It is well known that if $X$ is a Polish space and $\mathcal{F} \subset \mathcal{M}\_+(X)$ (the set of finite positive Radon measures on $X$) is uniformly tight and bounded in mass, it is relatively compact w.r.t. to the weak topology, i.e. the coarsest topology on $\mathcal{M}\_+(X)$ w.r.t. the maps $\mu \mapsto \int\_X \varphi \text{ d} \mu$ are continuous for every $\varphi \in C\_b(X)$, the continuous bounded real functions on $X$.


I am intersted in a similar statement but in the general case of a Hausdorff topological space $X$.


A possible way to introduce a topology on $\mathcal{M}\_+(X)$, when $X$ is a Hausdorff topological space, is to consider the coarsest topology on $\mathcal{M}\_+(X)$ w.r.t. the maps $\mu \mapsto \int\_X \varphi \text{ d} \mu$ are lower semi continuous for every $\varphi \in LSC\_b(X)$, the lower semi continuous and bounded real functions on $X$.


In the book of Schwarz "Radon measures on arbitrary topological spcaes" it is proven that, in this topology, uniform tightness and boundedness in mass together again imply relative compactness.


I am wondering, how much can I enrich the topology on $\mathcal{M}\_+(X)$ and still have that Prokhorov theorem holds?


For example, if $X$ is a Hausdorff topological space and I endow $\mathcal{M}\_+(X)$ with the coarsest topology w.r.t the maps $\mu \mapsto \int\_X\varphi \text{ d}\mu$ are continuous for every $\varphi \in LSC\_b(X) \cup USC\_b(X)$, what happens?


Here $LSC\_b(X)$ (resp. $USC\_b(X)$) is the set of the lower (resp. upper) semi continuous bounded real functions on $X$. 



Prokhorov theorem on non Polish spaces
This seems to be a pure general topology problem. Enriching the topology will necessarily destroy the Prohorov property whenever the enrichment matters. If you have two nested Hausdorff topologies and a set is relatively compact in the finer topology, then the closure is the same under both topologies and the trace topologies coincide on the closure.


To see this, let $\tau$ and $\tau'$ be Hausdorff topologies on $X$ such that $\tau\subseteq\tau'$ and let $R\subseteq X$ be relatively compact in the topology $\tau'$. Trivially, $R$ is also relatively compact in the coarser topology $\tau$ and the $\tau'$-closure of $R$ is a subset of the $\tau$-closure of $R$. Let $x$ be a point in the $\tau$-closure of $R$ and $\langle x\_\alpha\rangle$ be a net in $R$ converging to $x$ under $\tau$. Since $R$ is relatively compact under $\tau'$, a subnet will $\tau'$ converge to a point $x'$. But this subnet will also converge to $x'$ under $\tau$. Since $\tau$ is Hausdorff , $x=x'$ and the closure of $R$ is the same under both topologies. Let $C$ be this compact closure. The identity is a continuous function from $(C,\tau')$ to $(C,\tau)$ and will therefore map compact subsets of $C$ to compact subsets of $C$. But these are exactly the closed subsets of $C$ under both topologies, so both topologies must coincide on $C$.


