Hausdorff measure and dimension

Let us talk about Hausdorff measure $\mathcal{H}^s$ named after Felix Hausdorff, which comes up in image processing often. These measures allow us to measure very small subsets of $\mathbb{R}^n$. The idea is that $A$ is an $s-$ dimensional subset of $\mathbb{R}^n$ if $0<\mathcal{H}^s(A)<\infty$. We will follow the notation and treatment of Evans and Gariepy.

Felix Hausdorff

Definitions:
(i) Let $A\subset \mathbb{R}^n$, $0\leq s < \infty$, $0<\delta\leq \infty$. Define $$\mathcal{H}^s_{\delta}(A)=\inf \Big\{ \sum_{j=1}^{\infty} \alpha(s) \Big( \frac{\mbox{diam}(C_j)}{2} \Big)^s \, \Big\vert \,A\subset \cup_{j=1}^{\infty}C_j, \mbox{diam}(C_j)\leq \delta \Big\},$$ where $\{C_j\}\subset \mathbb{R}^n$ is the covering of the set $A$ and $\alpha(s)=\frac{\pi^{\frac{s}{2}}}{\Gamma(\frac{s}{2}+1)}.$ Here $\Gamma(s)=\int_0^{\infty}e^{-x}x^{s-1}\,dx,\,\,(0< s<\infty),$ is the usual gamma function. (ii) The $s-$dimensional Hausdorff measure of $A$ is denoted by $\mathcal{H}^s(A)$ and it is defined as the limit:
$$\mathcal{H}^s(A)=\lim_{\delta\rightarrow 0}\mathcal{H}^s_{\delta}(A)=\sup_{\delta >0} \mathcal{H}^s_{\delta}(A).$$

The book: 'Measure theory and fine properties of functions' by Evans and Gariepy includes the normalizing constant $\alpha(s)$ in the definition, whereas some other authors do not include this constant. The constant $\alpha(s)$ is included because if $s$ is an integer, then $\mathcal{H}^s(A)$ is equal to the $s-$dimensional surface area for nice sets (graph of Lipschitz function).


Some properties of Hausdorff measure:
1.$\mathcal{H}^s$ is a Borel regular measure. Refer to E.G. for the proof.

2. Let us consider a Hausdorff measure $\mathcal{H}^0$ of a singleton $\{a\}\in \mathbb{R}^n$. First note that $\alpha(0)=1$. So, $\mathcal{H}^0(\{a\})=\mathcal{H}^0_{\delta}(\{a\})=1$. Thus, $\mathcal{H}^0$ is a counting measure.

3. $\mathcal{H}^1=\mathcal{L}^1$ on $\mathbb{R}^1$, i.e. $\mathcal{H}^1$ measure of a line segment is its length.

4. Since $\mathbb{R}^n$ is not $\sigma-$finite, $\mathcal{H}^s$ is not a Radon measure for $0\leq s < n$.

5. If $s > n$ then $\mathcal{H}^s\equiv 0$ on $\mathbb{R}^n$.
To see this is easy. Let us consider a unit cube $Q$ in $\mathbb{R}^n$, fix an integer $m \geq 1$. The unit cube can be decomposed into $m^n$ cubes with sides $1/m$ and diameter ${n^{1/2}}/{m}$. Therefore,
$$\mathcal{H}^s_{{\sqrt{n}}/{m}}(Q)\leq \sum_{i=1}^{m^n} \alpha(s) (n^{1/2}/m)^s = \alpha(s) n^{s/2} m^{n-s}.$$ If $s >n$, then $m^{n-s}\rightarrow 0$ as $m\rightarrow \infty$ i.e. as the size of the covering vanishes. Thus, $\mathcal{H}^s(Q)=0$, so $\mathcal{H}^s(\mathbb{R}^n)=0$ for $s >n$.

6. Scaling: $\mathcal{H}^s(\lambda A)=\lambda^s\mathcal{H}^s(A)$ for all $\lambda > 0$, $A \subset \mathbb{R}^n$. (Easy to prove.)

7. Invariance under affine transformation: $\mathcal{H}^s(L(A))=\mathcal{H}^s(A)$ for each affine isometry $L:\mathbb{R}^n\rightarrow \mathbb{R}^n, \, A \subset \mathbb{R}^n$.

8. $\mathcal{H}^n=\mathcal{L}^n$ on $\mathbb{R}^n$.


Hausdorff-Besicovitch dimension:
The hausdorff dimension of a set $A\subset \mathbb{R}^n$ is defined to be

$$\mathcal{H}_{\mbox{dim}}(A)=\inf\{ 0\leq s < \infty | \mathcal{H}^s(A)=0 \}.$$

The following assertions follow from the definition:
Countable sets have Hausdorff dimension $0$. The Euclidean space has Hausdorff dimension $n$.
The circle has Hausdorff dimension $1$.
The Cantor set (a zero-dimensional topological space) is a union of two copies of itself, each copy shrunk by a factor 1/3; this fact can be used to prove that its Hausdorff dimension is ${\ln 2}/{\ln 3}$.