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.

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