# Inequality between Hausdorff measure and spherical Hausdorff measure

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?

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 $\left( \dfrac{2n}{n+1} \right)^{k/2}$.
• Thank you your answer. I think you meant to write $\text{diam} B_j=2\text{diam }E_j.$ Sep 23, 2015 at 8:07
• 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 $\mathcal{S}^1(E)=1$. Jan 24, 2020 at 12:53
• Why would you think that $S^1(E) = 1$? Jan 24, 2020 at 15:10