Let $(\Omega,\Sigma)$ be a measurable space. Denote by $ba(\Sigma)$ the set of all bounded and finitely additive measures on $(\Omega,\Sigma)$ (see http://en.wikipedia.org/wiki/Ba_space for a definition). Is the set of all probability measures $\mathcal{M}_1(\Sigma)\subseteq ba(\Sigma)$ weak*closed? The weak*topology on $ba(\Sigma)$ is the weakest topology such that the maps $l_Z:ba(\Sigma)\rightarrow \mathbb{R}$, mapping $\mu\mapsto \int_\Omega Z d\mu$, are continuous for all bounded and measurable maps $Z:\Omega\rightarrow \mathbb{R}$.

1$\begingroup$ You really need to put these questions on math.stackexchange. $\endgroup$– Nik WeaverJun 13, 2012 at 14:26
1 Answer
No. The most elementary case is $\Omega=\mathbb N$ and $\Sigma$ the power set. A nonfixed ultrafilter is a good example of a finitelyadditive but not countablyadditive measure: each set has either measure $0$ or $1$. But such a thing is a limit of fixed ultrafilters, that is, countably additive zeroone measures. Indeed, it is the limit along that very ultrafilter.
Nik seems to think this is too elementary. But there is no reason this should be clear to a nonspecialist.

1$\begingroup$ It's like asking whether the unit ball of $l^1$ is weak* closed in its double dual. $\endgroup$ Jun 13, 2012 at 15:02

1$\begingroup$ @Nik: since it is weak* sequentially closed, even that is an interesting question. $\endgroup$ Jun 13, 2012 at 15:46

3$\begingroup$ Nik: yes, it is exactly like that in the sense that neither is clear to a nonspecialist. $\endgroup$ Jun 13, 2012 at 15:47

2$\begingroup$ Annoyingly, this is exactly the same as an answer I gave over at math.stackexchange: math.stackexchange.com/questions/157795/… I had already suggested to Peter (who may or may not also be Andy) that it was polite to point out when you are crossposting... $\endgroup$ Jun 13, 2012 at 19:25

2$\begingroup$ @Peter: your comment is inappropriate. This is a professional forum and we don't talk that way here. $\endgroup$ Jun 14, 2012 at 13:22