most recent 30 from http://mathoverflow.net 2013-05-23T16:01:46Z http://mathoverflow.net/feeds/question/73046 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73046/turning-a-measurable-function-to-a-bijection BBB 2011-08-17T11:25:20Z 2011-08-17T11:34:23Z

<p>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 $g|_A=f|_A$?</p>

http://mathoverflow.net/questions/73046/turning-a-measurable-function-to-a-bijection/73047#73047 Tapio Rajala 2011-08-17T11:34:23Z 2011-08-17T11:34:23Z

<p>No, consider $f(x) = 2x \mod 1$.</p>