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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..

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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

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    Thanks for the details. I just checked the link. But I'm quite not sure how exactly it is $\mu^\ast f(B_k) \leq L^d \mu B_k$2012-10-08