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Let A be a Lebesgue measurable set. Let f: $\mathbb{R} \rightarrow \mathbb{R}$ be a function of the class $C^1$;

Is this true that f(A) is lebesgue measurable?

I know that this is true when f is injective.

But I don't know that when f is not injective.

Could you teach me it is true or not?

Sorry for my poor English.

0428 jutaro
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  • To elaborate on my comment: You can express every measurable set as $A =\bigcup K_n \cup N$ with compact sets $K_n$ and a null-set $N$. Then, $f(A)=\bigcup f(K_n) \cup f(N)$ is measurable if $f$ maps nullsets to nullsets, since images of compact sets are compact (because $f$ is continuous). – PhoemueX Jun 03 '15 at 00:33
  • Thanks for answering my question. But still I can't understand. are $K_n$ a compact sets? I think $K_n$ are just closed sets, not always compact (i.e. not always bounded). I can understand your answer except this part. – 0428 jutaro Jun 03 '15 at 02:59
  • If you find closed sets $K_n$ satisfying the desired equation, use $K_{n,k}=K_n \cap [-k,k]$. These will be countably many compact sets, which still satisfy the desired identity. – PhoemueX Jun 03 '15 at 12:27
  • Oh, That's true! Thank you very much! I understood. I greatly appreciate your answer! – 0428 jutaro Jun 03 '15 at 13:12

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