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

Question

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.

Answer 1

$\newcommand\F{\mathcal F}\newcommand\G{\mathcal G}\newcommand\ep{\epsilon}\newcommand\si{\sigma}$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.