# a finitely additive measure that is continuous at $\phi$ is $\sigma$-additive

Let $(\Omega,M)$ a $\sigma$-algebra of events, and let $P$ be a finitely additive measure.

We say that $P$ is continuous at an event $A\in M$ if $A_n,B_n\in M$ are sequences of events such that $A_{n+1}\subset A_n$ , $B_n \subset B_{n+1}$ and $\cup B_n=A=\cap A_n$ then $\lim_{n\to \infty}P(A_n)=\lim_{n\to \infty} P(B_n)$.

It's well known that if $P$ is $\sigma$-additive then it's continuous at each point. Here we have a reciprocal, but I can't prove it.

Prove that if $P$ is a finitely additive measure that is continuous at $\phi$, then $\sigma$-additive, and thus is continuous at each event.

Clearly continuous at $\phi$ only consider decreasing sequences $A_n$.

This is my attempt: Let $C_n$ a disjoint sequence of events. Let $C=\cup_{k=1}^{\infty}C_k$, $A_n= \cup_{k=n+1}^{\infty} C_k$, $B_n= \cup_{k=1}^{n} C_k$.

Obviously $C=A_n \cup B_n$ , we want to prove that $P(C)=\sum_{k=1}^{\infty}{P(C_k)}$. Note that using finitely additivity $P(C)=P(A_n)+P(B_n)=P(A_n)+\sum_{k=1}^{n}{P(C_k)}$. Since $A_n \to \phi$ decreasing, $P(A_n)\to 0$, thus given $\epsilon>0$ there exist $N_{\epsilon}$ such that, $n \ge N_{\epsilon} \Rightarrow 0\le P(A_n)< \epsilon$ . Fixing some $M$ such that $M> N_\epsilon$ $P(C)=P(A_M)+\sum_{k=1}^{M}{P(C_k)}<\epsilon+\sum_{k=1}^{M}{P(C_k)}<\epsilon+\sum_{k=1}^{\infty}{P(C_k)}$

And since this inequality it's for every $\epsilon>0$ we conclude that $P(C) \le \sum_{k=1}^{\infty}{P(C_k)}$ i.e is $\sigma$-subadditive , but I don't know how to continue, and if I have some mistakes in my proof, because I used only properties of measure, not even probability.

$\phi$ denotes the empty set

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Most people write $\emptyset$ (\emptyset) for the empty set instead of $\phi$. –  Nate Eldredge Sep 13 '13 at 15:43

$$\sum_{k=1}^MP(C_k)\uparrow\sum_{k=1}^{\infty}P(C_k)$$ Thus, for given $\epsilon>0$, there exists $N_{\epsilon}\in\mathbb{N}$, such that $\forall n\geq N_{\epsilon},~\sum_{k=1}^nP(C_k)+\epsilon\geq \sum_{k=1}^{\infty}P(C_k)$. Or $P\left(\bigcup_{k=1}^n C_k\right)+\epsilon\geq \sum_{k=1}^{\infty}P(C_k)$. Now $\bigcup_{k=1}^n C_k\subset C$. So $$P(C)+\epsilon\geq P\left(\bigcup_{k=1}^n C_k\right)+\epsilon\geq \sum_{k=1}^{\infty}P(C_k)$$
Here we have only used that $P$ is a finite measure. –  Abishanka Saha Sep 13 '13 at 16:12