In the book Principles of mathematical analysis by Walter Rudin these are the statements which I came across
- Corollary 5.12$\space\space\space$ If $f$ is differentiable on $[a,b]$ then $f'$ cannot have any simple discontinuities on $[a,b]$. But $f'$ may have discontinuties of the second kind.
- Theorem 6.20$\space\space\space$ Let $f$ $\in$ $\mathscr{R}$ on $[a,b].$ For $a\le x \le b,$ put $$F(x)=\int_a^xf(t)dt.$$ Then $F$ is continuous on $[a,b]$; furthermore, if $f$ is continuous at a point $x_0$ of $[a,b]$, then $F$ is differentiable at $x_0$, and $$F'(x_0)=f(x_0).$$
- Theorem 6.21$\space\space\space$ If $f$ $\in$ $\mathscr{R}$ on $[a,b]$ and if there is a differentiable function $F$ on $[a,b]$ such that $F'=f$, then $$\int_a^b{f(x)dx}=F(b)-F(a).$$
Now the problem is, we can have a differentiable function $F$ on $[a,b]$ such that $F'= f$ and $f$ is discontinuous, although the discontinuity will be of second kind according to the statement $1$. If these discontinuities are finite then $f\in\mathscr{R}.$ Let's assume a point $c\in[a,b]$ where $f$ is discontinuous (discontinuity of the second kind). Since, $f\in\mathscr{R}$ let's define a function $$G=\int_a^xf(t)dt\tag{1}$$ for $x\in[a,b]$. Then according to the statement $2$, $G(x)$ is clearly not differentiable at $c$ since $f$ is discontinuous at $c$. Now, from statement $3$ we can also say that $$\int_a^xf(t)dt=F(x)-F(a)\tag{2}$$ if $x\in[a,b]$. Thus from equations $(1)$ and $(2)$ we have $$F(x)-F(a)=G(x)$$ which would make $G(x)$ differentiable for all $x\in[a,b]$ but we know that $G(x)$ is not differentiable at $c$. So, this is a paradox which means I am wrong somewhere but try as I might I am not able to find it. It'd be a great help if someone can point out the error.
Edit: $\mathscr{R}$ here denotes the set of Riemann-integrable functions
Converse of a theorem is not true in general. Statement 2) does not tell you that if $f$ is not continuous at $c$ then $G(x)$ is not differentiable at $c$.
Example: Let $f(0)=1$ and $f(x)=0$ for $x\neq 0$. Then $G(x)=\int_0^{x} f(t)dt=0$ for all $x$ and $G$ is differentiable at $0$ even though $f$ is not continuous at $0$.