Let X be an infinite set. Let $\mathcal{A}$ be the collection of subsets A of X such that either A is finite, and then m(A):=0, or the complement of A is finite, and then m(A):=1.
Under what condition on X can $m$ be extended to a countably additive measure on a $\sigma$-algebra?
It can be shown that $m(\cdot)$ is finitely-additive. Let $\{A_n\}, A_n \in \mathcal{A}, n = 1,...,N$ is collection of pairwise-disjoint sets. If $\cup_{n=1}^NA_n$ is infinite, then there is infinite $A_m \in \{A_n\}_{n=1}^N$. $A_m \in \mathcal{A}$, then $A_m^c$ is finite.
$\left(\cup_{n=1}^N A_n\right)^c \subset A_m^c$, thus $\left(\cup_{n=1}^N A_n\right)^c$ is finite.
Elements of $\{A_n\}_{n=1}^N$ are pairwise-disjoint. If $\sum_{n = 1}^{N} m(A_n) = 0$, then $A_n$ - is finite and their $\cup_{n=1}^N A_n$ is finite. Thus $m(\cup_{n=1}^N A_n) = 0 = \sum_{n=1}^N m(A_n)$.
If $\sum_{n=1}^N m(A_n) \neq 0$, then there is m $\in \mathbb{N}$: $m(A_m) \neq 0$ and $A_m^c$ is finite. Since $A_n$ are disjoint, $\cup_{n = 1, n \neq m}^N A_n \subset A_m^c$, thus $\cup_{n = 1, n \neq m}^N A_n$ is finite. Hence, $m(\cup_{n = 1}^N A_n) = 1 = \sum_{n=1}^{N} m(A_n)$. This is why $m(\cdot)$ is finite-additive.
There is the theorem:
For each X and ring $\mathcal{A}$ defined on X, any countably-additive function $\mu$ from $\mathcal{A}$ into $[0,\infty]$ can be extended to a measure on the $\sigma$-algebra $\mathcal{S}$ generated by $\mathcal{A}$.
So it seems that we need to constraint X and as the result $m(\cdot)$ should became countably-additive. How can we do it?
