Polish spaces and standard Borel spaces


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1 APPENDIX A Polish spaces and standard Borel spaces We present here the basic theory of Polish spaces and standard Borel spaces. Standard references for this material are the books [143, 231]. A.1. Polish spaces A metric space (X, d), or the metric d itself, is said to be complete if every Cauchy sequence in X converges. A topological space X, or the topology itself, is said to be completely metrizable if there is a compatible complete metric on X. Note that a metric d is complete if and only if the metric d = d is complete. Thus every 1+d completely metrizable space admits a compatible complete metric which is bounded. DEFINITION A.1. A topological space X is called a Polish space if it is separable and completely metrizable. The following is obvious. PROPOSITION A.2. The family of completely metrizable topological spaces is closed under taking countable products and closed subspaces. The same is also true for the family of Polish spaces. PROPOSITION A.3. Let τ 0 be a Hausdorff topology on a set X, and for each n N let τ n be a completely metrizable topology on X which is finer than τ 0. Then the topology τ on X generated by n N τ n is completely metrizable. The same is also true if completely metrizable is replaced by Polish in both places. PROOF. The map (X, τ) (X, τ n) sending x to (x, x,... ) realizes (X, τ) as a closed subspace of (X, τ n). We can then apply Proposition A.2. THEOREM A.4. Let X be an uncountable Polish space. Then X has a closed subset Y which is a Cantor set. PROOF. Fix a compatible complete metric on X. Let Z be the subset of X consisting of all points x such that each neighbourhood of x is uncountable. Then Z is closed. For each x X \ Z, write r x for the supremum of all r (0, 1) such that the open ball centred at x with radius r is countable. Take a countable dense subset S of X \ Z. Then the open balls around x S with radius 1 2 r x for x S cover X \ Z. Thus X \ Z is countable. Therefore we may assume that Z = X. Then by a simple recursive procedure over n N we can construct pairwise disjoint closed balls 385
2 386 A. POLISH SPACES AND STANDARD BOREL SPACES B n,1, B n,2,..., B n,2 n so that each B n,k has a positive radius no larger than 1/n and contains two of the sets B n+1,1, B n+1,2,..., B n+1,2 n+1. Then clearly 2 n k=1 B n,k is a closed subset of X and is a Cantor set. COROLLARY A.5. Every uncountable Polish space has cardinality 2 ℵ 0, where ℵ 0 denotes the cardinality of N. PROOF. By Theorem A.4 it suffices to show that every Polish space X has cardinality at most 2 ℵ 0. Take a countable dense subset S of X. Then for each x X we can find a sequence {x n } in S converging to x. Since the set of sequences in S has cardinality at most 2 ℵ 0, we conclude that X has cardinality at most 2 ℵ 0. THEOREM A.6 (Kuratowski). Let X be a metric space and let Y be a complete metric space. Let A be a subset of X and let f : A Y be a continuous map. Then there exist a G δ subset B of X satisfying A B A and a continuous extension g : B Y of f. PROOF. Since A is G δ in X, we may assume that A = X. For each n N write B n for the union of all open sets U in X which satisfy diam(f(a U)) 1/n. Then B := n N B n is a G δ subset of X and A B by the continuity of f. It is clear that for each x B the intersection U f(a U) over all open neighbourhoods U of x is a singleton, say {y}. Set g(x) = y. Then it is readily checked that g is continuous and extends f. PROPOSITION A.7. Let X be a completely metrizable topological space and let Y be a nonempty subset of X. Then Y is a completely metrizable topological space if and only if Y is a G δ subset of X. The same is also true if completely metrizable topological space is replaced by Polish space in both places. PROOF. It suffices to prove the assertion for the case of completely metrizable spaces. The only if part follows from Theorem A.6. For the the converse, let d be a compatible complete metric on X. If Y is open, then d (y, z) := d(y, z) defines a compatible complete metric on Y. If Y is a G d(y,x\y ) d(z,x\y ) δ, say Y = U n for some open sets U n in X, then the map Y U n sending y to (y, y,... ) realizes Y as a closed subspace of U n, in which case we conclude by Proposition A.2 that Y is a completely metrizable space. PROPOSITION A.8. Every Polish space is homeomorphic to a G δ subset of [0, 1] N. PROOF. Let X be a Polish space and d a compatible metric on X, which we may assume to be bounded by 1 by replacing it with d if necessary. Take a dense 1+d sequence {p n } in X. Then the map X [0, 1] N sending x to (d(x, p n )) n N is clearly a homeomorphism of X onto its image. By Propositions A.2 and A.7, the image of X is a G δ subset of [0, 1] N.
3 A.2. STANDARD BOREL SPACES 387 EXAMPLE A.9. Clearly every Polish space is second countable, i.e., has a countable basis for the topology. Conversely, every second countable locally compact Hausdorff space is Polish. To see this, notice first that if we are given a basis for the topology of a locally compact Hausdorff space X then the elements in the basis with compact closure also form a basis. Thus any second countable locally compact Hausdorff space X has a countable basis consisting of open sets with compact closure, and hence can be written as n N K n where each K n is compact and contained in the interior of K n+1. It follows that the onepoint compactification X of X is also second countable. By Proposition A.7, if X is Polish then so is X. Take a countable basis U for the topology on X. Then U is an open cover of X. For each finite subcollection V of U which covers X, choose a partition of unity {f V,U } U V over X which is subordinate to V. Then, with the index V ranging over all finite subcollections of U which cover X, the map X V U V [0, 1] which sends x to {{f V,U(x)} U V } V is a homeomorphism of the compact space X onto a closed subset of V U V [0, 1]. By Proposition A.2, X is Polish. EXAMPLE A.10. Let X be a second countable locally compact Hausdorff space and let Y be a Polish space. Equip the space C(X, Y ) of continuous maps X Y with the compactopen topology, i.e., the topology generated by the sets O(K, U) consisting of continuous maps X Y sending K into U where K ranges over the compact subsets of X and U over the open subsets of Y. Then C(X, Y ) is Polish. To see this, write X as K n where each K n is compact and contained in the interior of K n+1 (see Example A.9), and let d be a bounded complete metric on Y. Then d (f, g) = 1 2 n max x K n d(f(x), g(x)) is a metric on C(X, Y ) inducing the compactopen topology. It is easy to see that d is complete. Thus it remains to prove that C(X, Y ) is separable. For this it suffices to show that C(K n, Y ) is separable for each n. By Proposition A.8, we can identify Y with a G δ subset of R N. Then C(K n, Y ) with the compactopen topology is clearly a subspace of C(K n, R N ). Thus it is enough to show that C(K n, R N ) is a separable metrizable space. Clearly C(K n, R N ) = (C(K n, R)) N as topological spaces. By the StoneWeierstrass theorem, C(K n, R) is Polish. It follows by Proposition A.2 that C(K n, R N ) = (C(K n, R)) N is Polish. A.2. Standard Borel spaces For a family τ of subsets of a set X, we denote by B(τ) the σalgebra generated by τ. A measurable space (X, B) is called a standard Borel space if it is isomorphic to (Y, B(τ)) for some Polish space (Y, τ). A subset of a topological space is said to be clopen if it is both closed and open.
4 388 A. POLISH SPACES AND STANDARD BOREL SPACES THEOREM A.11. Let (X, τ) be a Polish space. Let A B(τ). Then there exists a Polish topology τ A on X finer than τ such that B(τ A ) = B(τ) and A is clopen in (X, τ A ). PROOF. Denote by A the set of all A B(τ) satisfying the conclusion of the theorem. It suffices to show that A is a σalgebra containing τ. Let A τ. By Proposition A.7, both A and X \ A are Polish spaces. Denote by τ A the topology on the disjoint union of A and X \ A. Then clearly τ A is Polish and finer than τ, and also τ A B(τ), whence B(τ A ) = B(τ). Moreover, A is clopen in (X, τ A ). This proves that τ A. Now we show that A is a σalgebra. Let {A n } be a sequence in A. By Proposition A.3, the topology τ A on X generated by τ A n is Polish and finer than τ. Since each τ An has a countable basis and is contained in B(τ), it is readily checked that τ A B(τ). Thus B(τ A ) = B(τ). Note that n N A n is open in τ A. By the first paragraph, there exists a Polish topology τ A on X finer than τ A such that B(τ A ) = B(τ A) = B(τ) and A n is clopen in (X, τ A ). Therefore A is closed under taking countable unions. Finally, if A is a member of B(τ) which belongs to A, then clearly X \ A also belongs to A, and so we conclude that A is a σalgebra. The following corollary is a consequence of Theorem A.11 and Proposition A.3. COROLLARY A.12. Let (X, τ) be a Polish space and let Y be a second countable topological space. Let f : X Y be a Borel map. Then there exists a Polish topology τ f on X finer than τ such that B(τ f ) = B(τ) and f : (X, τ f ) Y is continuous. THEOREM A.13. Let (X, B) and (Y, C ) be measurable spaces. Let f : X Y and g : Y X be injections with f(b) C and g(c ) B such that each establishes an isomorphism of measurable spaces onto its image. Then there exist A B and B C such that f(a) = Y \B and g(b) = X \A. In particular, (X, B) and (Y, C ) are isomorphic. PROOF. Set B 1 = Y \ f(x), B = n 0 (f g)n B 1, and A = X \ g(b). Then g(b) = X \ A and f(a) = f(x \ g(b)) = f(x) \ (f g)b = Y \ (B 1 (f g)b) = Y \ B. As a direct consequence of Theorem A.11 we get the following. COROLLARY A.14. Let (X, B) be a standard Borel space and let Y be a nonempty set in B. Then (Y, B Y ) is also a standard Borel space. LEMMA A.15. Let {(X n, τ n )} be a sequence of second countable topological spaces. Then B( τ n) is exactly B(τ n), i.e., the smallest σalgebra on X n containing π 1 n (B(τ n )), where π n : m=1 X m X n is the coordinate projection map.
5 PROOF. Since π 1 n A.2. STANDARD BOREL SPACES 389 (B(τ n )) = B(πn 1 (τ n )) for each n N, we have ( B(τ n ) = B ) πn 1 (τ n ). For every n N, take a countable basis U n for τ n. Denote by U the family of subsets of X of the form n k=1 π 1 k (U k), where n N and U k U k for all k = 1,..., n. Then U is a countable basis for τ n. Consequently ( ) ( ) B τ n = B(U ) = B πn 1 (τ n ) = B(τ n ). LEMMA A.16. The closed unit interval [0, 1] equipped with the Borel structure associated to the standard topology is isomorphic to the Cantor set equipped with its Borel structure. PROOF. Define a map ϕ : {0, 1} N [0, 1] by 1 ϕ((a n )) = 2 a n. n Clearly ϕ is continuous and surjective. Denote by Y the set of points in [0, 1] with more than one preimage under ϕ. Then both Y and ϕ 1 (Y ) are countably infinite. It is readily checked that the restriction of ϕ to {0, 1} N \ ϕ 1 (Y ) is a homeomorphism onto [0, 1] \ Y, and hence a Borel isomorphism. Take a bijection ψ from ϕ 1 (Y ) onto Y. Define φ : {0, 1} N [0, 1] to be ϕ on {0, 1} N \ ϕ 1 (Y ) and ψ on ϕ 1 (Y ). Then φ is a Borel isomorphism. THEOREM A.17. Two standard Borel spaces are isomorphic if and only if they have the same cardinality, and any two uncountable standard Borel spaces are isomorphic. PROOF. For the nontrivial direction of the equivalence, observe first that two countable standard Borel spaces with the same cardinality are clearly isomorphic. Fixing a Cantor set (C, τ), it thus suffices to show that every uncountable standard Borel space (X, B) is isomorphic to (C, B(τ)). By Theorem A.13 it is then enough to show that there are injections f : C X and g : X C with f(b(τ)) B and g(b) B(τ) such that each establishes an isomorphism of measurable spaces onto its image. Fix a Polish topology τ X on X such that B = B(τ X ). By Theorem A.4 there exists a homeomorphism f from (C, τ) onto a closed subspace Z of (X, τ X ). Then f is an isomorphism from (C, B(τ)) onto (Z, B Z ). To construct g, we first apply Proposition A.8 to obtain an embedding ϕ : X [0, 1] N whose image is a G δ subset of [0, 1] N. Then ϕ is a Borel isomorphism from (X, B) onto ϕ(x). Since C N is homeomorphic to C, by Lemmas A.15 and A.16
6 390 A. POLISH SPACES AND STANDARD BOREL SPACES there is an isomorphism ψ from [0, 1] N, equipped the Borel structure associated to the product topology, to (C, B(τ)). Now take g to be ψ ϕ. A.3. Measures on standard Borel spaces PROPOSITION A.18. Let (X, τ) be a metrizable topological space and let µ be a finite measure on B(τ). Then for each A B(τ) one has µ(a) = sup{µ(f ) : F A and F is closed} = inf{µ(u) : A U and U is open}. PROOF. Denote by A the set of all A B(τ) satisfying the above conditions. It is easily checked that A is a σalgebra on X containing τ. Thus A = B(τ). A finite Borel measure µ on a topological space X is said to be regular if for every Borel set A X one has µ(a) = sup{µ(f ) : F A and F is compact} = inf{µ(u) : A U and U is open}. Together with Proposition A.18, the following proposition shows that every finite Borel measure on a Polish space is regular. PROPOSITION A.19. Let (X, τ) be a Polish space and let µ be a finite measure on B(τ). Then for each A B(τ) one has µ(a) = sup{µ(f ) : F A and F is compact}. PROOF. By Proposition A.18 it suffices to consider the case that A is closed. Thus we may assume that A = X. Fix a compatible complete metric on X. Let S be a countable dense subset of X. Let ε > 0. For each n N, the collection {B(x, 1 )} n x S of closed balls of radius 1 centred at points in S covers X, and so we can find a finite n set S n S such that µ(x \ x S n B(x, 1 )) < n ε/2n. Set F = x S n B(x, 1 ). n Then µ(x \ F ) µ(x \ x S n B(x, 1 )) < ε. Since F is complete and totally n bounded, it is compact. Let (X, B) be a measurable space such that {x} B for every x X. A measure µ on (X, B) is said to be atomless if µ({x}) = 0 for every x X. If (X, B) and (Y, C ) are measurable spaces, f : X Y is a measurable map, and µ is a measure on B, then we write f µ for the push forward measure on C, which is defined by f µ(a) = µ(f 1 (A)) for all A C. THEOREM A.20. Let µ be an atomless probability measure on a standard Borel space (X, B). Then there is a Borel isomorphism f : X [0, 1] for the Borel structure coming from the standard topology on [0, 1] such that f µ is Lebesgue measure on [0, 1].
7 A.4. BOREL MAPS AND DISJOINTNESS OF IMAGES 391 PROOF. Of course X must be uncountable. By Theorem A.17 we may assume that (X, B) is the unit interval [0, 1] with the Borel structure coming from its standard topology. Consider the function g : X [0, 1] defined by g(x) = µ([0, x]). This is an increasing function with g(1) = 1. Since µ is atomless, it is easy to see that g is continuous and g(0) = 0. Thus g is surjective. Clearly g µ([0, t]) = t for every t [0, 1]. It follows that g µ and Lebesgue measure coincide on open subsets of [0, 1]. By Proposition A.18 they coincide on all Borel subsets of [0, 1]. Clearly g 1 (t) is an interval, possibly of length 0, for each t [0, 1]. Write D for the set of all t [0, 1] such that g 1 (t) has positive length. Then D is countable. It is clear that g restricts to a Borel isomorphism from X \ g 1 (D) onto [0, 1] \ D. If D is empty then we may take f = g to get the conclusion of the theorem. Suppose then that D is nonempty. Choose a Borel set E [0, 1] with cardinality 2 ℵ 0 and Lebesgue measure 0 (for example, the middle thirds Cantor set). Then µ(g 1 (D E)) = 0, and g restricts to a measurepreserving Borel isomorphism from X \ g 1 (D E) onto [0, 1] \ (D E). By Corollary A.14, both D E equipped with the restriction of the Borel structure on [0, 1] and g 1 (D E) equipped with the restriction of B are standard Borel spaces. Thus by Theorem A.17 they are isomorphic. Define a map f : X [0, 1] so that f coincides with g on X \ g 1 (D E) and restricts to a Borel isomorphism from g 1 (D E) onto D E. Then f satisfies the requirements. A.4. Borel maps and disjointness of images Proposition A.22 below is useful for converting freeness for a p.m.p. action (Definition 1.2) into a property involving the disjointness of images of nonnull sets under the action of finitely many group elements (Proposition 1.4). This property gets applied in a crucial way in the proof of the Rokhlin lemma (Theorem 3.44 and, in the case of Zactions, Lemma 3.77). PROPOSITION A.21. Let (X, B) and (Y, C ) be standard Borel spaces, and let f, g : X Y be Borel maps. Then the set {x X : f(x) = g(x)} belongs to B. PROOF. Let τ be a Polish topology on Y such that C = B(τ). By Proposition A.2, Y Y is a Polish space. By Lemma A.15, the product Borel structure C 2 on Y Y is exactly the Borel structure generated by the product topology. In particular, the diagonal := {(y, y) : y Y } belongs to C 2. Observe also that the map h : (X, B) (Y Y, C 2 ) given by x (f(x), g(x)) is Borel. Thus the set h 1 ( ) = {x X : f(x) = g(x)} belongs to B. PROPOSITION A.22. Let (X, B) and (Y, C ) be standard Borel spaces and let µ be a measure on B. Let f, g : X Y be Borel maps such that the set {x X : f(x) = g(x)} has µmeasure 0. Let A be a member of B with µ(a) > 0. Then there exists a B B such that B A, µ(b) > 0, and f(b) g(b) =. PROOF. By Proposition A.21, the set Z := {x X : f(x) g(x)} belongs to B. Let τ X be a Polish topology on X with B(τ X ) = B and τ Y a Polish topology
8 392 A. POLISH SPACES AND STANDARD BOREL SPACES on Y with B(τ Y ) = C. By Corollary A.12 we may assume that both f and g are continuous. Then each z Z has a neighbourhood U such that f(u) g(u) =. Fix a metric on X inducing τ X. For each z Z let r z be the supremum of all r (0, 1) such that the open ball V r,z in X centred at z with radius r satisfies f(v r,z ) g(v r,z ) =. As (X, τ X ) is separable there is a countable dense subset S of Z. Then the collection {V z,rz/2} z S covers Z. As µ(a Z) = µ(a) > 0, we see that µ(a Z V w,rw/2) > 0 for some w Z. Then the set B := A Z V w,rw/2 satisfies the requirements.
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