Lebesgue measuarable sets under a differentiable bijection Let $U,V \subseteq \mathbb{R}^{n}$ be open and suppose $A\subseteq U$ are (Lebesgue) measurable. Suppose $\sigma \in C^{1} (U,V)$ be a bijective differentiable function. Then does it follow that $\sigma(A)$ is (Lebesgue) measurable?
I've tried work on it, but still stuck and cannot progress at all. I've tried to use continuity of $\sigma$ but then as $A$ is not an open set, I couldn't really use it. Should I use the deifinition of Lebesuge measurable set? But I think it will be more complex.. 

The answer is yes. Let me call your differentable bijection $f$...
Hint : Every Lebesgue measurable set is the union of a $F_{\sigma}$ and a set of measure zero. Now, use the fact that the image by $f$ of any $F_{\sigma}$ is Lebesgue measurable (why?) and that $f$ maps sets of measure zero to sets of measure zero...
EDIT
To show that $f$ maps sets of measure zero to sets of measure zero, note that $f$ is locally lipschitz, and you can proceed as in this question
