This question asks the following:
Let $\Omega$ be a countably infinite set and let $\mathcal{F}$ be a field that contains all subsets of $\Omega$ which are finite and also contains their complements. If $A$ is finite set $\mu(A)=0$, and if $A^C$ is finite, set $\mu(A)=1$.
(a) Show that $\mu$ is finitely additive, but is not countably additive on $\mathcal{F}$.
(b) Show that $\Omega$ is the limit of an increasing sequence of sets $A_n\in \mathcal{F}$ with $\mu(A_n)=0\ \forall n$, but $\mu(\Omega)=1$.
I did the second part by defining the sequence of sets $A_n=\{a_1,a_2,\cdots,a_n\},\ n\ge 1$, where $\Omega=\{a_1,a_2,\cdots\}$. Clearly $A_n\uparrow \Omega$, and $\mu(A_n)=0,\ \forall n$ since each of $A_n$ is finite, but $\mu(\Omega)=1$ since $\Omega^C=\emptyset$, which is finite.
However, for some reason I am not getting the first part. If I take $N$ sets $A_n\in \mathcal{F},\ 1\le n\le N$, if all of them are finite, the first part is obvious, however, I am having trouble when a finitely many of them are infinite and have finite complements. Let say, $A_1,A_2$ have finite complements, then $\mu(A_1)+\mu(A_2)=2$, but $\mu(A_1\cup A_2)=1$, as $A_1^C\cap A_2^C$, which contradicts the part (a). Am I missing something? Please help.