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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 in $(a,b)$ 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.)

kisten
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Potato
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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). –  Jun 06 '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. –  Jun 06 '12 at 17:45
  • Assume whatever you need. – Potato Jun 06 '12 at 17:50
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    Perhaps you could clarify the question so folks (like myself) don't waste time answering... – copper.hat Feb 18 '14 at 18:49

2 Answers2

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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.

Yai0Phah
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  • Thanks for solving this mystery! – Potato Feb 18 '14 at 18:05
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    Wonderful! But I suspect that $D^+R_n$ should be $D_+R_n$ in the sketch. Stein's hint was actually incomplete. $F'(x) \ge 0$ almost everywhere alone is not enough. We also need $D_+F(x) > -\infty$ everywhere. That's why we need to take $\phi_n = \min(n,F')$ instead of just $\phi = F'$. – Petra Axolotl May 19 '22 at 00:55
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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.

copper.hat
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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. – Yai0Phah 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. – Yai0Phah 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. – Yai0Phah 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
  • @FrankScience: I am just working from the OP's problem statement, I don't know what the original problem is. – copper.hat Feb 18 '14 at 17:05
  • Well, sorry, I was excited, but I have Stein's book and I know what's really asked to prove. I don't know why the OP (blindly) accepted this without clarifying anything, and in fact, I stumbled into this problem for long and thus googled around. – Yai0Phah Feb 18 '14 at 17:11
  • @FrankScience This question is almost two years old. I was not so adept at M.SE then, sadly. – Potato Feb 18 '14 at 23:10
  • @copper.hat I am sorry to bother you with this old question, but what problem did you solve exactly? The equivalence of two different definitions of AC? Standard $\iff$ the indefinite integral one? Or that if $f' \in L^1$ and exits everywhere then $f$ is AC? – Sorfosh Dec 19 '19 at 18:29
  • @Sorfosh: The OP wanted a proof that $F$ is absolutely continuous. – copper.hat Dec 19 '19 at 18:36
  • @Sorfosh: I believe the question and intent changed over the two years after I originally answered, so perhaps there is some confusion there. – copper.hat Dec 19 '19 at 18:41