Showing a measure is countably additive if it's finitely additive Let $\mu$ be a finitely additive nonnegative set function on some measurable space $(\Omega, \mathcal{F})$ with the continuity property $$B_n\in \mathcal{F},\, B_n\downarrow\emptyset, \, \mu(B_n)<\infty\Rightarrow \mu(B_n)\to 0 \tag{$*$} $$ 
is countably additive when $\mu(\Omega)<\infty$.
I am having difficulty using the property to show $\mu$ is countably additive. I would like to prove for any disjoint collection of sets $(A_i)$, $\lim \mu (\bigcup _{i=1} ^n A_n)=\mu(\bigcup _{i=1} ^\infty A_n)$. WLOG, (by rearranging indices) suppose $(A_n)$ is decreasing. Then $A_n\downarrow \emptyset$. By $(*)$ $\mu(A_n)\to 0$. I don't see how this helps or how this implies $\sum_i \mu (A_i)<\infty$.


*

*How can I show $\mu$ is countably additive?

*Doesn't the condition that $B_n\downarrow \emptyset$ automatically imply $\mu(B_n)\to 0$?

 A: Hint: note that 
$$\cup_0^\infty A_n = \cup_0^N A_n \cup \left( \cup_{N+1}^\infty A_n\right) $$
is a finite union and that you can apply the assumption to the sequence $(\cup_{N+1}^\infty A_n)_N$
A: *

*Take any $\{E_i\}$ a sequence of disjoint sets in $\mathcal{F}$, and $E\triangleq\bigcup\limits_{i=1}^{\infty}E_i$. We have to show that $\mu(E) = \sum\limits_{i=1}^{\infty}\mu(E_i)$.
Consider the sets $F_n = \bigcup\limits_{i=1}^n E_i$. You can observe that the sequence $(E\setminus F_n)\downarrow\emptyset$. Note that the finiteness of  $(E\setminus F_n)$ for every $n$ follows from the fact that $\mu(\Omega)<\infty$.
Now since $\mu$ is finitely additive $\mu(F_n)=\sum\limits_{i=1}^n \mu(E_i)$ .
$$ \mu(E)=\mu(E\setminus F_n)+\sum\limits_{i=1}^n \mu(E_i)$$
Take limits on both the sides;
$$\mu(E)=\lim_{n\to\infty}\mu(E\setminus F_n) + \sum\limits_{i=1}^\infty \mu(E_i)$$
By the property of $\mu$ mentioned in the problem, $\lim\limits_{n\to\infty}\mu(E\setminus F_n)=0$. And hence $\mu(E)=\sum\limits_{i=1}^\infty \mu(E_i)$.

*For your second doubt assume the measure (you can verify it is indeed a measure) $\mu:\mathcal{P}(\mathbb{R})\mapsto\mathbb{R}^*$ defined as 
$$\mu(E)=\begin{cases}
\infty & \text{if $E$ is infinite}\\n &\text{if $E$ is contains n elements }\end{cases}$$
Now consider the sets $F_n \triangleq\left(0,\frac{1}{n}\right]$. Note that though $F_n\downarrow\emptyset$, $\lim\limits_{n\to\infty}
\mu(F_n)=\infty$.    
