Absolute continuity on $[a,b]$ implies mapping of sets of measure zero to sets of measure zero

Question

I am reviewing some materials for real analysis and measure theory.

I want to show that if we have an absolutely continuous function on a closed interval, the function maps sets of measure zero to sets of measure zero.

I found the following that I have a few questions about, found at http://www.ms.uky.edu/~perry/676-s19/_assets/final-answers.pdf on the last page :

• Where the author writes that $m(O) < \delta$, is this because $m(O) < m(E) + \delta = \delta$ ?
• How does the author assert that $f$ is continuous on each interval $[a_k,b_k]$ ? Since $Z$ is contained in $O$, can't it be that $[a_k,b_k]$ is not necessarily contained in $[a,b]$ ?
• Why can we conclude from $m(f(O)) = 0$ that $m(f(Z)) = 0$ ?

Thank you ! This is my first time here.

Answer 1

• Where the author writes that $m(O) < \delta$, is this because $m(O) < m(E) + \delta = \delta$ ?

$\mathcal O$ is just chosen to be an open set containing $Z$ of measure less than $\delta$. There exist open sets of arbitrary small measure containing $Z$.

• How does the author assert that $f$ is continuous on each interval $[a_k,b_k]$ ? Since $Z$ is contained in $O$, can't it be that $[a_k,b_k]$ is not necessarily contained in $[a,b]$ ?

If $S \subseteq T \subseteq \mathbb R$, and $f$ is a continuous function on $T$, its restriction to $S$ is always continuous.

• Why can we conclude from $m(f(O)) = 0$ that $m(f(Z)) = 0$ ?

They proved that given any $\epsilon > 0$, there exists an open set $\mathcal O \subset [a,b]$ such that $m(f(Z)) \leq m(f(\mathcal O)) < \epsilon$. So for any $\epsilon >0$, we have $m(f(Z)) < \epsilon$. This can only happen if $m(f(Z)) = 0$.