# Integrable derivative implies absolutely continuous

Consider Lebesgue integrals over the real line. I have the following problem:

Problem: Suppose $F(x)$ is a continuous function in $[a,b]$, and $F'(x)$ exists everywhere and is integrable. Show $F(x)$ is absolutely continuous.

The hint in the book suggested showing that $F'(x)\ge 0$ a.e. implies that $F(x)$ is increasing. I did that, but I do not see how it helps with this problem. How can this hint be used to solve the problem?

(I am aware other proofs exist.)

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What do you already know about Lebesgue integral, absolutely continuous $(AC)$ functions and functions of bounded variation $(BV)$? It is known that $f\in BV$ iff $f$ is the difference of two increasing functions. If you know that and if you know under which conditions a $BV$ function is $AC$ then this might help (I don't know whether this is the direction to go, just my first thoughts about the question). – user20266 Jun 6 '12 at 17:39
I do know where a proof of this can be found and I do understand your intention to use the hint. My 1st comment was the attempt to ask you what kind of results are admissible to proof this. I don't know the Stein/Shakarchi source you are citing, though. – user20266 Jun 6 '12 at 17:45
Assume whatever you need. – Potato Jun 6 '12 at 17:50
Perhaps you could clarify the question so folks (like myself) don't waste time answering... – copper.hat Feb 18 '14 at 18:49

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.

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Thanks for solving this mystery! – Potato Feb 18 '14 at 18:05

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.

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Why is it true that $\lvert f(u_k)-f(l_k)\rvert\le\int_{I_k}\lvert F'\rvert$? The fundamental theorem of calculus, when $f$ is differentiable everywhere and $f'\in L^1$, is just the goal to prove, not assumed to use. – Frank Science Feb 18 '14 at 16:46
@FrankScience: The goal is to show that $F$ is absolutely continuous. We are given that $F'(x)$ exists everywhere and $F'$ is integrable. Then $F(u_k) - F(l_k) = \int_{I_k} F'(t) dt$ from which the estimate follows. – copper.hat Feb 18 '14 at 16:54
But $F(u_k)-F(l_k)=\int_{l_k}^{u_k}F'$ isn't verified in this case. In fact, the whole problem on Stein is: to show $F$ is AC, and that Barrow's formula is right. – Frank Science Feb 18 '14 at 16:59
Another note: if $f\in L^1$, then $\int_a^xf$ is AC. It's much easier than the original problem. – Frank Science Feb 18 '14 at 17:01
@FrankScience: I'm confused, that is the result I am using... – copper.hat Feb 18 '14 at 17:04