Absolutely continuous functions

Solution 1:

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 $\alpha$.

Solution 2:

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).