I apologize if that question has been asked already. I can't figure out this problem:

Let a function $f:[a,b]\rightarrow \mathbb{R}$ which be a Lipshitz function with a constant c, prove that it maps a set of Lebesgue measure zero onto a set of Lebesgue measure zero and a Lebesgue measurable set onto a lebesgue measurable set.

Now, I managed to do the first part, but the part about measurable sets is not as easy for me. If I take an open subset of $[a,b]$ it will be open and bounded hence lebesgue measurable, but its image won't necessarily be open, it may be closed, though still bounded... would it necessarily be measurable? can't go further from here.

I studied baby Rudin and Royden&Fitzpatrick, these books define measurable sets in different ways, can't fidure out which one to apply here...


Let $E $ be measurable. By inner regularity, $$m (E)=\sup\{m (K):\ K\subset E, \ K \text { compact }\}. $$ So we can write $$E=E_0\cup\bigcup_nK_n, $$ an increasing union of compacts, and $E_0$ a null set. Now $$ f (E)=f (E_0)\cup\bigcup_nf (K_n), $$ a union of a null set and compacts, so measurable.

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  • $\begingroup$ Thank you for your answer. I have a few questions about it. Firstly, can we do it without involving the inner regularity? We haven't studied this property in class, however it is stated in Rudin's RCA. Secondly, if we avoided inner regularity could we still use this decomposition of $E$? $\endgroup$ – BoBoB Nov 21 '16 at 0:26
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    $\begingroup$ This is the argument that comes to my mind; right now, I cannot think of another. As for your second question, inner regularity is precisely the fact that measurable sets are of the form I wrote, so you cannot separate them. $\endgroup$ – Martin Argerami Nov 21 '16 at 0:36

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