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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 …

We say a subset $E$ of $\mathbb{N}$ is a set of returns if there is some measure preserving system $(X,\mathcal{B},\mu,T)$ and some $A\in\mathcal{B}$ with $\mu(A)>0$ such that $E=\{n\in\mathbb{N};\mu( …

Here is a way to do it without the axiom of choice, but it isn't a nice formula either. Consider a Cantor set $C\subseteq[0,1]$ with Hausdorff dimension $0$. Now consider a countable disjoint union $\ …

This is a follow-up to this question by Dominic van der Zypen. For each bijection $f:\mathbb{N}\to\mathbb{N}$, let $$\operatorname{rc}(f) := \liminf_{N\to\infty} \frac{\left|\left\{(m,n)\in\{1,\dots,N …

$\newcommand\N{\mathbb N}$No, it is not possible to have $\mu_{[\N]^2}\big({\operatorname{rev}(f)}\big) = 1$. Given the function $f$, we will say that $n\in\mathbb{N}$ is good if $f(n)<f(k)$ for more …

One has $\max\big\{\mu\big(\exp(\varphi)\big) :\varphi \in S_{\mathbb{N}}\big\}=1$, as mentioned by Emil Jeřábek in the comments. We can also create a function $\varphi$ with $\mu\big(\text{shr}(\varp …

My comment reposted as an answer: If the continuum hypothesis holds, then we can give a well order $\prec$ to $\mathbb{R}$ isomorphic to the first uncountable ordinal. And then for each $j\in[-1,1]$ w …

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 …

Perhaps this is not the strongest result one can get, but it is true that, if $M$ is a complete Riemannian manifold, then for almost all $p^*\in M$ the set $F_{p^*}$ you define in the question has mea …

More generally, for any closed subset $S$ of a complete manifold $M$, the set of points $x$ at whose minimal distance to $S$ is attained at more than point has measure $0$. Indeed, consider the distan …

Yes, the almost partition exists. Instead of letting $\mu(E)\geq\frac{\mu(\Omega)}{2}$, I let $\mu(E)\in(0,\mu(\Omega))$ be arbitrary and proved that you can divide $\Omega$ into $N$ open simply conne …

Frostman's lemma seems to work for this problem. Suppose that $H^n(K\times\mathbb{R})>0$. Then $H^n(K\times[0,1])>0$, so there is a measure $\mu$ in $\mathbb{R}^{n+1}$ with $\mu(K\times[0,1])>0$ and $ …

Yes, $\mathbb{R}^n\setminus E$ has to be path-connected. Let $x,y\in\mathbb{R}^n\setminus E$, we will prove that there are many paths going from $x$ to $y$ inside $\mathbb{R}^n\setminus E$. We can sup …

Here is a short proof supposing that $M$ is complete (if not the statement is false, see the last paragraph) using the idea from Leo Moos' answer of using Rademacher's theorem. Suppose $\mathcal{B}(p, …

Let $p,q\in M$, $p\neq q$, where $M$ is a Riemannian manifold. We will let the bisector of $p,q$ be $\mathcal{B}(p,q)=\{x\in M;d(p,x)=d(q,x)\}$. Does $\mathcal{B}(p,q)$ have measure $0$? I was thinkin …