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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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up vote 2 down vote accepted

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...


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 your help!! Btw, if I may, what is $F_{\sigma}$? – julypraise Oct 8 '12 at 17:55
You're welcome! An $F_{\sigma}$ is a countable union of closed sets. If you are still having difficulty with your question, I suggest you take a look at Rudin's Real and Complex Analysis, thm 2.20 and lemma 7.25 in the 3rd edition. – Kalim Oct 8 '12 at 17:59
Thank you very much for your reference too! – julypraise Oct 8 '12 at 18:04
Ah.. just one more question if you are still there. You said $f$ maps to sets of measure zero to those of measure zero. I'm not sure how to prove this part.. And just to check: since $f$ is continuous, it is Borel measurable and thus as $f$ is bijective the image of $F_{\sigma}$, which is a Borel set, is a measurable set. Right? – julypraise Oct 8 '12 at 19:16
@julypraise : I edited my answer to provide more details. – Kalim Oct 8 '12 at 20:34

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