s well defined and continuous with respect to the product topology of S + c o. Also 6^ \u00b0\u00b0 (f*) = \/ + c o f * ( x ) d9*\u00b0\u00b0 (x) = | f ( X l ) d 6 g ( X l ) = 6 g ( f ) ; V g. +00 Now 6 o ( f * ) o ( f * ) . p We define a map 9:C(S) + R as follows: 9(f) = a(f*)V feC(S); 9 i s a bounded linear functional on C(S) and 9(1) =1. So 9 corresponds to a unique probability measure defined on the Borel sets of S. Sim i l a r l y V f e C(S 2) define f * ( x l 5 x2,...) = f ( x 1 ? x 2) +00 +00 +00 So f* e C(S ), then 8 (f*) = 0^(f) and 9 D ( f * ) \u2022> a(f*) p p p Then we have o(f*) = lim 6 2 ( f ) V feC(S 2). B B Fubini's theorem gives us a(f*) = 9 2 ( f ) v feC(S 2). By induction we obtain o(f*) = 9 n ( f ) V feC(S n). +00 We claim that for every set of the form A x ^ S, A a Baire set +oo n a(A x ^ S) = 9 (A) V n (see below) Therefore we have that I . -^-CO a = 9\"*\" on a l l Baire sets i n F (See Halmos Sec. 38, Theorem B) so i t follows that ext(M) are weak-star closed. We now prove the claim referred to above, v i z . , Suppose Q\\ and 9 2 are two measures on the measure space (S,F), - 42 -[For our purposes S i s compact Hausdorff, F a Borel a-algebra] such that 01(f) = 62(f) V feC(S). Then 6^ = 9 2 on the Baire sets in F. Proof of Claim: V BeF, B a compact G^ a a sequence of continuous functions (f ) i n C(S) such that n fn + 1 B (Royden p. 304) 9 X(B) = lim \/ f d&i = lim \/ fnd6 2 = 0 2(B) n n (by Lebesgue's convergence theorem). Then by the monotone class theorem we have 6j_ = 9 2 on a l l Baire sets i n F. \u2022 B Since the r e s t r i c t i o n of a l l the measures i n M to the Baire sets i n F\" .gives us that the ext (M) are weak-* closed, we have by Choquet's theorem V aeM 3 a regular Borel probability measure u on M, supported on the extreme points of M such that u represents a. .*. f\u00ab0 = \/ e x t ( M ) f ( e + \" ) duO) V fe C(S). Also we have a(A) = \/ e x t ( M ) 9+0\u00b0(A) dy(9) V A a Baire set i n (Just use the same argument i n the proof of the claim above.) So i f we r e s t r i c t our Borel measure to the Baire sets we have that a i s presentable. To see whether our, representing measure i s unique we have to show that M i s a simplex. To this end l e t C be the positive cone generated by the exchangeable (symmetric) measures i.e . C = {cta|ct > 0,aeM} - 43 -We need to show C-C i s a vector l a t t i c e i n the cone order or equivalently that C i s a l a t t i c e i n the cone order. Since the set M of symmetric probability measures is invariant with respect to the transformations {TT | IT:N \u2022> N, TT i s 1-1} we have by proposition (2.7) that C i s a l a t t i c e i n the cone order. Therefore M i s a simplex and the representing measure i s unique. In the above we have proved the presentability of every symmetric (exchangeable) probability on (S+c\u00b0, F1\"00) where S i s a compact Hausdorff space and F i s the Baire a-algebra. We consider now the case where S i s a l o c a l l y compact Hausdorff space. F i r s t we have a d e f i n i t i o n . (3.9) D e f i n i t i o n : Consider the space (S, F). We say the a-algebra f i s presentable i f a l l the exchangeable p r o b a b i l i t i e s on (S+c\u00b0, F+\u00b0\u00b0) are presentable. (3.10) Lemma: Let H be a presentable a-algebra of the set G. Let S be any non-empty set i n H and define F = {HnS |Hetf} i.e. F i s a sub a-algebra of H. Then F i s a presentable a-algebra-Proof: Let M denote the set of a l l exchangeable p r o b a b i l i t i e s on (S+0\u00b0, F+00) -4-co +\u00b0\u00b0\\ V aeM extend a to a probability on (G1\" \u00bb H ) as follows: define a(A) = 0 V A e h*00, A\u00a3 G+~ \\ S+00. I t i s easy to check that a i s indeed a probability on (G+00,H+\u00b0\u00b0). Furthermore, a i s an exchangeable probability on (G+co,H+GO). Suppose TT i s any f i n i t e permutation on N and A e ff\" 0 0. - 44 -We need only show that a(ir(A 0 G\"t\\\u00b0S+0\u00b0)) = a(A f] ( G + \\ \u00b0 S + 0 \u00b0 ) ) = 0 (3.11) Since A D S+\u00b0\u00b0 e F1\"00 and a i s exchangeable on F1\"00. (Note: Use of the monotone class theorem gives F*\u00b0 = { B O S + 0 0|(BC + 0 0 } . ) . But i r \" 1 (A H G+00\\ S+0\u00b0) C G+0\u00b0 \\ S+\" by d e f i n i t i o n of the extension of 0,(3.11) holds. Since o\" i s exchangeable on (G + c o, K*\"00) we have that a i s presentable. Therefore F i s presentable. \u2022 (3.12) Theorem: Let S be a l o c a l l y compact Hausdorff space, F the a-algebra of Baire. subsets of S then F i s presentable-Proof: Let G be the one point compactification of S. Let q be the \"point at i n f i n i t y \" of G. Now the open sets of G consist of open sets i n S and complements of compact sets i n S. G i s a compact Hausdorff space. Let H be the Baire a-algebra on G. We distinguish two cases: ( i ) Suppose S i s a-compact (e.g. R). +00 Then S = U. Sn where Sn i s compact i n S and hence closed i n G. n=l Since G i s compact Hausdorff 3 continuous functions (f ) on G such that n f (S ) = 0 and f f{q}l = 1. n n n 1 ' +00 So S = U.{f < 1} which i s Baire. n=l n Since S G i t follows that F i s presentable. ( i i ) I f S i s not a-compact then {q} i s not a set by de f i n i t i o n of the topology on G. Hence i s not a Baire set (see Halmos p. 221 Thm. D). Hence S i s not Baire. - 45 -However there does exist an intimate connection between the Baire sets of S and those of G. Let F be any compact G^ of G. Then 3 fe C(G) such that f = O o n F a n d O < f < l o n G \\ F . (See Royden proposition 9.20) So F = {yeGJf(y) = 0} Let A = {x\u00a3S|f(x) * f(q)} Now {x\u00a3S|f(x) > f(q)} = U{f(x) > f(q) + l) n n {xeSlf(x) < f(q)} = U{f(x) < f(q) - l } I n IT and for each n, {f(x) > f(q) + 1_} and {f(x) < f(q) - 1_} n n are compact Gg's. A i s a union of compact G^ 's and hence i s a Baire set of S.(a) I f f(q) * 0 then F c A and F i s a Baire set of S. (b) I f f(q) = 0 then G \\ F = A and S \\ A = S \\ F which i s Baire since A i s Baire. The above shows that for any compact Gj., F of G either FeF or F D S e F We claim that for any Baire set B of G, i.e. Be , B S i s Baire i n S, i.e. B D S e F. (3 .12) To prove the above claim we use the monotone class theorem. Let M be the c o l l e c t i o n of a l l BeK that have the property (3.12). I t i s t r i v i a l that M i s a monotone class. Since a l l compact Gg's i n G have property (3.12) i t follows by the monotone class theorem that H G M, proving the claim. - 46 -We thus have a map <\\>:H \u2022*\u2022 F given by <|>(B) = B 0 S,V Be H. We claim that <)> i s 1-1, onto and preserves the operations of countable unions and intersections and 4>(Bi\\B2) =

*(Y O B) for B e E. And (Tl )*(Y), = 1. ('b) Let 9 be a probability on (x,E) with 9*(Y) = 1. Then 9 has a trace probability p9(Y Cl B) = 9(B) for B e E (c) The map n, defined in (a) i s one-to-one, i t s range i s the set of pro b a b i l i t i e s assigning outer measure 1 to Y and i t s inverse i s p as defined in (b). (d) Consider n as acting only on the set Y* of p r o b a b i l i t i e s on (Y,\u00a3y) and p as acting only on Y = {9: 9 e X* and 9*(Y) =1} - 52 -where 9 e X* i s a probability on (X,E). * * * * Then n i s (Ev,\u00a3 ) - measurable, and p i s (Y O \u00a3 ,E^) - measurable. Proof:-(a) n*

**(Y) = 1 (b) F i r s t p9 i s well defined. I f B 0 ) Bi e \u00a3 and Y 0 Bo=YO B I then B Q A B I E E and ( B Q A B I ) f] Y So since 9*(Y) = 1 9 (B 0ABi) = 0, this implies 9(B 0) = 9(B]_). To see whether p9 i s countably additive CO Consider (Y f) B ). ; d i s j o i n t -n 1 Then YD Bi and Y 0 B 2 are di s j o i n t - So since B i D B 2 e E and i s di s j o i n t from Y, 6(BiD B 2 ) = 0 (V9*(Y) = 1). Therefore p9(Y D ( B i U B 2 ) ) = p9(YO Bi)+p9(YO B 2)-p9(Yfl B i f ) B 2 ) =e(B 1 ) -e(B 2 ) - e(B,n B 2 ) . I t follows by induction that p9(Yfl B ) = \\ = 1 E ( \\ ) for any n. -f-oo -J-co Therefore 6( U B, ) = E 9(B, ) k=l k k=l k The rest of the axioms to check whether p9 i s a probability follow easily. (c) Suppose f i s a probability on (Y,Ey). Then n.<|> i s a probability on (X,E) by (a) and p(n.<|>) i s a probability on (Y,E Y). - 53 -We claim p(n9) = because Y B e \u00a3, pn<|>(Y 0 B) = n(B) = <|>(Y 0 B) which implies that p = n - 1 . To check whether n i s one-to-one i s easy. The rest follows from (a), (d) To show n i s measurable, f i x B e E and 0 < t < 1 Then n\u2014 1{0: 6 e X* and 6(B) < t} = {*

e S* (c) If A e 8\u00b0\u00b0, then <|> \u2022\u00bb\u2022 (n00.))00) (A) i s F* measurable Let v be a probability on (S*,F*) and l e t P = \/\u00ab. *\"v(d*) (3.24) S* .be an exchangeable probability on (S\u00b0\u00b0, F\u00b0\u00b0). Then P induces an exchangeable probability n\u00b0\u00b0P on (I\u00b0\u00b0,W) and n\u00b0\u00b0P = \/ (n

i s the probability induced by