If $f'(x)$ is piecewise continuous in $[0,1]$, does it implie that $f$ is also piecewise continuous in $[0,1]$? My intuition tells me that it should be so, but I see many sentences like "Let $f$ and $f'$ be piecewise continuous."
Edit: $f$ is piecewise continuous in $[0,1]$ if there is a partition $$0=x_0<x_1<\cdots<x_n=1$$ such that each restriction $f|_{(x_{i-1},x_i)}$ is continuous, and the one-sided limits $$\lim_{x\to x_{i-1}^+}f(x)\qquad\mbox{and}\qquad\lim_{x\to x_i^-}f(x)$$ are finite.
It's not an answer. I am simply sharing my thoughts. Could you think about the following situation?
In that case, it's clear that $f$ is, in fact, continuous everywhere on $[0, 1$ because integration is a smooth operator ie takes an integrable function and return continuous function. Please don't get me wrong.