I am looking for a counterexample to the following:
Let $f$ be [Riemann] integrable $[a,b]$, let $c \in (a,b)$, and let $$F(x) = \int_a^xf$$ for $a\leq x\leq b$.
Find $f,c$ such that $f$ is differentiable at $c$, but $F'$ is not continuous at $c$.
I've been working on this with multiple people for quite some time, and though the solution says a counterexample exists, we have not been able to find one.
As I commented, I expect that the problem statement should have been
Finding examples for that, while not trivial, is not exceedingly difficult.
If we require that the integrand $f$ is differentiable at $c$, it becomes impossible to find an example. It is already impossible if we merely require $f$ to be continuous at $c$. We can write
$$F(x) = F(c) + f(c)(x-c) + \int_c^x \bigl(f(t) - f(c)\bigr)\,dt.$$
Let $\varepsilon > 0$ be given. By the continuity of $f$ at $c$, there is a $\delta > 0$ such that $\lvert f(t) - f(c)\rvert < \varepsilon$ for all $t \in (c-\delta, c+\delta)$. Then, for $x,y \in (c-\delta, c+\delta)$ with $x < y$ we have
$$F(y) - F(x) = f(c)(y-x) + \int_x^y \bigl(f(t) - f(c)\bigr)\,dt,$$
and so
$$\Biggl\lvert \frac{F(y) - F(x)}{y-x} - f(c)\Biggr\rvert \leqslant \frac{1}{y-x}\int_x^y \lvert f(t) - f(c)\rvert\,dt < \varepsilon.$$
Thus at all points of $(c-\delta, c+\delta)$ where $F$ is differentiable, we have
$$\lvert F'(x) - f(c)\rvert \leqslant \varepsilon.$$
So $F'$ is continuous at $c$.
Note: There can be points arbitrarily close to $c$ where $F$ is not differentiable, but that doesn't affect the continuity of $F'$ at $c$, for that one considers only the domain of $F'$.
If it was meant that there is no neighbourhood $(c-\eta, c+\eta)$ of $c$ such that $F$ is differentiable at all points of that neighbourhood, just take an $f$ that has a sequence of jump discontinuities converging to $c$. Such functions are easy to find even when one requires differentiability at $c$.