Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've read in a working paper (bottom of page 9) that the following is a "standard result":

Let $A$ be a compact metric space and $T$ be a Polish space. Let $\rho$ be a Borel probability measure on $T$. Let $\mathcal{M}^\rho(T\times A)$ be the set of Borel probability measures on $T\times A$ such that the marginal on $T$ is equal to $\rho$. Then $\mathcal{M}^\rho(T\times A)$ is a compact set in the narrow topology on the space of probability measures.

Could anyone tell me how to show this or give me a reference? The narrow topology is the same as the weak* topology or the topology of weak convergence of measures.

share|cite|improve this question
up vote 2 down vote accepted

As $T$ is separable and complete, by Ulam's theorem, for each $j\geq 1$, we can find a compact subset of $T$, say $K_j$, such that $\rho(K_j)\geq 1-j^{-1}$. If $P\in \mathcal M^{\rho}(T\times A)$, then $$P(K_j\times A)=\rho(K_j)\geq 1-j^{-1}.$$ As $K_j\times A$ is a compact subset of $T\times A$, the set $\mathcal M^{\rho}(T\times A)$ is uniformly tight. By of Prokhorov theorem, as $T\times A$ is separable and complete, $\mathcal M^{\rho}(T\times A)$ has a compact closure in the narrow topology. As this one is metrizable, to see that $\mathcal M^{\rho}(T\times A)$ is closed, we just have to check sequential closeness. Let $\{ P_n\}\subset \mathcal M^{\rho}(T\times A)$ converging in law to $P$. We just have to show that for each open set $O$ of $S$, $P(O\times A)=\rho(O)$. We have by portmanteau theorem, $$P(O\times A)\leq \liminf_n P_n(O\times A)=\rho(O).$$ As a closed set in a metric space is a countable intersection of open sets, we have $P(O^c\times A)\leq \rho(O^c)$, so $\rho(O)=P(O\times A)$ for all open set $O$.

Theorem.(Ulam) Let $(S,d)$ a complete separable metric space. Then each Borel probability measure on $S$ is tight.

Proof: Let $\{x_n\}$ a sequence dense in $S$, and $\varepsilon>0$. For each $j$, let $N_j$ integer such that $P\left(\bigcup_{l=1}^{N_j}B(x_l,j^{-1})\right)\geq 1-2^{-1}\varepsilon$. Then $K:=\bigcap_{j\geq 1}\overline{\bigcup_{l=1}^{N_j}B(x_l,j^{-1})}$ is pre-compact and closed, hence compact, and $P(K)\geq 1-\varepsilon$.

Theorem.(Prokhorov) Let $(S,d)$ a separable metric space, and $\mathcal P$ a family of Borel probability measures. If $\mathcal P$ is uniformly tight, that is, for all $\varepsilon>0$, we can find a compact subset $K$ of $S$ such that for each element $P$ of $\mathcal P$, $P(K)\geq 1-\varepsilon$, then $\mathcal P$ is relatively compact for the narrow topology.

share|cite|improve this answer
Thank you very much! – Michael Greinecker Nov 2 '12 at 12:13
You are welcome! – Davide Giraudo Nov 2 '12 at 12:14
In the first version, I forget to check that $\mathcal M^{\rho}$ is closed for the narrow topology. – Davide Giraudo Nov 2 '12 at 18:53
The Prokhorov theorem as stated is valid without the assumption that $S$ is separable. This assumption is needed for the converse statement, namely that a relatively compact set is tight. – Ahriman Nov 2 '12 at 19:40
@Ahriman Yes, the fact that $T\times A$ is Polish gives us that the narrow topology is metrizable (that what is needed here). And you are right, this direction of Prokhorov theorem doesn't need separability. – Davide Giraudo Nov 2 '12 at 19:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.