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If $f$ is a continuous function on $X$ and E a Lebesgue measurable set, can we conclude that $f^{-1}(E)$ is measurable?

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I presume $f : X \to Y$ is a continuous function between topological spaces equipped with their Borel $\sigma$-algebras? –  Qiaochu Yuan Sep 28 '12 at 2:35
    
I know that "The inverse image of any Borel set is measurable". But I want to see if the general case (any measurable) holds. –  Anita Sep 28 '12 at 2:37
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No. See here for the standard counterexample: math.stackexchange.com/questions/104519/… –  Chris Janjigian Sep 28 '12 at 2:39
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What do you mean by "the general case" here? What is the codomain of $f$ and what $\sigma$-algebra are you using on it? –  Qiaochu Yuan Sep 28 '12 at 2:45
    
Actually the example in the link provided by Chris is used to proof that "measurable" is really more general than "Borel": there is a measurable set $B$ such that its inverse image by a continuous map is not measurable so is not Borel. The same function in that post serves to show that measurability is not a topological property. –  leo Sep 28 '12 at 4:57
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