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Appendix Polish Spaces and Standard Borel Spaces

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Appendix Polish Spaces and Standard Borel Spaces Let X be a given set. A collection ~ of subsets is called a a-field if (a) any union of countably many members of ~ is a member of~, (b) the complement of any member of f!J belongs to f!J and (c) cJ> E f!J. Given a family Y of subsets of X, there exists the smallest a-field f!J.,/ of subsets of X containing Y, which will be called the a-field generated by Y. A Borel space is a set X equipped with a specified a-field f!J. The members of ~ are called Borel sets of X. When X is a topological space, the Borel space X generated by the topology means the Borel space X equipped with the a-field generated by the family of open (hence closed) subsets of X. Given a Borel space X, a collection of Borel subsets is said to be generating ifthe a-field of Borel subsets is generated by the family; it is said to be separating if any distinct pair of points of X are separated by a member of the family. A Borel space is said to be countably generated (resp. separated) if it admits a countable generating and separating (resp. separating) family of Borel sets. A (positive) measure on a Borel space X means an extended positive real valued countably additive set function J1. on the Borel sets in X. Given two Borel spaces X 1 and X 2, a map f of X 1 into X 2 is said to be Borel if the inverse image f -1(S) of every Borel set S in X 2 is Borel in X 1. If f is bijective and f - 1 is Borel also, then f is called a Borel isomorphism, and in this case X 1 and X 2 are said to be (Borel) iso­ morphic. If X 1 and X 2 are topological spaces, then any continuous map f of X 1 into X 2 is Borel, i.e., f - 1(S) is Borel in X 1 for every Borel set S in X 2. We note, however, that the image f(S) of a Borel set S in X 1 need not be a Borel set in X 2. A topological space is called Polish if it is homeomorphic to a separable complete metric space. Theorem A.t. In a Polish space, a subspace, with respect to the relative topology, is a Polish space if and only if it is a Gd-subset of the whole Polish 375 376 Appendix: Polish Spaces and Standard Borel Spaces space. A topological space is a Polish space if and only if it is homeomorphic to a Ga-subset of the cube IN, where 1= [0,1] and N denotes the set of all positive integers. PROOF. Let X be a Polish space and Y be a subspace. Suppose that Y is a Polish space. Let d be a complete metric on Y compatible with the topology in Y. Let Y be the closure of Y in X. For each n :::: 1,2, ... , we denote by y" the set of all those points x E Y such that there exists an open neighborhood U of x with diameter U n Ys lin. It is clear that Y" is an open subset of Y with respect to the relative topology of Y, and that Y" ~ Y. Hence we have n:; 1 Y" ~ Y. Suppose that x En:; 1 Y". For each n, there exists an open neighborhood Un of x such that the diameter of Un n Y s lin. Replacing Un by U 1 n ... n Un, we may assume that {Un} is decreasing. It then fol­ lows that { Un n Y} is a sequence of decreasing open subsets of Y whose diameters tend to zero. Hence n:;l {Un n Y} is a singleton, say {y}. Since {Un} can be chosen to be a fundamental system of neighborhoods of x, we have x = y E Y. Thus Y = n:; 1 Y". Since y" is open in Y, there exists an open set v" in X such that y" = Y n v". On the other hand, Y is a Gd-set in X, being closed, so that there exists a sequence of open sets {~} in X such that Y = n:; 1 ~. Thus Y = nn.m (v" n Wm ) is a Gd-set in X. Suppose that Y is a Ga-set of X. Let {Gn} be a sequence of open sets in X such that nGn = Y. Let d be a complete metric in X compatible with the topology in X. For each x E Y and n = 1,2, ... , let fn(x) = 1Id(x,G~) > 0. We set ~ 1 jJ,,(x) - J,,(y)j do(x,y) = d(x,y) + .'~'t 2n 1 + jJ,,(x) - J,,(y)j for each (x,y) E Y x Y. Then it is not difficult to prove that do is a complete metric in Y which is compatible with the topology in Y. Being a subspace of a separable metric space, Y is separable. Thus Y is a Polish space. Suppose that X is a Polish space with a complete metric d. Let {an} be a dense sequence in X. For a point x E X, we set d(an,x) } [ ]N q>(x) = { 1 + d(an,x) E 0,1 . It is not too hard to show that q> is a homeomorphism of X into the compact metrizable space [0,1 ]N. By the first half of our assertion, q>(X) is a Ga-set in [0,1 ]N. Q.E.D. Let N denote the set of all positive integers as usual. Set A = NN and define a metric d in A by Appendix: Polish Spaces and Standard Borel Spaces 377 It then follows that A is separable and complete with respect to the metric d. Let X be a separable complete metric space with a metric d; hence a Polish space. For each subset A c X, let c:5(A) be the diameter of A, i.e., c:5(A) = sup{d(x,y):x,y E A}. Let {F(n):n E N} be a closed covering of X with c:5(F(n)) ~ 1/2. Suppose that a closed covering {F(nb' .. ,nk):n 1 , ••• ,nk E N} of X has been defined for k = 1, ... ,m with the properties 00 F(nb ... ,n) = U F(n 1,n2 , ••• ,nj,n), n=1 j = 1,2, ... ,m - 1. Then there exists a closed covering {F(nb'" ,nm,n):n E N} of F(n 1 , ••• ,nm) such that c:5(F(nl," . ,nm,n)) ~ tc:5(F(n 1, • •• ,nm)). Now for any sequence {nd E A, {F(nl,'" ,nk):k E N} is a decreasing se­ quence of closed sets and limk_oo c:5(F(nl,' .. ,nk)) = O. Since X is complete, nr=1 F(n 1 ,··· ,nk) is a singleton which we denote by {cp({nk})}' Thus we obtain a mapping cp of A into X. It is easy to show that cp is continuous and surjective. Thus we obtain the following: Lemma A.2. If X is a Polish space, then there exists a continuous map from NN = A onto X. A Souslin space (or sometimes analytic space) means a metrizable space which is a continuous image of a Polish space. A subset of a topological space is said to be a Souslin set if it is a Souslin space as a topological space. By Lemma A.2, any Souslin space is a continuous image of A. Theorem A.3 (Separation of Souslin Sets). Let X be a metrizable space. If {Xn} is a disjoint countable family of Souslin subsets of X, then there exists a disjoint family {Bn} of Borel subsets of X such that Xn c Bn, n = 1,2, .... PROOF. First, we claim that if {An} and {A;"} are sequences of subsets of X such that for any n, m there exists a Borel set Bn.mwith the property Bn.m :;) An and Bn.m n A;" = 0, then there exists a Borel set B such that B :;) U:= 1 An and B n (U~= 1A;") = 0· Indeed, the Borel set B = U:= 1(n~= 1Bn.m) does the job. We now prove the theorem for two disjoint Souslin sets Yand Y' in X. Let f and f' be continuous maps of A onto Yand Y', respectively. For each (n 1, ••• ,nk) E,N x .~. x N:, k times we set 378 Appendix: Polish Spaces and Standard Borel Spaces We then have 00 A(nl, ... ,nk) = U A(nlon2,'" ,nk,n). n= 1 Set Y(nb ... ,nk) = f(A(nb ... ,nk», Y'(ml, ... ,mk) = f(A(mlo .. ,mk»· We now assume that there exists no Borel set B such that B ~ Y and B n Y' = 0. Jt then follows from the first argument that for some nl and mb Y(nl) and Y'(ml) are not separated by a Borel set. If Y(n 1, ... ,nk) and Y'(mb ... ,mk) are not separated by a Borel set, then for some nk+ 1 and mk+ 10 Y(nb'" ,nkonk+ d and Y'(ml,'" ,mbmk+ d are not separated by a Borel set. By induction, we obtain two sequences {nil and {mj} such that Y(nb' .. ,nk) and Y'(mb'" ,mk) are not separated by a Borel set. Set n = {nil E A and m = {mil E A and y = f(n) E Y, y' = f'(m) E Y'.
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