A function $f:[a,b]\rightarrow \mathbb{C}$ is said to be absolutely continuous if for each $\epsilon>0$, there exists $\delta>0$ such that for every mutually disjoint finite sequence of closed sub-intervals $\{[a_1,b_1],...,[a_n,b_n]\}$ of $[a,b]$ satisfying $\sum_{i=1}^n |b_i-a_i|<\delta$, $\sum_{i=1}^n |f(b_i)-f(a_i)|<\epsilon$ holds.
Let $f:[a,b]\rightarrow \mathbb{C}$ be an absolutely continuous function and $\epsilon>0$.
Then, how do I prove that there exists $\delta>0$ such that for every mutually disjoint finite sequence of open sub-intervals $\{(a_1,b_1),...,(a_n,b_n)\}$ of $[a,b]$ satisfying $\sum_{i=1}^n |b_i-a_i|<\delta$, $\sum_{i=1}^n |f(b_i)-f(a_i)|<\epsilon$ holds ?