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.
 A: If $f(x)=0$ for $x<0$ and 1 for $x\geq0$ then $F(x)=0$ for $x<0$ and $x$ for $x\geq 0$. The left hand derivative of F at 0 is 0 and the right hand derivative is 1 so F is not differentiable at 0.
A: 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 $x=0$
A: I don't know how to rigorously prove what I will now state, but I feel like more could be said about this than just "no, it does not".
Let $f\colon[a,b]\longrightarrow\mathbb{R}$ be a bounded and piece-wise continuous function. Then, this function is $\mathcal{R}$-integrable and
$$F(x) \equiv k+\int_{a}^xf(t)\space\mathrm{d}t$$
is, $\forall k\in\mathbb{R}$, a uniformly continuous function on $[a,b]$ piece-wise differentiable on $(a,b)$. The points at which $F$ is not differentiable are all the points at which $f$ has a non-removable discontinuity.
$F'(x)$ will only equal $f(x)$ for those point at which $f(x)$ is continuous, it will not be defined for points with non-removable discontinuities and it will equal the limit of the function at that point for removable discontinuities. In fact, I'm sure
$$F'(x) \equiv \lim_{t\to x}f(t)$$
holds true $\forall x\in[a,b]$ (for non-removable discontinuities the limit does not exist, and for points at which $f$ is continuous the limit equals $f(x)$).
This "antiderivative" can be computed for any piecewise-continuous function by just computing the antiderivative of each continuous piece and, then, making all their integration constants be such, that the resulting piecewise-defined function is continuous everywhere. This will make all of them depend on just one constant, as it should be.
With this "antiderivative", the Barrow rule holds true even for piecewise-continuous functions.
