Does derivative of antiderivative always equal the integrand function? For some reals $a<b$ we consider $F:(a,b)\to\mathbb{R}$ where $F(x):=\int\limits_{a}^xf(t)dt$ and assume that $F$ is continously differentiable. (We don't know if $f$ is continuous)
Can we conclude that $F'(x)=f(x)$ for all $x\in (a,b)$?

For some $x_0\in(a,b)$ we choose
$$f(t)=\begin{cases}0, & t\in(a,b)\setminus\{x_0\}\\2,& t=x_0.\end{cases}$$
and $g(t):=\hat{0}$. Due to properties of the Riemann integral
$$\int\limits_a^{x_0}f(t)dt=\int\limits_a^{x_0}g(t)dt=0.$$
Then
\begin{align*}
&\left|\frac{\int\limits_{a}^xf(t)~dt-\int\limits_{a}^xf(t)~dt}{x_0-x}\right|=\left|\frac{\int\limits_{x_0}^xf(t)~dt}{x_0-x}\right|=\left|\frac{\int\limits_{x_0}^xg(t)~dt}{x_0-x}\right|=0,\text{ where }x\neq x_0\\
&\implies F'(x_0)=0.
\end{align*}
As $x_0$ was chosen arbitrarily $F'(x)=0$ for all $x\in(a,b)$. Hence $F'$ is continuous but $F'\neq f$.

Is this correct?
 A: A more general result (that doesn't rely on the Riemann integral) along these lines is that approximately continuous functions (i.e. $f\in {\cal C}_{ap}$) need not be derivatives.  Then letting $F(x) := \int^x_0 f(t)\, dt$ it is usually easy to show that $f(y) \not= F'(y)$ for points $y$ of approximate continuity.
A function $f:{\mathbb R} \rightarrow {\mathbb R}$ is approximately continuous at $x_0$ if there is a set $E\subset {\mathbb R}$ for which $\mathrm{\mathop{dens}}(E,x_0) = 1$ and $f$ restricted to $E$ is continuous.  Here, $\mathrm{\mathop{dens}}(E,x_0) = \lim_{h\rightarrow 0} \frac{\mu (E\cap [x_0 -h, x_0+h])}{2h}$ is the usual (symmetric) density.
Now, let $\{a_n\}$ and $\{b_n\}$, $n\in \mathbb N$, be two sequences with $a_{n+1} < b_{n+1} < a_n$ so that both sequences decrease to $0$ and the intervals $(a_n,b_n)$ are pairwise disjoint.  Let also $\sum_k b_k < \infty$, and let $\mathrm{\mathop{dens}}(\cup_n(a_n,b_n),0) =1$.  Define a function $f$ so that:

*

*$f(0)=0$

*$f(x) = 0 \ \ \forall x \in \cup_n(a_n,b_n)$

*$f$ is continuous everywhere except at $0$

*$\int_{b_{n+1}}^{a_n} f(t) \, dt = b_n$
and extend $f$ to $[-b_1,0]$ by setting $f(x)=f(-x)$.  Then $f$ is approximately continuous on $[-b_1, b_1]$.  For each $n\in {\mathbb N}$ we also have that
$$\frac{1}{b_n} \int_0^{b_n} f(t) \, dt = \frac{1}{b_n} \sum_{k=n}^\infty b_k = 1 + \sum_{k=n+1}^\infty \frac{b_k}{b_n}$$
and this sum converges by hypothesis.  So $\frac{d}{dt}\int^{b_n}_0 f(t) \, dt \not= f(b_n)$, as you desired.
The usual concrete example for this is $F(x) = x^{-2}\sin(x^{-2})$, $F(0)=0$.
