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

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

• You really need to put these questions on math.stackexchange. – Nik Weaver Jun 13 '12 at 14:26

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.
• It's like asking whether the unit ball of $l^1$ is weak* closed in its double dual. – Nik Weaver Jun 13 '12 at 15:02