Integrals of a function which has finitely many discontinuities are not differentiable at the discontinuities I've tried solving an exercise stated below.
$$\text{Suppose that $f\in\mathcal R$ on $[a,b]$ and define $F(x)=\int_{a}^{x}{f(t)dt}$.}\\\text{If $x$ is a point at which $f$ is not continuous, is it still possible that $F'(x)=f(x)$?}$$
So I approached in two ways. Let $k\in\{\text{discontinuities of $f$}\}$.
One is to show $F$ is continuous at $x=k$.
And the other is to show that $\lim\limits_{h\to0}\frac{F(k+h)-F(k)}{h}$ exists.
First step, $^\exists M\gt0$ satisfying $\vert f(x)\vert\leq M$ for all $x\in [a,b]$, since $f\in\mathcal R$.
(I'll assume that $f$ is defined at all the point on $[a,b]$.)
So, $\vert F(y)-F(x)\vert=\vert\int_{x}^{y}f\left(t\right)dt\vert\leq M\left(y-x\right)$ for $a\leq x\leq y\leq b$.
For any $\epsilon\gt0$, $\vert F(y)-F(x)\vert\lt\epsilon$ where $\vert y-x\vert\lt\delta=\epsilon/M$ .
Hence, the continuity of $F$ is clear.
Now I'm suffering to show that the second step is false.
For a jump discontinuity case, I've tried to
Consider $f(x)=
\begin{cases}
x^2/2+2, &\text{if $x\lt 2$} \\
x^2/2+6, &\text{if $x\geq 2$}
\end{cases}$.
In this case $k=2$. Then, $$\left\vert \frac{F(2+h)-F(2-h)}{2h}-f(2)\right\vert=\left\vert\frac{\int_{2-h}^{2}\left(t^2/2+2\right)dt+\int_{2}^{2+h}\left(t^2/2+6\right)dt}{2h}-8\right\vert\gt\epsilon=1\\\text{for any $0\lt h\lt\delta=g(\epsilon).$ (I'll skip the part of evaluating improper integrals.)}$$
Thus $F$ is not differentiable at $x=k$ where $k$ is a jump discontinuity, and it says roughly that $F$ is changing sharply at $x=k$, right?
But how about removable discontinuities? Any hint will be helpful.

Hint: Consider the function $f(x) = \sin(1/x), x\ne 0, f(0)=0.$
