Hausdorff measure and densities#

Recall the definition of Hausdorff measure from Example 5:

Definition 20 (Hausdorff measure)

For $$X$$ a metric space, $$s\geq 0$$ and $$A\subset X$$, let

$\mathcal H^s_\delta(A)= \inf \sum_{S\in G}\operatorname{diam}(S),$

where the infimum is taken over all countable

$G\subset \{S\in F: \operatorname{diam}(S)\}<\delta$

such that

$A \subset \bigcup_{S\in G}S.$

Finally, define $$\mathcal H^s(A)=\sup_{\delta>0}\psi_\delta(A)$$.

We are interested in the measure $$\mathcal H^s|_S$$, for $$S\subset X$$ some $$\mathcal H^s$$-measurable set with $$\mathcal H^s(S)<\infty$$. In particular, we require some counterpart to the Lebesgue density theorem, but, of course, $$\mathcal H^s|_X$$ may not be locally finite.

Definition 21 (Hausdorff density)

Let $$X$$ be a metric space, $$A\subset X$$ and $$s\geq 0$$. The upper and lower Hausdorff densities of $$A$$ are

$\Theta^{*,s}(A,x)=\limsup_{r\to 0} \frac{\mathcal H^s(A\cap B(x,r))}{(2r)^s}$

and

$\Theta^{s}_*(A,x)=\liminf_{r\to 0} \frac{\mathcal H^s(A\cap B(x,r))}{(2r)^s}.$

Lemma 5

Let $$X$$ be a metric space, $$s\geq 0$$ and $$A\subset X$$ with $$\mathcal H^s(A)<\infty$$. Then

$2^{-s} \leq \Theta^{*,s}(A,x) \leq 1$

for $$\mathcal H^s$$-a.e. $$x\in A$$.

Proof. The set of $$x\in A$$ with $$\Theta^{*,s}(A,x)<2^{-s}$$ is a countable union countable of the sets

$S_\delta:= \{x\in A : \mathcal H^s(A\cap B(x,r)) < (1-\delta) r^s\ \forall 0<r< \delta\}.$

Thus, for the first inequality, it suffices to show that $$\mathcal H^s(S_\delta)=0$$ for all $$\delta>0$$.

Fix $$\delta,\epsilon>0$$. We may cover $$S_\delta$$ by sets $$E_1,E_2,\ldots$$ such that, for each $$i\in \mathbb N$$, $$\operatorname{diam}E_i<\epsilon$$, $$S_\delta\cap E_i\neq \emptyset$$ and

$\sum_{i\in\mathbb N} \operatorname{diam}E_i^s \leq \mathcal H^s(S_\delta)+\epsilon.$

For each $$i\in \mathbb N$$ let $$x_i\in S_\delta \cap E_i$$ and set $$r_i =\operatorname{diam}E_i$$. Then

$\begin{split}\begin{gathered} \mathcal H^s(S_\delta) \leq \sum_{i\in\mathbb N}\mathcal H^s(S_\delta \cap E_i) \leq \sum_{i\in\mathbb N} \mathcal H^s(A\cap B(x_i,r_i)) \\ \leq (1-\delta) \sum_{i\in\mathbb N} \operatorname{diam}E_i^s \leq (1-\delta)(\mathcal H^s(S_\delta)+\epsilon). \end{gathered}\end{split}$

Since $$\epsilon>0$$ is arbitrary and $$\delta>0$$, this implies $$\mathcal H^s(S_\delta)=0$$, as required.

For the second inequality, since $$\mathcal H^s$$ is Borel regular (see Example 42), it suffices to assume that $$A$$ is Borel. As before, given $$\delta>0$$, it suffices to prove that

$S:= \{x\in A: \Theta^{*,s}(A,x)>1+\delta\}$

satisfies $$\mathcal H^s(S)=0$$. Fix $$\epsilon>0$$ and let $$U\supset S$$ be open with

$\mathcal H^s(A \cap U)\leq \mathcal H^s(S)+\epsilon$

(which exists by the outer regularity of the measure $$\mathcal H^s|_A$$). Let $$\mathcal B_\epsilon$$ be the collection of balls $$B$$ centred at a point of $$S$$ with $$\operatorname{rad}B<\epsilon$$ such that $$B\subset U$$ and

(21)#$\mathcal H^s(A \cap B) > (1+\delta)(2\operatorname{rad}B)^s.$

This is a Vitali cover of $$S$$. Let $$\mathcal B'_\epsilon$$ be obtained from Proposition 1.

Since $$\mathcal H^s(S)<\infty$$, $$S$$ is separable (see Example 44) and so $$\mathcal B'_\epsilon =\{B_1,B_2,\ldots\}$$ is countable and the conclusion of Proposition 1 states that

$S\setminus \bigcup_{i\in\mathbb N} B_i \subset \bigcup_{i>n}5B_i$

for each $$n\in\mathbb N$$. Since $$\operatorname{diam}B_i<\epsilon$$ for each $$i\in \mathbb N$$, the $$B_i$$ and $$5B_i$$ may be used to estimate $$\mathcal H^s_{10\epsilon}(S)$$. For each $$n\in\mathbb N$$ we obtain

$\begin{split}\mathcal H^s_{10\epsilon}(S) &\leq \sum_{i\in\mathbb N}(2\operatorname{rad}B_i)^s + \sum_{i>n} (10\operatorname{rad}B_i)^s\\ &\leq \sum_{i\in\mathbb N} \frac{\mathcal H^s(A\cap B_i)}{1+\delta} + 5^s\sum_{i>n} \frac{\mathcal H^s(A\cap B_i)}{1+\delta}\end{split}$

where the second inequality follows by (21). Since the $$B_i$$ are disjoint and $$\mathcal H^s(A)<\infty$$, the second term converges to 0 as $$n \to\infty$$. Since the $$B_i$$ are subsets of $$U$$ we obtain

$\mathcal H^s_{10\epsilon}(S) \leq \frac{\mathcal H^s(A \cap U)}{1+\delta} \leq \frac{\mathcal H^s(S)+\epsilon}{1+\delta}.$

Since $$\epsilon>0$$ is arbitrary, this implies $$\mathcal H^s(S) \leq \mathcal H^s(S)/(1+\delta)$$ and hence $$\mathcal H^s(S)=0$$, as required.

Lemma 6

Let $$X$$ be a metric space, $$s\geq 0$$ and let $$A\subset X$$ be $$\mathcal H^s$$-measurable with $$\mathcal H^s(A)<\infty$$. Then

$\Theta^{*,s}(A,x) =0$

for $$\mathcal H^s$$-a.e. $$x\not\in A$$.

Proof. It suffices to show that, for $$t>0$$, the set

$S=\{x\in X\setminus A: \Theta^{*,n}(A,x)>t\}$

satisfies $$\mathcal H^s(S)=0$$. Fix $$\epsilon>0$$. Since $$A$$ is $$\mathcal H^s$$-measurable, $$\mathcal H^s|_A$$ is Borel regular. Therefore, since $$\mathcal H^s|_A(S)=0$$, there exists an open $$U\supset S$$ with

$\mathcal H^s(A\cap U)=\mathcal H^s|_A(U)<\epsilon.$

For each $$x\in S$$ and $$\delta>0$$ there exists a ball $$B$$ centred on $$x$$ with $$\operatorname{rad}B<\delta$$ such that

$\frac{\mathcal H^s(A\cap B)}{(2\operatorname{rad}B)^s}>t.$

By Lemma 4 there exists a disjoint collection $$\mathcal B$$ of such balls such that

$S\subset \bigcup_{B\in\mathcal B}5B.$

Since $$\mathcal H^s(A)<\infty$$, $$A$$ is separable and each of these balls contains a point of $$A$$, $$\mathcal B$$ is countable. Therefore

$t\mathcal H^s_{5\delta}(S) \leq t \sum_{B\in\mathcal B} (2\operatorname{rad}5B)^s < 5^s \sum_{B\in\mathcal B} \mathcal H^s(A\cap B) \leq 5^s \mathcal H^s(A\cap U) < 5^s \epsilon.$

Since $$\delta,\epsilon>0$$ are arbitrary, this completes the proof.

Exercises#

Example 41

Let $$\mathcal V\subset [0,1]$$ be a Vitali set as constructed in Exercise 8.

1. Show that, for any Borel $$B\subset \mathcal V$$, $$\mathcal L^1(B)=0$$.

2. Deduce that $$\mathcal L^1([0,1]\setminus \mathcal V)=1$$ and hence, if $$C$$ is a Borel set with

$[0,1]\setminus \mathcal V\subset C \subset [0,1],$

then $$\mathcal L^1(C)=1$$.

3. Hence show that $$\mathcal L^1(\mathcal V \cap C)=\mathcal L^1(\mathcal V)>0$$.

Note however that we cannot deduce the value of $$\mathcal L^1(\mathcal V)$$ from our construction in Example 8. Indeed, for any $$\epsilon>0$$, that construction may produce a $$\mathcal V\subset [0,\epsilon]$$.

Example 42

Let $$X$$ be a metric space and $$s\geq 0$$.

1. Show that $$\mathcal H^s$$ is Borel regular. Hint: first show that in the definition of $$\mathcal H^s$$, we may take $$F$$ to be the collection of closed sets.

2. We are usually interested in $$\mathcal H^s|_A$$ for some $$A\subset X$$. Show that for any $$A\subset X$$, $$\mathcal H^s|_A$$ is a Borel measure.

3. Now assume that $$A\subset X$$ is $$\mathcal H^s$$-measurable with $$\mathcal H^s(A)<\infty$$. Show that $$\mathcal H^s|_A$$ is Borel regular. Hint: show that there exist Borel sets $$B\supset A \supset B'$$ with $$\mathcal H^s(B\setminus B')=0$$.

4. Show that $$\mathcal H^s|_A$$ may not be Borel regular if $$A$$ is not $$\mathcal H^s$$ measurable. Hint: consider Example 41.

Example 43

In this exercise we construct the four corner Cantor set. Let $$K_0 = [0,1]^2$$. Let $$K_1$$ be the “four corners” of $$K_0$$ of side length 1/4. That is

$K_1 = [0,1/4]^2 \cup [3/4,1]^2 \cup [0,1/4]\times [3/4,1] \cup [3/4,1]\cup [0,1/4].$

Inductively, $$K_n$$ is constructed by taking the four corners of side length $$1/4^n$$ of all the squares of $$K_{n-1}$$. Finally let $$K=\bigcap_{n\in\mathbb N}K_n$$, a compact set. Show that $$0<\mathcal H^1(K)<\infty$$.

Example 44

For $$s\geq 0$$ let $$X$$ be a metric space with $$\mathcal H^s(X)<\infty$$. Show that $$X$$ is separable.

Example 45

Show that Lemma 5 and Lemma 6 may be false if $$A$$ has only $$\sigma$$-finite $$\mathcal H^s$$ measure.