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?

**Answer**

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 $\alpha$.

**Attribution***Source : Link , Question Author : Vishesh , Answer Author : BaronVT*