Let $I$ be a closed interval and $f:I\rightarrow \mathbb{R}$ be a continuous function which maps measure zero sets to measure zero sets.
If $f$ is monotonically increasing, $f$ must be absolutely continuous.
However, is the monotone assumption necessary?
Is there a continuous function mapping measure zero sets to measure zero sets but not absolutely continuous?
Best Answer
There is a theorem that if a function is continuous, bounded variation, and has the "Luzin N property" (i.e. maps measure zero sets to measure zero sets) then it is absolutely continuous. One way to get bounded variation is to assume monotonicity but this is not the only way. However, if you don't have bounded variation then you will not have absolute continuity. In particular:
$$f : [0,1] \to \mathbb{R},f(x)=\begin{cases} x \sin(1/x) & x \neq 0 \\ 0 & x=0 \end{cases}$$
is a continuous function with the Luzin N property which is not absolutely continuous. To see that you can simply bound its variation below by $\sum_{k=0}^\infty |f(x_k)-f(x_{k-1})|$ where $x_{-1}=1,x_k=\frac{1}{\frac{\pi}{2}+k \pi}$.