The problem reads:
Suppose that $F$ is continuous on $[a,b]$, $F'(x)$ exists for every $x\in(a,b)$, and $F'(x)$ is integrable. Then $F$ is absolutely continuous and $$F(b)-F(a)=\int_a^b F'(x)\,dx.$$ [Hint: Assume $F'(x)\geq 0$ for a.e. $x$. We want to conclude that $F(b)\geq F(a)$. Let $E$ be the set of measure zero of those $x$ such that $F'(x)<0$. Then according to Exercise 25, there is a function $\Phi$ which is increasing, absolutely continuous, and for which $(D^+\Phi)(x)=+\infty$, $x\in E$. Consider $F+\delta\Phi$ for each $\delta$ and apply the result (a) in Exercise 23.]
Result (a) in Exercise 23 is that if $F$ is continuous on $[a,b]$ and $(D^+F)(x)\geq 0$ for every $x\in[a,b]$ then $F$ is increasing on $[a,b]$. S&S have defined $$(D^+F)(x)=\limsup_{h\to 0^+} \frac{F(x+h)-F(x)}{h}.$$
My question: I am able to prove what they suggest in the hint, but I don't see how it connects to the original problem. Why can I assume that $F'(x)\geq 0$ for a.e. $x$? And even then, knowing $F(b)-F(a)\geq 0$ is a far cry from knowing its exact value. Rudin proves this theorem in Chapter 7 of his "Real and Complex Analysis" by an entirely different proof, but I would like to understand what S&S are suggesting here. Any reference or hint is welcome.