The problem states the following.
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
$$F(b)-F(a)=\int_a^b F'(x)dx.$$
It comes with the following hint:
Assume $F'(x)\ge0$ for almost every $x$. We want to conclude that $F(b)\ge F(a)$. Let $E$ be the set of measure 0 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 $D^+\Phi=+\infty$ on $E$, where $D^+$ is the right upper Dini derivative. Consider $F+\delta\Phi$, for each $\delta$ and apply the results (a) in Exercise 23 (which says $D^+f\ge0\implies F$ increasing).
While I can find the proof of the statement in other sources (e.g. Theorem 7.21 in Rudin's Real and Complex Analysis), I cannot make this particular hint work. A natural attempt is to define
$$G(x)=F(a)+\int_a^x F'(t)dt$$
and consider $F-G$. Then this function is merely differentiable almost everywhere with derivative almost everywhere zero. A priori we cannot conclude $F-G=0$ due to the counterexample of the Cantor-Lebesgue function. Thus it is desirable to apply the hinted lemma, which, although not explicitly stated, seems to require the function in question to be differentiable everywhere, and is not directly applicable to $F-G$.
Therefore I am asking for a proof of the statement along the lines of the provided hint.