# When are Differentiation and Integration Inverse Operations?

Edit: Naively, for F differentiable:

$$\int F'=F$$

But this is not always true, as in the case where F is the Cantor function. Then we see that

$$F'=0$$ so that $$\int F'=\int 0 =C \neq F$$ , for C a Real constant.

I believe adding the condition that F be Absolutely Continuous is sufficient, though

not clear that it is necessary or somehow minimal. I understand that characterizing all

functions that can be a derivative is an open problem ( a necessary condition being satisfying

the Darboux property), but I believe my question is different.

• For $F$ continuous, it doesn't even make sense to say $\int F'$. You need $F$ to be differentiable, or at least differentiable almost everywhere. It also depends if you mean the Riemann or Lebesgue integral. This page addresses the Lebesgue integrability for absolutely continuous functions en.wikipedia.org/wiki/Absolute_continuity. Commented Nov 10, 2021 at 21:47
• @ElliotG: Thanks, but I am also trying to figure out when $\int F'=F$ For F the Cantor function (Not Absolutely Continuous), we get that $F'=0$ so that $\int F'=\int 0 \neq F$. When do we have $\int F'=F$. And I think AC implies differentiable. And if /when $F$ is continuous , we get $G= \int_0^x F(t)dt$ gives us G'=F.
– MSIS
Commented Nov 10, 2021 at 21:47
• AC does not imply differentiable, and the Cantor function is also not differentiable; it's differentiable almost everywhere. The derivative is only defined on $[0,1]$ minus the Cantor set, but it turns out that, no matter how you extend $F'$ to the interval, $F'$ is still Riemann integrable and $\int F'=0$. Commented Nov 10, 2021 at 21:54
• It is true that $F$ is absolutely continuous if and only if $F$ is differentiable almost everywhere, $F'$ is Lebesgue integrable, and $F(x)=F(a)+\int_a^xF'$. I'm not sure if that answers your question though. Commented Nov 10, 2021 at 22:08
• What's also interesting is that $F$ being differentiable is not enough to assume $F'$ Riemann integrable, even if $F'$ is bounded. But "differentiable" and "absolutely continuous" are apparently not comparable. Commented Nov 10, 2021 at 22:11

Ok I think this answers the question so I'm writing as an answer.

According to the Wikipedia page for absolute continuity, the following are equivalent for a function $$f\colon[a,b]\to\Bbb R$$:

1. $$f$$ is absolutely continuous
2. $$f$$ is differentiable almost everywhere, the derivative $$f'\colon S\to\Bbb R$$ is Lebesgue integrable, and $$f(x)=f(a)+\int_a^xf'$$ for all $$x\in[a,b]$$. Here, $$S$$ is a subset of $$[a,b]$$, $$[a,b]\setminus S$$ has measure zero, and $$\int$$ means the Lebesgue integral. (If you like, you can consider $$f'$$ to be a function on $$[a,b]$$ by extending to $$0$$ on $$[a,b]\setminus S$$.)

One might ask if the Riemann integral works instead, but the question is ill-posed because the Riemann integral can't ignore sets of measure zero. It turns out this problem can't be overcame:

Theorem $$1$$: If $$f\colon[a,b]\to\Bbb R$$ is differentiable and $$f'$$ is bounded, then $$f$$ is absolutely continuous.

Theorem $$2$$: There exists a function $$f\colon[a,b]\to\Bbb R$$ which is differentiable, $$f'$$ is bounded, and $$f'$$ is not Riemann integrable (on any subinterval of $$[a,b]$$).

Putting these together, we get an absolutely continuous (and differentiable) function $$f$$ for which $$f$$ is not, in any sense, equal to $$\int f'$$ (if $$\int$$ is the Riemann integral).

I don't have a source for Theorems $$1$$ and $$2$$ off the top of my head other than my undergraduate thesis (Theorem $$1$$ is Theorem $$3.6$$; Theorem $$2$$ is basically all of section $$2$$).

• Thank you. Please give me some time to read it; will be pretty busy for a while.
– MSIS
Commented Nov 13, 2021 at 22:24