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}

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