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> Let $X$ be a countable and infinite set and $$\mathcal{F}:=\{A\subset X;\#A<\infty\ \text{or}\ \#A^C<\infty\}.$$
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> (a) Show that $\mathcal{F}$ is an algebra.
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> (b) Given $A\in\mathcal{F}$, defines $$\mu(A):=\left\lbrace\begin{array}{c}0,\ \text{if}\ A\ \text{is finite}\\1,\ \text{if}\ A^C\ \text{is finite}\end{array}\right..$$ Show that $\mu$ is finitelly additive, but isn't necessarily countable additive.
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I'm with some problems in (b). Taking the finite and disjoint family $\{A\_j\}\_{j=1}^n\subset\mathcal{F}$, if $A\_j$ is finite, for each $j\in\{1,...,n\}$, is easy to see that $\mu$ is finitelly additive, but i'm stacked in the other cases. Since $\{A\_j\}$ is 2-2 disjoint, if we have $A\_{j\_0}^C$ finite, for some $j\_0\in\Bbb{N}$, we get $$A\_k\subset A\_{j\_0}^C, \forall k\in\{1,...,n\}\setminus\{j\_0\},$$ hence $$\bigcup\_{\begin{array}{c}j=1\\j\ne j\_0\end{array}}^{n}A\_j\subset A\_{j\_0}^C,$$ so $\bigcup\_{\begin{array}{c}j=1\\j\ne j\_0\end{array}}^{n}A\_j$ is finite.

So, how to conclude what we want? This looks very confused to me.