Let $f$ be an absolutely continuously monotone function on $[0,1]$. Suppose $E$ has measure zero. How do I go about showing that the measure of $f(E)$ is zero?
Thanks.
Edit: Would this work?
Since $f$ is absolutely continuous, for every $\varepsilon \gt 0$, there is a $\delta \gt 0$ such that for a family of non-overlapping intervals $\{[x_i,y_i]\}_i$ of $[0,1]$, we have $$ \sum_i(y_i - x_i)\lt \delta ~\implies ~ \sum_i |f(y_i)-f(x_i)|\lt \varepsilon.$$
Let $(x_k,y_k)\subset [x_i,y_i]$ cover $E$. then $E\subset \bigcup (x_k,y_k)$(disjoint) and $\sum(y_k-x_k)\lt \delta.$
Also, $$f(E)\subset f\left(\bigcup (x_i,y_i)\right)=\bigcup\left(f(x_k),f(y_k)\right)~,$$
and
$$ \mu(f(E))\leq \mu \left(\bigcup\left(f(x_k),f(y_k)\right)\right)=\sum_k \left|f(y_k)-f(x_k)\right|\lt \varepsilon.$$
Thus , $\mu(f(E))=0$.