2
$\begingroup$

I have a doubt on spherical Hausdorff measure. Given $k, \delta \in (0,\infty)$, the $\delta$-Hausdorff premeasure is defined for $E\subset \mathbb R^n$ as: $$\mathcal H^k_\delta(E):=\inf\{\sum_j\alpha_k \frac{\text{ diam}E_j^k}{2^k}: E\subset \bigcup E_j, \text{ diam}E_j\leq \delta\},$$ where $\alpha_k=\frac{2 \pi^\frac{k}{2}}{k\Gamma(\frac{m}{2})}.$

The spherical Hausdorff $\delta$-premeasure $\mathcal S^k_\delta$ differs from the last one in the fact that coverings made only by balls are allowed.

With this definition it is clear that $\mathcal H^k_\delta \leq \mathcal S^k_\delta$. But it is also true that $\mathcal S^k_\delta\leq 2^k \mathcal H^k_\delta.$ My question is: why is this true?

I've though that, for any $\epsilon>0$, there exists a $\delta$-covering $\{E_j\}$ of $E$ such that $$\sum_j \alpha _k\frac{\text{ diam}E_j^k}{2^k}\leq\sum_j \alpha^k \text{ diam}E_j^k\leq 2^k\mathcal H^k_\delta(E)+2^k\epsilon.$$ Now, I would finished the proof if I could assume that the covering $\{E_j\}$ is made by balls, because I would have $$\mathcal S^k_\delta(E)\leq\sum_j\alpha_k \frac{\text{ diam}E_j^k}{2^k}\leq\sum_j \alpha^k \text{ diam}E_j^k\leq 2^k\mathcal H^k_\delta(E)+2^k\epsilon,$$ and by the arbitrariety of $\epsilon$ it would follow $\mathcal S^k_\delta\leq 2^k \mathcal H^k_\delta.$ Can I assume this?

Also, why and where spherical Hausdorff measure is useful?

$\endgroup$

1 Answer 1

4
$\begingroup$

It is important to be able to sort through families of sets that cover other sets. When the covering is a family of balls, there are many theorems available. When the covering is arbitrary it can be much more challenging to sort. The Hausdorff measure and spherical Hausdorff measure (in integer dimensions) coincide on rectifiable sets, so in many important cases one is just as good as the other.

If $\{E_j\}$ is a covering of $E$ with sets whose diameter does not exceed $\delta$, you can select points $x_j \in E_j$ and consider the balls $\newcommand{\diam}{\mathrm{diam}\, }B_j = B(x_j,\diam E_j)$. Then $E_j \subset B_j$ and $\diam B_j = 2\diam E_j$ so that $\{B_j\}$ covers $E$, $\diam B_j \le 2\delta$ for all $j$, and $$S_{2\delta}^k(E) \le \sum_j \alpha_j \frac{(\diam B_j)^k}{2^k} = 2^k \sum_j \alpha_j\frac{(\diam E_j)^k}{2^k}.$$ Take the infimum over all such families $\{E_k\}$ to get $$S_{2\delta}^k(E) \le 2^k H_\delta^k(E).$$

In fact, the constant $2^k$ can be improved to $\left( \dfrac{2n}{n+1} \right)^{k/2}$.

$\endgroup$
6
  • $\begingroup$ Thank you your answer. I think you meant to write $\text{diam} B_j=2\text{diam }E_j.$ $\endgroup$
    – batman
    Sep 23, 2015 at 8:07
  • $\begingroup$ I did. Thanks for pointing out the error. $\endgroup$
    – Umberto P.
    Sep 23, 2015 at 13:11
  • $\begingroup$ I don't understand why $B_j$ exists here. Say $E=[0,1]\cap\mathbb{Q}$ in $\mathbb{R}$, then one of the covering could be singletons. And thus $\mathcal{H}^1(E)=0$ but $\mathcal{S}^1(E)=1$. $\endgroup$
    – Edward
    Jan 24, 2020 at 12:53
  • $\begingroup$ Why would you think that $S^1(E) = 1$? $\endgroup$
    – Umberto P.
    Jan 24, 2020 at 15:10
  • $\begingroup$ Intuition, since open balls interests with rational numbers. I am suspicious about it now. But have no clue about how to prove it to be zero. $\endgroup$
    – Edward
    Jan 24, 2020 at 16:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .