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Here is a way to implement Christian Remling's comment. The construction is similar to the Cantor step function. Define a function $f:[0,1]\to[0,1]$ as follows. For $x\in[0,1]$ let $x=\sum_{n=2}^\inft …

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 …

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 …

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 …

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 …

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 …