# Different definition of continuity

Condition: $$f:I\to\mathbb R$$ is continuous. For any countable sequence of pairwise disjoint sub-intervals $$(x_k, y_k)$$ of $$I$$, we have $$\forall\epsilon\exists\delta$$ such that $$\sum_{k} |y_{k} - x_{k}| < \delta$$ implies $$\sum_{k} |f(y_{k}) - f(x_{k})| < \epsilon.$$

Is this condition a necessary or sufficient condition of absolution continuity? Note that the order of the logic identifiers has changed.

A function $$f: I \to \mathbb{R}$$ is absolutely continuous on an interval $$I$$ if for every $$\epsilon > 0$$ there is a $$\delta > 0$$ such that whenever a finite sequence of pairwise disjoint sub-intervals $$(x_k, y_k)$$ of $$I$$ satisfies $$\sum_{k} |y_{k} - x_{k}| < \delta$$ then $$\sum_{k} |f(y_{k}) - f(x_{k})| < \epsilon$$

Given any function $$f:I\to\mathbb R$$ (not necessarily continuous), the condition: For any countable sequence of pairwise disjoint sub-intervals $$(x_k, y_k)$$ of $$I$$, we have $$\forall\epsilon\exists\delta$$ such that $$\sum_{k} |y_{k} - x_{k}| < \delta$$ implies $$\sum_{k} |f(y_{k}) - f(x_{k})| < \epsilon$$ is trivially true.
Proof: Given any countable sequence of pairwise disjoint sub-intervals $$(x_k, y_k)$$ of $$I$$, just choose $$\delta = \frac{1}{2} \sum_{k} |y_{k} - x_{k}|$$. Then the condition $$\sum_{k} |y_{k} - x_{k}| < \delta$$ will be false and so the implication "$$\sum_{k} |y_{k} - x_{k}| < \delta$$ implies $$\sum_{k} |f(y_{k}) - f(x_{k})| < \epsilon$$" will be trivially true.
Yes, they are equivalent. Suppose you choose $$\delta$$ according to the usual definition of absolute continuity with $$\epsilon$$ repalced by $$\epsilon /2$$. If $$(a_k.b_k)$$ is a disjoint sequence of interval with total length less than $$\delta$$ then $$\sum\limits_{k=1}^{N} |f(b_k)-f(a_k)| < \epsilon /2$$ for each $$N$$. Let $$N \to \infty$$ to complete the proof.
• Many thanks for your teaching! So you proved that the definition "$\forall\epsilon\exists\delta(\forall \text{finite subintervals we have} (\sum|y_k-x_k|<\delta \Rightarrow \sum|f(y_k)-f(x_k)|<\epsilon))$" is equivalent to "$\forall\epsilon\exists\delta(\forall \text{countable subintervals we have} (\sum|y_k-x_k|<\delta \Rightarrow \sum|f(y_k)-f(x_k)|<\epsilon))$". However, my first condition means "$\forall \text{countable subintervals}(\forall\epsilon\exists\delta \text{we have} (\sum|y_k-x_k|<\delta \Rightarrow \sum|f(y_k)-f(x_k)|<\epsilon))$". Not sure my understanding is correct, though. – High GPA Mar 25 '19 at 22:05