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 $X$.

  • $\begingroup$ 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? $\endgroup$ – Dieter Kadelka May 4 '20 at 8:39
  • $\begingroup$ 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! $\endgroup$ – Bremen000 May 4 '20 at 8:48
  • $\begingroup$ 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. $\endgroup$ – Dieter Kadelka May 4 '20 at 9:05
  • $\begingroup$ I mean for functions that are upper or lower semi continuous. Edited again, now it should be clear! $\endgroup$ – Bremen000 May 4 '20 at 9:59
  • $\begingroup$ 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. $\endgroup$ – Nate Eldredge May 4 '20 at 12:29

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 $C$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.