Sum of Countable Increasing Absolutely Continuous Functions is Absolutely Continuous

Question

I am trying to show that $f(x) = \sum_n f_n(x)$ is absolutely continous on $[a,b]$, where $f_n(x)$ is an increasing absolutely continuous function on $[a,b]$ for each $n$ and $\sum_n f_n(x)$ converges.

My thoughts: choose $\delta_k$ as a response to the $\epsilon/2^k$ challenge regarding the criterion for the absolute continuity of $f_k$ on $[a, b]$. Then $\min \{\delta_k\}$ would be the response the $\sum_k \epsilon/2^k = \epsilon$ challenge regarding the criterion for the absolute continuity of $f = \sum_n f_n(x)$ on $[a, b]$. This is certainly true for finite sums, but is this still valid for countable sums? Also I realized that I did not use the condition that each $f_n$ is increasing anythere, which sugguests that this approach might not work.

Can someone give me a hint? Thanks in advance!

Answer 1

If you are familiar with the fundamental theorem of Calculus (FTC-Lebesgue)in the sense of Lebesgue the problem has a rather simple solution.

The FTC-Lebesgue implies that $f_n$ (by virtue of being absolutely continuous) is differentiable almost surely, $f'_n\in L_1([a,b])$ and $f_n(x)=f(a)+\int^x_af'_n$ for all $a\leq x\leq b$. The monotonicity assumption implies that $f'_n\geq0$ when it exists.

Monotone convergence yields $\sum_n\int_{(a,x]}f'_n=\int_{(a,x]}\sum_n f'_n$. Putting this together $$\sum_n(f_n(x)-f_n(a))=\sum_n\int_{(a,x]}f'_n=\int_{(a,x]}\sum_nf'_n$$ Thus, if $F=\sum_nf_n$ (which by assumption exists on $[a,b]$ then $$F(x)=F(a)-\int_{(a,x]}\sum_nf'_n$$ and so, $F$ is itself ansulutely continuous.