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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.

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    $\begingroup$ No, $F$ might be nondifferentiable at the point of discontinuity. However, if one-sided limit of $f$ exists at a point, then one-sided derivative of $F$ will exist and those will be equal. The proof is nearly the same as that of FTC. $\endgroup$ – Wojowu Mar 2 '18 at 9:28
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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.

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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 $x=0$

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