Let $f : [0,1] \to \mathbb{R} $ be an absolutely continuous function.
I am trying to prove (or disprove) that, for any null set $N \subset \mathbb{R}$, then $f^{-1} (N) \cap \{ t \in [0,1] \mid f'(t) \neq 0 \}$ is a null set.
In the literature I found several references showing that, when $f$ is nondecreasing, then the result is true. See here for example (and references therein).
However I was unable to adapt the proof to the non-monotonic case. Furthermore, to try to construct a counterexample, I found an absolutely continuous function which is nowhere monotone (here and here). Unfortunately I was not able to go any further.
Can someone please help me? Thank you.
This is true, but the proof is not simple. Let $I=[0,1]$, $E:=f^{-1}\left( N\right)$, and $E^{\ast}:=\left\{ x\in E:\,\left\vert f^{\prime}\left( x\right) \right\vert >0\right\} $.
We need to prove that $\mathcal{L}^{1}\left( E^{\ast }\right) =0$. For every integer $k\in\mathbb{N}$ let $$ E_{k}^{\ast}:=\left\{ x\in E^{\ast}:\,\left\vert f\left( x\right) -f\left( y\right) \right\vert \geq\frac{\left\vert x-y\right\vert }{k}\text{ for all }y\in\left( x-\tfrac{1}{k},x+\tfrac{1}{k}\right) \cap I\right\} . $$ Note that $$ E^{\ast}=\bigcup_{k=1}^{\infty}E_{k}^{\ast}. $$ Hence, if we fix $k$ and we let $F:=J\cap E_{k}^{\ast}$, where $J$ is an interval of length less than $\frac{1}{k}$, then to prove that $\mathcal{L}^{1}\left( E^{\ast }\right) =0$, it suffices to show that $\mathcal{L}^{1}\left( F\right) =0$.
Since $\mathcal{L}^{1}\left( f\left( E\right) \right) =0$ and $F\subset E$, for every $\varepsilon>0$ we may find a sequence of intervals $\left\{ J_{n}\right\} $ such that $$ f\left( F\right) \subset\bigcup_{n=1}^{\infty}J_{n},\quad\sum_{n=1}^{\infty }\mathcal{L}^{1}\left( J_{n}\right) <\varepsilon. $$ Let $E_{n}:=f^{-1}\left( J_{n}\right) \cap F$. Since $\left\{ E_{n}\right\} $ covers $F$, we have \begin{align*} \mathcal{L}_{o}^{1}\left( F\right) & \leq\sum_{n=1}^{\infty}% \mathcal{L}_{o}^{1}\left( E_{n}\right) \leq\sum_{n=1}^{\infty}\sup_{x,y\in E_{n}}\left\vert x-y\right\vert \\ & \leq\sum_{n=1}^{\infty}k\sup_{x,y\in E_{n}}\left\vert f\left( x\right) -f\left( y\right) \right\vert, \end{align*} where we have used the fact that $E_{n}\subset J\cap E_{k}^{\ast}.$ Since $f\left( E_{n}\right) \subset J_{n}$, we have $$ \sup_{x,y\in E_{n}}\left\vert f\left( x\right) -f\left( y\right) \right\vert \leq\mathcal{L}^{1}\left( J_{n}\right) , $$ and so $$\mathcal{L}_{o}^{1}\left( F\right) \le \sum_{n=1}^{\infty}k\sup_{x,y\in E_{n}}\left\vert f\left( x\right) -f\left( y\right) \right\vert\leq k\sum_{n=1}^{\infty}\mathcal{L}^{1}\left( J_{n}\right) <k\varepsilon. $$ It now suffices to let $\varepsilon\rightarrow0^{+}$.