Let X be any countable! set and and let F be the cofinite set, i.e., $A \in F $ if A or $A^{c}$ is finite (this is an algebra).
Then show that the set function $\mu: F \rightarrow [0,\infty)$ defined as $\mu(A)=0$ if A is finite $\mu(A)=1$ if $A^{c}$ is finite is finitely additive.
I have started the argument by letting $A=\sqcup_i A_i$. If all $A_i$ are finite, then $\mu(A)=0=\sum_i\mu(A_i)$ and finite additivity follows. If there is at least one $A_i$ not finite then $\sqcup A_i$ is not finite. But $\sqcup A_i \in F$, which implies $(\sqcup_i A_i)^{c}$ is finite. But then $\mu(\sqcup_i A_i)=1 \neq \sum_i\mu(A_i)$.
Could anyone let me know where am I going wrong with the second part of the argument and how to finish this off? I can imagine that finite additivity of $\mu$ relies on the fact that X is countable.
Best Answer
If there is at least one infinite $A_i$ then it is the only one: Let $j \neq i$, then since $A_i \cap A_j = \emptyset$ we have $A_j \subseteq A_i^C$, so $A_j$ is finite. Hence in the sum $\sum_i \mu(A_i)$ there is exactly one $1$, giving $$\sum_i \mu(A_i) = \mu\left(\bigcup_i A_i\right) = 1.$$