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Does there exist a function $f : \mathbb{R} \to \mathbb{R} \hspace{.1 in}$ such that for all $B$, if $B \subsetneq \mathbb{R}$ and $B$ is a nonempty Borel set, then $\lbrace x \in \mathbb{R} : f(x) \in B \rbrace$ is nonmeasurable?

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up vote 20 down vote accepted

There are continuum many pairwise disjoint subsets of [0,1] each having Lebesgue outer measure 1. (By transfinite recursion of length continuum.) Assume that they are $\{A_x:0\leq x\leq 1\}$ and their union is [0,1]. Then set $f:[0,1]\to[0,1]$ so that $f(y)=x$ is $y\in A_x$. Now the inverse image of any $X\subseteq [0,1]$, $X\neq\emptyset, [0,1]$ has innermeasure 0, outer measure 1, hence $f^{-1}[X]$ is nonmeasurable.

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Peter, you mean "hence, $f^{-1}(X)$ is not measurable" in your last sentence. –  Joel David Hamkins Sep 23 '10 at 15:57
Joel: thanks, I edited accordingly. –  Péter Komjáth Sep 24 '10 at 4:31
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