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Questions about Hausdorff measures, their variants (such as spherical Hausdorff measures) and generalisations.
5
votes
If $\mathcal{H}^{n-1}(K)=0$ then $\mathcal{H}^n(K\times \mathbb{R})=0$
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 $ …
11
votes
Accepted
If $\mathcal{H}^{n-1}(E)=0$ then $\mathbb{R}^n\setminus E$ is connected
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 …
4
votes
Accepted
Is there an explicit, everywhere surjective $f:\mathbb{R}\to\mathbb{R}$ whose graph has zero...
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 $\ …