# Space of probability measures “complete”? (In the other sense)

I want to consider a space of probability measures on some set $X$, such that the space of measures is complete, not in the sense of complete probability measures (though probably that too), but as in the standard analysis meaning. I haven't studied far enough in probability measures, so I can't tell -- what conditions do I need on my space such that it is complete? And how should I metrize it?

Thanks!

Edit: A reference/citation to a text would also be super helpful!

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For any measurable space $(X,\mathscr B)$ let's denote by $\mathscr M$ the linear space of all bounded (signed) measures. This is known to be a Banach space w.r.t. the total variation norm $$\|\mu\| = \sup_{B\in \mathscr B}(|\mu(B)|+|\mu(B^c)|).$$ so that this space is complete. Define $f:\mathscr M\to\Bbb R$ as $f(\mu) = \mu(X)$. This map is clearly continuous: $$|f(\mu) - f(\nu)|\leq \|\mu - \nu\|$$ and thus $\mathscr M_1:=\{\mu:f(\mu) = 1\}$ is a closed subspace of $\mathscr M$. The space of all non-negative measures $\mathscr M_+:=\{\mu:\mu\geq 0\}$ can be characterized as $$\mathscr M_+ = \{\mu:\|\mu\| - f(\mu) = 0\}$$ and thus is a closed subspace of $\mathscr M$ as well. The space of all probability measures $\mathscr P = \mathscr M_1\cap\mathscr M_+$ is thus a closed subspace of a complete space, hence complete itself.

Besides of the total variation distance which can be introduced regardless the structure of the underlying measurable space, there are other sorts of metric spaces of measures. The relation of completeness for them are more involved.

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Great, thanks, that's very helpful! Do you happen to know of a good text/reference with a statement that I can use as a citation as well? –  usul Jan 31 '13 at 19:20
@usul: of the fact that $\mathscr P$ is complete under the total variation distance? –  Ilya Jan 31 '13 at 19:22
yeah unless this is considered obvious/trivial, I guess. Or a chapter where this is developed if you know of one offhand, but it's not a big deal (your explanation is very clear). –  usul Feb 1 '13 at 0:15
@usul unfortunately, I don't –  Ilya Feb 5 '13 at 16:35

This is related to Prokhorov's theorem and tightness. If you have a seperable metric space $(X,d)$ that is complete, then $(\mathcal{P},d_P)$ is complete, where $\mathcal{P}(X)$ is the set of all Borel probability measures and $d_P$ is the Prokhorov metric.

You can find a good detailed discussion of this here., section 9 pg 26.

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Great, thanks, and thanks for the useful links! –  usul Jan 31 '13 at 19:20