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.
 A: 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.
A: 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 $$\sum_{n \in \cup_k A_k} a_n - \epsilon \leqslant a_{i_1} + \ldots + a_{i_N} \leqslant \sum_{n \in \cup_{k \leqslant m} A_k} a_n \leqslant \sum_i \sum_{n \in A_i} a_n$$
