Let $f$ be an asbolutely continuous, non-decreasing function on a finite, closed interval $[a,b]$. Let $E$ be a set of Lebesgue measure zero in $[a,b]$. Prove that the measurable set $f(E)$ has Lebesgue measure zero.
Attempt
Fix $\varepsilon >0$. Then there exists $\delta$ such that $\sum|f(b_i)-f(b_i) < \varepsilon$ when $\sum|b_i-a_i|<\delta$ is a finite collection of intervals. Clearly, since $E$ is of measure zero we can parition in $E$ into finitely many disjoint sub-intervals of measure $ < \delta$ for all $\delta >
0$. Thus, $\lambda([f(a),f(b)]) \leq \sum|f(b_i)-f(a_i)| < \varepsilon$.
Edit
Fix $\varepsilon >0$. Then there exists $\delta$ such that $\sum|f(b_i)-f(b_i) < \varepsilon$ when $\sum|b_i-a_i|<\delta$ is a finite collection of intervals.
Since $E$ is of measure zero we can write $E=\cup_{i=1}^\infty E_i$ where $\sum \lambda (E_i) < \delta$. However, since $[a,b]$ is compact we need only finitely many $E_i$ andso $E = \cup^k E_i$ and $\sum^k \lambda (E_i) <\delta \implies$ giving the result. Note: we may assume $E_i = (a_i,b_i)$.
Best Answer
You may modify your reasoning slightly like:
Fix an $\epsilon>0$, then there exists a $\delta>0$ such that $\displaystyle\sum|f(b_{i})-f(a_{i})|<\epsilon$ whenever $\displaystyle\sum|b_{i}-a_{i}|<\delta$, where $\{[a_{i},b_{i}]\}$ is a collection of non-overlapping intervals in $[a,b]$. Here the indexes $i$ may be countably infinite.
Now, as $E$ is of measure zero, there exists a collection $\{I_{i}\}$ of intervals such that $E\subseteq\displaystyle\bigcup_{i}I_{i}$ and $\displaystyle\sum_{i}|I_{i}|<\delta$. By absorbing the overlapping intervals, we know that the resulting collection is non-overlapping and must have measure no more than $\displaystyle\sum_{i}|I_{i}|$, so we may assume that $\{I_{i}\}$ is non-overlapping. Express each $I_{i}$ as $[a_{i},b_{i}]$, then $\displaystyle\sum_{i}|b_{i}-a_{i}|<\delta$. Since $f$ is non-decreasing, then $J_{i}:=[f(a_{i}),f(b_{i})]$ is an interval and $\displaystyle\sum_{i}|J_{i}|=\sum_{i}|f(b_{i})-f(a_{i})|<\epsilon$.
The result follows if we are able to show that the collection $\{J_{i}\}$ is a cover for $f(E)$. For any $x\in E$, choose an $i$ with $x\in I_{i}$, then $a_{i}\leq x\leq b_{i}$, so $f(a_{i})\leq f(x)\leq f(b_{i})$, and hence $f(x)\in[f(a_{i}),f(b_{i})]=J_{i}$, we are done.