# Finitely additive “measure” that is not upper-continuous

Let $X$ be a set and $\mathcal{A}\subset 2^X$ an algebra of subsets (i.e closed under finite intersection, union and complements) containing $\emptyset$. Suppose $$\mu: \mathcal{A}\rightarrow [0,1]$$

is finitely additive (for disjoint sets) with $\mu(X)=1$. Is it possible that for a nested family $A_1 \supset A_2 \supset...$ of elements of $\mathcal{A}$ with $\cap_n A_n =\emptyset$, we have $\lim_n\mu(A_n)\neq 0$?

Remark: I'm pretty sure there is such a measure. The reason is that the hypothesis of upper-continuity is part of a well-known extension theorem used to extend $\mu$ to the $\sigma$ algebra generated by $\mathcal{A}$.

• Which extension procedure is that? Also yes, you're right. There should be such a measure. The limit of the measure converges to zero if and only if the measure is countably additive. – Theoretical Economist Dec 9 '16 at 2:20
• It says that if $\mu$ and $\mathcal{A}$ are as above, and if the measure of a nested intersection is always the limit of the measures, then there exists a measure on the sigma algebra generated by $\mathcal{A}$ which agrees with $\mu$ on $\mathcal{A}$. – Tim kinsella Dec 9 '16 at 2:42

Let $X=\mathbb N$ and let $\mathcal A=2^\mathbb N.$ Let $\mathcal U$ be a nonprincipal ultrafilter on $\mathbb N$ (this requires the axiom of choice). Define $\mu:\mathcal A\to\{0,1\}$ by setting $\mu(A)=1$ for $A\in\mathcal U$ and $\mu(A)=0$ for $A\notin\mathcal U.$ Then $\mu$ is a finitely additive probability measure, and $\mu(A)=0$ for all finite $A.$ Let $A_n=\{n,n+1,n+2,\dots\}$; then $A_1\supset A_2\supset A_3\supset\cdots$ and $\bigcap_n A_n=\emptyset$ and $\lim_n\mu(A_n)=1.$
On $\mathbb{N}$, let $\mathcal{A}$ be the collection of finite or cofinite sets. Define $\mu$ to be $0$ on finite sets and $1$ on cofinite sets. This is essentially the same example that bof gave. I'm not sure why you need ultrafilters.