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

$\begingroup$ Are there any additional conditions on f that will make it true? $\endgroup$ – BBB Aug 18 '11 at 9:57

$\begingroup$ All the conditions I can think of are essentially tautological. $\endgroup$ – Tapio Rajala Aug 18 '11 at 10:46