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Let $\{p_j\}_{j = 1}^{\infty}$ be a set of nonnegative values such that they all sum to $1$, let $\mathcal{B}$ be a $\sigma$-algebra of subsets of the countable sample space $S = \{s_1, s_2, \dots\}$, and let $\{A_i\}_{i=1}^{\infty}$ be an infinite sequence of disjoint subsets in $\mathcal{B}$.

We define $$\mathbf{1}_{A_i}(s_j) = \begin{cases} 1, & s_j \in A_i \\ 0, & s_j \notin A_i\text{.} \end{cases}$$ What justifies the equivalence $$\sum\limits_{j=1}^{\infty}\sum\limits_{i=1}^{\infty}p_j\cdot \mathbf{1}_{A_i}(s_j) = \sum\limits_{i=1}^{\infty}\sum\limits_{j=1}^{\infty}p_j\cdot \mathbf{1}_{A_i}(s_j)\text{?}$$ This answer talks a bit in brief about this, but the given link has a 404 error.

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    $\begingroup$ One can always interchange summation signs when every entry is nonnegative. $\endgroup$ – Did Jun 13 '15 at 20:02
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If $a_{mn} \geq 0$, then

$$\sum_m \sum_n a_{mn} = \sum_n \sum_m a_{mn}$$

by Tonelli's theorem for non-negative functions

Convergence is not necessary. See: 1 2 3 4

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