If $f$ is differentiable almost everywhere, it implies $f'$ is measurable. How to prove this?
$f$ may not be continuous.
I'm thinking to use $$f_n(x) = \frac{f(x+1/n) - f(x)}{1/n}$$ to approximate the "derivative" f' (almost everywhere). However, I can't convince me if $f_n$ is measurable to reach the conclusion.
I gather you only need to prove that $f$ is Lebesgue measurable, which it is because it's continuous almost-everywhere (every point where $f$ is differentiable is a continuity point).
In point of fact, let $f$ be a.e. continuous and let $C$ be the set of continuity points of $f$. $$f^{-1}(-\infty,a)=(f^{-1}(-\infty,a)\cap C)\cup (f^{-1}(-\infty, a)\setminus C)$$
By pointwise continuity, for all $x\in f^{-1}(-\infty,a)\cap C$ there is an open set $U\subseteq \Bbb R$ such that $x\in U\subseteq f^{-1}(-\infty,a)$. Taking the union of all those sets over all the points of $f^{-1}(-\infty,a)\cap C$ we obtain an open set $V$ such that $f^{-1}(-\infty,a)\cap C\subseteq V\subseteq f^{-1}(-\infty,a)$. Therefore $$f^{-1}(-\infty,a)=V\cup (f^{-1}(-\infty,a)\setminus C)$$
which is Lebesgue measurable because $V$ is open and $f^{-1}(-\infty,a)\setminus C\subseteq \Bbb R\setminus C$ is a null set.