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I've searched quite a lot in this site and found many questions that mention this result, but not a single proof. I tried proving it myself and failed.

So, I ask it here: how can one prove that if $f:[a, b]\rightarrow \mathbb{R}$ is absolutely continuous, then for every set $A \subseteq[a,b]$ with lebesgue measure $0$, $f(A)$ also has lebesgue measure $0$?

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  • $\begingroup$ How are you defining absolute continuity in the first place? There are several equivalent definitions that you could choose. $\endgroup$
    – Ian
    Commented Aug 19, 2016 at 19:46
  • $\begingroup$ You can use the $\epsilon - \delta$ definition, or the fundamental-theorem-of-calculus-definition. I know that these two are equivilant. $\endgroup$
    – 35T41
    Commented Aug 19, 2016 at 19:49

1 Answer 1

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Let $N \subset (a,b)$ be a null set and let $\epsilon > 0$.

Select $\delta > 0$ corresponding to the definition of absolute continuity and let $O \subset [a,b]$ be an open set with $m(O) < \delta$. There exists a countable family $\{(a_k,b_k)\}$ of disjoint open intervals with $O = \bigcup_k (a_k,b_k)$.

Since $f$ is continuous on each $[a_k,b_k]$ there exist points $c_k,d_k \in [a_k,b_k]$ on which $f$ attains its infimum and supremum, respectively. This means that $f([a_k,b_k]) = [f(c_k),f(d_k)]$ so that $$m^*(f((a_k,b_k))) = |f(d_k) - f(c_k)|.$$

Since $\sum_k |c_k - d_k| \le \sum_k |a_k - b_k| = m(O) < \delta$ you have $$\sum_k |f(d_k) - f(c_k)| < \epsilon$$ by absolute continuity. Consequently $$m^*(f(O)) = m^*( f(\bigcup_k (a_k,b_k)) \le \sum_k m^*(f((a_k,b_k)) < \epsilon.$$

In particular, if $O$ is an open set with $N \subset O \subset (a,b)$ and $m(O) < \delta$ then $$m^*(f(N)) \le m^*(f(O)) < \epsilon.$$

Since this is valid for any $\epsilon > 0$ it follows that $m^*(f(N)) = 0$ as desired.

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  • $\begingroup$ Wow, thank you so much! I've been stuck on that one for 2 days. This simple reduction to open sets escaped my attension. $\endgroup$
    – 35T41
    Commented Aug 19, 2016 at 20:01
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    $\begingroup$ Once again Littlewood's principles come to the rescue! - en.wikipedia.org/wiki/… $\endgroup$
    – Umberto P.
    Commented Aug 19, 2016 at 20:04
  • $\begingroup$ Can I argue $f(N)$ is measurable in this way? There is a $\delta_n$ such that $\sum |f(b_i)-f(a_i)|<1/n$ if $\sum |b_i-a_i|<\delta_n$. For each $\delta_n$, there is a collection of open interval such that $I_{n,i}=(a_{n,i},b_{n,i})$ which cover $N$ and $m(\bigcup_i I_{n,i}) \leq \delta_n$ and thus $\sum_i f(I_{n,i}) <1/n$. Then $m(\bigcap_n \bigcup_i f(I_{n,i}))=0$. Since $f(N) \subset \bigcap_n \bigcup_i f(I_{n,i}) $. So $f(N)$ is measurable as Lebesgue measure is complete. $\endgroup$
    – mnmn1993
    Commented Nov 22, 2017 at 8:47

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