Consider a function $g:\mathcal{I}\to\mathbb{R}$ defined in an open interval $\mathcal{I}$ of the real line and therein differentiable almost everywhere.
Is it true that all points in which $g$ is differentiable have a neighborhood in which the derivative of $g$ is defined everywhere?
No. Define a function $f:(-1,1) \to \mathbb R$ by $$f(x) = \max\{2^{-n}:n \in \mathbb N, 2^{-n} \le \lvert x\rvert\}.$$ Then $f$ has countably many discontinuities, at the points $\pm 2^{-n}$, but is continuous at $0$. Hence the function $g:(-1,1) \to \mathbb R,$ $$g(x) = \int_0^x f(t)\ dt$$ is differentiable almost everywhere, including at $0$, but not on any open interval containing $0$.