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
- $F$ 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 $0$)
- 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.
Best Answer
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. $\diamondsuit$