About absolute continuity $\Rightarrow$ null set maps to null set 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 $0$?
 A: 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.
