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 $Z:\Omega\rightarrow \mathbb{R}$.



Is the set of all probability measures weak* closed?
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.


