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 
  $$\sum_{k} |f(y_{k}) - f(x_{k})| < \epsilon$$

 A: 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. 
A: 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.
So this conditionis not sufficient for absolute continuity or even continuity.
