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This is a crosspost from MSE. Can we find for any given $\varepsilon>0$ an open subset $A\subseteq[0,1]^2$ with measure $>\frac{1}{100}$ such that, for any smooth curve $\gamma:[0,1]\to\mathbb{R}^2$ o …

Firstly, a set $P$ of positive measure need not contain anything of the form $A\times B$, for example consider for some $k\in\mathbb{R}\setminus\{0\}$ the set $P=\{(x,y)\in\mathbb{R}^2;x-ky\not\in\mat …

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, 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 …

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( …

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 …

Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function such that the set $A:=\{x\in\mathbb{R};f'(x)\not\in\{1,-1\}\}$ has measure $0$. Does this imply that $f'$ is constant? Context: I was think …

For any smooth Riemannian manifold $(M,g)$ there is a countable disjoint union of balls with complement of measure $0$. Let $\mu$ be Riemannian measure and for each $p\in M$ let $B_p$ be a small preco …

Given a compactly supported probability measure $m$ on $\mathbb{R}^n$, we can define its average distance to a point $x$ as $\int_\mathbb{R^n}d(x,y)dm(y)$. In this question I found that for a given me …

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 …

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 …

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 …

$\newcommand\N{\mathbb N}$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 …

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 …