# Hausdorff measure

Jordan Bell
October 29, 2014

## 1 Outer measures and metric outer measures

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

 $\nu\left(\bigcup_{j=1}^{\infty}A_{j}\right)\leq\sum_{j=1}^{\infty}\nu(A_{j}).$

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

 $\nu(E)=\nu(E\cap A)+\nu(E\cap A^{c}),\qquad E\in\mathscr{P}(X).$ (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 theorem11 1 Gerald B. Folland, Real Analysis, second ed., p. 29, Theorem 1.11. states that the collection $\mathscr{M}$ of $\nu$-measurable sets is a $\sigma$-algebra and that the restriction of $\nu$ to $\mathscr{M}$ 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.22 2 Gerald B. Folland, Real Analysis, second ed., p. 349, Proposition 11.16. 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 $\mathscr{M}$ of $\nu$-measurable sets is a $\sigma$-algebra, and hence to prove that $\mathscr{M}$ contains the Borel $\sigma$-algebra it suffices to prove that $\mathscr{M}$ 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,

 $\nu(E)=\nu((E\cap F)\cup(E\cap F^{c}))\leq\nu(E\cap F)+\nu(E\cap F^{c}).$

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

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

which satisfies $\rho(E_{n},F)\geq n^{-1}$. Because $\rho(E_{n},E\cap F)\geq\rho(E_{n},F)\geq n^{-1}$, 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

 $E\setminus F=\bigcup_{n=1}^{\infty}E_{n}.$ (3)

Therefore

 $E=(E\cap F)\cup(E\cap F^{c})=(E\cap F)\cup\bigcup_{n=1}^{\infty}E_{n}=\bigcup_% {n=1}^{\infty}((E\cap F)\cup E_{n}),$

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

 $\nu(E)\geq\nu((E\cap F)\cup E_{n})=\nu(E\cap F)+\nu(E_{n}).$

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

 $\lim_{n\to\infty}\nu(E_{n})=\nu(E\cap F^{c}).$

Let $D_{n}=E_{n+1}\setminus E_{n}$. For $x\in D_{n+1}$ and $y\in X$ satisfying $\rho(x,y)<((n+1)n)^{-1}$, we have

 $\rho(y,F)\leq\rho(x,y)+\rho(x,F)<\frac{1}{n(n+1)}+\frac{1}{n+1}=\frac{1}{n},$

which implies that $y\not\in E_{n}$. Thus,

 $\rho(D_{n+1},E_{n})\geq\frac{1}{n(n+1)}.$ (4)

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

 $\displaystyle\nu(E_{2n+1})$ $\displaystyle=$ $\displaystyle\nu(D_{2n}\cup E_{2n})$ $\displaystyle\geq$ $\displaystyle\nu(D_{2n}\cup E_{2n-1})$ $\displaystyle=$ $\displaystyle\nu(D_{2n})+\nu(E_{2n-1})$ $\displaystyle\geq$ $\displaystyle\cdots$ $\displaystyle=$ $\displaystyle\nu(D_{2n})+\nu(D_{2n-2})+\cdots+\nu(D_{2})+\nu(E_{1})$ $\displaystyle\geq$ $\displaystyle\sum_{j=1}^{n}\nu(D_{2j}),$

and

 $\displaystyle\nu(E_{2n})$ $\displaystyle=$ $\displaystyle\nu(D_{2n-1}\cup E_{2n-1})$ $\displaystyle\geq$ $\displaystyle\nu(D_{2n-1}\cup E_{2n-2})$ $\displaystyle=$ $\displaystyle\nu(D_{2n-1})+\nu(E_{2n-2})$ $\displaystyle\geq$ $\displaystyle\cdots$ $\displaystyle=$ $\displaystyle\nu(D_{2n-1})+\nu(D_{2n-3})+\cdots+\nu(D_{3})+\nu(D_{1})+\nu(E_{0})$ $\displaystyle=$ $\displaystyle\sum_{j=1}^{n}\nu(D_{2j-1}).$

But $E_{n}\subset E$ so $\nu(E_{n})\leq\nu(E)$, and hence each of the series $\sum_{j=1}^{\infty}\nu(D_{2j})$ and $\sum_{j=1}^{\infty}\nu(D_{2j-1})$ converges to a value $\leq\nu(E)$. Thus the series $\sum_{j=1}^{\infty}\nu(D_{j})$ converges to a value $\leq 2\nu(E)$. But for any $n$,

 $\nu(E\setminus F)=\nu\left(E_{n}\cup\bigcup_{j=n}^{\infty}D_{j}\right)\leq\nu(% E_{n})+\sum_{j=n}^{\infty}\nu(D_{j}).$

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

 $\nu(E\setminus F)\leq\liminf_{n\to\infty}\nu(E_{n})\leq\limsup_{n\to\infty}\nu% (E_{n})\leq\nu(E\setminus F);$

the last inequality is due to (3), which tells us $\nu(E_{n})\leq\nu(E\setminus F)$. Therefore,

 $\lim_{n\to\infty}\nu(E_{n})=\nu(E\setminus F)=\nu(E\cap F^{c}),$

which completes the proof. ∎

We shall use the following.33 3 Gerald B. Folland, Real Analysis, second ed., p. 29, Proposition 1.10.

###### Lemma 2.

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

 $\nu(A)=\inf\left\{\sum_{j=1}^{\infty}d(E_{j}):E_{j}\in\mathscr{E}\textrm{ and % }A\subset\bigcup_{j=1}^{\infty}E_{j}\right\},\qquad A\in\mathscr{P}(X)$

is an outer measure.

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

## 2 Hausdorff measure

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

 $H_{p,\delta}(A)=\inf\left\{\sum_{j=1}^{\infty}d(E_{j}):\textrm{E_{j}\in% \mathscr{E} and A\subset\bigcup_{j=1}^{\infty}E_{j}}\right\},\qquad A\in% \mathscr{P}(X)$

is an outer measure. If $\delta_{1}\leq\delta_{2}$ then $H_{p,\delta_{1}}(A)\geq H_{p,\delta_{2}}(A)$, from which it follows that for each $A\in\mathscr{P}(X)$, as $\delta$ tends to $0$, $H_{p,\delta}(A)$ tends to some element of $[0,\infty]$. We define $H_{p}=\lim_{\delta\to 0}H_{p,\delta}$ and show that this is a metric outer measure.44 4 Gerald B. Folland, Real Analysis, second ed., p. 350, Proposition 11.17.

###### Theorem 3.

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

 $H_{p}(A)=\lim_{\delta\to 0}H_{p,\delta}(A),\qquad A\in\mathscr{P}(X).$

is a metric outer measure.

###### Proof.

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

 $H_{p}(A)=\lim_{\delta\to 0}H_{p,\delta}(A)\leq\lim_{\delta\to 0}H_{p,\delta}(B% )=H_{p}(B).$

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

 $\displaystyle H_{p}\left(\bigcup_{j=1}^{\infty}A_{j}\right)$ $\displaystyle=\lim_{\delta\to 0}H_{p,\delta}\left(\bigcup_{j=1}^{\infty}A_{j}\right)$ $\displaystyle\leq\lim_{\delta\to 0}\sum_{j=1}^{\infty}H_{p,\delta}(A_{j})$ $\displaystyle=\sum_{j=1}^{\infty}\lim_{\delta\to 0}H_{p,\delta}(A_{j})$ $\displaystyle=\sum_{j=1}^{\infty}H_{p}(A_{j}).$

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 $H_{p}(A\cup B)\geq H_{p}(A)+H_{p}(B)$. Let $0<\delta<\rho(A,B)$ and let $\mathscr{E}$ be the collection of those subsets of $X$ with diameter $\leq\delta$ together with the set $X$. If there is no covering of $A\cup B$ by countably many elements of $\mathscr{E}$, then $H_{p}(A\cup B)\geq H_{p,\delta}(A\cup B)=\infty$. Otherwise, let $\{E_{j}\}\subset\mathscr{E}$ be a covering of $A\cup B$. For each $j$, because $\mathrm{diam}\,E_{j}\leq\delta<\rho(A,B)$, it follows that $E_{j}$ does not intersect both $A$ and $B$. Write

 $\mathscr{E}=\{E_{a_{j}}\}\cup\{E_{b_{j}}\},$

where $E_{a_{j}}\cap B=\emptyset$ and $E_{b_{j}}\cap A=\emptyset$. Then $A\subset\bigcup E_{a_{j}}$ and $B\subset\bigcup E_{b_{j}}$, so

 $\sum_{j=1}^{\infty}(\mathrm{diam}\,E_{j})^{p}=\sum_{j=1}^{\infty}(\mathrm{diam% }\,E_{a_{j}})^{p}+\sum_{j=1}^{\infty}(\mathrm{diam}\,E_{j_{b}})^{p}\geq H_{p,% \delta}(A)+H_{p,\delta}(B).$

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

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

The above inequality is true for any $0<\delta<\rho(A,B)$, and taking $\delta\to 0$ yields

 $H_{p}(A\cup B)\geq H_{p}(A)+H_{p}(B),$

completing the proof. ∎

We call the metric outer measure $H_{p}:\mathscr{P}(X)\to[0,\infty]$ 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 $\mathscr{B}_{X}$ 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 $X=\mathbb{R}^{n}$, $H_{p}$ is invariant under translations.

We will use the following inequality when talking about Hausdorff measure on $\mathbb{R}^{n}$.55 5 Gerald B. Folland, Real Analysis, second ed., p. 350, Proposition 11.18.

###### Lemma 4.

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

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

then for any $A\in\mathscr{P}(Y)$,

 $H_{p}(f(A))\leq C^{p}H_{p}(g(A)).$
###### Proof.

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

 $\sum(\mathrm{diam}\,E_{j})^{p}\leq H_{p}(g(A))+\epsilon.$

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 $g(y),g(z)\in E_{j}$,

 $\rho(u,v)=\rho(f(y),f(z))\leq C\rho(g(y),g(z))\leq C\mathrm{diam}\,E_{j},$

hence

 $\mathrm{diam}\,f(g^{-1}(E_{j}))\leq C\mathrm{diam}\,E_{j}.$

Since the sets $f(g^{-1}(E_{j}))$ cover $f(A)$ and each has diameter $\leq C\mathrm{diam}\,E_{j}\leq\delta$,

 $H_{p,\delta}(f(A))\leq\sum(\mathrm{diam}\,f(g^{-1}(E_{j})))^{p}\leq\sum C^{p}(% \mathrm{diam}\,E_{j})^{p}\leq C^{p}(H_{p}(g(A))+\epsilon).$

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

 $H_{p}(f(A))\leq C^{p}(H_{p}(g(A))+\epsilon).$

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

 $H_{p}(f(A))\leq C^{p}H_{p}(g(A)).$

## 3 Hausdorff dimension

###### Theorem 5.

If $H_{p}(A)<\infty$ then $H_{q}(A)=0$ for all $q>p$.

###### Proof.

Let $\delta>0$. Then $H_{p,\delta}(A)\leq H_{p}(A)<\infty$ Let $\{E_{j}\}$ be countably many sets each with diameter $\leq\delta$ such that $A\subset\bigcup E_{j}$ and

 $\sum(\mathrm{diam}\,E_{j})^{p}\leq H_{p,\delta}(A)+1\leq H_{p}(A)+1.$

This gives us

 $\displaystyle H_{q.\delta}(A)\leq\sum(\mathrm{diam}\,E_{j})^{q}$ $\displaystyle=\sum(\mathrm{diam}\,E_{j})^{q-p}(\mathrm{diam}\,E_{j})^{p}$ $\displaystyle\leq\delta^{q-p}\sum(\mathrm{diam}\,E_{j})^{p}$ $\displaystyle\leq\delta^{q-p}(H_{p}(A)+1).$

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

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

 $\inf\{q\geq 0:H_{q}(A)=0\}.$

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

## 4 Radon measures and Haar measures

Before speaking about Hausdorff measure on $\mathbb{R}^{n}$, 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 $E$,

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

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

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

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.66 6 Gerald B. Folland, Real Analysis, second ed., p. 217, Theorem 7.8.

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 $E\in\mathscr{B}_{G}$,

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

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 $\mu=c\nu$.77 7 Gerald B. Folland, Real Analysis, second ed., p. 344, Theorem 11.9.

## 5 Hausdorff measure in Rn

Let $m_{n}$ denote Lebesgue measure on $\mathbb{R}^{n}$.

###### Lemma 6.

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

 $H_{n}(E)\geq 2^{n}m_{n}(E).$
###### Proof.

Let $\epsilon>0$ and let $\{E_{j}\}$ be countably many closed sets that cover $E$ and such that

 $\sum(\mathrm{diam}\,E_{j})^{n}\leq H_{n}(E)+\epsilon.$

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

 $m_{n}(A)\leq\left(\frac{\mathrm{diam}\,A}{2}\right)^{n}.$

Using this gives

 $\sum 2^{n}m_{n}(E_{j})\leq H_{n}(E)+\epsilon.$

But because the sets $E_{j}$ cover $E$ we have $m_{n}(E)\leq m_{n}(\bigcup E_{j})\leq\sum m_{n}(E_{j})$, so we get

 $m_{n}(E)\leq\frac{H_{n}(E)+\epsilon}{2^{n}}.$

This expression does not involve the sets $E_{j}$ (which depend on $\epsilon$), and since this expression is true for any $\epsilon>0$, taking $\epsilon\to 0$ yields

 $m_{n}(E)\leq\frac{H_{n}(E)}{2^{n}}.$

Let

 $Q=\left\{x\in\mathbb{R}^{n}:|x_{1}|\leq\frac{1}{2},\ldots,|x_{n}|\leq\frac{1}{% 2}\right\}.$
###### Lemma 7.

$0.

###### Proof.

For any $m\geq 1$, the cube $Q$ can be covered by $m^{n}$ cubes $q_{1},\ldots,q_{m^{n}}$ of side length $\frac{1}{m}$. Let $0<\delta<1$ and let $m>\frac{1}{\delta}$. The distance from the center of $q_{j}$ to one of the vertices of $q_{j}$ is

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

Inscribe $q_{j}$ in a closed ball $b_{j}$ with the same center as $q_{j}$ and radius $r$. These balls cover $Q$. Hence

 $H_{p,\delta}(Q)\leq\sum_{j=1}^{m^{n}}(\mathrm{diam}\,b_{j})^{n}=\sum_{j=1}^{m^% {n}}(2r)^{n}=(2r)^{n}\cdot m^{n}=n^{n/2}.$

Taking $\delta\to 0$ gives $H_{p}(Q)\leq n^{n/2}<\infty$.

On the other hand, by Lemma 6,

 $H_{n}(Q)\geq 2^{n}m_{n}(Q)=2^{n}>0.$

###### Theorem 8.

There is some constant $c_{n}>0$ such that

 $H_{n}=c_{n}m_{n}.$
###### Proof.

$\mathbb{R}^{n}$ is a locally compact Hausdorff space in which every open set in $\mathbb{R}^{n}$ 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 $\mathbb{R}^{n}$, there is some $r>0$ such that $K\subset rQ$. By Lemma 4 and Lemma 7 we get $H_{n}(rQ)<\infty$, so $H_{n}(K)<\infty$. 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 $\mathbb{R}^{n}\to\mathbb{R}^{n}$, so $H_{n}$ is invariant under translations. Thus $H_{n}$ is a left Haar measure on $\mathbb{R}^{n}$. But Lebesgue measure $m_{n}$ is also a left Haar measure on $\mathbb{R}^{n}$, so there is some $c_{n}>0$ such that

 $H_{n}=c_{n}m_{n},$

proving the claim. ∎