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Hausdorff measure

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hbghlyj posted 2022-5-4 18:18 |Read mode
A Course in Metric Geometry
The following theorem provides a basis for measure theory in Euclidean spaces.
Theorem 1.7.5 (Lebesgue). There exists a unique Borel measure $m_{n}$ over $\mathbb{R}^{n}$ which is invariant under parallel translations and such that $m_{n}\left([0,1]^{n}\right)=1$.

The measure $m_{n}$ is called Lebesgue measure. The uniqueness part of the theorem implies that every translation-invariant Borel measure in $\mathbb{R}^{n}$ with a finite value for a cube is a constant multiple of $m_{n}$.
Exercise 1.7.6. Prove that
1. Lebesgue measure is invariant under isometries of $\mathbb{R}^{n}$.
2. If $L: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ is a linear map, then $m_{n}(L(A))=|\operatorname{det} L| \cdot m_{n}(A)$ for any measurable set $A \subset \mathbb{R}^{n}$. In particular, a homothety with coefficient $C$ multiplies the Lebesgue measure by $C^{n}$.
Hint: Use the uniqueness part of Theorem 1.7.5.
1.7.2. Hausdorff measure. To motivate the definition of Hausdorff measures, let us recall the main idea of the construction behind the proof of Theorem 1.7.5. It begins with choosing a class of simple sets (such as balls or cubes). Then a set (from an appropriate $\sigma$-algebra) is covered by simple sets; then the Lebesgue measure of the set is defined as the infimum of total measures of such covers. Speaking about “the total measure of a cover” one means here that certain measure is already assigned to simple sets.
To define Hausdorff $n$-dimensional measure on a metric space, one could proceed along the same lines: cover a set by metric balls such that all their radii are less than $ε$. For each ball, consider a Euclidean ball of the same radius and add their volumes for all balls from the cover: this will be the total measure of the cover. Taking its infimum over all covers and passing to the limit as $ε$ approaches zero, one gets a version of Hausdorff measure of the set. Instead of adding volumes of Euclidean balls of the same radii, one could simply add the radii of balls from the cover raised to the power $n$: the result is the same up to a constant multiplier. It turns out that arbitrary sets and diameters are technically more convenient to use than metric balls and radii.
Now we pass to formal definitions.
Definition 1.7.7. Let $X$ be a metric space and $d$ be a nonnegative real number.

For a finite or countable covering $\left\{S_{i}\right\}_{i \in I}$ of $X$ (that is, a collection of sets such that $\left.X \subset \bigcup S_{i}\right)$, define its $d$-weight $w_{d}\left(\left\{S_{i}\right\}\right)$ by the formula
$$
w_{d}\left(\left\{S_{i}\right\}\right)=\sum_{i}\left(\operatorname{diam} S_{i}\right)^{d}
$$
If $d=0$, substitute each (if any) $0^{0}$ term in the formula by 1.
For an $\varepsilon>0$ define $\mu_{d, \varepsilon}(X)$ by
$$
\mu_{d, \varepsilon}(X)=\inf \left\{w_{d}\left(\left\{S_{i}\right\}\right): \operatorname{diam}\left(S_{i}\right)<\varepsilon \text { for all } i\right\} \text {. }
$$
The infimum is taken over all finite or countable coverings of $X$ by sets of diameter $<\varepsilon$; if no such covering exists, then the infimum is $+\infty$.
The $d$-dimensional Hausdorff measure of $X$ is defined by the formula
$$
\mu_{d}(X)=C(d) \cdot \lim _{\varepsilon \rightarrow 0} \mu_{d, \varepsilon}(X)
$$
where $C(d)$ is a positive normalization constant. This constant is introduced for only one reason: for integer $d$ it is convenient to choose $C(d)$ so that the $d$-dimensional Hausdorff measure of a unit cube in $\mathbb{R}^{n}$ equals 1. In fact, almost nothing depends on the actual value of $C(d)$.

It may be unclear from the definition what the value of $\mu_{d}(\emptyset)$ is. We explicitly define $\mu_{d}(\emptyset)=0$ for all $d \geq 0$.

Clearly $\mu_{d, \varepsilon}(X)$ is a nonincreasing function of $\varepsilon$. Since such a function has a (possibly infinite) limit as $\varepsilon \rightarrow 0, \mu_{d}(X)$ is well defined for any metric space $X$. It may be either a nonnegative real number or $+\infty$.

Though we have defined Hausdorff measure for a metric space, this notion will often be applied to subsets of metric spaces. In such cases, a subset should be considered as a metric space with the restricted metric. (Note that the definition can be read verbatim if $X$ is a subset of a larger metric space; it does not matter whether covering sets $S_i$ are actually contained in $X$.)
The following proposition summarizes the properties of Hausdorff measure that follow immediately from the definition.
Proposition 1.7.8. Let $X$ and $Y$ be metric spaces, and let $A$ and $B$ be subsets of $X$. Then
(1) If $A \subset B$, then $\mu_{d}(A) \leq \mu_{d}(B)$.
(2) $\mu_{d}\left(\bigcup A_{i}\right) \leq \sum \mu_{d}\left(A_{i}\right)$ for any finite or countable collection of sets $A_{i} \subset X$.
(3) If $\operatorname{dist}(A, B)>0$, then $\mu_{d}(A \cup B)=\mu_{d}(A)+\mu_{d}(B)$.
(4) If $f: X \rightarrow Y$ is a Lipschitz map with a Lipschitz constant $C$, then $\mu_{d}(f(X)) \leq C^{d} \cdot \mu_{d}(X) .$
(5) If $f: X \rightarrow Y$ is a $C$-homothety, i.e., $\left|f\left(x_{1}\right) f\left(x_{2}\right)\right|=C\left|c_{1} x_{2}\right|$ for all $x_{1}, x_{2} \in X$, then $\mu_{d}(f(X))=C^{d} \cdot \mu_{d}(X)$.

According to Carathéodory's criterion, ([Fe], 2.3.1(9)), any nonnegative function on the Borel $\sigma$-algebra of $X$ possessing the properties 1-3 from Proposition 1.7.8 is actually a measure. Thus we obtain

Theorem 1.7.9. For any metric space $X$ and any $d \geq 0, \mu_{d}$ is a measure on the Borel $\sigma$-algebra of $X$.

Exercise 1.7.10. Prove that 0-dimensional Hausdorff measure of a set is its cardinality. In other words, $\mu_{0}(X)$ is a number of points in $X$ if $X$ is a finite set, and $\mu_{0}(X)=\infty$ if $X$ is an infinite set.

Exercise 1.7.11. Let $X$ and $Y$ be metric spaces and $f: X \rightarrow Y$ a locally Lipschitz map with dilatation $\leq C$. Prove that $\mu_{d}(f(X)) \leq C^{d} \cdot \mu_{d}(X)$ assuming that (a) $X$ is compact; (b) $X$ has a countable topological base.
1.7.3. Hausdorff measure in $\mathbb{R}^{n}$. Let $I$ denote the interval $[0,1]$ of $\mathbb{R}$. Then $I^{n}=[0,1]^{n}$ is the unit cube in $\mathbb{R}^{n}$.
Theorem 1.7.12. $0<\mu_{n}\left(I^{n}\right)<\infty$.
Exercise 1.7.13. Prove the theorem.
Now we can define the normalization constant $C(n)$ from the definition of Hausdorff measure. Namely, choose $C(n)$ so that $\mu_{n}\left(I^{n}\right)=1$. The existence of such a constant follows from Theorem 1.7.12. Theorem 1.7.5 then implies that Hausdorff measure $\mu_{n}$ on $\mathbb{R}^{n}$ coincides with the standard $n$-dimensional volume $m_{n}$.

In most cases, the actual value of $C(n)$ is not important. However it is an interesting fact that $C(n)$ equals the volume of the Euclidean $n$-ball of diameter 1. The proof is based on the following theorem.

Theorem 1.7.14 (Vitali's Covering Theorem). Let $X$ be a bounded set in $\mathbb{R}^{n}$ and let $\mathfrak{B}$ be a collection of closed balls in $\mathbb{R}^{n}$ such that for every $x \in X$ and $\varepsilon>0$ there is a ball $B \in \mathfrak{B}$ such that $x \in B$ and $\operatorname{diam}(B)<\varepsilon$. Then $\mathfrak{B}$ contains a finite or countable subcollection $\left\{B_{i}\right\}$ of disjoint balls which covers $X$ up to a set of zero measure, i.e., such that $B_{i} \cap B_{j}=\emptyset$ if $i \neq j$ and $\mu_{n}\left(X \backslash \bigcup_{i} B_{i}\right)=0$.

Proof. We may assume that every ball $B \in \mathfrak{B}$ contains at least one point of $X$ and exclude the balls with radius greater than 1. Then all these balls are contained in the 2-neighborhood of $X$ which is bounded and hence has finite volume. We construct a sequence $\left\{B_{i}\right\}_{i=1}^{\infty}$ of balls by induction. If $B_{1}, \ldots, B_{m}$ are already constructed, we choose the next ball $B_{m+1}$ as follows. Let $\mathfrak{B}_{m}$ denote the set of balls from the collection that do not intersect any of $B_{1}, \ldots, B_{m}$. If $\mathfrak{B}_{m}$ is empty, then $B_{1} \cup \cdots \cup B_{m}$ covers the entire set $X$ and the proof is finished (this follows from the condition that every point is covered by balls of arbitrarily small radii). If $\mathfrak{B}_{m}$ is not empty, choose $B_{m+1}$ to be any element of $\mathfrak{B}_{m}$ with
$$
\operatorname{diam}\left(B_{m+1}\right)>\frac{1}{2} \sup \left\{\operatorname{diam}(B): B \in \mathfrak{B}_{m}\right\} .
$$
The balls $B_{i}$ are disjoint by the construction. We will now show that they cover $X$ up to a set of zero measure. Fix an $\varepsilon>0$. Since the balls are disjoint and are contained in a set of finite volume, we have $\sum_{i=0}^{\infty} \mu_{n}\left(B_{i}\right)<\infty$. Hence there is an index $m$ such that $\sum_{i=m+1}^{\infty} \mu_{n}\left(B_{i}\right)<\varepsilon$. Let $x \in X \backslash \bigcup_{i} B_{i}$ and let $B$ be any ball from the collection that contains $x$ and does not intersect the balls $B_{1}, \ldots, B_{m}$. Note that $B$ must intersect $\bigcup_{i} B_{i}$ because otherwise $B \in \mathfrak{B}_{m}$ for all $m$ which contradicts that $\mu_{n}\left(B_{i}\right) \rightarrow 0$. Let $k$ be the minimal index such that $B \cap B_{k} \neq \emptyset$. Then $B \in \mathfrak{B}_{m-1}$ and hence $\operatorname{diam}\left(B_{k}\right)>\frac{1}{2} \operatorname{diam}(B)$ by (1.1). It follows that the distance from $x$ to the center of $B_{k}$ is not greater than 5 times the radius of $B_{k}$. Hence $x$ belongs to the ball with the same center as $B_{k}$ and radius 5 times larger. We denote this ball by $5 B_{k}$.

We have just proved that every $x \in X \backslash \bigcup_{i} B_{i}$ belongs to a ball $5 B_{k}$ for some $k>m$. Thus $X \backslash \bigcup_{i} B_{i} \sup \bigcup_{i=m+1}^{\infty}\left(5 B_{i}\right)$; hence
$$
\mu_{n}\left(X \backslash \bigcup_{i} B_{i}\right) \leq \sum_{i=m+1}^{\infty} \mu_{n}\left(5 B_{i}\right)=5^{n} \sum_{i=m+1}^{\infty} \mu_{n}\left(B_{i}\right)<5^{n} \varepsilon
$$
Since $\varepsilon$ is arbitrary, it follows that $\mu_{n}\left(X \backslash \bigcup_{i} B_{i}\right)=0$.
Corollary 1.7.15. The normalization constant for the $n$-dimensional Hausdorff measure equals the volume of the Euclidean $n$-ball of diameter 1.

Proof. Let $C_{n}$ denote the constant from the formulation. Then the volume of a Euclidean $n$-ball equals $C_{n} d^{n}$ where $d$ is its diameter. Let $\mu_{n}^{\prime}$ be the $n$-dimensional Hausdorff measure with normalization constant $C_{n}$. We have to prove that $\mu_{n}^{\prime}=\mu_{n}$, i.e., that $\mu_{n}^{\prime}\left(I^{n}\right)=1$.
1. $\mu_{n}^{\prime}\left(I^{n}\right) \leq 1$. To prove this, apply Theorem $1.7 .14$ to the set $\mathfrak{B}$ of all closed Euclidean balls contained in $I^{n}$. This yields a countable collection $\left\{B_{i}\right\}$ of such balls such that the set $Y=I^{n} \backslash \bigcup B_{i}$ has zero measure. Hence $\mu_{n}^{\prime}\left(I^{n}\right) \leq \mu_{n}^{\prime}(Y)+\sum C_{n} \operatorname{diam}\left(B_{i}\right)^{n}=0+\sum m_{n}\left(B_{i}\right) \leq m_{n}\left(I^{n}\right)=1$.
2. $\mu_{n}^{\prime}\left(I^{n}\right) \geq$ 1. By a well-known Bieberbach inequality (cf. e.g. [BZ], Theorem 11.2.1), a Euclidean ball has the maximal volume among the sets with the same diameter. Hence $m_{n}(S) \leq C_{n} \operatorname{diam}(S)^{n}$ for any bounded set $S \subset \mathbb{R}^{n}$. Now if $\left\{S_{i}\right\}_{i=1}^{\infty}$ is a covering of $I^{n}$, then $1=m_{n}\left(I^{n}\right) \leq \sum m_{n}\left(S_{i}\right) \leq$ $\sum C_{n} \operatorname{diam}(S)^{n}$. The statement follows.
1.7.4. Hausdorff dimension. The next theorem tells us how the Hausdorff measure of a fixed set depends on dimension. Briefly, the measure is zero or infinite for all dimensions except at most one. More precisely, there is a "critical dimension" below which the measure is infinity and above which the measure is zero. This dimension is an important characteristic of a metric space, called the Hausdorff dimension. Warning: at the critical dimension, all three possibilities (the measure is zero, positive number or $+\infty$) may take place.

Theorem 1.7.16. For a metric space $X$ there exists a $d_{0} \in[0,+\infty]$ such that $\mu_{d}(X)=0$ for all $d>d_{0}$ and $\mu_{d}(X)=\infty$ for all $d<d_{0}$.

Proof. Define $d_{0}=\inf \left\{d \geq 0: \mu_{d}(X) \neq \infty\right\}$. Trivially $\mu_{d}(X)=\infty$ for all $d<d_{0}$. If $d>d_{0}$, there is a $d^{\prime}<d$ such that $\mu_{d^{\prime}}(X)=M<\infty$. Therefore for any $\varepsilon>0$ there exists a covering $\left\{S_{i}\right\}$ of $X$ such that $\operatorname{diam} S_{i}<\varepsilon$ for all $i$ and $\sum\left(\operatorname{diam} S_{i}\right)^{d^{\prime}}<2 M$. Then
$$
\sum\left(\operatorname{diam} S_{i}\right)^{d} \leq \varepsilon^{d-d^{\prime}} \cdot \sum\left(\operatorname{diam} S_{i}\right)^{d^{\prime}} \leq 2 \varepsilon^{d-d^{\prime}} M
$$
Hence $\mu_{d, \varepsilon}(X) \leq 2 \varepsilon^{d-d'} M$. Since $\varepsilon^{d-d'} \rightarrow 0$ as $\varepsilon \rightarrow 0$, we have $\mu_{d}(X)=0$.

Definition 1.7.17. The value $d_{0}$ from Theorem 1.7.16 is called the Hausdorff dimension of $X$ and denoted by $\operatorname{dim}_{H}(X)$.
Remark 1.7.18. Hausdorff dimension is not necessarily integer.

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original poster hbghlyj posted 2022-5-4 18:19
en.wikipedia.org/wiki/List_of_fractals_by_Hausdorff_dimension
225px-Quadriccross[1].gif
Quadric cross的Hausdorff维数为${\displaystyle \log _{\sqrt {5}}\left({\frac {10}{3}}\right)}$

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