Derivative and Antiderivative: Suppose $F: [a,b]\rightarrow \mathbb{R}$ is differentiable on $[a,b]$. Suppose $F: [a,b]\rightarrow \mathbb{R}$ is differentiable on $[a,b]$.
Does it follow that $\int_{a}^{b} F' = F(b)-F(a)$?
This is an interesting question. Don't we have to assume that $F'(x)$ is always equal to $f(x)$ for this to work? In other words, does $F(x)$ being differentiable always guarantee that $F'(x)$ is integrable?
Thanks!
 A: There are several questions in the post that are intertwined, and I'll try to list a bunch of definitions/conclusions below in the hope to answer your question (all integrations below are Lebesgue integration):
Firstly, if $F$ is an antiderivative of $f$, then the derivative of $F$ is $f$ a.e.

Suppose $f:\mathbb R \mapsto \mathbb R$ is integrable and $a \in \mathbb R$. Define $F(x) = \int_a^xf(y)\,dy$ as the antiderivative of $f$. Then $F$ is differentiable almost everywhere and $F'(x)=f(x)$ a.e.


Secondly however, if $f$ is the derivative of $F$ a.e., $F$ is not necessarily the antiderivative of $f$. One good-enough relationship is based on functions of bounded variation. (But notice that we need more condition in order to guarantee the equality)

Functions of bounded variation are differentiable and can be written as the difference of two increasing functions. If $F$ is increasing, then $F'$ exists a.e., and $\int_a^bF'(x)\,dx \le F(b) - F(a)$; 

Notice that here the equality is not guaranteed.
Example: let $F$ be the Cantor-Lebesgue function, then it is differentiable with $F'(x)= 0$ a.e., and thus 
$$1=F(1)-F(0) \color{red}> 0 = \int_0^1F'(x) \, dx $$

Finally, here is one stricter condition under which the equality is achieved.

If $F$ is absolutely continuous, then $F'$ exists a.e. and $\int_a^bF'(x)\,dx = F(b) - F(a)$. 

One could consider absolutely continuous as a special case of bounded variation, because it could be proved that if $f$ is absolutely continuous, then it is of bounded variation.
