# Hausdorff Measure and Capacity

Definition 1 Let ${E \subset {\mathbb R}^n, 0 \le d < \infty, 0 < \delta \le \infty}$. Define

$\displaystyle \mathcal{H}^d_\delta(E) : = \inf \bigg\lbrace (1/2)^d \alpha(d) \sum_{i=1}^\infty diam(A_i)^d | E\subset \bigcup_{i=1}^\infty A_i, diam(A_i)\le \delta \bigg\rbrace,$

where ${\alpha(d) := \frac{\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2} + 1)}}$.

$\displaystyle \mathcal{H}^d (E) := \lim_{\delta \rightarrow 0} \mathcal{H}^d_\delta(E) = \sup_{\delta >0} \mathcal{H}^d_\delta(E)$

Then ${\mathcal{H}^d}$ is called d-dimensional Hausdorff measure in ${{\mathbb R}^n}$.

Remark 1 Hausdorff measure is a Borel outer measure, it is not Radon for ${0 \le d < n}$ because ${{\mathbb R}^n}$ is not ${\sigma-}$finite with respect to ${\mathcal{H}^d}$. And it is a measure if restricted on Lebesgue measurable sets (by Caratheodory condition: We say a set ${E}$ satisfies Caratheodory condition, if for any set ${A \subset {\mathbb R}^n}$, ${\mathcal{L}^n(A) = \mathcal{L}^n(A \cap E) + \mathcal{L}^n(A \setminus E)}$). Moreover, note that ${\alpha(d)}$ gives the volume of unit ball in ${d}$ dimension if ${d}$ is an integer, we naturally have (not trivially) ${\mathcal{H}^n = \mathcal{L}^n }$, where ${\mathcal{L}^n}$ is the n-dimensional Lebesgue measure.