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Extra resources for Measures On Infinite Dimensional Spaces

Example text

X,B) is a-compact. Let is compact regular, Oi be the family of which satisfy: inf{d^(K,B)| K is compact and K C B } = 0. is closed under countable intersection and countable union, so that in order to prove Oi m 0nm 0 1 = 'Ey it is sufficient to show that contains all open sets and all closed sets. 40 Evidently, a compact set is a Borel set, and belongs to ¿71. A closed set F OL belongs to y because F pressed as a countable union of compact sets: 00 can be ex­ assuming X= 00 \ J (FnK ) . 4) = {'X € Xj d(x,0°) > 1/n}.

B* all open sets, therefore is a a-algebra and contains contains all Borel sets. 1» two Suslin sets ly ^ B DS, Corollary. (X,t ), mutually disjoint In a topological space S and S’ ^B’D S ’, B If both S are separated by Borel sets, name­ and and B* S^ are Borel sets and B n B ’=0. are Suslin sets, then S is a Borel set. Proof (of Lemma). First, note that if ted by B’ B^j^ S ’=0SjJj m=l and for each pair are separated by B and S^ and S^J^ (n,m), then B ’, where and B*=l~) ( / ^ B ’ ). In other words, if S m=l n=l separated by Borel sets, then some S^ and are separa­ S= Q S^ n=l B= Q ( n=l m=l and S^ S’ and ) are not are not sepa­ rated by Borel sets.

3) f“'^(P)e(7t. 1. In a locally compact space X, a compact set is always contained in a compact Baire set. 4) KCO, 0 be an open neigh­ is compact. Since U(x). K is Thus, denot­ 0, we have is open and 0 is compact. Here, we shall use the following lemma without proof. 2. A compact space disjoint closed sets function Proof f on X P^ X and is a normal space, namely for P 2 , there exists a continuous such that f=l on P^ and f=0 on P2 . g. N. Bourbaki "Topologie générale” Chap. 9, §4. Since f of on 0).

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