Is the following set compact w.r.t. the Wasserstein distance?

Question

Fix a finite first moment probability measure $q\in\mathcal{P}_1(\mathbb R ^d)$, and real numbers $K,M,R$. Consider the following set:

$$A:=\left\{p\in\mathcal{P}_1(\mathbb R ^d): \int |x|dp\leq K, \int x dp=M, \mathcal{W}_1(p,q)\leq R \right\}$$

Is it a compact set in $(\mathcal{P}_1(\mathbb R ^d),\mathcal W _1)$? It is not hard to check that this set is closed and that it is tight (weakly) so to show compactness in the Wasserstein space we can equivalently show either of the following two:

1. $\lim_{R\to\infty} \sup_{p\in A}\int_{|x|>R}|x| dp(x)=0$
2. $\exists g:\mathbb R_+\to\mathbb R_+\text{ monotonically divergent s.t. } \sup_{p\in A}\int_{|x|>R}|x| g(|x|) dp(x)=0$

I haven't had much luck in showing this, so any help would be greatly beneficial!

I did manage to show that the limit in 1. is at most $R$, but that's not really that useful.

PS: I do not actually know that this set is compact, I am merely hoping that it is for my purposes.

Many thanks in advance!

Answer 1

Unfortunately not. Take $q = \delta_0$ and $p_n = (1-n^{-1})\delta_0 + n^{-1}\delta_n$. Then $p_n \in A$ (with say $K = M = R = 1$) and $p_n \to \delta_0$ weakly but not in $\mathcal{P}_1$...