Fundamental Theorem of Calculus For Piecewise Continuous Functions Recall that the first fundamental theorem of calculus says that if $f$ is continuous real values function defined on $[a,b]$. Then the function defined by $F(x)= \int_ a^x f(t) dt$ is differentiable and $F’(x)= f(x) $ 

Does the same result hold if $f$ is merely piecewise continuous instead of continuous?

Edit: The above question is motivated from a set of notes. The exact statement written in notes is: Let $f$ be continuous on $ \mathbb R$ with period 2l, then  $\int_ l^x f’(t) dt= f(x) dx$
P.S: I don’t know about measure theory so please avoid to use it.

The answer is no. 
Consider the function $f(x)$ defined on $[-1,1]$ as $ f(x)=1$ for $-1\le x\le 0$ and $f(x)=-1$ for $0<x\le 1$
Then  $F(x)= \int _{-1}^x f(t)dt =x+1$ if $x \le 0$ and $F(x)= \int _{-1}^x f(t)dt =1-x$ if $x > 0$
Note that F(x) is a tent map which is not differentiable at $x=0$
