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I have seen two definitions of Absolutely Continuous Functions, my main query is whether the collection of subintervals is supposed to be finite / infinite, and whether the two definitions coincide.

Definition 1 (Wikipedia): A real-valued function $f$ on a closed, bounded interval $[a,b]$ is said to be absolutely continuous on $[a,b]$ provided for each $\epsilon>0$, there is a $\delta>0$ such that for every finite disjoint collection $\{(a_k, b_k)\}_{k=1}^n$ of open intervals in $(a,b)$, if $\sum_{k=1}^n(b_k-a_k)<\delta$, then $\sum_{k=1}^n |f(b_k)-f(a_k)|<\epsilon$.

Definition 2 (Wheedon): $f$ is said to be absolutely continuous on $[a,b]$ if given $\epsilon>0$, there exists $\delta>0$ such that for any collection $\{[a_i, b_i]\}$ (finite or not) of nonoverlapping subintervals of $[a,b]$, $\sum|f(b_i)-f(a_i)|<\epsilon$ if $\sum(b_i-a_i)<\delta$.

On the surface, it seems that definition 2 is more general than definition 1. They may well be equivalent, but how do we see that?

Thanks!

yoyostein
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    They are equivalent. From definition 1, you get $\sum \lvert f(b_i) - f(a_i)\rvert \leqslant \epsilon$ directly for an infinite family of disjoint intervals $(a_i,b_i)$. – Daniel Fischer Oct 04 '16 at 11:51
  • Thanks. How does that work? By taking limits? – yoyostein Oct 04 '16 at 12:08
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    Yes. For a family ${ t_{\alpha} : \alpha \in A}$ of non-negative numbers, $\sum\limits_{\alpha\in A} t_{\alpha}$ is defined as $$\sup \Biggl{ \sum_{\alpha \in F} t_{\alpha} : F \subset A \text{ is finite}\Biggr},$$ so that works for arbitrary index sets. If we a priori have a countable collection, indexed by $\mathbb{N}$, we have an ordinary series, which we learned to handle in the first semesters. – Daniel Fischer Oct 04 '16 at 12:12

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