Is the derivative of the Riemann integral continuous wherever it exists?

Question

Let $f$ be Riemann integrable on $[a,b]$ and let $F(x)$ be its integral on $[a,x]$. (Note the minimal assumption of integrability, not continuity.)

Must $F’$ be continuous wherever it’s defined? (Nota bene: here I mean continuous in the general sense of continuous on its domain. On this definition, it’s possible for a function to be continuous at a point without there being a deleted interval of the point on which the function is defined.)

Context

The claim is plausible on these heuristic grounds: $F$ ignores trouble at discontinuities (which happen on a set of measure zero), so even if $f$ fails to be continuous somewhere, $F’$ should be continuous there (if it exists). In other words, $F’$ should be “at least as nice” as $f$ (again, given that it exists).

As an example, take an $f$ continuous except at $c$, where it has a removable discontinuity. Let $g$ be the everywhere continuous function that removes the discontinuity. Since $f=g$ almost everywhere, their integrals $F$ and $G$ are everywhere the same. By the FTC, $G’=g$; combining with the previous sentence we have $F’=g$. Thus $F’$ will be continuous at $c$ even though $f$ isn’t: it plugs the hole.

This is just intuition. Any proof will need to deal with two cases: one at points where $f$ is continuous and the other at points where $f$ is discontinuous.

In the former case we know by the FTC that $F’$ exists at $c$ (and moreover $F’(c)=f(c)$). Can we prove that it’s continuous there?

In the latter case, we could break into subcases depending on the type of discontinuity. If $f$ has a removable discontinuity we can reduce to the previous case. If $f$ has a jump discontinuity, I think we can show $F’$ doesn’t exist. And if $f$ has an essential (oscillatory) discontinuity, I also think we can show $F’$ doesn’t exist. (I haven’t produced a rigorous argument but can’t find counterexamples. I’m stuck with $f(x)=\sin(1/x)$ on $[-1,0)\cup(0,1]$ and $0$ at the origin. Will $F’$ exist at the origin?)

If all that’s right, the proof depends essentially on whether the following implication holds: $f$ continuous at $c$ implies $F’$ continuous at $c$.

If we strengthen the hypothesis by assuming $f$ is not just continuous but differentiable at $c$, it does indeed follow that $F’$ is continuous at $c$. The question is whether we can relax the hypothesis to bare continuity of $f$ at $c$.

EDIT

Paramanand Singh points out in comments that $F'$ can indeed exist and be discontinuous where $f$ has an oscillatory discontinuity. (The right example, as I guessed, was $f(x)=\sin(1/x)$ on $[-1,0)\cup(0,1]$ and $0$ at the origin, but it turns out that $F'$ exists here, contrary to my hunch.)

So the answer to the question in the title is no. But this question is still open: if $f$ is continuous somewhere, must $F'$ be continuous there?

Answer 1

Yes, if $f$ is continuous at $c,$ then $F'$ is continuous at $c$ within the domain of $F'.$ (Somewhat humorously, the proof of this is easier than my previous proof in the case where $f'(c)$ exists. See the link the OP provides.)

Proof: Suppose $x_1> x_2 > \cdots \to c$ and $F'(x_n)$ exists for each $n.$ WLOG, $[x_n,x_n+1/n]\subset (c,b]$ for all $n.$

Fix $n$ for the moment. Then

$$F'(x_n)-f(c) = \lim_{h\to 0^+}\frac{1}{h}\int_{x_n}^{x_n+h} (f(x)-f(c))\,dx.$$

We can assume $0<h<1/n$ above. Thus

$$\tag 1| F'(x_n)-f(c)| \le \sup_{[x_n,x_n+1/n]}|f-f(c)|.$$

Now unfreeze $n.$ Because $f$ is continuous at $c,$ the right side of $(1)$ $\to 0$ as $n\to \infty.$ Hence so does the left side. Since $F'(c)=f(c),$ we have shown the desired continuity of $F'$ at $c.$