Integrable derivative implies absolutely continuous

Solution 1:

A proof using Vitali-Caratheodory's theorem could be found in Papa Rudin, Chapter 7, Theorem 7.21.

But it's unrelated to the hint given by Stein. If you want one of this kind, well, there's one (I spent so much time in looking for this!):

Natanson, Theory of Functions of a Real Variable, volume 1, Chapter IX, section 7, theorem 1. The hint given by Stein, is just the two lemmas in that proof!

The sketch of the proof:

Let $\phi_n=\min(n,F')$, and $R_n(x)=F(x)-\int_a^x\phi_n(t)dt$, we have $R_n'=F'-\phi_n\ge0$ a.e. and $D^+R_n\ge F'-n>-\infty$, thus (by hint) $R_n(b)\ge R_n(a)$ and $F(b)-F(a)\ge\int_a^b\phi_n$, hency by domintated convergence theorem, $F(b)-F(a)\ge\int_a^b F'$. Replace $F$ with $-F$, we'll obtain the result.

Solution 2:

I'm not sure how monotonicity helps you here.

I think the result follows most naturally from the fundamental theorem of calculus: $$F(x) = F(a) + \int_a^x F'(t) \, dt,$$

and the fact that since if $f'$ is integrable, then the measure $\mu(A) = \int_A |f'(t)| \, dt$ is absolutely continuous wrt the Lebesgue measure. It is fairly straightforward to establish that $\forall \epsilon >0$, $\exists \delta>0$ such that if $m(A) < \delta$, then $\mu(A) < \epsilon$ (first take $|f'|$ bounded, then use the dominated convergence theorem).

Then if $I_k = [l_k, u_k]$ are a collection of disjoint intervals with $\sum_{i=1}^n m(I_k) < \delta$, we have $\sum_{k=1}^n |F(u_k)-F(l_k)| \leq \sum_{i=1}^n \int_{I_k} |F'(t)| \, dt = \int_{I_1 \cup ... \cup I_n} |F'(t)| \, dt < \epsilon$.

Another (similar) approach would be to invoke the Banach Zarecki Theorem.