Rigorously Proving Countable Additivity of Discrete Measure I'm trying to show, in full rigor, that a measure on a discrete probability space is necessarily countably additive. That is, with countable $\Omega$ and $p: \sum_{x\in \Omega} p(x) = 1$ and $p(x) \ge 0$ for all $x$, we define:
$P(A) = \sum_{x\in A} p(x)$
Now my hand-wavy solution is, with the $A_i$ disjoint:
$P(\bigcup A_i) =\sum_{x\in \bigcup A_i} p(x) = \sum_i \sum_{x\in A_i} p(x) = \sum_i P(A_i) $
But it feels like the middle equality is doing some work that I can't necessarily justify. Specifically, if the collection of $i$ is countably infinite, and each $A_i$ is countably infinite.
 A: What make things work are some basic properties of absolutely convergent series (in the present case, it suffices to consider convergent series with nonnegative terms) and combinations of such properties:

*

*adding and removing parenthesis,


Theorem 1. Let $p:\mathbb{N}\rightarrow\mathbb{N}$ be a strictly monotone increasing function. Given a series $\sum_na_n$ define
$$\begin{align}
b_1&=a_1+\ldots + a_{p(1)},\\
b_n&=a_{p(n)+1}+\ldots a_{p(n+1)}
\end{align}$$
ans consider the series $\sum_nb_n$. If $\sum_na_n$ converges to $s$, then $\sum_nb_n$ also converges to $s$.



*rearrangements,


Theorem: Suppose $g:\mathbb{N}\rightarrow\mathbb{N}$ is a bijective function. For any series $\sum_na_n$ define $b_n=a_{f(n)}$ and consider the series $\sum_nb_n$. If the series $\sum_na_n$ converges absolutely and $\sum_na_n=s$, then $\sum_nb_n$ also converges absolutely and $\sum_nb_n=s$.



*forming subsequences,


Theorem: Suppose $f:\mathbb{N}\rightarrow\mathbb{N}$ is injective. For any series $\sum_na_n$ define $b_n=a_{f(n)}$. Is $\sum_na_n$ converges absolutely, then $\sum_nb_n$ converges absolutely.



*partitioning or subgrouping series.


Theorem: Suppose $\{f_k:k\in\mathbb{N}\}$ is a sequence of injective functions from $\mathbb{N}$ into $\mathbb{N}$ such that $\{f_k(\mathbb{N}):k\in\mathbb{N}\}$ forms a partition of $\mathbb{N}$ (i.e. the sets $f_n(\mathbb{N})$ are pairwise disjoint and their union is $\mathbb{N}$). Given a series $\sum_na_n$ define
$b_k(n)=a_{f_k(n)}$, and
$$s_k=\sum_nb_k(n)$$
If $\sum_na_n$ is absolutely convergent and $\sum_na_n=s$, then  the series $\sum_ks_k$ is also absolutely convergent, and $\sum_ks_k=s$.

I omit proofs  of these results as they can be found in many Calculus and Analysis textbooks that cover convergence of series. A nice reference is Apostol, T. Mathematical Analysis, 2nd edition,  Addison Wesley,  1974 pp. 187, 196-199.
