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Let $E\subset (0,1)$ is Lebesgue measurable, $f:(0,1)\to\mathbb{R}$ is of class $C^1$ and strictly increasing. Show that $f(E)$ is Lebesgue measurable.

My attempt: Since every Lebesgue measurable set $E$ can be decomposed as $F \cup N$, where $F$ is an $F_\sigma$-set and $N$ has Lebesgue measure zero. Since $f$ is a homeomorphism between the interval $(0,1)$ and its image, $f(E)=f(F)\cup f(N)$ and $f(F)$ is also an $F_\sigma$-set. Therefore, it remains to show that $f(N)$ is also Lebesgue measurable.

However, since $f$ is defined on an open interval, it may diverge to $\pm\infty$ at the endpoint. That means I cannot assume that $f$ is absolutely continuous and I cannot claim directly that $f(N)$ also has Lebesgue measure zero.

What should I do in this case?

Any hints or advices will help a lot!

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  • $\begingroup$ Hint: can you show that $A\subset\Bbb R$ is Lebesgue measurable if and only if $A\cap[a,b]$ is Lebesgue measurable for every $[a,b]\subset\Bbb R$? Does that help here? $\endgroup$ Mar 26, 2018 at 7:23
  • $\begingroup$ note that every $F_{\sigma}$ set is a countable union of compacts, so $f(F)$ also is, even if $f$ is only assumed to be continuous. So you can get away with the strictly increasing condition. $\endgroup$
    – orangeskid
    Mar 26, 2018 at 8:17

1 Answer 1

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Let $f_{n}=f\big|_{[1/n,1-1/n]}$ for $n\geq 2$ and $E_{n}=E\cap[1/n,1-1/n]$, and we have $f(E)=\displaystyle\bigcup_{n\geq 2}f_{n}(E_{n})$ and each $f_{n}(E_{n})$ is Lebesgue measurable by the reasoning.

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  • $\begingroup$ Thanks for the answer! Here each $f_n$ is absolutely continuous on the given interval $[1/n,1-1/n]$, so we can argue that $f_n(E_n)$ is measurable for each $n$, right? $\endgroup$
    – bellcircle
    Mar 26, 2018 at 7:28

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