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The answer is yes under the additional assumption that the set is compact and I do not know what happens in the general case. The result is a consequence of the following one, see [1] and references …

Theorem 1. The Hausdorff dimension of the graph $\Gamma_f$ of $f$ equals $1$. Proof. Take a partition of $[0,1]$ by intervals of length $1/n$. Since the function is increasing you can cover the g …

In fact one can say quite a lot of regularity of the boundary of a convex set. Assume that $X\subset\mathbb{R}^n$ is a bounded convex set with non-empty interior. Convex functions are locally Lipschi …

Unfortunately, you can always find a Lipschitz map $$ f:\mathbb{R}^m\to\mathbb{R}^{m-k} \quad \text{and} \quad y\in\mathbb{R}^{m-k} $$ such that $\partial f^{-1}(y)$ has positive $m$-dimensional measu …

If $s>1$, then clearly $H^s(B_x)=0$ so there is nothing to do. If $s=1$, $H^1$ is just the Lebesgue measure so measurability follows. If $0<s<1$ the situation is a way more complicated, but the answer …

Answer to Q2 is yes. If $\mathcal{H}^1(C_g)<\infty$, then the graph has finite length (it is known that for a one-to-one curve $\mathcal{H}^1$ coincides with the length). However that implies that $g$ …

The answer is no. Any two Riemannian metrics when restricted to a compact set are bi-Lipschitz equivalent and bi-LIpschitz homeomorphism preserves the Hausdorff dimension.

This is true and it follows from the following result: Lemma. Suppset that $f:W\to\mathbb{R}$ is convex and $L$-Lipschitz, where $W\subset\mathbb{R}^n$ is convex. Then, $$ \tilde{f}(x)=\inf_{z\in W}\ …

Let $s>1$ be aby number. The unit interval $[0,1]$ with the metric $d(x,y)=|x-y|^{1/s}$ has poisitive and finite $s$-dimensional Hausdorff measure. While, it is an abstract metric space, it can be emb …