# Can we replace finite with countably infinite sequences in the definition of absolute continuous functions?

Absolute Continuous functions require a finite sequence of subintervals of the domain. Can we replace this finite sequence with a countably infinite one in this definition?

Please note that an absolutely continuous function preserves null sets. To prove this result, I choose an open set G that contains a given null set N. Then eventually, I proved that the image of N has measure zero. In this proof, at some point, I chose a finite sequence from the countable sequence of intervals that covers G. Moreover, the converse is also true ie any continuous function with a bounded variation that preserves the null set is absolutely continuous. Based on my own shallow observation, I feel like we can replace this finite sequence criterion with a countably infinite one. Is this true? Could you give me an easy counter-example where my guess is false? Thank you for your time.

Definition 1 Let $$f:\mathbb{R}\to \mathbb{R}$$ be a function. We say that $$f$$ is finitely absolutely continuous if for every $$\epsilon>0$$ there exist $$\delta>0$$ such that for every finite collection of disjoint intervals $$((\alpha_i,\beta_i))_{i=1}^{n}$$ with $$\sum_{i=1}^{n}|\beta_i-\alpha_i|<\delta$$ we have $$\sum_{i=1}^{n}|f(\beta_i)-f(\alpha_i)|<\epsilon$$.
Definition 2 Let $$f:\mathbb{R}\to \mathbb{R}$$ be a function. We say that $$f$$ is countably absolutely continuous if for every $$\epsilon>0$$ there exist $$\delta>0$$ such that for every countable collection and disjoint intervals $$((\alpha_i,\beta_i))_{i=1}^{\infty}$$ with $$\sum_{i=1}^{\infty}|\beta_i-\alpha_i|<\delta$$ we have $$\sum_{i=1}^{\infty}|f(\beta_i)-f(\alpha_i)|<\epsilon$$.
(Def 1 $$\implies$$ Def 2) Let $$\epsilon>0$$. Let $$\delta>0$$ be the positive real number that corresponds to $$\epsilon/2$$ that satisfies the condition described in Definition 1. Suppose we are given a countable collection of disjoint intervals $$((\alpha_i,\beta_i))_{i=1}^{\infty}$$ with $$\sum_{i=1}^{\infty}|\beta_i-\alpha_i|<\delta.$$ Then, for every fixed $$N$$ we have that the collection $$((\alpha_i,\beta_i))_{i=1}^{N}$$ constitutes a finite collection of disjoint intervals with $$\sum_{i=1}^{N}|\beta_i-\alpha_i|\leq \sum_{i=1}^{\infty}|\beta_i-\alpha_i|<\delta.$$ Then, we can apply the result of the 1st definition to this collection to get $$\sum_{i=1}^{N}|f(\beta_i)-f(\alpha_i)|<\frac{\epsilon}{2}.$$ As the above inequality holds for every $$N$$, to the limit $$N\to \infty$$ we get $$\sum_{i=1}^{\infty}|f(\beta_i)-f(\alpha_i)|\leq \frac{\epsilon}{2}<\epsilon.$$ And we proved the 2nd definition.
(Def 2 $$\implies$$ Def 1) I think this claim is rather obvious, as if have it for countably infinite disjoint intervals, we can adjust an infinite number of disjoint singleton sets at every finite collection without changing its total length.