Show that the measure is sigma-additive Let $X = \{ x_{1}, x_{2}, \ldots \}$ be any countable set and let $p_{1}, p_{2}, \ldots $ be any positive numbers such that
$$\sum_{n = 1}^{\infty} p_{n} = 1.$$
On the power set $\mathscr{P}(X)$ of $X$, define a measure $\mu$ by the formula 
$$\mu(A) = \sum_{x_{n} \in A} p_{n} \quad (A\subseteq X),$$
where the sum is over all $n$ such that $x_{n} \in A$. Prove that $\mu$ is a $\sigma$-additive measure, with $\mu(X) = 1$.
My book defines a measure $\mu$ by
(1) the domain of definition of $\mu$ is a semi-ring of sets;
(2) $\mu$ is real and non-negative;
(3) $\mu$ is additive in the sense that if $A = \cup_{i = 1}^{n} A_{i}$ for pairwise disjoint sets $A_{i}$, then $\mu(A) = \sum_{i = 1}^{n} \mu(A_{i})$.
And a measure $\mu$ is $\sigma$-additive if
(4) $\mu(A) = \sum_{i = 1}^{\infty} \mu(A_{i})$ for $A = \cup_{i = 1}^{\infty} A_{i}$ for pairwise disjoint $A_{i}$.
Now, showing that $\mu(X) = 1$ seems easy since $\mu(X) = \sum_{x_{n} \in X} p_{n} = \sum_{n = 1}^{\infty} p_{n} = 1$, since we are summing over all $x_{n} \in X$ and there are countably many points in $X$. Also, $\mathscr{P}(X)$ is a semi-ring and $\mu$ is real and non-negative since $p_{1}, p_{2}, \ldots$ are positive reals.
So it remains to be shown that $\mu$ is $\sigma$-additive. For this, let $A \in \mathscr{P}(X)$ such that 
$$A = \bigcup_{n = 1}^{\infty} A_{n},$$
for pairwise disjoint sets in $\mathscr{P}(X)$. Then, we want to show that
$$\mu (A) = \mu (\bigcup_{n = 1}^{\infty} A_{n}) = \sum_{x_{n} \in \bigcup_{n = 1}^{\infty} A_{n}} p_{n},$$
where $x_{n} \in \bigcup_{n = 1}^{\infty} A_{n}$ means that there is $n_{0}$ such that $x_{n} \in A_{n_{0}}$; we want to show that the above is equal to
$$\sum_{n= 1}^{\infty} \mu(A_{n}) = \sum_{n= 1}^{\infty} \sum_{x_{n} \in A_{n}} p_{n}.$$
And this is where I am stuck.
 A: First prove that $\mu$ is additive.  
Let $$A = \bigcup_{n = 1}^m A_{n},$$ for pairwise disjoint sets in $\mathscr{P}(X)$.
We have 
$$ \mu (A) = \mu \left (\bigcup_{n = 1}^m A_{n} \right ) = \sum_{x_{k} \in \bigcup_{n = 1}^m A_{n}} p_{k} = \sum_{n=1}^m\sum_{x_{k} \in A_{n}} p_{k} =  \sum_{n=1}^m\mu(A_n)$$
Now let us prove that $\mu$ is $\sigma$-additive. 
Let $$A = \bigcup_{n = 1}^{\infty} A_{n},$$ for pairwise disjoint sets in $\mathscr{P}(X)$. For al $m$, we have (using the monotonicity of $\mu$): 
$$ \mu (A) = \mu \left (\bigcup_{n = 1}^{\infty} A_{n} \right ) \geqslant  \mu \left (\bigcup_{n = 1}^m A_{n} \right ) =  \sum_{n=1}^m\mu(A_n)$$
Since $ \left ( \sum_{n=1}^m\mu(A_n) \right)_m$ is a non-decreasing sequence of real number and it is limited above by $ \mu(A)$, we can conclude that  
$$ \mu (A) = \mu \left (\bigcup_{n = 1}^{\infty} A_{n} \right ) \geqslant   \lim_{m \to \infty}  \sum_{n=1}^m\mu(A_n)= \sum_{n=1}^{\infty}\mu(A_n)$$
To complete the proof, note that for all $x_k \in A$, there is $n$ such that $x_k\in A_n$, so we can conclude that 
$$ \mu (A) = \sum_{x_{k} \in A} p_{k} \leqslant  \sum_{n=1}^{\infty}\sum_{x_{k} \in A_{n}} p_{k} =  \sum_{n=1}^{\infty}\mu(A_n)$$
