Proving that one can split a convergent series into a (countably) sum of “subseries”

Let $I\subseteq \mathbb{N}$, $\left(A_{i}\right)_{i\in I}$ be a sequence of disjoint subsets of $\mathbb{N}$. If $a:\bigcup_{i\in I}A_{i}\rightarrow\mathbb{R}^{+}$ is a sequence of positive real numbers in $\bigcup_{i\in I}A_{i}$ such that $\sum_{n\in\bigcup_{i\in I}A_{i}}a_{n}$ converges, does then the equality $$\sum_{n\in\bigcup_{i\in I}A_{i}}a_{n}=\sum_{i\in I}\,\sum_{n\in A_{i}}a_{n}$$ hold ? If $I$ is finite I managed to prove it (by using the fact that a sum $\sum_{i\in I} A_i$ can be represented as $\sup\{\sum_{i\in F} \mid F\subseteq I, F \ \text{finite}\}$), but for countable $I$ I don't know what to do.

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You do want to require that the sets $A_i$ be pairwise disjoint. – Brian M. Scott Dec 23 '12 at 16:36
@BrianM.Scott Ah,yes,thankyou. – resu Dec 23 '12 at 17:43

For $I$ countable (just take it to be $\mathbb{N}$ for notation) if you have (*) the case of $I$ finite we have pretty easily that $$\sum_{i} \sum_{n \in A_i} a_n \leqslant \sum_{n \in \cup_i A_i} a_n$$Indeed, the LHS is $\lim_j \sum_{k \leqslant j} \sum_{n \in A_k} a_n$, so we have for fixed $j$ that $$\sum_{k \leqslant j} \sum_{n \in A_k} a_n =^{(*)} \sum_{n \in \cup_{k \leqslant j} A_{k}} a_n \leqslant \sum_{n \in \cup_k A_{k} }a_n$$

Now in the other direction, it suffices to prove that $$\sum_{n \in \cup_k A_k} a_n \leqslant \sum_k \sum_{n \in A_k} a_n + \epsilon$$for any $\epsilon > 0$. For the sum on the left, for that same $\epsilon$ we can get within the total sum by just adding up finitely many terms, and they appear in finite time on the RHS.

In more detail, we have $\sum_{n \in \cup_k A_k} a_n - \epsilon \leqslant a_{i_1} + \ldots + a_{i_N}$, where $\cup_k A_k = i_1 < i_2 < \ldots$ (we obtain an appropriate $N$ from the definition of convergence). Now $i_1$ through $i_N$ all lie in $A_1 \cup \ldots \cup A_m$ for some $m$, and we have $$\sum_{n \in \cup_k A_k} a_n - \epsilon \leqslant a_{i_1} + \ldots + a_{i_N} \leqslant \sum_{n \in \cup_{k \leqslant m} A_k} a_n \leqslant \sum_i \sum_{n \in A_i} a_n$$

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I know the proof of the rearrangement theorem; you don't have to prove it here (seeing that you have begun it's proof); I'm more interested in seeing how you use that to circumvent the Dominated Convergence Theorem. – resu Dec 23 '12 at 18:26
I pondered some time over the second inequality and I don't quite get it: Could you please structure the proof a little bit more ? (Specific questions: I assume, that $k$ ranges over $I$ as well ? And why do we have $\sum_{n \in \cup_k A_k} a_n - \epsilon \leqslant a_{i_1} + \ldots + a_{i_n}$ ? I'm slightly confused by the $n$ which appears on both sides; why is the sum on the right finite ?) – resu Dec 23 '12 at 19:26
Hey resu thanks for the reply. I had carelessly used $n$ for two different things, I now call one of them $N$. Do let me know if you have further questions. – uncookedfalcon Dec 23 '12 at 20:32
excellent answer! – resu Dec 31 '12 at 17:58

This is a special case of the countable additivity of integration over disjoint countable collections for the Lebesgue integral. The following theorem is from Royden,

Theorem 20 (the Countable Additivity of Integration) Let $f$ be integrable over $E$ and $\{E_n\}_{n=1}^\infty$ a disjoint countable collection of measurable subsets of $E$ whose union is $E$. Then $$\int_E f = \sum_{n=1}^\infty \int_{E_n} f$$

The proof of the above theorem is an immediate application of the Dominated Convergence Theorem. Royden develops the theory in terms of the Lebesgue measure, but the proof of this theorem relies only on the Dominated Convergence Theorem, which is of course true for integrals with respect to general measures.

As for your question, in case either side is infinite the result follows from a simple monotonicity argument. Otherwise, take $\nu$ to be the counting measure on $\mathbb{N}$ and $E = \cup_{i \in I} A_i$, $E_n = A_n$ and the result follows.

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Woah that is a little bit too heavy for me, since I don't know that much integration theory (although I think I will appreciate your answer once I do); I thought there would be a more direct and elementary proof. BTW, meanwhile I managed to prove it for finite $I$ (and arbitrary $\bigcup_{i\in I}A_{i}$) - but for countable $I$ I still don't know what to do. – resu Dec 23 '12 at 18:18
kk I wrote up how to get to $I$ countable from $I$ finite just playing with $\epsilon$s – uncookedfalcon Dec 23 '12 at 18:46