real analysis - A finitely additive measure is a measure if and only if we have continuity from below

A finitely additive measure $\mu$ is a measure if and only if it is continuous from below.

I want to know how I should proceed in proving this statement. My idea is to first assume we have a finitely additive measure $\mu$ and then prove that it is continuous from below. Then for the converse assume we have an increasing sequence of sets in $M$ and show that $\mu$ is finitely additive.

I am not sure if this is the correct approach I should take. Any suggestions is greatly appreciated.

Answer

Note that the only difference between a measure and a finitely additive measure is that a finitely additive measure may not satisfy countable additivity. Moreover, if $\mu$ is in fact a measure then it is a standard fact that $\mu$ satisfies continuity from below. So what the problem is really asking is to show that a finitely additive measure which satisfies continuity from below is countably additive, hence is a measure.

To show this, let $\{E_n\}_{n=1}^{\infty}$ be a sequence of disjoint sets in the $\sigma$-algebra $\mathcal{M}$. Define
$$ F_k=\bigcup_{n=1}^kE_n$$
and observe that $F_1\subset F_2\subset\cdots$ and that
$$ \bigcup_{n=1}^{\infty}E_n=\bigcup_{k=1}^{\infty}F_k$$
Therefore by finite additivity and continuity from below,
$$\mu\Big(\bigcup_{n=1}^{\infty}E_n\Big)=\mu\Big(\bigcup_{k=1}^{\infty}F_k\Big)=\lim_{k\to\infty}\mu(F_k)=\lim_{k\to\infty}\sum_{n=1}^k\mu(E_n)=\sum_{n=1}^{\infty}\mu(E_n)$$

Therefore $\mu$ is countably additive, hence is a measure.