This is how the function is constructed in my notes:
Let $C$ be a Cantor set with $\lambda(C)>0$ and $C = [0,1]\setminus U$ where $U = \bigcup\limits_{k=1}^\infty I_k$ is the disjoint union of open intervals $I_1,I_2,...$ Choose intervals $J_k \subsetneq I_k$ so that $J_k$ has the same center as $I_k$ with $|J_k| = \frac{1}{2}|I_k|$. Finally, choose continuous functions $f_k:[0,1]\rightarrow [0,1]$ such that $f_k(x) = 0$ outside $J_k$ and $f_k(x) = 1$ at the midpoint of $J_k$. Define $f:[0,1] \rightarrow [0,1]$ by $f(x) = \sum\limits_{k=1}^\infty f_k(x)$. Then $f(x)$ is continuous on $U$, discontinuous on $C$, and not integrable (by Lebesgue's criterion).
If we define $g:[0,1]\rightarrow [0,1]$ by $g(x) = \sum\limits_{k=1}^\infty {\displaystyle \int_{0}^{x}} f_k(t)\, dt$, then $g$ is differentiable with $g' =f$ on $[0,1]$.
I can follow everything here, except I don't see why $g' =f$. I believe that $\sum\limits_{k=1}^\infty f_k(x)$ doesn't converge uniformly to $f$ so I don't think I can switch the differentiation and summation in $\frac{d}{dx}g(x) = \frac{d}{dx}\sum\limits_{k=1}^\infty {\displaystyle \int_{0}^{x}} f_k(t)\, dt$.
I'm not sure where to go from here. Maybe $g$ isn't actually differentiable?