We know that a function $f$ is Lebesgue-integrable on an open interval $I$. If $f'(x)$ exists almost everywhere on $I$, prove that $f'$ is measurable on $I$.
This appears as the 35th exercise of Chapter 10 in Apostol's book of Mathematical Analysis. It hard to relate the step-function approximation of $f$ to $f'$. I tried to prove that $f'$ can be approximated by step functions. If $f$ is Lebesgue-integrable then there is a sequence of step functions $\{s_n\}$ converges to $f$ almost everywhere. Maybe it's possible to construct another sequence from $\{s_n\}$ to approximate $f'$, using the same partition scheme. The difficulty of this approach is that we don't know anything further about $f'$, like whether $f'$ is continuous at every subinterval. Maybe it is a not so smart idea. Since if we make further assumptions about $f'$, for example, $f'$ is only discontinuous at a finite number of points, the added assumption is enough to prove that $f'$ is measurable.
Hint: for $x \in I$ and $n \in \mathbb N$ let
$f_n(x)=\frac{f(x+1/n)-f(x)}{1/n}$. Then $f_n$ is measurable and
$f_n(x) \to f'(x)$ almost everywhere on $I$.
Can you proceed ?