let $f(x)$ be a bounded measurable function defined on $\mathbb{R}$, then define
$$F(x)=\int_0^xf(t)dt,\ \ x\in\mathbb{R}$$
We can see that $F(x)$ is a absolutely continuous function and by some classical results we know $F$ is a.s. differentiable and the integral of its derivative in an interval $[a, b]$ equals to the difference $F(b)-F(a)$. I would like to know whether we have
$$F'(x)=f(x),\ \ a.s.$$
Since we have
$$\int_0^xf(t)dt=\int_0^xF'(t)dt,\ \ \forall x\in\mathbb{R}$$
for two measurable functions. How could we conclude that
$$F'(x)=f(x),\ \ a.s.$$
Many thanks!
First of all, let us define two measures $\mu(A) = \int_A f \;\mathrm d\lambda$ and $\nu(A) = \int_A F' \;\mathrm d\lambda$ where $\lambda$ is the Lebesgue measure. Let us further restrict our attention to an arbitrary bounded interval $[a,b]$. Since due to the equality you have in OP $\mu$ and $\nu$ coincide on $[a,x)$, then they coincide for any Borel subset of $[a,b]$. Hence $$ \int_A f\;\mathrm d\lambda = \int_A F'\;\mathrm d\lambda\qquad \forall \text{ Borel }A\subseteq[a,b] $$ and as a result, $f = F'$ on $[a,b]$ $\lambda$-a.e. Due to the point that $[a,b]$ is arbitrary, we obtain that $f = F'$ $\lambda$-a.e. on $\Bbb R$.