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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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The answer is no, here is a counterexample. Consider $$f(x) =\left\{\begin{array}{lc}2x,& x\le 1/2 \\ 2x-1, & x>1/2 \end{array}\right. $$ Then for all $y$ in $(0,1)$, $f^{-1}(y)$ has cardinal $2$ and hence has measure zero. Now assume that for some measurable $A \subset (0,1)$, $g:=f_{|A}$ is injective. Letting $A_1=A \cap (0,1/2)$, $A_2=A \cap (1/2,1)$, since $g$ is injective $x \in A_1 \implies 1/2+x \notin A_2$. This implies $\mu(A_2^c \cap [1/2,1]) \geq \mu(A_1)$, and $\mu(A) = \mu(A_1)+\mu(A_2) \leq 1/2$.

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What additional conditions on f will make it true? –  Arnold Aug 18 '11 at 10:00

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