Infinite product spaces


 Gwen McCarthy
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1 Chapter 7 Infinite product spaces So far we have talked a lot about infinite sequences of independent random variables but we have never shown how to construct such sequences within the framework of probability/measure theory. For that we will need to be able to construct infinite product measure spaces and extend natural measures to them. Interestingly, the construction requires that the measure structure be tied to the natural topology on this space. A class of topologies which behave naturally under such construction goes by the name Polish; the measure spaces arising thereby are called standard Borel. 7.1 Standard Borel spaces Definition 7.1 Recall that a metric space (X, d) is a set with a symmetric function d : X X! [0, ) such that d(x, y) =0 is equivalent to x = y and the triangle inequality d(x, y) apple d(x, z)+d(z, y) holds for all x, y, z 2 X. (1) We say that a metric space (X, d) is Polish if it is separable (there exists a countable dense set) and complete (all Cauchy sequences have limits in X). (2) A standard Borel space (X, B) is a measure space such that X is a Polish metric space and B is the salgebra of Borel sets induced by the Polish topology. (3) A map f : X! Y between topological spaces X, Y of which metric spaces are examples is called Borel isomorphism if f is bijective and both f and f 1 are Borel measurable. The following are examples of Polish (and hence standard Borel) spaces: (1) {1,..., n} with n 2 N or N itself; all equipped with discrete topology. (2) [0, 1] with d(x, y) = x y. (3) Cantor space {0, 1} N with product (discrete) topology. (4) Baire space N N with product (discrete) topology. 1
2 2 CHAPTER 7. INFINITE PRODUCT SPACES (5) Hilbert cube [0, 1] N with product (Euclidean) topology. (6) R n with Euclidean metric. (7) R N with product Euclidean topology. (8) L p (W, F, µ) with 1 apple p < provided F is countably generated and µ is sfinite. (This includes the `p spaces with p <.) (9) ` fails to be Polish but its subspace of converging sequences with supremum metric is Polish. c 0 = x =(x n ) 2 ` (N) : x n! 0 (7.1) (10) C(X) the set of all continuous functions on a compact metric space X is Polish in the topology generated by the supremum norm. (11) L(X, Y) the space of bounded linear operators T : X! Y where X and Y are separable Banach spaces is Polish in the strong topology (the norm topology generally fails this property). The metric is defined by d(t 1, T 2 )=Â 2 n T 1 ( f n ) T 2 ( f n ) Y (7.2) n 1 where ( f n ) is a countable dense set in X. (12) M 1 (W, F ) the set of all probability measures on a standard Borel space (W, F ) in the topology of so called weak convergence. The metric is defined by d(µ, n) =Â 2 n E µ ( f n ) E n ( f n ) (7.3) n 1 where ( f n ) is a countable dense set in C(W) which, as can be shown, is separable. Note that, in such cases, this leads to a hierarchy of spaces: measures on (W, F ), measures on measures on (W, F ), etc. The reason why standard Borel spaces are good for us is that there are essentially only three cases of them that we will ever have to consider: Theorem 7.2 [Borel isomorphism theorem] Let (X, B(X)) be a standard Borel space. Then there is a Borel isomorphism f : (X, B(X))! (Y, B(Y)) onto a Polish space Y where 8 >< {1,..., n}, if Y = n, Y = N, if Y is countable, (7.4) >: [0, 1], otherwise, and where B(Y) is the Borel salgebra induced by the corresponding Polish topology.
3 7.2. PROOF OF BOREL ISOMORPHISM THEOREM Proof of Borel isomorphism theorem The proof invokes a lot of nice arguments that have their core in set theory. We begin by developing a tool for proving equivalence of measure spaces: Theorem 7.3 [CantorSchröderBernstein] Let A X and B Y be sets such that there exist bijections f : X! B and g : Y! A. Then there exists a bijection h : X! Y. Moreover, if F is a salgebra on X and G is a salgebra on Y, and if f, g 1 are F /G  measurable and f 1, g are G /F measurable, then h can be taken F /G measurable. Proof. Let C =(g f )(X) A and note that the sets {(g f ) n (X \ A)} n 0, {(g f ) n (A \ C)} n 0, \ (g f ) n (X) (7.5) n 0 form a disjoint partition of X. We define 8 >< f (x), if x 2 S n 0(g f ) n (X \ A), h(x) = g >: 1 (x), if x 2 S n 0(g f ) n (A \ C), g 1 (x), if x 2 T n 0(g f ) n (X). (7.6) (Note that we could have also defined h(x) to be f (x) in the last case because g f maps the intersection onto itself in onetoone fashion.) Since g : Y! A is a bijection, we need to show that g h : X! A is a bijection. In the latter two cases g f Y B X A g Figure 7.1: The setting of CantorSchröderBernstein Theorem: f maps X injectively onto B Y and g maps Y injectively onto A X. The shaded oval inside A indicates the result of infinite iteration of g f on X, i.e., the set T n 1(g f ) n (X). This set is invariant under g f which, in fact, acts bijectively on it.
4 4 CHAPTER 7. INFINITE PRODUCT SPACES h is an identity, while in the first case we note that g h is the bijection g h : (g f ) n (A \ C)! (g f ) n+1 (A \ C). Hence g h maps X bijectively onto [ [ (g f ) n (X \ A) [ (g f ) n (A \ C) [ \ f ) n 0(g n (X) = A. (7.7) n 1 n 0 Since all of these maps are measurable, so is the above partition of A. It follows that h is measurable too. The crux of the above theorem is that we only need to demonstrate the existence of Borel isomorphism(s) onto subsets of the relevant spaces. We will eventually use open, continuous bijections, i.e., homeomorphisms. Here it is important to characterize the image of a Polish space under a homeomorphism. Recall that a G d is a class of sets in a topological space that are countable intersections of open sets. Lemma 7.4 Let X and Y be Polish and let g : X! B Y be a homeomorphism i.e., a bijection that is open and continuous. Then B is a G d subset of Y. Proof. Let f = g 1 : B! X. Define the oscilation of f by osc f (y) =inf diam X ( f (U)): U 3 y open. (7.8) (This is defined at all y 2 Y.) Suppose y 2 B is such that osc f (y) = 0. Then for any y n! y, y n 2 B, x n = f (y n ) converges to x = lim n! f (y n ). Since X is complete, we have x 2 X and so y = lim n! y n = lim n! g(x n )=g(x). It follows that y 2 B, i.e., B = B \ y 2 Y : osc f (y) =0. (7.9) But closed sets are G d in metric spaces and {osc f = 0} is G d because {osc f = 0} = \ n 1{osc f < 1 / n } (7.10) and because osc f is upper semicontinuous, nonnegative and so continuous on {osc f = 0}. Hence B is also a G d set. The actual use of Theorem 7.3 is enabled by the following observation: Lemma 7.5 The spaces [0, 1], C =[0, 1] N, N = N N and [0, 1] N, equipped with their natural Polish topologies, are Borel isomorphic. Proof. This results from the string of open, continuous (and thus Borel) injections: [0, 1],! C,! N,! [0, 1] N,! [0, 1] (7.11) The first map is the injection defined by writing each number in [0, 1] in dyadic form. The second map is a simple inclusion. The third map is induced by the continued fraction expansion: if x 2 [0, 1] \ Q, there exists a unique sequence (a n ) 2 {0, 1,... } N such that 1 x = (7.12) 1 a 1 + a a
5 7.2. PROOF OF BOREL ISOMORPHISM THEOREM 5 and noting that, since a n = b 1 x a n 1 c, if x is irrational two distinct sequences will lead to two distinct numbers. The fourth injection is a diagonal map: Write an element x =(x 1, x 2,...) 2 [0, 1] N by means of a matrix a m,n 2{0, 1} such that Now consider the diagonal map x m = Â n 1 f(x) =Â a m,n, m 1. (7.13) 2n k 1 Â k 1 m=1 a m,m k. (7.14) 21+ +k 2+m It is easy to check that this map is continuous, onetoone and the inverse is continuous. By Lemma 7.4, the images of these maps are Borel and so all maps are Borel measurable. By Theorem 7.3, all of these spaces are Borel isomorphic. Recall the wellknown fact which enables the definition of dimension that [0, 1] and [0, 1] 2 are not homeomorphic. However, by the arguments in the above lemma, as standard Borel spaces they are indistinguishable. Having proved the Borelequivalence of examples 24 in our list above, let us now address the embedding of a general Polish space. Here it is convenient to work with Hilbert cube: Proposition 7.6 [Universality of Hilbert cube] to a G d subset of [0, 1] N. Every Polish space is homeomorphic Proof. Let (X, d) be a Polish space and let us assume without loss of generality d that d 1 (otherwise maps (X, d)! (X, 1+d ) by identity map). Let (x n) be a countable dense set in X. Define the map f : X! [0, 1] N by x 2 X 7! f (x) = d(x, x n ) n 1 2 [0, 1] N. (7.15) As is easy to check, f is continuous and onetoone and so we have f 1 : f (X)! X. We claim that f 1 is also continuous. Indeed, suppose that (z n ) is a sequence such that f (z n )! f (z) in f (X). Let e > 0 and find n 1 such that d(x n, z) < e. Then d(z m, z) apple d(z m, x n )+d(x n, z) = f (z m ) n + d(z, x n )! n! f (z) n + d(z, x n ) < 2e. (7.16) Hence z m! z or, in explicit terms, f 1 ( f (z m ))! f 1 ( f (z)), and so f 1 is continuous on f (X). Thus f : X! f (X) is a homeomorphism and so, by Lemma 7.4, f (X) is a G d subset of [0, 1] N. The previous proposition gives the desired embedding of X into the Hilbert cube and, by Lemma 7.5, a Borel isomorphism into the Cantor space. It remains to construct an embedding of the Cantor space into a general (uncountable) Polish space.
6 6 CHAPTER 7. INFINITE PRODUCT SPACES Theorem 7.7 [CantorBendixon] Let X be a second countable topological space (i.e., with countable basis of the topology). Then there exists unique subsets P and O with P perfect (i.e., having no isolated points) and O countable open such that X = P [ O and P \ O =. (7.17) Proof. Let {U n } be a countable basis (of open sets) and let O = [ U n : U n countable. (7.18) Then O is open and countable while P = X \ O is closed. Let x 2 P. If U 3 x is open, then U being the union of a subcollection from {U n } is necessarily uncountable, because otherwise x would end up in O. This means that P \ U \{x} 6=, i.e., x is not isolated. If P 0 and O 0 are two other such sets, then U \ P 0 6= implies that U \ P 0 is uncountable, for any U open. Hence P 0 \ O =, i.e., P 0 P. On the other hand, O 0 is countable open and so it appears in the union defining O. Hence O 0 O. If P 0 [ O 0 = X, then we must have P 0 = P and O 0 = O. The relevant consequence of this general theorem is as follows: Corollary 7.8 [Cantor space to Polish space] Every uncountable Polish space contains a homeomorphic copy of C (and so it has cardinality 0). Proof. Let X be an uncountable Polish space. Every separable metric space is second countable, so let X = P \ O be as in the previous theorem. Then P being a closed subset of X is Polish too, so we may as well assume that X = P. We will use the standard treelike construction to identify a copy of C in X. Let (x n ) be a countable dense set (of distinct points) in P. As is easily checked, there exist two functions f 0 and f 1 from {x 1,...} to itself such that and max d( f 0 (x n )), x n ), d( f 1 (x n )), x n ) apple 2 n (7.19) min d( f 0 (x n )), x n ), d( f 1 (x n )), x n ) d f 0 (x n ), f 1 (x n ) > 0. (7.20) We now define a collection of balls B s, indexed by s{0, 1} n, centered at points of {x 1, x 2,...} as follows: Set B? to ball of radius 1 / 2 about x 1. If B s is defined for all s{0, 1} n, and x s are their centers, let r n be the minimum of f 0 ( x s ) and f 1 ( x s ) for all s 2{0, 1} n. Then let B s0 be the ball centered at f 0 ( x s ) of radius r n 4, and similarly let B s1 be the corresponding ball centered at f 1 ( x s ). These balls are disjoint with B s0, B s1 B s and, by induction, the same is true for {B s : s{0, 1} n } for all n 1. Now let s =(s 1, s 2,...) 2{0, 1} N. By construction, the intersection T n 1 B (s1,...,s n ) is nonempty and, by the completeness of X, it contains exactly one point y s. Moreover, if s and s agree in the first n digits and s n+1 = 0 and s 0 n+1 = 1, then y s 2 B (s1,...,s n,0) and y s 2 B (s1,...,s n,1) and so r n 2 apple d(y s, y s ) apple 2 n (7.21)
7 7.3. PRODUCT MEASURE SPACES 7 Thus the map s 7! y n is onetoone and continuous. To show that the inverse map is continuous, let us note that if d(y s, y s ) < e, then s n = sn 0 for all n for which r n/ 2 > e. So the smaller e, the larger part of s and s must be the same. Hence, the inverse map is continuous. Now we are finally ready to put all pieces together: Proof of Theorem 7.2. If X is finite or countable, then the desired Borel isomorphism is constructed directly. So let us assume that X is uncountable. Then Corollary 7.8 says there is a homeomorphism C,! X while Proposition 7.6 says there exists a homeomorphism X,! [0, 1] N. Both of these image into a G d set by Lemma 7.4 and so the inverse is Borel measurable. Since [0, 1] N is Borel isomorphic to [0, 1] N, we thus we have Borelembeddings C,! X and X,! C. From Theorem 7.3 it follows that X is Borel isomorphic to C and thus to any of the spaces listed in Lemma Product measure spaces Definition 7.9 [General product spaces] Let (W a, F a ) a2i be any collection of measurable spaces. Then W = a2i W a can be equipped with the product salgebra F = N a2i F a which is defined by n F = s A a : A a 2 F a &#{a2i: A a 6= W a } < o. (7.22) a2i If W a = W 0 and F a = F 0 then we write W = W I 0 and F = F I 0. Next we note that standard Borel spaces behave well under taking countable products: Lemma 7.10 Let I be a finite or countably infinite set and let (W a, F a ) a2i be a family of standard Borel spaces. Let (W, F ) be the product measure space defined above. Then (W, F ) is also standard Borel. Moreover, if f a : W a! [0, 1] are Borel isomorphisms, then f = a2i f a is a Borel isomorphism of W onto [0, 1] I. Our goal is to show the existence of product measures. We begin with finite products. The following is a restatement of Lemma 4.13: Theorem 7.11 Let (W 1, F 1, µ 1 ) and (W 2, F 2, µ 2 ) be probability spaces. Then there exists a unique probability measure µ 1 µ 2 on F 1 F 2 such that for all A 1 2 F 1, A 2 2 F 2, µ 1 µ 2 (A 1 A 2 )=µ 1 (A 1 )µ 2 (A 2 ). (7.23) The reason why this is not a trivial application of Carathéodory s extension theorem is the fact that (7.26) defines µ 1 µ 2 only on a structure called semialgebra and not algebra. Definition 7.12 A nonempty collection S P(W) is called semialgebra if
8 8 CHAPTER 7. INFINITE PRODUCT SPACES (1) A, B 2 S implies A \ B 2 S. (2) If A 2 S, then there exist A i 2 S disjoint such that A c = S n i=1 A i. It is easy to verify that semialgebras on W automatically contain W and. Lemma 7.13 The following are semialgebras: (1) S = {(a, b] : apple a apple b apple } (2) S = {A 1 A n : A i 2 A i } where A i are algebras. (3) S = A a : A a 2 A a &#{a: A a 6= W a } < provided A a s are algebras. a2i Proof. Let us focus on (2): Since all A i are psystems, S is closed under intersections. To show the complementation property in the definition of semialgebra, let D be a collection of all sets of the form B 1 B n, where B i is either A i or A c i, but such that A 1 A n is not included in D. Then (A 1 A n ) c = [ From here the complementation property follows by noting that D is finite and D S. Next we will show that via finite disjoint unions semialgebras give rise to algebras: Lemma 7.14 Let S be a semialgebra and let S be the set of finite unions of sets A i 2 S. Then S is an algebra. Proof. First we show that S is closed under intersections. Indeed, A is the (finite) union of some A i 2 S and B is the (finite) union of some B j 2 S. Therefore, [ A \ B = i [ A i \ j B2B B. B j = [ A i \ B j. i,j Since A i, B j 2 S, then A i \ B j 2 S and A \ B is the finite union of sets from S. Consequently, A \ B 2 S. Next we will show that S is closed under taking the complement. Let A be as above. Then [ c A c = A i = \ A c i = \ [ B i,j, i i i j where B i,j 2 S are such that A c i is the union of B i,j see the definition of semialgebra. The union on the extreme right is finite and, therefore, it belongs to S. But S is closed under finite intersections, so A c 2 S as well. Now we will learn how (and under what conditions) a set function on a semialgebra can be extended to a measure on the associated algebra:
9 7.3. PRODUCT MEASURE SPACES 9 Theorem 7.15 Let S be a semialgebra and let µ : S! [0, 1] be a set function. Suppose the properties hold: (1) Finite additivity: A, A i 2 S,A= S n i=1 A i disjoint ) µ(a) =Â n i=1 µ(a i), (2) Countable subadditivity: A, A i 2 S,A= S i=1 A i disjoint ) µ(a) apple Â i=1 µ(a i). Then there exists a unique extension µ : S! [0, 1] of µ, which is a measure on S. Proof. Note that (1) immediately implies that µ( ) =0. The proof comes in four steps: Definition of µ: For A 2 S, then A = S n i=1 S i for some S i 2 S. Then we let µ(a) =Âi=1 n µ(s i). This definition does not depend on the representation of A; if A = S m j=1 T i for some T j 2 S, then the fact that S is a psystem and finite additivity of µ ensure that n Â µ(s i )= i=1 n [ Â m µ S i \ T j = i=1 j=1 n Â i=1 m Â j=1 µ(s i \ T j )= m Â µ(t j ). j=1 Finite additivity of µ: By the very definition, µ is also finitely additive, because if A n is given as the disjoint union S i S i,n, then S n A n = S i,n S i,n and thus µ( S n A n )= Â i,n µ(s i,n )=Â n µ(a n ). (All indices are finite.) Countable subadditivity of µ: We claim that µ is countably subadditive, whenever the disjoint union lies in S. Let A i 2 S and A = S k 1 A k 2 S. Writing A k = S n k applej<n k+1 S j note the efficient way to index the S j s contributing to A k we have A = S j 1 S j where now S j 2 S. But A 2 S implies that A = S n i=1 T i and thus T i = S j 1 S j \ T i all sets again disjoint. Now countable subadditivity of µ gives us that µ(t i ) apple Â µ(s j \ T i ) j 1 and thus µ(a) apple n Â Â i=1 j 1 µ(s j \ T i )=Â so µ is countably subadditive as well. n Â j 1 i=1 µ(s j \ T i )=Âµ(S j )=Â µ(a k ), j 1 k 1 Countable superadditivity of µ: To prove the opposite inequality we note that A can be written as the disjoint union B n [ S n k=1 A k, where B n = S k>n A k is also in S because S is an algebra. Since we already showed that µ is finitely additive, we have µ(a) = µ(b n )+ n Â k=1 µ(a k ) Letting n!, the opposite inequality follows. n Â µ(a k ). k=1
10 10 CHAPTER 7. INFINITE PRODUCT SPACES Thus we have a well defined function µ : S! [0, 1] which is finitely additive on S and countably additive on disjoint unions that belong to S. These are the properties required to be a measure on an algebra. Note that the combination of Lemma 7.13, Lemma 7.14 and the construction of the measure µ from Theorem 7.15 provide an alternative proof of the fact that LebesgueStieltjes measures on R are defined by the corresponding distribution function. Proof of Theorem The set S = {A 1 A 2 : A i 2 F i } is a semialgebra by Lemma 7.13 and s(s ) is the product salgebra F 1 F 2. We will show that the set function µ : S! [0, 1] defined by µ(a 1 A 2 )=µ 1 (A 1 )µ 2 (A 2 ), A i 2 F i, is countably additive on S this implies the desired properties (12) in Theorem Let A 2 F 1 and B 2 F 2 and suppose that A B is the disjoint union S j A j B j, where A j 2 F 1 and B j 2 F 2, and where j is either finite or countable index. The proof uses integration: For each x 2 A, let I(x) ={j : A j 3 x}. Now since {x} B j A j B j whenever j 2I(x), the sets {x} B j with j 2I(x) are disjoint and {x} S j2i(x) B j = {x} B. It follows that B is the disjoint union B = [ j2i(x) B j and thus 1 A (x)µ 2 (B) =Â 1 Aj (x)µ 2 (B j ) j for any x 2 W 1. Both sides are F 1 measurable, and integrating with respect to µ 1 we get µ 1 (A)µ 2 (B) =Â µ 1 (A j )µ 2 (B j ). j This shows that µ(a B) equals the sum of µ(a j B j ) and so µ is countably additive. Lemma 7.14, Theorem 7.15 and Carathéodory s extension theorem then guarantee that µ has a unique extension to F 1 F 2. General finite product spaces are handled by induction using the fact that no no 1 F i = F i F n (7.24) i=1 Note that this proves the associativity of the product of measures: i=1 We leave the details of these calculations implicit. (µ 1 µ 2 ) µ 3 = µ 1 (µ 2 µ 3 ). (7.25)
11 7.4. KOMOGOROV S EXTENSION THEOREM Komogorov s extension theorem We now address the corresponding extension of measures to infinite product spaces: Theorem 7.16 [Kolmogorov s extension theorem] Let M be a compact metric space, B = B(M) the Borel salgebra on M and, for each n 1, let µ n be a probability measure on (M n, B n ). We suppose these measures are consistent in the following sense: 8n 1, 8A 1,...,A n 2 B : µ n+1 (A 1 A n M) =µ n (A 1 A n ). Then there exists a unique probability measure µ on (M N, B N ) such that µ(a 1 A n R R...)=µ n (A 1 A n ) (7.26) holds for any n 1 and any Borel sets A i 2R. As many extension arguments from measure theory, the proof boils down to proving that if a sequence of sets decreases to an empty set, then the measures decrease to zero. In our case this will be ensure through the following lemma: Lemma 7.17 Let W be a metric space and let A be an algebra of subsets of W (which is not necessarily a salgebra). Suppose that µ : A! [0, 1] is a finitelyadditive set function which is regular in the following sense: For each set B 2 A and each e > 0, there exists a compact set C 2 A such that C B (7.27) and µ(b \ C) apple e. (7.28) Then for each decreasing sequence B n 2 A with B n # we have µ(b n ) # 0. Proof. Let B n be a decreasing sequence of sets with lim n! µ(b n )=d > 0. We will show that T n B n 6=. By the assumptions, we can find compact sets C n 2 A such that C n B n and µ(b n \ C n ) < d2 n 1. Now since B n B n+1, we have B n \ n[ (B k \ C k ) k=1 n\ C k. Indeed, if x 2 B n then x 2 B k for all k apple n and then x can only belong to the set on the left if x 2 C k for all k apple n. Using finite additivity of µ, it follows that k=1 \ n µ k=1 C k µ(b n ) n Â µ(b k \ C k ) d Â µ(b k \ C k ) k=1 k 1 d 2. Consequently, since finite additivity implies that µ( ) =0, the intersection T n k=1 C k is nonempty for any finite n. But the sets C n are compact and thus T k=1 C k 6=. The inclusions C n B n imply that also T n B n 6=. Lemma 7.17 links measure with topology (which is the reason why we often consider standard Borel spaces). Next we observe:
12 12 CHAPTER 7. INFINITE PRODUCT SPACES Lemma 7.18 Let (M, r) be a compact metric space. Equip M N with the metric d(x, y) =Â r(x n, y n )2 n (7.29) n 1 whenever x =(x n ) 2 M N and y =(y n ) 2 M N. Then (M N, d) is a compact metric space. Moreover, if C 1,...,C n are compact sets in (M, r), then C 1 C n M M is compact in (M N, d). Proof. The first part is a reformulation of Tychonoff s theorem for product spaces. The second part is checked directly. Now we are finally ready to prove Kolmogorov s theorem: Proof of Theorem Consider the set S = A 1 A n M : A i compact in M 8i (7.30) Then, as is easy to check, S is a semialgebra. Let µ : S! [0, 1] be defined by µ(a 1 A n M )=µ n (A 1 A n ) (7.31) The consistency condition for µ n shows that this gives the same number regardless of the representation of the set A 1 A n M. Let S be the set of all finite disjoint unions of elements from S. By Lemma 7.14, S is an algebra; we extend µ to a set function µ : S! [0, 1] by additivity. A similar argument to that used in the proof of Theorem 7.15 now shows that µ is welldefined and finitely additive. It remains to show that µ is countably additive. To that end, let A 1, A 2, 2S be disjoint with A = S n 1 A n 2 S. Define D N = S n>n A n and note that D N 2 S as well. Then A 1,...,A n, B n are disjoint and so finite additivity of µ implies µ(a) = µ(a 1 )+ + µ(a n )+ µ(d n ) (7.32) We thus need to show that µ(d n )! 0 as n!. Here we use the topology: By Lemma 7.17 it suffices to show 8e > 0 8B 2 S 9C 2 S : C B compact & µ(b \ C) < e. (7.33) But B 2 S is a finite disjoint union of B i 2 S and so it clearly suffices to prove this for B 2 S. However, every B 2 S is compact by Lemma 7.18 and so we can take C = B above and the regularity property holds trivially. By Lemma 7.17, we have µ(d n )! 0 and µ is thus countably additive on S. The Carathéodory extension theorem now implies the existence of a unique extension of µ to s(s )= B N. The proof in the textbook bundles Lemma 7.17 with the proof of Tychonoff s theorem. We prefer to separate these to highlight the role of Borel regularity (Lemma 7.17) which is a property of separate interest. Note also that the consistency condition holds only for probability; extensions of other measures are not meaningful. In fact, it is known that one cannot construct an infinite product of Lebesgue measures on R.
13 7.5. TWO ZEROONE LAWS Two zeroone laws We will demonstrate the usefulness of the infinite product spaces by proving two ZeroOne Laws. We demonstrate these on an example of percolation. Example 7.19 Percolation : Let p 2 [0, 1] and let B denote the set of edges of Z d. (We regard the latter as a graph with vertices at points of R d with integer coordinates and edges between any two points with Euclidean distance one.) Let (h b ) b2b be i.i.d. Bernoulli(p). We may view each h =(h b ) as a subgraph of (Z d, B) with edge set {b 2 B : h b = 1}. The question is: Does this graph have an infinite connected component? We define E = {h : h has an infinite connected component}. (7.34) We refer to edges with h b = 1 as occupied and those with h b = 0 as vacant. To set the notation, we will denote by P p the law of h with parameter p. First we have to check that E is an event: Lemma 7.20 E is measurable (w.r.t. the product salgebra on {0, 1} B ). Proof. Let F denote the product salgebra on {0, 1} B. We have to show E 2 F. For each n 1 and each x 2 Z d, consider the event E n (x) = x connected to (x +[ n, n] d ) c in h (7.35) that x is connected by a path of occupied edges to the boundary of a box of side 2n centered at x. Since E n (x) depends only on a finite number of h b s, it is measurable. But E = \ [ E n (x) (7.36) n 1 x2z d and so E 2 F as well. Next we observe that E does not depend on the status on any given finite number of edges. Indeed, if an edge is occupied, making it vacant may increase the number of infinite component but it won t destroy them while. Similarly, making a vacant edge occupied may connect two infinite components together but if there is none, it will not create one. In a more technical language, E is a tail event according to the following definition: Definition 7.21 Let X 1, X 2,... be random variables. Then T = T n 1 s(x n, X n+1,...) is the tail salgebra and events from T are called tail events. Here is our first zeroone law: Theorem 7.22 [Komogorov s ZeroOne Law] Let X 1, X 2,... be independent and let T = T n 1 s(x n, X n+1,...) be the tail salgebra. Then 8A 2 T : P(A) 2{0, 1}. (7.37)
14 14 CHAPTER 7. INFINITE PRODUCT SPACES Figure 7.2: The infinite connected component in box centered at the origin for percolation with parameter p = The finite components are not depicted. Proof. We will show that T is independent of F. First pick n 1 and note A 2 s(x 1,...,X n ) & B 2 s(x n+1, X n+2,...) ) A, B independent (7.38) But T s(x n+1, X n+2,...) and so A 2 s(x 1,...,X n ) & B 2 T ) A, B independent (7.39) This now holds for all n 1 and so A 2 [ n 1 s(x 1,...,X n ) & B 2 T ) A, B independent (7.40) But S n 1 s(x 1,...,X n ) is a psystem and F = s( S n 1 s(x 1,...,X n )) and so Theorem 4.6 implies A 2 F & B 2 T ) A, B independent (7.41) Let now A 2 T. Then A is independent of itself which yields This permits only P(A) =0 or P(A) =1. Here is a consequence of this law for percolation: P(A) =P(A \ A) =P(A) 2. (7.42)
15 7.5. TWO ZEROONE LAWS 15 Corollary 7.23 For each d 1 there exists p c 2 [0, 1] such that ( P p (E )= 0, if p < p c, 1, if p > p c, (7.43) At p = p c we have P pc (E ) 2{0, 1}. Proof. We have already shown that E 2 T and so P p (E ) 2{0, 1} for all p. It is also clear that the larger p the more edges there are and so P p (E ) should be increasing. However, to prove this we have to work a bit. The key idea is that we can actually realize percolation for all p s on the same probability space. Consider i.i.d. random variables U =(U b ) b2b whose law is uniform on [0, 1] and define h (p) b = 1 {Ub applep}. (7.44) Clearly, h (p) =(h (p) b ) are i.i.d. Bernoulli(p) so they realize a sample of percolation at parameter p. Now let E (p) ={U : h (p) contains infinite connected component}. Since p 7! h (p) b increases, if E (p) occurs, then so does E (p 0 ) for p 0 > p. In other words p 0 > p ) E (p) E (p 0 ) (7.45) This implies P p (E )=P E (p) apple P E (p 0 ) = P p 0(E ), (7.46) i.e., p 7! P p (E ) is nondecreasing. Since it takes values zero or one, there exists a unique point where P p (E ) jumps from zero to one. This defines p c. Let us remark what really happens: It is known that p c ( = 1, d = 1, 2 (0, 1), d 2. (7.47) At d = 2 it is known that p c = 2 (Kesten s Theorem) but the values for d 3 are not known explicitly (and presumably are not of any special form). Concerning P pc (E ), it is widely believed that this probability is zero there is no infinite cluster at the critical point but the proof exists only in d = 2 and d 19. Our next object of interest is the random variable: N = number of infinite connected components in h (7.48) The values N can take are {0, 1,... }[{ }. The random variable N definitely depends on any finite number of edges and so N is not T measurable. (The reason why we care is if it were tail measurable then it would have to be constant a.s.) However, N is clearly translation invariant in the following sense: Definition 7.24 Let t x : {0, 1} B!{0, 1} B be the map defined by (t x h) b = h b+x (7.49) We call t x the translation by x. An event A is translation invariant if t 1 x (A) =A for all x 2 Z d. A random variable N is translation invariant if N t x = N.
16 16 CHAPTER 7. INFINITE PRODUCT SPACES First we note that the measure P p is translation invariant: Lemma 7.25 Let P p be the law of Bernoulli(p) on B. Then P p is translation invariant, i.e., P tx 1 = P p for all x 2 Z d. Proof. This is verified similarly as translation invariance of the Lebesgue measure. First, a direct calculation shows that P p (tx 1 (A)) = P p (A) for all cylinder events. But the class of sets in F satisfying this identity is a lsystem, and since cylinder sets form a psystem, we have invariance of P p on the salgebra generated by cylinder events. This is all of F. Next we will prove that every translation invariant event in F is a zeroone event: Theorem 7.26 [Ergodicity of i.i.d. measures] leftshift acting on the sequence (X n ) via Let (X n ) n2z be i.i.d. and let t be the t(, X 1, X 0, X 1, )=(, X 0, X 1, X 2, ). (7.50) Then 8A 2 s(x n : n 2 Z) : t 1 (A) =A ) P(A) 2{0, 1}. (7.51) Proof. We proceed in a way similar to Komogorov s law; the goal is to show that a translation invariant set is independent of itself. However, since not all translation invariant events are tail events, we have use approximation to derive the result. Let F = s(x n : n 2 Z), F n = s(x n,, X n ) and let A 2 F be of positive measure. The construction of Carathéodory extension of the measure to F shows that there exists a sequence A n 2 F n such that A n approximates A in the sense P(A4A n )! n! 0, (7.52) where A4A n denotes symmetric difference. Suppose now that t 1 (A) =A and recall that P is tinvariant. This means that t 1 (A n ) approximates t 1 (A) as well as A n, P (A4t 1 (A n ) = P(A4A n ). (7.53) Iterating this 2n + 1 times we get a set t (2n+1) (A n ) which is in s(x 3n 1,...,X n 1 ) and is thus independent of A n. Thus, we may approximate A equally well by two disjoint sets A n and t (2n+1) (A n ). Now we will derive the independence of A of itself: The key will be the identity P(A \ B) apple P(C \ D)+P(A4C)+P(B4D) (7.54) which we will apply to B = A, C = A n and D = t (2n+1) (A n ). This yields P(A) =P(A \ A) apple P A n \ t (2n+1) (A n ) + P(A4A n )+P A4t (2n+1) (A n ). (7.55) Since A n and t (2n+1) (A n ) are independent have equal probability, (7.53) gives us P(A) apple P(A n ) 2 + 2P(A4A n ). (7.56)
17 7.5. TWO ZEROONE LAWS 17 However P(A n ) apple P(A)+P(A4A n ) and so (7.52) gives P(A) apple P(A) 2. (7.57) This is only possible if P(A) =0 or P(A) =1. Going back to percolation, the above zeroone law tells us: Corollary 7.27 one. One of the events {N = 0}, {N = 1}, and {N = } has probability Proof. All of these events are translation invariant (or tail) and so they have probability zero or one. We have to show that P(N = k) =0 for all k 2{2, 3,... }. Again, if this is not the case then P(N = k) =1 because {N = k} is translation invariant; for simplicity let us focus on k = 2. If P(N = 2) =1 then there exists n 1 such that the event ( ) box [ n, n] d is connected to infinity by two disjoint paths that are C n = not connected to each other in the complement of the box [ n, n] d (7.58) occurs with probability at least 1 / 2. Indeed, {N = 2} S n 1 C n and C n is increasing. However, the event C n is independent of the edges inside the box [ n, n] d and so with positive probability, the two disjoint paths get connected inside the box. This means that N = 1 with positive probability on the event {N = 2}\C n, i.e., P(N = 1) > 0. This contradicts P(N = 2) =1. Let us conclude by stating what really happens: we have P(N = ) =0 so we either have no infinite cluster or one infinite cluster a.s. The most beautiful version of this argument comes in a theorem of Burton and Keane which combines rather soft arguments not dissimilar to what we have seen throughout this section.
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