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.