Q:

Algebras: Finite Additivity $\implies$ Countable Additivity (Idea?)

Given an algebra $\mathcal{A}$.
Consider a set function $\mu:\mathcal{A}\to\mathbb{R}_+$:
Then countable additivity follows from finite additivity:
$$\mu(A+B)=\mu(A)+\mu(B)\implies\mu(A)=\sum_{n\in\mathbb{N}}\mu(A_n)$$
Can you explain the idea behind the proof?
(I always forget the precise steps probably because I haven't understand it yet.)

A:

I think there is something missing in yiur question. 
To see this, consider the algebra(!) of all subsets $A \subset \Bbb{N}$ which are finite or for which $A^c$ is finite. 
Define $\mu(A)=0$ if $A$ is finite and $\mu(A)=1$ otherwise. It is easy to verify that $\mu$ is finitely additive, but not countably additive. 

