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

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

\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.
• thank you so in the end you'll see that $\sum_k^n |f(b_k) - f(a_k)| < \epsilon$ if $\sum_k^n [b_k - a_k] < \min\{\delta^+, \delta^-\}$? Oct 13, 2020 at 19:44
• So $m$ depends on $a_k, b_k$. It's not clear how you choose $\delta >0$ so that the inequality holds for all choices of $(a_k, b_k)$. Oct 13, 2020 at 20:06
• @ArcticChar Yes, it is ok for $m$ to depend on $a_k, b_k$. Once $m$ is fixed, applying the definition of absolute continuity of $f_m$ will allow you to make the middle term small no matter the choice of $a_k, b_k$. Oct 13, 2020 at 20:26
• Can you write down precisely the definition of $\delta$? I really can't see how it is independent of $(a_k, b_k)$... Oct 13, 2020 at 20:32
• @ArcticChar Suppose you want the middle term to be smaller than some number $\epsilon / 3$. The fact that $f_m$ is abs. cont. gives you a $\delta$ that guarantees this for any $a_k, b_k$ satisfying $\sum_k (b_k-a_k) < \delta$. Oct 13, 2020 at 20:33