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 $\mu(B_n)\to 0$?
Best Answer
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 $(\cup_{N+1}^\infty A_n)_N$