## mu is countably additive if and only if it satisfies the Axiom of Continuity

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Let $\mu$ be a finite, non-negative, finitely additive set function on a measurable space $(\Omega, \mathcal{A})$. Show that $\mu$ is countably additive if and only if it satisfies the Axiom of Continuity: For $E_n\in\mathcal{A}, E_n\downarrow\emptyset \implies \mu(E_n)\to 0$.

(=>) Assume $\mu$ is countably additive. Let $E_n\in\mathcal{A}$, $E_n\downarrow\emptyset$. Then,

$\displaystyle \lim_n \mu (E_n)=\mu (\cap_{n=1}^\infty E_n)=\mu (\emptyset)$.

Suppose $\mu(\emptyset)=c$. Then $\mu(\emptyset)=\mu(\cup_{n=1}^\infty \emptyset)=\sum_{n=1}^\infty c$ implies $c=0$.

(<=) Assume $\mu$ satisfies Axiom of Continuity. Let $A_n\in\mathcal{A}$ be mutually disjoint sets. Define $E_n=\cup_{i=1}^\infty A_i\setminus \cup_{i=1}^n A_i$.

Then $E_n\downarrow\emptyset$. $\lim_n \mu(E_n)=0$, $\lim_n \mu(\cup_{i=1}^\infty A_i)-\mu (\cup_{i=1}^n A_i)=0$. $\lim_n \mu(\cup_{i=1}^n A_i)=\mu (\cup_{i=1}^\infty A_i)$.

Therefore

\begin{aligned}\mu(\cup_{i=1}^\infty A_i)&=\lim_n \mu(\cup_{i=1}^n A_i)\\ &=\lim_n \sum_{i=1}^n \mu (A_i)\\ &=\sum_{i=1}^\infty \mu(A_i) \end{aligned}