Proving $f(x)$ is absolutely continuous on $[a,b].$

Question

Can someone please help me prove the following? I am having difficulties proving it.

Let $f_n(x) (n=1,2,\cdots)$ be increasing absolutely continuous functions on $[a,b].$ Assume $f(x) = \sum_{n=1}^\infty f_n(x)$ converges on $[a,b],$ prove that $f(x)$ is absolutely continuous on $[a,b].$

$\textbf{My idea:}$ 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 $(a,b),$ $$\vert \sum_{k=1}^n [f(b_k) – f(a_k)]\vert < \epsilon, \text{ if } \sum_{k=1}^n [b_k – a_k] < \delta.$$

Answer 1

\begin{align}|\sum_k (f(b_k)-f(a_k))| &\le \sum_k |f(b_k) - f(a_k)| \\ &\le \sum_k |f(b_k) - f_m(b_k)| + \sum_k |f_m(b_k) - f_m(a_k)| + \sum_k |f_m(a_k) - f(a_k)| \end{align}

Choose $m$ large enough so the first and third terms are small. Choose $\delta$ small enough so that the middle term is small.