I have been having an argument with my friend. My friend claims that the following statement is true.
Let $a<b$ be real numbers. Suppose that there is a function $F:[a,b]\rightarrow\mathbb{R}$ which is a function such that it is differentiable everywhere, i.e., there is a function $f:[a,b]\rightarrow\mathbb{R}$ such that $f(x)=F'(x)$ for all $x\in [a,b]$. Then it is true that
\begin{equation} \displaystyle\int_{a}^b f(x)\,dx = F(b)-F(a). \end{equation}
Here $f$ is assumed to be integrable, in whatever integrals.
I've been trying to disprove this statement. He convinces me by using a result in Tao's Analysis I that this is true in the case of Riemann integrable $f$. However, that does not imply that this statement is still true if $f$ is just Lebesgue integrable. I've been searching for a function that is differentiable everywhere, but its derivative does not satisfy the above equation.
My idea is to use an example of a function $F$ which is differentiable everywhere on $[a,b]$, but not absolutely continuous. This implies that it does not satisfy the above equation. My only problem is, this (desired) conclusion is too strong. We can infer only that
$f$ is not Lebesgue integrable or not satisfy the equation.
while my desired conclusion is
$f$ is Lebesgue integrable, but not satisfy the equation.
Can anyone suggest me a function $F$ which is differentiable everywhere (on a compact interval) such that its derivative is Lebesgue integrable, but not satisfy the above equation?
Edited : Fixed some confusion when I first typed it.
I think the statement your friend claims is true. According to this page, (https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus)
I quote from that page: If a real function $F$ on $[a, b]$ admits a derivative $f(x)$ at every point $x$ of $[a, b]$ and if this derivative $f$ is Lebesgue integrable on $[a, b]$, then $$ \int_a^b f(t) dt = F(b) - F(a) $$ The Wikipedia page cites Rudin's Real and Complex Analysis for this result.