I know that given a measurable space $(\Omega, \mathcal F)$, if a finitely additive set function $\mu$ is finite and continuous from above at $\varnothing$, then $\mu$ is a measure. But what if $\mu (\Omega)=\infty$? Could you please prove it or give a counterexample?
Finitely additivity and continuity from above at $\varnothing$ implies countably additivity.
measure-theory
Best Answer
Here is a counter-example.
Let $\Omega = \Bbb N$ and $\mathcal F= 2^{\Bbb N}$. Define $\mu$ on $\mathcal F$ by: $\mu(E)= 0$ if $E$ is finite and $\mu(E)= +\infty$ if $E$ is infinite.
Note that $\mu(\Omega) = +\infty$.
It is easy to see that $\mu$ is finitely additive and $\mu$ is continuous from above at $\emptyset$. BUT $\mu$ is not $\sigma$-additive.
Remark. Recall the definition: A set function $\mu$ is continuous from above at $\emptyset$ if given any decreasing sequence $E_1 \supseteq E_2\supseteq E_3 \supseteq ... $ of sets in $\mathcal F$ such that $\bigcap_n E_n = \emptyset$, if there is at least one $n$ such that $\mu(E_n)< +\infty$, then $\lim_{n \to +\infty} \mu(E_n) =0$.