Topological necessary and sufficient condition for tightness - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T01:33:25Z http://mathoverflow.net/feeds/question/99497 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99497/topological-necessary-and-sufficient-condition-for-tightness Topological necessary and sufficient condition for tightness Davide Giraudo 2012-06-13T20:18:53Z 2012-07-14T00:32:29Z <p>Recall the definition of tightness for a probability measure $\mathbb P$ on the Borel $\sigma$-algebra of a metric space $(S,d)$:</p> <blockquote> <p>For each $\varepsilon>0$, we can find a compact subset $K$ of $X$ such that $\mathbb P(K)\geq 1-\varepsilon$. </p> </blockquote> <p>The question is: is there a "nice" topological characterization of metric spaces such that each Borel probability measure is tight?</p> <p>In Billingsley's book <em>Convergence of probability measures</em>, 1968, it's said that it's an open problem. I wish know whether some progress have been done so far.</p> <p>Call EPT a metric space on which each Borel probability measure is tight. Some remarks:</p> <ul> <li>By Ulam's theorem, each separable metric space topologically complete (Polish is a shorter term) is EPT. </li> <li>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. </li> <li>If we assume the metric space separable, we have the answer from Dudley's book <em>Real Analysis and Probability</em>: 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 $\widehat S$).</li> </ul> http://mathoverflow.net/questions/99497/topological-necessary-and-sufficient-condition-for-tightness/102194#102194 Answer by heinrichvw for Topological necessary and sufficient condition for tightness heinrichvw 2012-07-14T00:32:29Z 2012-07-14T00:32:29Z <p>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. </p>