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Math Help - borel measurability

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    borel measurability

    if (S,d) is a metric space and B(d) is the sigma algebra on S generated by all open sets, how would i prove that every continuous function f:S->R is (B(d),B(R))-measurable?
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    Quote Originally Posted by squarerootof2 View Post
    if (S,d) is a metric space and B(d) is the sigma algebra on S generated by all open sets, how would i prove that every continuous function f:S->R is (B(d),B(R))-measurable?
    If U is open in R, then f^{-1}(U) is open, hence Borel, in S. Now suppose that V_n\ (n=1,2,3,\ldots) are sets in R such that f^{-1}(V_n) is Borel for each n. Then f^{-1}\Bigl(\bigcup_nV_n\Bigr) = \bigcup_nf^{-1}(V_n), which is a countable union of Borel sets and in therefore Borel in S. By a similar argument, f^{-1}\Bigl(\bigcap_nV_n\Bigr) is Borel.

    That shows that the class of subsets V ⊆ R for which f^{-1}(V) is Borel in S is a σ-algebra containing all the open sets in R. Therefore this class contains all the Borel sets in R. Thus f is Borel measurable.
    Last edited by Opalg; May 27th 2008 at 10:08 AM.
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