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 $Z:\Omega\rightarrow \mathbb{R}$.
Take the 2minute tour
×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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. 

