Chapter 7 Infinite product spaces

More documents

Recommendations

Info

$Math 33B/1 S00 K. Solutions to Midterm Exam G1, Version B If you ...$

$Math 3C Homework 3 Solutions$

6 CHAPTER 7. INFINITE PRODUCT SPACES Theorem 7.7 [Cantor-Bendixon] 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 {Un} be a countable basis (of open sets) and let O = [ Un : Un 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 {Un} 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 2 @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 “tree-like” construction to identify a copy of C in X. Let (xn) be a countable dense set (of distinct points) in P. As is easily checked, there exist two functions f0 and f1 from {x1,...} to itself such that and max d( f0(xn)), xn), d( f1(xn)), xn) apple 2 n min d( f0(xn)), xn), d( f1(xn)), xn) (7.19) d f0(xn), f1(xn) > 0. (7.20) We now define a collection of balls Bs, indexed by s{0, 1} n , centered at points of {x1, x2,...} as follows: Set B? to ball of radius 1/ 2 about x1. If Bs is defined for all s{0, 1} n , and ˜xs are their centers, let rn be the minimum of f0( ˜xs) and f1( ˜xs) for all s 2{0, 1} n . Then let Bs0 be the ball centered at f0( ˜xs) of radius rn 4 , and similarly let Bs1 be the corresponding ball centered at f1( ˜xs). These balls are disjoint with Bs0, Bs1 ⇢ Bs and, by induction, the same is true for {Bs : s{0, 1} n } for all n 1. Now let s =(s1, s2,...) 2{0, 1} N . By construction, the intersection T n 1 B (s1,...,sn) is non-empty and, by the completeness of X, it contains exactly one point ys. Moreover, if s and ˜s agree in the first n digits and sn+1 = 0 and s0 then ys 2 B (s1,...,sn,0) and y ˜s 2 B (s1,...,sn,1) and so n+1 = 1, rn 2 apple d(ys, y ˜s) apple 2 n (7.21)
7.3. PRODUCT MEASURE SPACES 7 Thus the map s 7! yn is one-to-one and continuous. To show that the inverse map is continuous, let us note that if d(ys, y ˜s) < e, then sn = s 0 n for all n for which rn/ 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 Gd-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 Borel-embeddings 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 7.5. 7.3 Product measure spaces Definition 7.9 [General product spaces] Let (Wa, Fa)a2I be any collection of measurable spaces. Then W = ⇥a2I Wa can be equipped with the product s-algebra F = N a2I Fa which is defined by ✓n o F = s ⇥ Aa : Aa 2 Fa &#{a 2I: Aa 6= Wa} < • a2I ◆ . (7.22) If Wa = W0 and Fa = F0 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 (Wa, Fa)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 fa : Wa ! [0, 1] are Borel isomorphisms, then f = ⇥a2I fa 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 (W1, F1, µ1) and (W2, F2, µ2) be probability spaces. Then there exists a unique probability measure µ1 ⇥ µ2 on F1 ⌦ F2 such that for all A1 2 F1, A2 2 F2, µ1 ⇥ µ2(A1 ⇥ A2) =µ1(A1)µ2(A2). (7.23) The reason why this is not a trivial application of Carathéodory’s extension theorem is the fact that (7.26) defines µ1 ⇥ µ2 only on a structure called semialgebra and not algebra. Definition 7.12 A non-empty collection S ⇢ P(W) is called semialgebra if
Page 1 and 2: Chapter 7 Infinite product spaces S
Page 3 and 4: 7.2. PROOF OF BOREL ISOMORPHISM THE
Page 5: 7.2. PROOF OF BOREL ISOMORPHISM THE
Page 9 and 10: 7.3. PRODUCT MEASURE SPACES 9 Theor
Page 11 and 12: 7.4. KOMOGOROV’S EXTENSION THEORE
Page 13 and 14: 7.5. TWO ZERO-ONE LAWS 13 7.5 Two z
Page 15 and 16: 7.5. TWO ZERO-ONE LAWS 15 Corollary
Page 17: 7.5. TWO ZERO-ONE LAWS 17 However P

Chapter 7 Infinite product spaces

Create successful ePaper yourself

Delete template?

Save as template?