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This is a follow-up to this 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 $\sigma$-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). As was already seen in the original question, this set is not compact, but I am hoping that it is still $\sigma$-compact.

Many thanks in advance!

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It seems $A$ is not $\sigma$-compact. Check my argument as I have never worked with Wasserstein distances. Let $d=1$, let $q$ be the measure supported in $\{0\}$, so that $W_1(q,\mu)=\int|x|d\mu$ for all $\mu$, and take $K,R=2$, $M=1$.

$A$ is closed in $\mathcal{P}_1(\mathbb{R}^d)$, so it is a complete metric space. So by the Baire category theorem, it will be enough to prove that every compact subset of $A$ has a dense complement.

And for any compact $K\subseteq\mathcal{P}_1(\mathbb{R}^d)$, any $\mu\in A$ and for any $\varepsilon>0$, you can find a measure in $A\setminus K$ at distance $\leq10\varepsilon$ of $A$ by taking a big enough element of the sequence $\mu_n=\left(1-\varepsilon\right)\mu+\varepsilon\left(1-\frac{2}{n}\right)\delta_1+\frac{\varepsilon}{2n}\left(\delta_n+\delta_{-n+2}\right)$ (similar to this answer), where $\delta_n$ is the measure supported in $\{n\}$. Indeed, for any big $N$ we have $\lim_{n\to\infty}W_1(\mu_N,\mu_n)\geq\frac{\varepsilon}{2}$ (so the entire sequence $(\mu_n)_n$ cannot be contained in a compact set), and the measures $\mu_n$ have average $\int xd\mu_n=1$ (as they are an affine combination of measures with average $1$) and satisfy

$$\int|x|d\mu_n\leq(1-\varepsilon)2+\varepsilon+\frac{\varepsilon}{2n}(2n)\leq2.$$

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  • $\begingroup$ Thanks for the answer! Could you elaborate a bit on how exactly you use the Baire category theorem? I don't really have a background in topology unfortunately :\ And how does the second paragraph show that compact subsets of $A$ have dense complements? Sorry for being a bit slow... $\endgroup$
    – J.R.
    Commented Jun 7 at 12:04
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    $\begingroup$ A form of the Baire category theorem says that any complete metric space is a Baire space, that is, any countable intersection of dense open sets is nonempty, which implies that no countable union of compacts whose complements are dense can be the whole space. To prove that a compact subset $K$ of $A$ has dense complement, it is enough to show that for any point $\mu$ in $A$ there are points in $A\setminus K$ as close as you want to $\mu$. So it would be enough to check that no compact can contain infinitely many of the measures $\mu_n$, and that $\lim_{n\to\infty}W_1(\mu,\mu_n)=\varepsilon$. $\endgroup$
    – Saúl RM
    Commented Jun 7 at 12:36
  • $\begingroup$ I thought I understood, but now I'm not so sure. The proposed $\mu_n$ are no longer in $A$, since they do not have mean $M$ $\endgroup$
    – J.R.
    Commented Jun 7 at 13:39
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    $\begingroup$ I am applying the Baire category theorem to the topological space $A$, not to $\mathcal{P}_1(\mathbb{R}^d)$. So the interior of $A$ is the entire $A$. The only thing one needs to check to apply the Baire category theorem to the space $A$ is that it is closed in $\mathcal{P}_1(\mathbb{R}^d)$, so that is is a complete metric space $\endgroup$
    – Saúl RM
    Commented Jun 8 at 14:15
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    $\begingroup$ Okay thanks, I'll mull it over $\endgroup$
    – J.R.
    Commented Jun 8 at 14:17

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