TITLE: Topological necessary and sufficient condition for tightness
QUESTION [8 upvotes]: Recall the definition of tightness for a probability measure $\mathbb P$ on the Borel $\sigma$-algebra of a metric space $(S,d)$:

For each $\varepsilon>0$, we can find a compact subset $K$ of $X$ such that $\mathbb P(K)\geq 1-\varepsilon$. 

The question is: is there a "nice" topological characterization of metric spaces such that each Borel probability measure is tight?
In Billingsley's book Convergence of probability measures, 1968, it's said that it's an open problem. I wish know whether some progress have been done so far.
Call EPT a metric space on which each Borel probability measure is tight. 
Some remarks:

By Ulam's theorem, each separable metric space topologically complete (Polish is a shorter term) is EPT. 
A necessary and sufficient condition that each probability has a separable support is that each subset of $D$ of $S$ which is discrete have a non-measurable cardinal (i.e. we can't find a probability measure $\mu$ on $2^D$ such that $\mu(\{x\})=0$ for each $x\in D$). Hence for each EPT space, each discrete subset have a non-measurable cardinal.  
If we assume the metric space separable, we have the answer from Dudley's book Real Analysis and Probability: each probability measure on $S$ is tight if and only if $S$ is universally measurable (that is, if $\widehat S$ is the metric completion of $S$, then $S$ is $\mathbb P$-measurable for each probability measure $\mathbb P$ on $\widehat S$).

REPLY [6 votes]: You run into questions of set theory: If $X$ has the discrete topology (which is metric) 
then your condition means: Every probability measure on the power set $2^X$ is discrete. 
This is related to the 'measurability' of the cardinal $card(X)$. For general metric spaces
the maximal cardinality of discrete subsets is the key.