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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High order BV function has continuous representative

As known, bounded variation (BV) functions are not continuous, but evidently higher order BV functions on corresponding dimension have continuous representative. As a consequence, {W^{n,1}({\mathbb R}^n) \hookrightarrow C^0({\mathbb R}^n)}.

Definition A function {f \in BV_n({\mathbb R}^n)}, if {f \in W^{n-1,1}({\mathbb R}^n)}, and the nth order distributional derivative {D^n f} is a finite Radon measure.

Theorem If {f \in BV_n({\mathbb R}^n)}, then {f} has a continuous representative.

Proof: Note that {C_c^\infty({\mathbb R}^n)} is dense in {BV_n({\mathbb R}^n)}, i.e. for any {f \in BV_n({\mathbb R}^n)}, there exists a sequence {f_k \in C_c^\infty({\mathbb R}^n)} such that {\lim_{k \rightarrow \infty} f_k = f} in {W^{n-1,1}}, and {\lim_{k\rightarrow \infty} \| \nabla^n f_k\|_{L^1({\mathbb R}^n)} = \|D^n f \| ({\mathbb R}^n)}. (See for example, book by Evans and Gariepy). So it suffice to show for all {f \in C_c^\infty({\mathbb R}^n)},

\displaystyle \|f\|_{L^\infty({\mathbb R}^n)} \le \| \nabla^n f\|_{L^1({\mathbb R}^n)}

Indeed,

\displaystyle f(x_1, \cdots , x_n) = \int_{-\infty}^{x_1} \partial_1 f(s_1,x_2, \cdots , x_n)\,ds_1 = \int_{-\infty}^{x_1} \cdots \int_{-\infty}^{x_n} \partial_1 \cdots \partial_n f

This gives us the desire inequality. Then by density result we know that {BV_n({\mathbb R}^n)} admits continuous representative for every element. \Box

Corollary {W^{n,1}({\mathbb R}^n) \hookrightarrow C^0({\mathbb R}^n)}.

Remark If the dimension is greater than the order of BV function, then the above theorem will fail. For example, let {f(x) = |x|^{-\frac{1}{2}}} be defined in a neighborhood {\Omega} around 0, with smooth boundary in {{\mathbb R}^3}. {|\nabla f| \sim |x|^{-\frac{3}{2}}} and {|\nabla^2 f| \sim |x|^{-\frac{5}{2}}}. So {f \in W^{2,1}(\Omega)}. Then we can extend {f} on {{\mathbb R}^3}, since {\partial \Omega} is smooth (See for example, PDE book by Evans). But then {f \rightarrow \infty} as {|x| \rightarrow 0}, which doesn’t admit a continuous representative.