# Prokhorov theorem on non Polish spaces

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 $$X$$.

• I have problems with your text. What do you mean by "continuous on varying of $\phi$? The last sentence is dubious. Why should $\mu \o \int \phi d\mu$ be continuous for semicontinuous $\phi$? Do you know the notes on p. 454 and 455 in Bogachev, Measure Theory 2, Springer 2007? – Dieter Kadelka May 4 '20 at 8:39
• I have edited the text, I hope it is clear now. I am just considering different initial topologies on $\mathcal{M}_+(X)$. I've just read those notes: I think I will surely find something useful in the papers suggested there! – Bremen000 May 4 '20 at 8:48
• Much clearer now! But I still have a problem with your last sentence. Being both u.s.c. and l.s.c. implies continuity of a function. – Dieter Kadelka May 4 '20 at 9:05
• I mean for functions that are upper or lower semi continuous. Edited again, now it should be clear! – Bremen000 May 4 '20 at 9:59
• Your topology at the end is finer than the weak topology, and may be much much finer. For instance, take $X = [0,1]$; since the functions $1_{\{x\}}$ are USC, I think the set of Dirac masses is discrete in this topology, and certainly not relatively compact even though uniformly tight. – Nate Eldredge May 4 '20 at 12:29

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 $$C$$.