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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 …

This question is motivated by section 15.1 (Codes) of Alon and Spencer's The probabilistic method. Fix $\alpha<\frac{1}{2}$ and for each $n\in\mathbb{N}$ let $\{0,1\}^n$ be the length $n$ binary strin …

The statement is false in general, I added a counterexample at the end of my answer to show that some separability condition like countable separability (or the stronger condition of being Lebesgue fr …

I found an answer to Question 1 little after asking it, but as it was part of the material I was planning to upload to arXiv (and since the proof is long so I prefer to just cite it), I decided to wai …

Let $G$ be a countable amenable group. We consider sequences $(z_g)_{g\in G}$ of complex numbers with $|z_g|=1$ for all $g\in G$. I will say $(z_g)_{g\in G}$ is background noise for a (left-)Følner se …

Let $(X,\mathcal{B},\mu)$ be an arbitrary finitely additive probability measure space, let $a>0$ and let $(A_i)_{i\in I}$ be an uncountable family of subsets with measure $\geq a$. Is there an uncount …

This is false for any fixed $\theta\in(\frac{1}{2},1)$, we will use the balls of radius $1$ in $\mathbb{R}^d$ as a counterexample. Given $\theta$, if we want $ \theta K \subseteq \operatorname{conv}\{ …

Those constants don't exist for any $d\geq4$, here is an idea of why. For each $\varepsilon>0$ let $A_\varepsilon=\{(x_1,\dots,x_d)\in[-1,1]^d;\lvert (d-1)x_d-\sum_{i=1}^{d-1} x_i\rvert\leq\varepsilon …

As you say, it is not difficult to prove that $g=0$ when there are finitely many fluctuations in sign. More precisely, suppose we have points $0=a_0,a_1,\dots,a_n=M$ such that $g\geq0$ or $g\leq0$ in …