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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.

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    $\begingroup$ 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. $\endgroup$
    – pancini
    Commented Nov 10, 2021 at 21:47
  • $\begingroup$ @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. $\endgroup$
    – MSIS
    Commented Nov 10, 2021 at 21:47
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    $\begingroup$ 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 $\int F'=0$. $\endgroup$
    – pancini
    Commented Nov 10, 2021 at 21:54
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    $\begingroup$ 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. $\endgroup$
    – pancini
    Commented Nov 10, 2021 at 22:08
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    $\begingroup$ 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. $\endgroup$
    – pancini
    Commented Nov 10, 2021 at 22:11

1 Answer 1

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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 $2$).

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

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