Let $f:S\rightarrow\mathbb{R}$, then
$\sum_{s\in S} f(s)=\infty$ means For all $M\in\mathbb{R}$, there is a finite set $T\subseteq S$ such that, for all finite sets $T'\subseteq S$ that contain $T$ one has $\sum_{s\in T'} f(s)>M$
$\sum_{s\in S} f(s)=a$ means For all $\epsilon>0$, there is a finite set $T\subseteq S$ such that, for all finite sets $T'\subseteq S$ that contain $T$, one has $|a-\sum_{s\in T'} f(s)|<\epsilon$
Let $C$ be a collection of pairwise-disjoint subsets $U \subseteq S$ whose union is $S$. Prove that $\sum_{U\in C}\left(\sum_{s\in U}|f(s)|\right)$ converges if and only if $\sum_{s\in U}|f(s)|$ converges.
$\Rightarrow$
Assume $\sum_{U\in C}\left(\sum_{s\in U}|f(s)|\right)$ converges. Then $\sum_{s\in U_1}|f(s)|+\sum_{s\in U_2}|f(s)|+\dots +\sum_{s\in U_n}|f(s)|$ converges. So no $\sum_{s\in U_i}|f(s)|=\infty$, thus $\sum_{s\in U}|f(s)|$ converges.
Is this portion of the proof valid?
$\Leftarrow $
Assume $\sum_{s\in U}|f(s)|$ converges.
I'm not sure where to go with this part. This seems intuitive to me if I'm understanding correctly. I'm basically trying to prove that it doesn't matter how $S$ is partitioned, it will still converge. But I am not sure how to prove that.