Absolutely continuous functions send a set of positive derivative and positive measure to a set of positive measure In P49 of Ziemer's "Weakly Differentiable Functions", it was used that "by classical considerations", if $u$ is an absolutely continuous function on $\mathbb R$, and let \begin{align}P:=\{x\in\mathbb R: u'(x) \mbox{ exists and } u'(x)\neq 0\}.\end{align}
If $S\subset P$ such that $\mu(u(S))=0$, then $\mu(S)=0$, where $\mu$ is the Lebesgue measure on $\mathbb R$.
Why is this true? I don't know what the classical considerations he mentioned are.
 A: Perhaps I find an answer:
Lemma: For $A\subset \mathbb R$, we have $ \mu(u(A))\le \int_A |u'| \,d\mu$. And if $u$ is increasing, we have $\int_A u'\,d\mu= \mu(u(A))$.
Proof by approximation of $A$ by open intervals.
As for the problem I mentioned above, without loss of generality we assume $\mu(S)>0$ with $S\subset \{x:u'(x)>1\}$. Write $u=f-g$ with $f,g$ increasing and $f'=u'\mathbb 1_{u'>0}$ a.e. By the lemma above we have $\mu (f(E))\ge \mu(E)$ for any $E\subset S$. We choose a Lebesgue density point $x$ of $S$ and of $|u'|\mathbb 1_{S^c}$ at the same time. Then for a small enough nbhd $I$ of $x$, we have \begin{align*}\int_I |g'|\le \int_I|u'|1_{S^c}<<\mu(S\cap I)\approx \mu(I).\end{align*}
Then \begin{align*}
\mu(u(I\cap S))&\ge\mu(u(I)) -\int_I |u'|\mathbb 1_{S^c}\ge \mu(f(I))-\int_I |g'|\,d\mu-\int_I |u'|\mathbb 1_{S^c}\\
&\ge \mu(I\cap S)-\int_I |g'|\,d\mu-\int_I |u'|\mathbb 1_{S^c}>0
\end{align*}
Is this argument correct?
(Previously I tried to give a lower bound for $\mu\Big((f-g)(I\cap S)\Big)$ in this way:
\begin{align*}
\mu\Big((f-g)(I\cap S)\Big)\ge\mu\Big(f(I\cap S)\Big)-\int_I |g'|.
\end{align*}
But unfortunately, this fails for general absolutely continuous functions $f$ and $g$. Say, even a very small $g$ can lead to a very large difference between $\mu(f(E))$ and $\mu((f-g)(E))$ for a measurable set $E\subset I$.)
