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.
Best Answer
I don't know whether this is exactly the solution they had in mind, because we need something that's sort of not quite the same as the hint. But it's very much like the hint. (And I'm glad you asked the question, because I never remember anything except things I've figured out myself - now I finally know a proof of this fact.)
The hint amounts to this:
Hint If $F$ is differentiable and $F'\ge0$ almost everywhere then $F$ is increasing.
We need the fact that $F$ is differentiable everywhere to show that $D^+(F+\delta\Phi)\ge 0$ everywhere; for example we know the following is false:
False Fact If $F$ is continuous, differentiable almost everywhere, and satisfies $F'\ge0$ almost everywhere then $F$ is increasing.
We're going to use a stronger version of the hint:
Lemma Suppose $F$ is continuous, $F$ is differentiable almost everywhere, $F'\ge0$ almost everywhere, and $D^+F>-\infty$ everywhere. Then $F$ is increasing.
Proof: The fact that $D^+F>-\infty$ shows that $D^+(F+\delta\Phi)\ge0$ everywhere, for every $\delta>0$. So $F+\delta\Phi$ is increasing, and letting $\delta\to0$ shows that $F$ is increasing. QED.
Solution to the exercise: First, of course we may assume that $F$ is real-valued. Define $$G(x)=\int_a^x(F'(t))^+\,dt.$$Then $G$ is absolutely continuous and $G'=(F')^+$ almost everywhere. Let $$H=G-F.$$Then $H'=(F')^+-F'=(F')^-\ge0$ almost everywhere. And since $G$ is increasing and $F$ is differentiable it follows that $D^+H>-\infty$ everywhere. So the lemma shows that $H$ is increasing.
Now $H(b)\ge H(a)$ shows that $$F(b)-F(a)\le G(b)-G(a)=\int_a^b(F')^+.$$Applying this with $-F$ in place of $F$ shows that $$F(a)-F(b)\le\int_a^b(F')^-.$$But $|F(b)-F(a)|$ is either $F(b)-F(a)$ oor $F(a)-F(b)$. So we have $$|F(b)-F(a)|\le\int_a^b|F'|.$$
And now we're done. If $a_1<b_1<\dots<a_n<b_n$ then the same argument shows that $$\sum|F(b_n)-F(a_n)|\le\sum\int_{a_n}^{b_n}|F'|.$$Since $F'\in L^1$ this shows that $F$ is absolutely continuous. QED