# Is the separability of the space needed in the proof of the Prohorov's theorem?

The Section 5 of the book:

Billingsley, P., Convergence of Probability Measures, 1999,

studies Prohorov's theorem. A short reminder is given below.

Let $$\Pi$$ be a family of probability measures on $$(S,\mathcal{F})$$. We call $$\Pi$$ relatively compact if every sequence of elements of $$\Pi$$ contains a weakly convergent subsequence. The family $$\Pi$$ is tight if for every $$\epsilon$$ there is a compact set $$K$$ such that $$P(K)>1-\epsilon$$ for every $$P$$ in $$\Pi$$.

The direct half of the Prohorov's theorem is given in the Theorem 5.1: If $$\Pi$$ is tight, then it is relatively compact.

The converse half of Prohorov's theorem is given in the Theorem 5.2: Supose that $$S$$ is separable and complete. If $$\Pi$$ is relatively compact, then it is tight.

My question: In the proof of the Theorem 5.2 (i.e. relatively compact $$\Rightarrow$$ tight), we use separability and completness of the space $$S$$. On the other hand, in the proof of the Theorem 5.1 (i.e. tight $$\Rightarrow$$ relatively compact), I know that we do not need completness of $$S$$, but I do not know if we do need separability. I didn't find the place where separability is used in the proof of the Theorem 5.1. So my question is do I need or not the separability of the space $$S$$ in the direct part of the Prohorov's theorem?

Remarks:

• I know the proofs of the same theorem that use separability (e.g. Note).
• Prohorov's theorem in most books is given as one theorem on Polish spaces, so they assume separability in both halfs. It goes like this usually: Let $$S$$ be a Polish space and $$\Pi$$ a collection of probability measures on $$S$$. Than $$\Pi$$ is tight if and only if it is relatively compact.

The reason I am asking is that I would like to use the direct half of Prohorov's theorem on the problem I am currently working. Space $$S$$ in my case is complete but not separable.

Help with this would be great and needed. Thanks in advance.

• Note that if $\Pi = \{\mu\}$ then $\Pi$ is compact but without additional assumptions $\mu$ is not tight! – Dieter Kadelka Feb 12 '20 at 17:06
• @DieterKadelka, thank you for the comment. This is true, and this would be useful to remember that in the Theorem 5.2 (relative compact $\Rightarrow$ tight). This explains why do we need separability and completness there (if I am not mistaken). – Mark Feb 13 '20 at 9:52

You are correct that separability is not needed. However, there is also not really any loss of generality in assuming it. For suppose that $$\Pi$$ is tight. Then for every $$n$$ there exists a compact set $$K_n$$ such that $$\mu(K_n) > 1-\frac{1}{n}$$ for all $$\mu \in \Pi$$. So if we set $$S_0 = \bigcup_{n=1}^\infty K_n$$, then $$S_0$$ is separable and $$\mu(S_0) = 1$$ for all $$\mu \in \Pi$$. We can now view $$\Pi$$ as a set of probability measures on $$S_0$$, and it is still tight (since the $$K_n$$ are also compact in $$S_0$$). The separable case of the theorem then implies that $$\Pi$$ is weakly relatively compact in $$\mathcal{P}(S_0)$$, i.e. every sequence in $$\Pi$$ has a subsequence converging weakly in $$\mathcal{P}(S_0)$$, and you can easily check that such a subsequence also converges weakly in $$\mathcal{P}(S)$$. So $$\Pi$$ is weakly relatively compact in $$\mathcal{P}(S)$$, as desired.

In other words, once you have a tight family, then all those measures live on a separable subset of $$S$$ anyway, so the rest of the space is irrelevant and might as well not be there.

• Nate Eldredge, thank you for the answer. The last sentence explains it all. So in the problem I am working currently $S$ is $C([0,T];\mathcal{M})$ or $L^{p}(0,T;\mathcal{M})$ or something similar. Here $\mathcal{M}$ represents space of Radon measures or some subspace of it or some other similar space of measures, and I use weak convergence of measures. So if I am not mistaken, I could apply the direct half of the Prohorov's theorem (i.e. Theorem 5.1) in those cases? – Mark Feb 13 '20 at 9:59
• @Mark: Hm, could you explain more precisely what spaces you have in mind? If $\mathcal{M}$ is really all the Radon measures on some space, with the weak topology, then that is not metrizable, and so neither is $C([0,T],\mathcal{M})$. On the other hand, if $\mathcal{M}$ is just the probability measures, then it is Polish and so is $C([0,T];\mathcal{M})$; in particular it's separable and the issue doesn't arise. And in either case I don't know how to define $L^p([0,T];\mathcal{M})$ because the weak topology on $\mathcal{M}$ isn't a normed space. – Nate Eldredge Feb 13 '20 at 13:43
• Nate Eldredge, I work on a problem where I have one process that is in $L^{\infty}\cap BV$ and the other process that is in $C([0,T];H^k), k>2, k-integer$. I should compare them on the space that contains both of them. And later I should apply Prokorov's theorem on that new space. I got the recommendation of a one of my coworker to try to use some spaces such as $C([0,T];\mathcal{M})$ where $\mathcal{M}$ is some subspace of the space of Radon measures, $C([0,T];\mathcal{D^{'}})$ or $L^p([0,T];\mathcal{M})$. – Mark Feb 13 '20 at 14:49
• Also it is important to note, that the two processes I mentioned above are stochastic. If $\mathcal{M}$ is the space of probability measures, this maybe could work on the probability laws of the two processes I mentioned but not on the processes itself. In this case I could use that $\mathcal{M}$ is the space of the probability measures with the finite first moment and than use with it Wasserstein metric $W_1$. Also as I recall the space of distributions $\mathcal{D}^{'}$ isn't metrizable too, so that the space $C([0,T];\mathcal{D}^{'})$ shouldn't work. – Mark Feb 13 '20 at 14:56
• With regard to your question about Prohorov spaces, quite a lot of work was done on this in the 70’s—I’m not sure about more recent work. It seems to have more to do with completeness properties than with separability. Topologically complete spaces have this property but there are separable metrisable which fail. The references that I know are D. Preiss (Z. Wahrscheinlichkeitstheorie 34 (1973) 109-126) and F. Topsøe (Math. Scand. 34 (1974) 187-210). There is also material in the two volume “Measure Theory” by V. Bogachev. – user131781 Feb 15 '20 at 11:20

Separability is not necessary. In fact, tightness of a family of Borel probability measures implies relative compactness in the vague/weak-* topology on any completely regular space. For instance, this can be found in volume 4 of Fremlin's Measure Theory. Specifically Proposition 437U (b) shows that tight families are compact in the narrow topology, and 437K (c) shows that for completely regular spaces, the narrow topology agrees with the weak-* topology.

My original answer below answers the wrong question - the question is about whether tight implies relatively compact, rather than the other way.

Let $$\kappa$$ be a real-valued measurable cardinal, and $$\mu : \mathcal{P}(\kappa) \rightarrow [0,1]$$ a probability measure vanishing on singletons. Consider $$\kappa$$ to be a discrete metric space. Then the 1-element family $$\{\mu\}$$ is compact, because it is a singleton, but it is not tight because all compact subsets of $$\kappa$$ are finite sets, so have measure zero.

You say that completeness is not necessary, but (unless you are making an extra assumption) it is, for essentially the same reason - there are separable metric spaces with Borel measures on them that are not tight.

• @NateEldredge You are quite right - how silly of me! – Robert Furber Feb 12 '20 at 18:14
• Robert Furber, thank you for the answer. The spaces $S$ that I have in my problem are for example $C([0,T];\mathcal{M})$ or $L^{p}(0,T;\mathcal{M})$ where $\mathcal{M}$ represents space of Radon measures or some subset of it, and I use weak convergence. If I understood correctly I could apply Theorem 5.1 on it? – Mark Feb 13 '20 at 9:49
• This is just an addition to the answers below but might be of interest. There is a natural lc topology on $C^b(S)$, the strict topology (R.C. Buck), for which the dual is the space of bounded (signed) tight meaures and the the uniformly tight sets are just the equicontinuous ones. Hence they are automatically relatively weakly compact. The converse implication is much more delicate but it is true for paracompact, locally compact spaces. Of course, there are stronger results for certain non-locally compact spaces if one confines attention to probability measures. – user131781 Feb 13 '20 at 15:59
• @RobertFurber Your answer is for a general completely regular space, so you have to be careful about the definition of a relatively compact set of probability measures. The definition given by OP is equivalent to the usual one (used by Fremlin) when $S$ is metrizable but not in general. – user95282 Feb 13 '20 at 20:44
• @user131781, thanks for the interest in this. Could you give an examples for your last sentence (if one confines attention to probability measures)? – Mark Feb 14 '20 at 12:21