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I was finding a proof of the following result which is a characterisation of functions which are derivative of some function i.e. posseses an antiderivative.

Theorem

Let $I$ be an interval. $f:\mathbb {R\to R}$ be the derivative of some function on $I$ only if the set of all points of discontinuities of $f$ is $F_\sigma$ set of first category.

I want to find a concrete and detailed proof of the theorem.I have tried to prove this on my own,but I do not think this is trivial.Can someone provide me a detailed proof of this theorem?

Kishalay Sarkar
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  • Well the result you seek is actually the converse of this. It is a established theorem that discontinuities of a derivative form an $F_{\sigma} $ meager set. See https://math.stackexchange.com/a/112133/72031 You have the misconception that this result works in reverse like "iff". – Paramanand Singh Jul 18 '20 at 13:24
  • @NateEldredge Sorry,See the edit. – Kishalay Sarkar Jul 18 '20 at 13:40
  • @ParamanandSingh Sorry,I have made an edit...see it – Kishalay Sarkar Jul 18 '20 at 13:40
  • The linked answer has all the references with detailed proofs. You should have a look at them. – Paramanand Singh Jul 18 '20 at 13:44
  • Also the result in linked answer is that if $f$ is differentiable everywhere then discontinuities of $f'$ form an $F_{\sigma} $ meager set. And conversely if $D$ is an $F_\sigma$ meager set then there is a function $f$ which is everywhere differentiable and $D$ is the set of discontinuities of $f'$. I don't know if this is equivalent to the result you seek. – Paramanand Singh Jul 18 '20 at 13:48
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    Your problem seems to be related to characterizing functions which possess an anti-derivative. I don't know if there is any definitive result of this type. – Paramanand Singh Jul 18 '20 at 13:50
  • With the edit I think that $f(x) = -x^{-2} \cos(x^{-1})$, with $f(0)=0$, on $I=(-1,1)$, is a counterexample. Its only discontinuity is at $x=0$ and is of the second kind, and the set ${0}$ is $F_\sigma$ and first category. But if $g' = f$ then we have $g(x) = \sin(1/x)+c$ for all $x>0$ and then $g$ cannot even be continuous at $x=0$. – Nate Eldredge Jul 18 '20 at 13:59
  • @NateEldredge I have again made an edit. – Kishalay Sarkar Jul 18 '20 at 14:11

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