Let $\Omega$ be a nonempty set, $\{x_n\}_{n}$ a sequence of distinct elements of $\Omega$, and $\{a_n\}_{n}$ a sequence of nonnegative real numbers. For E $\subset$ $\Omega$, define $\mu$($E$) $=$ $\sum_{x_n\in E} a_n$.
I'm trying to show $\mu$ is a measure on $\mathcal P$($\Omega$).
I've already shown that condition (1) holds: Since $\{a_n\}_{n}$ is a sequence of nonnegative real numbers, we have $\mu$($E$) $=$ $\sum_{x_n\in E} a_n$. $\ge$ $0$ for all $E$ $\in$ $\mathcal P$($\Omega$).
However, I'm stuck showing that conditions (2) and (3) hold.
That is, $\mu$($\emptyset$) $=$ $0$, and for a sequence $\{E_n\}_{n=1}^{\infty}$ of pairwise disjoint sets $\mu$(${\bigcup_{n=1}^{\infty}E_n}$)$=$$\sum_{n=1}^{\infty}$ $\mu$($E_n$)
Any suggestions?
You have actually already asked this question here. The connection with your previous question is that we are taking $\lambda$ to be counting measure, and if $f:\Bbb \{x_n\}_n\to[0,\infty)$ is a sequence, i.e., a measurable function, with $f(x_n) = a_n$, then $\int_E f\,d\lambda = \sum_{x_n\in E} a_n$ by definition.
For (2), the empty sum is $0$ by convention. For (3), use the monotone convergence theorem.