Finding an example the fundamental theorem of calculus fails when $f$ is not continuous. Is the following statement true or not?

If $f$ is Riemann integrable on $[a,b]$, then the function $F(x)=\int_a^xf(t)dt$ is differentiable on $(a,b)$ and $F'(x)=f(x)$ for $x\in (a,b)$

I think it is not. Since $F$ is differentiable at the point where $f$ is continuous, but I can not think a counter example, is anyone has a counterexample? Thanks a lot.
 A: False: Take $f(x)=sgn(x)$ (sign function) on the interval $[-1,1]$ (that is, $a=-1$). Then
$$F(x)=|x|-1$$
and is not differentiable at $x=0$. The theorem as stated is true if $f$ is a continuous function however. Indeed, that is the full statement of the fundamental theorem of calculus (at least the first part).
A: Start with an arbitrary continuous function $f$ on $[a, b] $ (eg $f(x) =x$). Take any point $c\in(a, b) $ and define a function $g$ on $[a, b] $ such that $g(x) =f(x) $ for all $x\neq c$ and define $g(c)$ to be any number different from $f(c)$ (eg $g(c) =f(c) +1$).
Then $g$ is Riemann integrable on $[a, b] $ and moreover $$F(x) =\int_a^x f(t) \, dt=\int_a^x g(t) \, dt\, \forall x\in[a, b] $$ and by continuity of $f$ we have $F'(c) =f(c) \neq g(c) $. The function $g$ now serves as a simple and obvious counterexample you seek.
The above example gives you a case when $F$ remains differentiable at $c$ but it's derivative at $c$ does not match the value of integrand $g$ at $c$.
Other answers have tried to construct an example where $F$ is not differentiable at $c$. For such examples it is sufficient to consider integrands which have jump discontinuity at $c$ and then it will be guaranteed that the integral function $F$ will not be differentiable at $c$.
A: Let $[z]$ denote integer GIF, then $f(x)=x+x[x], x\in [0,2)$ is  discontinuous at $x=1$, If $g(x)=\int_{0}^{x} t(1+[t]) dt$ then $g(x)=\frac{x^2}{2}, x \in [0,1]; g(x)=x^2-\frac{1}{2} \in x \in [1,2)\implies g'(x)=x, x\in [0,2) \implies g'(0.3)=0.3, g'(1.3)=1.3$
But by fundamental theorem of Integration, we get
$$g'(x)=x(1+[x]) \implies g'(0.3)=0.3, g'(1.3)=2.6$$
Hence, there is a contradiction for $x=1.3$
