Hausdorff measure

1 Outer measures and metric outer measures

Suppose that $X$ is a set. A function $\nu :𝒫(X)\to [0,\mathrm{\infty }]$ is said to be an outer measure if (i) $\nu (\mathrm{\varnothing })=0$, (ii) $\nu (A)\le \nu (B)$ when $A\subset B$, and, (iii) for any countable collection $\{A_j\}\subset 𝒫(X)$,

$$\nu \left(\bigcup _{j=1}^{\mathrm{\infty }}{A}_j\right)\le \sum _{j=1}^{\mathrm{\infty }}\nu (A_j).$$

We say that a subset $A$ of $X$ is $\nu$-measurable if

$$\nu (E)=\nu (E\cap A)+\nu (E\cap A^c),E\in 𝒫(X).$$ (1)

Here, instead of taking a $\sigma$-algebra as given and then defining a measure on this $\sigma$-algebra (namely, on the measurable sets), we take an outer measure as given and then define measurable sets using this outer measure. Carathéodory’s theorem states that the collection $ℳ$ of $\nu$-measurable sets is a $\sigma$-algebra and that the restriction of $\nu$ to $ℳ$ is a complete measure.

Suppose that $(X,\rho )$ is a metric space. An outer measure $\nu$ on $X$ is said to be a metric outer measure if

$$\rho (A,B)=inf\{\rho (a,b):a\in A,b\in B\}>0$$

implies that

$$\nu (A\cup B)=\nu (A)+\nu (B).$$

We prove that the Borel sets are $\nu$-measurable. That is, we prove that the Borel $\sigma$-algebra is contained in the $\sigma$-algebra of $\nu$-measurable sets.

Theorem 1.

If $\nu$ is a metric outer measure on a metric space $(X,\rho )$, then every Borel set is $\nu$-measurable.

Proof.

Because $\nu$ is an outer measure, by Carathéodory’s theorem the collection $ℳ$ of $\nu$-measurable sets is a $\sigma$-algebra, and hence to prove that $ℳ$ contains the Borel $\sigma$-algebra it suffices to prove that $ℳ$ contains all the closed sets. Let $F$ be a closed set in $X$, and let $E$ be a subset of $X$. Because $\nu$ is an outer measure,

$$\nu (E)=\nu ((E\cap F)\cup (E\cap F^c))\le \nu (E\cap F)+\nu (E\cap F^c).$$

In the case $\nu (E)=\mathrm{\infty }$, certainly $\nu (E)\ge \nu (E\cap F)+\nu (E\cap F^c)$. In the case , for each $n$ let

$$E_n=\{x\in E\setminus F:\rho (x,F)\ge n^{-1}\},$$

which satisfies $\rho (E_n,F)\ge n^{-1}$. Because $\rho (E_n,E\cap F)\ge \rho (E_n,F)\ge n^{-1}$, the fact that $\nu$ is a metric outer measure tells us that

$$\nu ((E\cap F)\cup E_n)=\nu (E\cap F)+\nu (E_n).$$ (2)

Because $F$ is closed, for any $x\in E\setminus F$ we have $\rho (x,F)>0$, and hence

$$E\setminus F=\bigcup _{n=1}^{\mathrm{\infty }}{E}_n.$$ (3)

Therefore

$$E=(E\cap F)\cup (E\cap F^c)=(E\cap F)\cup \bigcup _{n=1}^{\mathrm{\infty }}{E}_n=\bigcup _{n=1}^{\mathrm{\infty }}((E\cap F)\cup E_n),$$

hence for each $n$, using this and (2) we have

$$\nu (E)\ge \nu ((E\cap F)\cup E_n)=\nu (E\cap F)+\nu (E_n).$$

To prove that $\nu (E)\ge \nu (E\cap F)+\nu (E\cap F^c)$, it now suffices to prove that

$$\underset{n\to \mathrm{\infty }}{lim}\nu (E_n)=\nu (E\cap F^c).$$

Let $D_n=E_{n+1}\setminus E_n$. For $x\in D_{n+1}$ and $y\in X$ satisfying , we have

which implies that $y\notin E_n$. Thus,

$$\rho (D_{n+1},E_n)\ge \frac{1}{n(n+1)}.$$ (4)

For any $n$, using (4) and the fact that $\nu$ is a metric outer measure,

$\nu (E_{2n+1})$ $=$ $\nu (D_{2n}\cup E_{2n})$

$\ge$ $\nu (D_{2n}\cup E_{2n-1})$

$=$ $\nu (D_{2n})+\nu (E_{2n-1})$

$\ge$ $\mathrm{\cdots }$

$=$ $\nu (D_{2n})+\nu (D_{2n-2})+\mathrm{\cdots }+\nu (D_2)+\nu (E_1)$

$\ge$ ${\displaystyle \sum _{j=1}^n}\nu (D_{2j}),$

and

$\nu (E_{2n})$ $=$ $\nu (D_{2n-1}\cup E_{2n-1})$

$\ge$ $\nu (D_{2n-1}\cup E_{2n-2})$

$=$ $\nu (D_{2n-1})+\nu (E_{2n-2})$

$\ge$ $\mathrm{\cdots }$

$=$ $\nu (D_{2n-1})+\nu (D_{2n-3})+\mathrm{\cdots }+\nu (D_3)+\nu (D_1)+\nu (E_0)$

$=$ ${\displaystyle \sum _{j=1}^n}\nu (D_{2j-1}).$

But $E_n\subset E$ so $\nu (E_n)\le \nu (E)$, and hence each of the series ${\sum }_{j=1}^{\mathrm{\infty }}\nu (D_{2j})$ and ${\sum }_{j=1}^{\mathrm{\infty }}\nu (D_{2j-1})$ converges to a value $\le \nu (E)$. Thus the series ${\sum }_{j=1}^{\mathrm{\infty }}\nu (D_j)$ converges to a value $\le 2\nu (E)$. But for any $n$,

$$\nu (E\setminus F)=\nu \left(E_n\cup \bigcup _{j=n}^{\mathrm{\infty }}{D}_j\right)\le \nu (E_n)+\sum _{j=n}^{\mathrm{\infty }}\nu (D_j).$$

Because the series ${\sum }_{j=1}^{\mathrm{\infty }}\nu (D_j)$ converges, the sum on the right-hand side of the above tends to $0$ as $n\to \mathrm{\infty }$, so

$$\nu (E\setminus F)\le \underset{n\to \mathrm{\infty }}{lim\; inf}\nu (E_n)\le \underset{n\to \mathrm{\infty }}{lim\; sup}\nu (E_n)\le \nu (E\setminus F);$$

the last inequality is due to (3), which tells us $\nu (E_n)\le \nu (E\setminus F)$. Therefore,

$$\underset{n\to \mathrm{\infty }}{lim}\nu (E_n)=\nu (E\setminus F)=\nu (E\cap F^c),$$

which completes the proof. ∎

We shall use the following.

Lemma 2.

Let $(X,\rho )$ be a metric space. Suppose that $ℰ\subset 𝒫(X)$ satisfies $\mathrm{\varnothing },X\in ℰ$ and that $d:ℰ\to [0,\mathrm{\infty }]$ satisfies $d(\mathrm{\varnothing })=0$. Then the function $\nu :𝒫(X)\to [0,\mathrm{\infty }]$ defined by

$$\nu (A)=inf\{\sum _{j=1}^{\mathrm{\infty }}d(E_j):E_j\in ℰ\text{ and }A\subset \bigcup _{j=1}^{\mathrm{\infty }}{E}_j\},A\in 𝒫(X)$$

is an outer measure.

We remark that if there is no covering of a set $A$ by countably many elements of $ℰ$ then $\nu (A)$ is an infinimum of an empty set and is thus equal to $\mathrm{\infty }$.

2 Hausdorff measure

Suppose that $(X,\rho )$ is a metric space and let $p\ge 0$, $\delta >0$. Let $ℰ$ be the collection of those subsets of $X$ with diameter $\le \delta$ together with the set $X$, and define $d(A)={(\mathrm{diam}A)}^p$. By Lemma 2, the function $H_{p,\delta }:𝒫(X)\to [0,\mathrm{\infty }]$ defined by

$$H_{p,\delta }(A)=inf\{\sum _{j=1}^{\mathrm{\infty }}d(E_j):E_j\in ℰ\text{ and }A\subset {\bigcup }_{j=1}^{\mathrm{\infty }}{E}_j\},A\in 𝒫(X)$$

is an outer measure. If ${\delta }_1\le {\delta }_2$ then $H_{p,{\delta }_1}(A)\ge H_{p,{\delta }_2}(A)$, from which it follows that for each $A\in 𝒫(X)$, as $\delta$ tends to $0$, $H_{p,\delta }(A)$ tends to some element of $[0,\mathrm{\infty }]$. We define $H_p={lim}_{\delta \to 0}{H}_{p,\delta }$ and show that this is a metric outer measure.

Theorem 3.

Suppose that $(X,\rho )$ is a metric space and let $p\ge 0$. Then $H_p:𝒫(X)\to [0,\mathrm{\infty }]$ defined by

$$H_p(A)=\underset{\delta \to 0}{lim}{H}_{p,\delta }(A),A\in 𝒫(X).$$

is a metric outer measure.

Proof.

First we establish that $H_p$ is an outer measure. It is apparent that $H_p(\mathrm{\varnothing })=0$. If $A\subset B$, then, using that $H_{p,\delta }$ is a metric outer measure,

$$H_p(A)=\underset{\delta \to 0}{lim}{H}_{p,\delta }(A)\le \underset{\delta \to 0}{lim}{H}_{p,\delta }(B)=H_p(B).$$

If $\{A_j\}\subset 𝒫(X)$ is countable then, using that $H_{p,\delta }$ is a metric outer measure,

$H_p\left({\displaystyle \bigcup _{j=1}^{\mathrm{\infty }}}{A}_j\right)$ $=\underset{\delta \to 0}{lim}{H}_{p,\delta }\left({\displaystyle \bigcup _{j=1}^{\mathrm{\infty }}}{A}_j\right)$

$\le \underset{\delta \to 0}{lim}{\displaystyle \sum _{j=1}^{\mathrm{\infty }}}{H}_{p,\delta }(A_j)$

$={\displaystyle \sum _{j=1}^{\mathrm{\infty }}}\underset{\delta \to 0}{lim}{H}_{p,\delta }(A_j)$

$={\displaystyle \sum _{j=1}^{\mathrm{\infty }}}{H}_p(A_j).$

Hence $H_p$ is an outer measure.

To obtain that $H_p$ is a metric outer measure, we must show that if $\rho (A,B)>0$ then $H_p(A\cup B)\ge H_p(A)+H_p(B)$. Let and let $ℰ$ be the collection of those subsets of $X$ with diameter $\le \delta$ together with the set $X$. If there is no covering of $A\cup B$ by countably many elements of $ℰ$, then $H_p(A\cup B)\ge H_{p,\delta }(A\cup B)=\mathrm{\infty }$. Otherwise, let $\{E_j\}\subset ℰ$ be a covering of $A\cup B$. For each $j$, because , it follows that $E_j$ does not intersect both $A$ and $B$. Write

$$ℰ=\{E_{a_j}\}\cup \{E_{b_j}\},$$

where $E_{a_j}\cap B=\mathrm{\varnothing }$ and $E_{b_j}\cap A=\mathrm{\varnothing }$. Then $A\subset \bigcup E_{a_j}$ and $B\subset \bigcup E_{b_j}$, so

$$\sum _{j=1}^{\mathrm{\infty }}{(\mathrm{diam}{E}_j)}^p=\sum _{j=1}^{\mathrm{\infty }}{(\mathrm{diam}{E}_{a_j})}^p+\sum _{j=1}^{\mathrm{\infty }}{(\mathrm{diam}{E}_{j_b})}^p\ge H_{p,\delta }(A)+H_{p,\delta }(B).$$

This is true for any covering of $A\cup B$ by countably many element of $ℰ$, so

$$H_{p,\delta }(A\cup B)\ge H_{p,\delta }(A)+H_{p,\delta }(B).$$

The above inequality is true for any , and taking $\delta \to 0$ yields

$$H_p(A\cup B)\ge H_p(A)+H_p(B),$$

completing the proof. ∎

We call the metric outer measure $H_p:𝒫(X)\to [0,\mathrm{\infty }]$ in the above theorem the $p$-dimensional Hausdorff outer measure. From Theorem 1 it follows that the restriction of $H_p$ to the Borel $\sigma$-algebra ${ℬ}_X$ of a metric space is a meausure. We call this restriction the $p$-dimensional Hausdorff measure, and denote it also by $H_p$.

It is straightforward to verify that if $T:X\to X$ is an isometric isomorphism then $H_p\circ T=H_p$. In particular, for $X={ℝ}^n$, $H_p$ is invariant under translations.

We will use the following inequality when talking about Hausdorff measure on ${ℝ}^n$.

Lemma 4.

Let $Y$ be a set and $(X,\rho )$ be a metric space. If $f,g:Y\to X$ satisfy

$$\rho (f(y),f(z))\le C\rho (g(y),g(z)),y,z\in Y,$$

then for any $A\in 𝒫(Y)$,

$$H_p(f(A))\le C^p{H}_p(g(A)).$$

Proof.

Take $\delta >0$ and $ϵ>0$. There are countably many sets $E_j$ that cover $g(A)$ each with diameter $\le C^{-1}\delta$ and such that

$$\sum {(\mathrm{diam}{E}_j)}^p\le H_p(g(A))+ϵ.$$

Let $a\in A$. There is some $j$ with $g(a)\in E_j$, so $a\in g^{-1}(E_j)$ and then $f(a)\in f(g^{-1}(E_j))$. Therefore the sets $f(g^{-1}(E_j))$ cover $f(A)$. For $u,v\in f(g^{-1}(E_j))$, there are $y,z\in g^{-1}(E_j)$ with $u=f(y),v=f(z)$. Because $g(y),g(z)\in E_j$,

$$\rho (u,v)=\rho (f(y),f(z))\le C\rho (g(y),g(z))\le C\mathrm{diam}{E}_j,$$

hence

$$\mathrm{diam}f(g^{-1}(E_j))\le C\mathrm{diam}{E}_j.$$

Since the sets $f(g^{-1}(E_j))$ cover $f(A)$ and each has diameter $\le C\mathrm{diam}{E}_j\le \delta$,

$$H_{p,\delta }(f(A))\le \sum {(\mathrm{diam}f(g^{-1}(E_j)))}^p\le \sum C^p{(\mathrm{diam}{E}_j)}^p\le C^p(H_p(g(A))+ϵ).$$

This is true for all $\delta >0$, so taking $\delta \to 0$,

$$H_p(f(A))\le C^p(H_p(g(A))+ϵ).$$

This is true for all $ϵ>0$, so taking $ϵ\to 0$,

$$H_p(f(A))\le C^p{H}_p(g(A)).$$

∎

3 Hausdorff dimension

Theorem 5.

If then $H_q(A)=0$ for all $q>p$.

Proof.

Let $\delta >0$. Then Let $\{E_j\}$ be countably many sets each with diameter $\le \delta$ such that $A\subset \bigcup E_j$ and

$$\sum {(\mathrm{diam}{E}_j)}^p\le H_{p,\delta }(A)+1\le H_p(A)+1.$$

This gives us

$H_{q.\delta }(A)\le {\displaystyle \sum {(\mathrm{diam}{E}_j)}^q}$ $={\displaystyle \sum {(\mathrm{diam}{E}_j)}^{q-p}{(\mathrm{diam}{E}_j)}^p}$

$\le {\delta }^{q-p}{\displaystyle \sum {(\mathrm{diam}{E}_j)}^p}$

$\le {\delta }^{q-p}(H_p(A)+1).$

This is true for any $\delta >0$ and $q-p>0$, so taking $\delta \to 0$ we obtain $H_q(A)=0$. ∎

For $A\in 𝒫(X)$, we define the Hausdorff dimension of $A$ to be

$$inf\{q\ge 0:H_q(A)=0\}.$$

If the set whose infimum we are taking is empty, then the Hausdorff dimension of $A$ is $\mathrm{\infty }$.

4 Radon measures and Haar measures

Before speaking about Hausdorff measure on ${ℝ}^n$, we remind ourselves of some material about Radon measures and Haar measures. Let $X$ be a locally compact Hausdorff space. A Borel measure $\mu$ on $X$ is said to be a Radon measure if (i) it is finite on each compact set, (ii) for any Borel set $E$,

$$\mu (E)=inf\{\mu (U):U\text{ open and }E\subset U\},$$

and (iii) for any open set $E$,

$$\mu (E)=sup\{\mu (K):K\text{ compact and }K\subset E\}.$$

It is a fact that if $X$ is a locally compact Hausdorff space in which every open set is $\sigma$-compact, then every Borel measure on $X$ that is finite on compact sets is a Radon measure.

Suppose that $G$ is a locally compact group. A Borel measure $\mu$ on $G$ is said to be left-invariant if for all $x\in G$ and $E\in {ℬ}_G$,

$$\mu (xE)=\mu (E).$$

A left Haar measure on $G$ is a nonzero left-invariant Radon measure on $G$. It is a fact that if $\mu$ and $\nu$ are left Haar measures on $G$ then there is some $c>0$ such that $\mu =c\nu$.

5 Hausdorff measure in Rⁿ

Let $m_n$ denote Lebesgue measure on ${ℝ}^n$.

Lemma 6.

If $E$ is a Borel set in ${ℝ}^n$, then

$$H_n(E)\ge 2^n{m}_n(E).$$

Proof.

Let $ϵ>0$ and let $\{E_j\}$ be countably many closed sets that cover $E$ and such that

$$\sum {(\mathrm{diam}{E}_j)}^n\le H_n(E)+ϵ.$$

The isodiametric inequality (which one proves using the Brunn-Minkowski inequality) states that if $A$ is a Borel set in ${ℝ}^n$, then

$$m_n(A)\le {\left(\frac{\mathrm{diam}A}{2}\right)}^n.$$

Using this gives

$$\sum 2^n{m}_n(E_j)\le H_n(E)+ϵ.$$

But because the sets $E_j$ cover $E$ we have $m_n(E)\le m_n(\bigcup E_j)\le \sum m_n(E_j)$, so we get

$$m_n(E)\le \frac{H_n(E)+ϵ}{2^n}.$$

This expression does not involve the sets $E_j$ (which depend on $ϵ$), and since this expression is true for any $ϵ>0$, taking $ϵ\to 0$ yields

$$m_n(E)\le \frac{H_n(E)}{2^n}.$$

∎

Let

$$Q=\{x\in {ℝ}^n:|x_1|\le \frac{1}{2},\mathrm{\dots },|x_n|\le \frac{1}{2}\}.$$

Lemma 7.

.

Proof.

For any $m\ge 1$, the cube $Q$ can be covered by $m^n$ cubes $q_1,\mathrm{\dots },q_{m^n}$ of side length $\frac{1}{m}$. Let and let $m>\frac{1}{\delta }$. The distance from the center of $q_j$ to one of the vertices of $q_j$ is

$$r=\sqrt{{\left(\frac{1}{2m}\right)}^2+\mathrm{\cdots }+{\left(\frac{1}{2m}\right)}^2}=\frac{\sqrt{n}}{2m}.$$

Inscribe $q_j$ in a closed ball $b_j$ with the same center as $q_j$ and radius $r$. These balls cover $Q$. Hence

$$H_{p,\delta }(Q)\le \sum _{j=1}^{m^n}{(\mathrm{diam}{b}_j)}^n=\sum _{j=1}^{m^n}{(2r)}^n={(2r)}^n\cdot m^n=n^{n/2}.$$

Taking $\delta \to 0$ gives .

On the other hand, by Lemma 6,

$$H_n(Q)\ge 2^n{m}_n(Q)=2^n>0.$$

∎

Theorem 8.

There is some constant $c_n>0$ such that

$$H_n=c_n{m}_n.$$

Proof.

${ℝ}^n$ is a locally compact Hausdorff space in which every open set in ${ℝ}^n$ is $\sigma$-compact. Therefore, to show that $H_n$ is a Radon measure it suffices to show that $H_n$ is finite on every compact set. If $K$ is a compact subset of ${ℝ}^n$, there is some $r>0$ such that $K\subset rQ$. By Lemma 4 and Lemma 7 we get , so . Therefore $H_n$ is a Radon measure.

Because $H_n(Q)>0$, $H_n$ is not the zero measure. Any translation is an isometric isomorphism ${ℝ}^n\to {ℝ}^n$, so $H_n$ is invariant under translations. Thus $H_n$ is a left Haar measure on ${ℝ}^n$. But Lebesgue measure $m_n$ is also a left Haar measure on ${ℝ}^n$, so there is some $c_n>0$ such that

$$H_n=c_n{m}_n,$$

proving the claim. ∎