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This might probably be classed as a soft question. But I would be very interested to know the motivation behind the definition of an absolutely continuous function. To state "A real valued function $ f $ on $ [a,b] $ is absolutely continuous on said interval if $\forall $$\epsilon >0 \ $, $\exists\delta>0\ $ such that $$ \sum^n_{k=1}|f(b_k) -f(a_k)|< \epsilon $$

for every $n$ disjoint subintervals $ \ (a_k,b_k) $ of $ \ [a,b] $, $k=1,\ldots,n$, such that $ \sum^n_k_=_1|b_k -a_k|< \delta $. Why the use of the disjoint sub-intervals? What purpose do they serve? Somehow the definition didn't seem natural to me.

What I mean is just as the notion of uniform continuity is motivated by the definition of continuity itself or as the concept of compactness serves to generalise the notion of finiteness, how can one look at Absolute Continuity in this respect?

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Yikes ! Your accept rate would make absolutely continuous functions go absolutely nuts. You don't like any of the answers you've received in this site to your questions? –  DonAntonio Sep 5 '12 at 4:07
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Well I am new to the ways of this site. I realise now that this has something to do with the earlier questions I asked. I only up voted some answers to my earlier questions, the thought of accepting answers totally escaped my mind. I really apologise for that. –  Vishesh Sep 5 '12 at 4:27
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Hi Vishesh. I would really recommend reading this (arxiv.org/pdf/1203.1462.pdf). The conditions of Absolute Continuity as they are defined above is essential to proving a very important theorem in Analysis: The Fundamental Theorem of Calculus for the Lebesgue Integral. –  Shankara Pailoor Sep 5 '12 at 5:42
    
I think there is a typo. Should be $|f(b_k) - f(a_k)|$. –  Tunococ Sep 5 '12 at 5:50
    
@Shankara.Thanks for the link.I guess I am being too narrow minded here and I am sorry for that, but I would be happy to know why this definition arose? What I mean is just as the notion of uniform continuity is motivated by the definition of continuity itself or even the concept of compactness which serves to generalise the notion of finiteness, how can one look at Absolute Continuity in this respect? –  Vishesh Sep 5 '12 at 7:04

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You can think of absolute continuity as motivated by / modeled on "differentiable almost everywhere" plus "satisfies Fundamental Theorem of Calculus".

Precisely, $f$ is absolutely continuous if and only if $f$ is differentiable almost everwhere and $f(x) = f(a) + \int_a^x f'(x) dx$ for all $x \in [a,b]$.

At first glance, it may seem like a.e.-differentiability should be a nice enough property to ensure FTC is true, but there are counterexamples (like the Cantor function). You can think of absolute continuity as a way of shoring up that kind of pathology, i.e. it eliminates so-called singular (in the measure-theory sense) functions.

The "disjoint" part of the definition serves to weaken the definition a little bit. If "disjoint" were omitted, you'd be describing a condition for Lipschitz functions, which is stronger than needed (if your aim is "abs. cont." $\iff$ "a.e.-diff + FTC"). That is, there are functions which are abs. continuous, but not Lipschitz (like $\sqrt{x}$ on $[0,1]$).

As an exercise, show that this definition fails for $\sqrt{x}$ on $[0,1]$ if the disjoint hypothesis is removed. Hint: take each of the intervals $[a_k,b_k]$ to be the $\textit{same}$ interval $[0,\alpha]$ for some appropriately small $\alpha$.

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Yes. "Absolutely continuous" would be just another obscure function property loved only by real analysts except for the theorem: $F$ is absolutely continuous on $[a,b]$ iff $F(x) = F(a)+\int_a^x F'(t)dt$ for all $x \in [a,b]$. –  GEdgar Sep 5 '12 at 12:40

The way I like to think of it is that it says that the image under $f$ of a sufficiently small finite collection of intervals is arbitrarily small (where "small" refers to total length).

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