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Let $f:(0,1)\rightarrow (0,1)$ be a borel measurable function such that for every $y$ in $(0,1)$ , $f^{-1}(y)$ is a borel set and $\mu(f^{-1}(y))=0$ and also $\mu (f((0,1)))=1$ where $\mu$ is the lebesgue measure . Is it possible to build a function $g:A\rightarrow A$ where $A\subseteq[0,1]$ is a borel set and $\mu (A)=1$, such that $g$ is a bimeasurable bijection and $g|_A=f|_A$?

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1 Answer

No, consider $f(x) = 2x \mod 1$.

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Are there any additional conditions on f that will make it true? –  BBB Aug 18 '11 at 9:57
    
All the conditions I can think of are essentially tautological. –  Tapio Rajala Aug 18 '11 at 10:46
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