Prove that if $A_n$, $n\in\mathbb{N}$, are mutually disjoint, then $P(\cup_n A_n)=\sum_n P(A_n)$ I gave this problem a try and I feel like I am missing something and I wanted to get some help on what I might be missing. 
Suppose $\Omega$ is a countable space and $p:\Omega\to[0,1]$ is such that $\sum_{\omega\in\Omega}p(\omega)=1$.
For $A\subset\Omega$ let $P(A)=\sum_{\omega\in A}p(\omega)$ with $P(\varnothing)=0$.  Prove that if $A_n$, $n\in\mathbb{N}$, are mutually disjoint, i.e.\ $A_n\cap A_m=\varnothing$ for $m\ne n$, then 
    $P(\cup_n A_n)=\sum_n P(A_n)$
My try:
To show this, notice that $P(\cup_n A_n)=$ (sum of the measures of the $A_n$) - (sum of the measures of the intersections of the $A_n$). Since the $A_n$ are mutually disjoint, we know that the sum of the intersections is 0. So, it follows that $P(\cup_n A_n)=\sum_n P(A_n)$.
 A: This is more a question about summation rather than probability.
We need to show that $\sum_{\omega \in A} p(\omega) = \sum_n \sum_{\omega \in A_n} p(\omega)$, where $A= \cup_n A_n$ and the $A_n$ are pairwise disjoint.
Note that all the index sets are subsets of $\Omega$ which is countable and $p$
is non negative.
It is not hard to show that
$\sum_{\omega \in B} p(\omega) = \sup_{I \subset B, \text{ finite}} \sum_{\omega \in I} p(\omega)$, and hence if $B \subset C$ then
$\sum_{\omega \in B} p(\omega) \le \sum_{\omega \in C} p(\omega)$.
Suppose $I \subset A$ is finite, then $I \cap A_n$ is finite and we have
$\sum_{\omega \in I} p(\omega) \le \sum_n \sum_{\omega \in I \cap A_n} p(\omega) \le \sum_n \sum_{\omega \in A_n} p(\omega)$ and hence 
$\sum_{\omega \in A} p(\omega) \le \sum_n \sum_{\omega \in A_n} p(\omega)$.
Now let $\epsilon>0$ and pick $N$ such that $\sum_{n \le N} \sum_{\omega \in A_n} p(\omega) > \sum_n \sum_{\omega \in A_n} p(\omega) - {1 \over 2} \epsilon$
Now choose finite $I_n \subset A_n$ such that
$\sum_{\omega \in I_n} p(\omega)  > \sum_{\omega \in A_n} p(\omega) - {1 \over 2^{n+1}} \epsilon$.
Then
\begin{eqnarray}
\sum_{\omega \in A} p(\omega) &\ge& \sum_{n \le N} \sum_{\omega \in I_n} p(\omega) \\
&\ge& \sum_{n \le N} (\sum_{\omega \in A_n} p(\omega) -{1 \over 2^{n+1}} \epsilon) \\
&\ge& \sum_{n \le N} \sum_{\omega \in A_n} p(\omega) -{1 \over 2} \epsilon \\
&\ge& \sum_n \sum_{\omega \in A_n} p(\omega) - \epsilon
\end{eqnarray}
from which we get the other direction.
A: EDIT I've realized that this proof is circular since I assumed at the beginning of the proof that $P$ is a measure. Nevertheless, I feel like leaving it here as an example of "math made difficult", in case anyone might find it amusing.

Let $A:=\bigcup_n A_n$. We shall consider the characteristic functions
$\chi_{A_n}$ and recall that $P$ is the measure whose density function is $p(\omega)$.
Your question can be interpreted as proving that
$$\begin{align}
P(A) &= \sum_{n=1}^\infty \int_\Omega \chi_{A_n}\,dP \\
&=\lim_{n\to\infty} \sum_{i=1}^n \int_\Omega \chi_{A_i}\,dP \\
&= \lim_{n\to\infty} \int_\Omega f_{n}\,dP 
\end{align}$$
where  $f_n :=\sum_{i=1}^n  \chi_{A_i}$.
By mutual disjoint-ness of $A_i$, we can see that $f_n=\chi_{\cup_{i=1}^n A_i}$ so $f_n\le 1$. It is also easy to verify that $f_n\to\chi_A$ pointwise, hence we can apply the Dominated Convergence Theorem to conclude that
$$
\lim_{n\to\infty} \int_\Omega f_{n}\,dP = \int_\Omega \chi_A dP = P(A)
$$
which proves the assertion.
Alternatively, if you want a more "elementary" method then I think you can fix an arbitrary enumeration of $A$ and try to invoke the Riemann rearrangement theorem. I haven't work out the full details but it should be doable.
