Lecture 9: Hausdorff measure

28/10/2024

Hausdorff measure is a way of measuring $d$-dimensional size of sets in $\R^n$ where $d<n$. This is done by covering with sets and taking infimum, like in the definition of the Lebesgue measure. For $E\subset\R^n$ denote by

$\mathrm{diam}(E) = \sup\{ |x-y| \ | \ x, y \in E\}$

the diameter with respect to the Euclidean distance. Let $d\in[0,\infty)$. For $\delta>0$ and $A\subset\R^n$ define

$\mathcal{H}^d_\delta(A)=\inf\{\sum_{k=1}^\infty \mathrm{diam}(E_k)^d \ | \ A \subset \bigcup_{k=1}^\infty E_k, \ \mathrm{diam}(E_k)\leq\delta\}.$

This is similar to the definition of the Lebesgue outer measure except we take diameter to the power $d$. This is related to the notion of the dimension. For example, if $E$ is a $d$-dimensional ball or a $d$-dimensional cube, then the $d$-dimensional volume of $E$ is proportional to $\mathrm{diam}(E)^d$.

This is an outer measure on $P(\R^n)$ (recall the general construction of outer measures from Lecture 6). In particular, it does lead to a $\sigma$-algebra and a measure on $\R^n$ by the Caratheodory theorem but it is not what we want. For example, we would want any rectangle in $\R^2$, which is a $2$-dimensional shape, to have infinite $1$-dimensional length (we can fit a curve of any length inside a rectangle) but $\mathcal{H}^1_\delta$ is finite on all compact sets.

Example. Let $A \subset \R^2$ be a square of unit side. For every $\delta>0$ we can cover it by approximately $2/\delta^2$ squares $E_k$ with side length $\delta/\sqrt{2}$ and diameter $\delta$. For $d=1$ the sum in the definition of the

$\sum_k \mathrm{diam}(E_k)^1 = 2/\delta^2 \cdot \delta = 2/\delta$

Therefore,

$\mathcal{H}^d_\delta(A) \leq 2/\delta$.

One can actually show that this is not only an upper bound but a good approximation of $\mathcal{H}^1_\delta(A)$. Therefore, for $\delta\to0$ we have $\mathcal{H}^1_\delta(A)\to\infty$ as desired. On the other hand, for $A$ a $1$-dimensional interval in $\R^2$, we see that the same procedure leads an approximation which converges to the length of $A$.

Therefore, we should take the limit as $\delta\to 0$. To that end, observe that for $\delta_1 \leq \delta_2$, if $\mathrm{diam}(E)\leq\delta_1$, then $\mathrm{diam}(E)\leq\delta_2$. Therefore, every $\delta_1$-covering is $\delta_2$-covering. If $S, T \subset \R$ are two sets and $S\subset T$, then $\inf T \leq \inf S$. We conclude that for

$\mathcal{H}^d_{\delta_2}(A) \leq \mathcal{H}^d_{\delta_1}(A)$ for $\delta_1 \leq \delta_2$

The $d$-dimensional Hausdorff outer measure is defined by