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Well known theorem of Ulam says, that each probability measure $\mu$ defined on Borel subsets of polish space $X$ satisfies the following condition: for each $\epsilon>0$ there is a compact subset $K$ of $X$, such that $\mu(K)>1-\epsilon$.

I wonder there are any reasonable condition on measure $\mu$ which would guarantee that for each $\epsilon>0$ there is an open set subset $U$ of $X$, such that $\mbox{cl}\,U$ is compact set and $\mu(\mbox{cl}\,U)>1-\epsilon$. Any idea? It would be very helpful for me.

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Think about Brownian Motion on ${\mathcal C}[0,1]$ that starts at 0. This is a probability measure on the polish space ${\mathcal C}[0,1]$. This is also a Banach space that is infinite dimensional. Such a space cannot have a precompact open subset -- normed linear spaces with this property are finite-dimensional. Therefore, I think you are out of luck here.

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    ok, but i rather think about a polish space that is not a vector space...2011-07-12
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    What conditions do you want to put on it?2011-07-12
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    I'm not sure, exactly i wonder if it at all possible to find such conditions, maybe something like a positivity on the open sets... There is a theorem for polish spaces, which says that for each $\sigma$-compact set $D$ there is a $\sigma$-compact set $C$ such that $D\subset C$ and Markov-Feller chain takes values a.c in the set $C$. The condition i asked about would allow me proof something like that with $C$ - locally compact. So in my case, the measure is a transition probability.2011-07-12
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    Local compactness will surely suffice, but I doubt that this is a condition you're willing to impose.2011-07-12