Question $10.34$, Chapter $10$, in Apostol's "Mathematical Analysis" ($2^{\text{nd}}$ edition):
If $f$ is Lebesgue-integrable on an open interval $I$ and if $f^{\prime}(x)$ exists almost everywhere on $I$, prove that $f^{\prime}$ is measurable on $I$.
In view of the answer here (as modified by Hagen von Eitzen's comment and adapted appropriately for open intervals), is the assumption that $f$ is Lebesgue-integrable (as opposed to just measurable) necessary to solve question $10.34$?
No, I would not say so. Let $A$ be the subset of $I$ where $f'$ exists. By definition, $A$ is the difference of $I$ and a set of measure zero, so it is measurable. Now set $Df_n(x) := \frac{f(x+n^{-1})-f(x)}{n^{-1}}$ for $x \in A$. Clearly, $Df_n$ are measurable and also note that $x \mapsto \inf_{n \in \mathbb{N}} Df_n(x)$ is measurable ($a \in \mathbb{R}$ arbitrary): $$ (\inf_{n \in \mathbb{N}} Df_n)^{-1}([-\infty, a]) = \bigcup_{n \in\mathbb{N}} \underbrace{Df_n^{-1} [-\infty, a)}_{\text{measurable}} $$ Analogously, $\sup Df_n$ is also measurable (for Details see Evans, Gariepy - Measure theory...). Then $$ \liminf_{n\rightarrow \infty} Df_n = \sup_{k \in \mathbb{N}} \inf_{n \geq k} Df_n $$ is measurable and hence $$ f'(x) = \liminf_{n \rightarrow\infty}\chi_A(x) Df_n(x) $$ is measurable.
Integrability was not needed here. It is enough to just demand measurability.