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

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?

• In case of oscillation at $c$ it is possible that $F'(c)$ exists. In particular if $f(x) =\sin(1/x)$ then $F'(0)=0$. The proof is tricky and available on mathse. Commented Mar 10, 2021 at 7:34
• And clearly in the example in above comment we see that $F'$ is discontinuous at $0$. Commented Mar 10, 2021 at 7:37
• On the other hand if $f$ is continuous at $c$ then perhaps (but not sure) we can use density arguments to show that $F'$ is also continuous at $c$. Commented Mar 10, 2021 at 7:50
• @ParamanandSingh: thanks! I've updated. That $\sin(1/x)$ example is indeed tricky, my gut said $F'$ wouldn't exist there. Commented Mar 10, 2021 at 16:23

## 1 Answer

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 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 $$c.$$