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 $Z:\Omega\rightarrow \mathbb{R}$.
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1$\begingroup$ You really need to put these questions on math.stackexchange. $\endgroup$– Nik WeaverJun 13, 2012 at 14:26
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No. The most elementary case is $\Omega=\mathbb N$ and $\Sigma$ the power set. A non-fixed ultrafilter is a good example of a finitely-additive but not countably-additive measure: each set has either measure $0$ or $1$. But such a thing is a limit of fixed ultrafilters, that is, countably additive zero-one 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 non-specialist.
answered Jun 13, 2012 at 14:49
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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
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1$\begingroup$ @Nik: since it is weak* sequentially closed, even that is an interesting question. $\endgroup$ Jun 13, 2012 at 15:46
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3$\begingroup$ Nik: yes, it is exactly like that in the sense that neither is clear to a non-specialist. $\endgroup$ Jun 13, 2012 at 15:47
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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 cross-posting... $\endgroup$ Jun 13, 2012 at 19:25
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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