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 $\varepsilon\rightarrow0^{+}$.
