How is it that a finitely additive probability measure on a field may not be countably subadditive? I know that the field must be countably additive and thus finite additivity does not suffice, but I'm struggling with the reasoning.

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What do you mean by "the field must be countably additive"? – Qiaochu Yuan Sep 27 '12 at 5:41

Revised: Let $\mathscr{F}=\{A\subseteq\Bbb N:A\text{ is finite or }\Bbb N\setminus A\text{ is finite}\}$, and define $\mu(A)=0$ if $A$ is finite and $\mu(A)=1$ if $\Bbb N\setminus A$ is finite.

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Keeping in mind that I'm reading chapter 3 of my first measure theoretic probability text, could you be a bit more specific? – rmh52 Sep 27 '12 at 3:38
For each $A\subseteq\Bbb R$ set $\mu(A)=0$ if there is some $x>0$ such that $A\subseteq[-x,x]$ and $\mu(A)=1$ otherwise. Check that this is finitely additive. Then set $A_n=[-n,n]$ for $n\in\Bbb N$, and consider whether $\mu$ is countably subadditive. – Brian M. Scott Sep 27 '12 at 3:41
I doesn't seem that such a measure would be a probability measure. If A is the interval from 1 to infinity and B from -1 to -infinity the measure of their union would be 1 while the sum of both would be 2... – rmh52 Sep 27 '12 at 3:45
@rmh52: Yes, it needs to be defined a little more carefully than I thought at first. Try the revised example instead; it avoids that problem. – Brian M. Scott Sep 27 '12 at 3:47
is script F supposed to be a sigma algebra? – esproff May 15 '13 at 4:02

Examples of measures which cannot be extended to be countably summable are: 1) a uniform distribution on the rational numbers in the interval $[0,1]$. This is easy to see: There are only countably many rationals, so any finite set of rationals will have measure zero (or the total probability will be infinite), yet all of them together have measure 1. This is inconsistent with countable additivity.

2) a uniform distribution on the integers, by the same argument, sinvce any finite interval of inyegers must have probability zero.

More interesting: On some set of increasing sequences of natural numbers, define the density to be the limit (if it does not exist, take the limsup) of $$\#\{n_k \colon n_k\le N\} / N$$ when $N$ goes to infinity. It can be shown that density defined in this way is onlyn a finitely additive measure, cannot be extended to countably additive.

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What do you mean by "a uniform distribution on the integers"? – Qiaochu Yuan Sep 27 '12 at 5:41
well, let us reformulate to "uniform on the natural numbers 0,1,2,3 ...".. Then, a subset such as the even numbers would have probability 1/2, $\{0,3,6,\dots \}$ would have probability 1/3, etc. for this distribution all finite subsets must have probability zero. – kjetil b halvorsen Sep 27 '12 at 5:45
You should be more careful about stating explicitly what sets you're allowing to be "measurable" here. – Qiaochu Yuan Sep 27 '12 at 6:30
Certainly, but I wrote a comment and not a thesis ... Anyhow, I don think measurability is such a problem with only finitely additive measures? – kjetil b halvorsen Sep 27 '12 at 6:57
@kjetilbhalvorsen Measurability is not a big problem if you are willing to extend a finitely additive measure using the Hahn-Banach theorem. But it prevents you from explicitely constructing a canonical finitely additive measure on a $\sigma$-algebra that is not countably additive. – Michael Greinecker Sep 27 '12 at 9:53