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One form of the FTC is the following:

if $f$ is continuous on an interval $I$ and $a$ a point of $I$, then the function $F$ defined on $I$ by $\displaystyle{ F(x)=\int_a^x f(t) \, dt }$ is differentiable on $I$ and satisfies $F'(x)=f(x)$ for any $x$ in $I$.

It is well-known that if $f$ is not continuous, then

  1. $F$ may not be differentiable (for example, for $f(x)=0$ when $x<0$ and $f(x)=1$ when $x\geqslant 1$, $F$ is not differentiable at $0$)
  2. and if $F$ is differentiable, we may not have $F'(x)=f(x)$ at the discontinuities of $f$. For example, if $f(x)=0$ for $x\neq 0$ and $f(0)=1$, then $F$ is constant equal to $0$, hence differentiable. But $F'(0)=0\neq f(0)$.

I am now looking for a converse of the above version of the FTC: if $F$ is an antiderivative of $f$, then $f$ is continuous on $I$. Is this statement true? I was not able to prove it, nor to find a counter-example.

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  • $\begingroup$ You need to say something about the kind of functions $f$ you are considering when you are integrating them. Since you don't want to assume $f$ is continuous, what are you willing to assume about $f$ so that $\int_a^x f(t)\,dt$ makes sense? Also mention your mathematical background: one or two semesters of calculus? Measure theory? Something in between? $\endgroup$ – KCd Oct 6 '14 at 6:22
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    $\begingroup$ There are differentiable functions with a non continuous derivative like $F(x)=x^2\sin(1/x)\chi_{\{x>0\}}(x)$. Thus $F'=f$ for an appropriate $f$ but $f$ is not continuous. Extension of the FTC usually consider so called absolutely continuous functions. This is handled in Rudin's book on real and complex analysis. $\endgroup$ – Quickbeam2k1 Oct 6 '14 at 6:23
  • $\begingroup$ There is some consolation in Darboux's theorem; see here: en.wikipedia.org/wiki/Darboux%27s_theorem_(analysis) $\endgroup$ – Guy Apr 26 '16 at 19:17
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You ask to prove that if $F$ is an antiderivative of $f$, defined over some interval $I$, then $f$ is continuous on $I$.

This is false.

For a simple counterexample, consider $f(x) = 0$ for $x \leq 0$ and $f(x) = 1$ for $x > 0$. Then this function is integrable on any closed interval. For instance, $$ F(x) = \int_{-1}^x f(t) dt = \begin{cases} 0 & x \leq 0 \\ x & x \geq 0 \end{cases}.$$ So an antiderivative $F$ exists and is continuous, but $f$ is clearly not continuous at $0$.

In this way, you can make integrable functions that are have countably many discontinuities. $\diamondsuit$

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