Let $F(x) = \int_{-\infty}^xf(t)dt,\;$. If $f(x)$ is discontinuous at $x=c$, then is $F(x)$ not differentiable at $x=c$

Question

There is a the fundamental theorem of calculus:

Define $F(x) = \int_{-\infty}^xf(t)~dt$ where $f(x)$ is and must be continuous at $[a,b]$, then $F'(x) = f(x)$

Is the following statement true?

Let $ f(x)$ be discontinuous at a point $x = c$, then $F(x)$ is not differentiable at $x=c$.

How do I prove that?

I tried to use method of reducing to absurdity to prove by assuming $F(x)$ is differentiable at $x =c$. What should I do next?

Answer 1

This is not true. Let $$f(x)=\begin{cases} 1 \text{ if } x=2\\ 0 \text{ else} \end{cases}$$

Then $F(x)\equiv 0$ and is differentiable at $2$, even though $f(x)$ isn't continuous at $2$.

Answer 2

You cannot prove it because it is false. Define$$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}0&\text{ if }x\leqslant0\\-\cos\left(\frac1x\right)+2x\sin\left(\frac1x\right)&\text{ otherwise.}\end{cases}\end{array}$$Then $f$ is discontinuous at $0$, but$$F(x)=\begin{cases}0&\text{ if }x\leqslant0\\x^2\sin\left(\frac1x\right)&\text{ otherwise,}\end{cases}$$which is differentiable at $0$.