Is there a function $F:[a,b]\to\mathbb{R}$ differentiable with $F' = f$, but $f$ is not Riemann integrable on $[a,b]$. $[a,b]$ is a bounded interval?
Motivation
Rudin page 152 Theorem 7.17: Suppose $\{f_n\}$ a sequence of functions, differentiable on $[a,b]$ and such that $\{f_n(x_0)\}$ converges for some point $x_0\in [a,b]$. If $\{f_n'\}$ converges uniformly on $[a,b]$ to $g$, then $\{f_n\}$ converges uniformly on $[a,b]$ to $f$ where $f' = g$.
In the remark, it says if the continuity of the function $f'_n$ is assumed in addition to the above hypothesis, then a much shorter proof can be based on fundamental theorem of calculus and the theorem that $\int_a^b f_n d\alpha\to\int_a^b f d\alpha$ if we have $f_n\to f$ uniformly.
I was thinking that we don't really need $f_n'$ to be continuous, we just need it to be Riemann integrable so we can apply Fundamental Thm of Calculus. From the remark I guess that there should be a derivative that is not integrable.
EDIT
Dominic Michaelis answered my question immediately. A further thought yields a more difficult question, is there a function as said above with $f'$ bounded on $[a,b]$?
Answer
Take $F[-1,1] \to \mathbb{R}$ with
$$F(x)= \begin{cases}
x^2 \sin\left(\frac{1}{x^2}\right) & x \neq 0\\
0 & x=0 \\
\end{cases}
$$
The derivative is not a regulated function, and not bounded at $0$. I don't know a definition of riemann integrable allowing those things.
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