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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 $\widehat S$).
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I asked the question at math.stackexchange (math.stackexchange.com/questions/157093/topological-n) but I didn't receive answer). –  Davide Giraudo Jun 13 '12 at 20:19
    
What is "topologically complete", did you want to say "complete"? –  Anton Petrunin Jun 13 '12 at 20:45
    
By topological complete, I mean homeomorphic to a complete metric space (then we apply Ulam's theorem to this metric space). –  Davide Giraudo Jun 13 '12 at 20:47
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Instead of "separable metric space topologically complete", you can just say "Polish space". –  George Lowther Jun 13 '12 at 22:49
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A completely regular space is called "Strongly measure compact" ir.library.osaka-u.ac.jp/metadb/up/LIBOJMK01/ojm15_01_14.pdf ... by definition iff every Baire measure is tight, but of course in metric space Baire=Borel. So also search papers with this phrase. –  Gerald Edgar Jul 14 '12 at 3:08

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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.

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