Absolutely continuous maps measurable sets to measurable sets

Question

Show that if $f:\mathbb{R} \rightarrow \mathbb{R}$ is absolutely continuous, then $f$ maps measurable sets to measurable sets.

Any ideas on how to do this?

Answer 1

Because Lebesgue measure is inner regular, given a Lebesgue measurable set $E\subset\mathbb{R}$, there exist a set $N$ of measure zero and a sequence of compact sets $\{K_n\}_{n\ge 1}$, such that $$E=N\cup(\cup_{n=1}^\infty K_n).$$ Since $f$ is continuous, $f(K_n)$ is compact for every $n\ge 1$; since $f$ is absolutely continuous, $f(N)$ is of measure zero. Therefore, $$f(E)=f(N)\cup\big(\cup_{n=1}^\infty f(K_n)\big)$$ is measurable.

Answer 2

I am just providing an already stated proof which is in a solution manual here

Theorem 11 part iv) from real analysis by Royden states for a measure set there exist a $F_\sigma$ set contained in E, for which $m(E\setminus F) =0$. lipchitz function maps a set of measure zero to a set of measure zero (proof) so $f(E\setminus F)$ is measureable and continious function maps $F_\sigma$ onto $F_\sigma$ set proof, $F_\sigma$ set is measurable. So $f(E) = f(E\setminus F)\cup f(F)$, since union of measurable sets is measurable $f(E)$ is measurable.