So, I'm trying to get my head around when you can have finitely but not countably additive probabilities.

The standard example of a finitely additive but not countably additive space is the following strange distribution over the natural numbers. All finite sets get measure 0, but the whole space gets measure 1. This is finitely additive but not countably so, since a finite union of finite sets is finite, but a countable union needn't be so.

So this got me thinking that if you had an atomless space, examples of this form would be harder to come by. Does atomlessness plus finite additivity guarantee countable additivity? If not, what is missing?

I know that Villegas (1964) shows that for a comparative probability structure to be countably additive, the important properties of the structure are atomlessness and a certain kind of continuity. But I don't know how relevant that point is to the current question.

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why did you roll back the edit? What do you mean by "measurement theory" if not "measure theory"? (I am aware of a different use of measurement theory that is tangentially related to data analysis, but your question certainly doesn't seem to be about that!) – Willie Wong Dec 11 '12 at 15:35
@WillieWong I had suggested a tag wiki for measurement theory. See for example this wikipedia page or the Krantz et al book Foundations of Measurement. The comments at the end about Villegas relate to measurement theory (not measure theory) in this sense. – Seamus Dec 11 '12 at 15:40
Okay, you are thinking of the same measurement theory as I am thinking of, but I still cannot see what your question has to do with measurement theory; your final paragraph really doesn't have much to do (mathematically) with the first three. – Willie Wong Dec 11 '12 at 15:58
@WillieWong Atomlessness and a kind of continuity are the extra that is required for a comparative probability relation to have a countably additive representation. That is, I guess, more suggestive than anything more concrete, but it is still relevant I think. – Seamus Dec 11 '12 at 16:15
The usual statement of continuity that guarantees $\sigma$-additiv. is kinda trivial: we just require upper-continuity that if $A_i\searrow \emptyset$ are sets that $\mu(A_i) \searrow 0$. If you think Villegas is relevant, can you give a link and/or a short description of the conditions that were used? – Willie Wong Dec 11 '12 at 16:33

Here’s an example.

Define an equivalence relation $\sim$ on $\wp(\Bbb N)$ by $A\sim B$ iff $A\,\triangle\, B$ is finite, where $\triangle$ is symmetric difference, and let $\mathscr{B}=\wp(\Bbb N)/\sim$. For $A\subseteq\Bbb N$ denote by $[A]$ the $\sim$-equivalence class of $A$. Let $\mathscr{U}$ be a free ultrafilter on $\Bbb N$. Note that for any $A\subseteq\Bbb N$, $A\in\mathscr{U}$ iff $[A]\subseteq\mathscr{U}$. Now define a $\{0,1\}$-valued measure $\mu$ on $\mathscr{B}$ by $\mu\big([A]\big)=1$ iff $A\in\mathscr{U}$. Then $\mathscr{B}$ is atomless, and $\mu$ is finitely additive. However, $\mu$ is not countably additive, since it is possible to partition $\Bbb N$ into countably infinitely many infinite sets, none of which is in $\mathscr{U}$.

Added: Michael Greinecker has pointed out that I’m using a notion of atomless that may be different from the one intended by Seamus. Here’s another example that may be preferable.

Let $d:\wp(\Bbb N)\to[0,1]$ be asymptotic density, and let $\mathscr{U}$ be a free ultrafilter on $\Bbb N$. For $A\subseteq\Bbb N$ let $$\mu(A)=\mathscr{U}\text{-}\lim_n\frac{|A\cap\{1,\dots,n\}|}n\;.$$ (For basic information on $\mathscr{U}$-limits see this answer by Martin Sleziak.) Then $\mu$ is a finitely additive non-atomic probability measure on $\wp(\Bbb N)$ such that $\mu(A)=d(A)$ whenever $A$ has an asymptotic density.

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OK. This is good. So atomlessness isn't sufficient. But is there something in the area that is sufficient? Presumably this sort of measure will have problems with limits? – Seamus Dec 11 '12 at 12:10
What notion of atomless do you use here? The usual one in measure theory is that every measurable set with positive measure contains a measurable subset of smaller positive measure. No $\{0,1\}$-valued set function can be atomless in this sense. – Michael Greinecker Dec 11 '12 at 12:12
@Michael: The usual notion for Boolean algebras. – Brian M. Scott Dec 11 '12 at 12:38
I meant atomless in the sense of the boolean algebra not having atoms. – Seamus Dec 11 '12 at 15:41
@Seamus: Okay; then my first example does what you want. – Brian M. Scott Dec 11 '12 at 15:42