# Tightness on a set $A$ implies tightness on a set $B$ where $A\subset B$?

From the book Billingsley - Convergence of probability measures, 1999, we have the following definitions of tightness and relative compactness and the Prohorov's theorem:

Tightness: Let $$\Pi$$ be a family of probability measures on $$(S,\mathcal{F})$$. 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 $$\Pi$$. Relative compactness: We call $$\Pi$$ relatively compact if every sequence of elements of $$\Pi$$ contains a weakly convergent subsequence. Prohorov's theorem: 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.

Remark: in the literature on the evolution stochastic partial differential equations it is usually said that $$\Pi$$ is tight or relatively compact in $$S$$.

My question: If I show tightness of family $$\Pi$$ in some set $$A$$, and $$A\subset B$$ (or $$A\hookrightarrow B$$ or $$A\hookrightarrow_c B$$, where $$\hookrightarrow$$ and $$\hookrightarrow_c$$ represent continuous and compact embeddings), is it true then that family $$\Pi$$ is tight in $$B$$ also?

In the concrete case: I am working with family $$\Pi$$ that is a family of probability laws of solutions of some stochastic partial differential problem. And I am mostly interested in the next three cases:

1. $$A=C([0,T];H^r(\mathbb{R}))$$, $$B=L^2(0,T;L^2(\mathbb{R}))$$, where $$H^r$$ is the Sobolev space of $$L^2$$-type and $$r\geq2$$ is integer.

2. $$A=C([0,T];H^r([-a,a]))$$, $$B=L^2(0,T;L^2(\mathbb{R}))$$, where $$[-a,a] \subset \mathbb{R}$$ is a bounded interval.

3. $$A=C([0,T];H^r([-a,a]))$$, $$B=C([0,T];H^r(\mathbb{R}))$$.

I think that in the three cases mentioned the answer should be yes.

My reasoning is: If we talk about the relative compactness (for the type of spaces I mentioned above) and if $$\Pi$$ is relatively compact in $$A$$ and $$A\subset B$$, then $$\Pi$$ should be relatively compact in $$B$$ also. If that is true, the same conclusion is valid in the cases $$A\hookrightarrow B$$ or $$A\hookrightarrow_c B$$. By the Prohorov's theorem tightness is equivalent to relatively compactness. So if the family $$\Pi$$ is tight in $$A$$ it should be tight in $$B$$ if I am not mistaken?

If this is correct, it should save me some time in the problem I am working on (I won't need to do tightness then). Help with this would be great and I am $$99\%$$ sure that the answer is yes. But that $$1\%$$ chance is bugging me. Thanks in advance.

• Since you mention continuous/compact embedding, I think you assume that $S$ is a separable (?) Banach space. Is this correct. It would help if you are a little bit more specific about your assumptions. Jan 13, 2021 at 14:24
• @DieterKadelka, I assume that space $S$ could be separable. But to be honest I didn't used it. I work on some problem that I approximated with the spde problem. I showed tightness of the family of probability laws of it's solutions in the space $A=C([0,T];H^r(\mathbb{R}))$. I noticed that it would be useful if I show it in the space bigger than $A$. In the concrete case I am interested in the cases $A \hookrightarrow B$, and the cases $A\subset B$, $A\hookrightarrow$ looked interesting also. I hope that this doesn't sound even more confusing. Thanks for your comment.
– Mark
Jan 13, 2021 at 19:49

$$\newcommand\F{\mathcal F}\newcommand\G{\mathcal G}\newcommand\ep{\epsilon}\newcommand\si{\sigma}$$The notion of "tightness of family $$\Pi$$ in some set $$A$$" is in general undefined. Instead, you can define it as follows:

Let $$\Pi$$ be a set of probability measures over $$(S,\F)$$, where $$S$$ is a topological space and $$\F$$ is a $$\si$$-algebra over $$S$$ containing all the compact subsets of $$S$$. The family $$\Pi$$ is called tight if for every real $$\ep>0$$ there is a compact set $$K\subseteq S$$ such that $$P(K)>1-\ep$$ for every $$P\in\Pi$$.

Now suppose we have measurable spaces $$(S,\F)$$ and $$(T,\G)$$, where $$S$$ and $$T$$ are topological spaces, $$\F$$ is a $$\si$$-algebra over $$S$$ containing all the compact subsets of $$S$$, and $$\G$$ is a $$\si$$-algebra over $$T$$ containing all the compact subsets of $$T$$. Let $$\Pi$$ be a tight set of probability measures over $$(S,\F)$$. Let $$f\colon S\to T$$ be a continuous measurable function. Let $$\Pi f^{-1}:=\{Pf^{-1}\colon P\in\Pi\}$$, the set of of all pushforward measures $$Pf^{-1}$$ for $$P\in\Pi$$ (so that $$(Pf^{-1})(B)=P(f^{-1}(B))$$ for all $$B\in\G$$). Then the set $$\Pi f^{-1}$$ is tight.

This follows immediately because, if $$K\subset S$$ is a compact, then $$f(K)$$ is also compact, since $$f$$ is continuous.

In each of your three concrete cases, the natural embedding of $$A$$ into $$B$$ is continuous and therefore preserves the tightness.

• Thank you for the answer. Your explanation is pretty good. This real makes my job easier because I don't need to show tightness again. I have showed tightness for a family of probability laws of the solutions of some system of spde in the space $A=C([0,T];H^r(\mathbb{R}))$. It is good to know that I don't need to do that again for the space $B=L^2(0,T;L^2(\mathbb{R}))$. Also if I am not mistaken space $B$ could also be for example $B=L^{\infty}(0,T;L^2(\mathbb{R}))$ or $B=L^2(0,T;L^{\infty}(\mathbb{R}))$ and similar because the emedding stays continuous?
– Mark
Jan 13, 2021 at 19:51
• @Mark : Yes, as long as you the continuity, the tightness is preserved. Jan 13, 2021 at 20:21
• Thank you for the follow up. This helps me a lot.
– Mark
Jan 13, 2021 at 20:22