Let $g:[0,T] \to \mathbb R$ be a bounded measurable function.
Let $f(t)=\int_0^t g(\tau) \ d\tau +a$ for some $a\ne 0$. Then, $f$ is absolutely continuous and almost everywhere differentiable.
Assume there exists $E \subset [0,T]$ with $\mu(E)>0$ and $f(t)=0$ for all $t \in E$.
The question: Is there a subset $E' \subset E$ such that $\mu(E')=\mu(E)$ and $f'(t)=0$ for all $t \in E'$? [Recall $f'(t)=g(t)$]
Yes, there is.
Let $E^*$ be the subset of $E$ where $f'$ exist. $\mu(E^*)=\mu(E)$ since $f'$ exist almost everywhere.
Pick any $x_0\in E^*$ which is not isolated, so we have a sequence $\{x_n\}_n\subset E^*$ that converges to $x_0$.
$f'(x_0)$ exists. Therefore,
$$f'(x_0)=\lim_{n\to\infty}\frac{f(x_n)-f(x_0)}{x_n-x_0}=\lim_{n\to\infty}\frac{0-0}{x_n-x_0}=0.$$
We only need $f$ to be constant here, doesn't have to be $0$.
Set of all isolated points in $E^*$ must be countable, thus has a measure of $0$.