# Equivalence of sigma algebras on the set of probability measures.

I was wondering if somebody could help me see/provide a reference to the following fact: Let $X$ be a metrizable set, $\mathcal{F}$ the corresponding Borel sigma-algebra on $X$, and $\triangle\left(X,\,\mathcal{F}\right)$ be the corresponding set of probability measures.

Suppose we endow $\triangle\left(X,\,\mathcal{F}\right)$ itself with the sigma-algebra generated by sets of the form

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ A\left(E,\, p\right)=\left\{ {\mu\in\triangle\left(X,\,\mathcal{F}\right)\,|\,\mu\left(E\right)\geq p}\right\} ,\,\,\,\,\,\,\,\,\,E\in F,\,\,\ p\in[0,\,1]$

Then the above sigma-algebra coincides with the Borel sigma-algebra generated by the weak * topology.

• Parts of this should be straightforward. If you can make any progress at all, describing it would help focus attention on what's nontrivial here. – Kevin Carlson Jan 27 '15 at 17:59
• @KevinCarlson I know that if the sets $E$ are taken to be closed, then the corresponding set of measures is closed, and these sets would in fact be included in the sigma-algebra generated by the weak * topology. I'm trying to use inner regularity to show this for arbitrary measurable $E$ but can't seem to get it quite right. In any case, that would only prove one inclusion, the other way seems even trickier... – Mark Jan 27 '15 at 18:18
• I've made some corrections to the notation to make it clearer. Thanks a lot. – Mark Jan 28 '15 at 3:12
• @KevinCarlson So, it can be shown that for each $E\in F$, the function $\mu\mapsto\mu\left(E\right)$ is measurable. (Aliprantis 1999) This immediately implies that the sigma algebra generated by the $A\left(E,\, p\right)$ is a subset of the Borel sigma-algebra. For the converse, it can be shown that basis elements in the weak topology can be described as sets of the form $A\left(E,\, p\right)$, but then the result doesn't follow because sigma-algebras are only closed under countable unions. – Mark Jan 29 '15 at 2:12