Preimage of a null set by a non-monotonic absolutely continuous function 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.
 A: 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 $\varepsilon\rightarrow0^{+}$.
