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## 数学代写|几何测度论代写geometric measure theory代考|Radon measures. Restriction, support

The results of Section $2.3$ motivate the following crucial definition. An outer measure $\mu$ is a Radon measure on $\mathbb{R}^n$ if it is a locally finite, Borel regular measure on $\mathbb{R}^n$. By Theorem $2.10$, if $\mu$ is a Radon measure on $\mathbb{R}^n$, then
\begin{aligned} \mu(E) &=\inf {\mu(A): E \subset A, A \text { is open }} \ &=\sup {\mu(K): K \subset E, K \text { is compact }} \end{aligned} for every Borel set $E \subset \mathbb{R}^n$. Thus, by Borel regularity, a Radon measure $\mu$ is characterized on $\mathcal{M}(\mu)$ by its behavior on compact (or open) sets.

Example 2.11 By Example 2.4, the Lebesgue measure, which is trivially locally finite, is a Radon measure. If $s \in[0, n)$, then $\mathcal{H}^s$ is not locally finite (as $\mathcal{H}^s(A)=\infty$ for every open set $A$ : see Chapter 3), and thus it is not a Radon measure. However, if $E$ is a Borel set with $\mathcal{H}^s(E)<\infty$, then the restriction $\mathcal{H}^s\left\llcorner E\right.$ of $\mathcal{H}^s$ to $E$ is a Radon measure on $\mathbb{R}^n$; see Proposition $2.13$ below.
We now notice that (2.6) and (2.7) have in fact a wider range of validity.
Proposition 2.12 If $\mu$ is a Radon measure, then (2.6) holds true for every $E \subset \mathbb{R}^n$, while (2.7) holds true for every $E \in \mathcal{M}(\mu)$.

Proof Step one: We prove (2.6) for $E \subset \mathbb{R}^n$. By Borel regularity, there exists a Borel set $F$ with $E \subset F$ and $\mu(E)=\mu(F)$. By (2.6) (applied to $F$ ),
\begin{aligned} \mu(E)=\mu(F) &=\inf {\mu(A): F \subset A, A \text { is open }} \ & \geq \inf {\mu(A): E \subset A, A \text { is open }} \geq \mu(E) . \end{aligned}
Step two: We prove (2.7) for $E \in \mathcal{M}(\mu)$. As (2.7) holds true on closed set in $\mathbb{R}^n$, it is enough to prove that
$$\mu(E)=\sup {\mu(C): C \subset E, C \text { is closed }} .$$

## 数学代写|几何测度论代写geometric measure theory代考|Hausdorff measures and the notion of dimension

We begin our discussion by introducing a measure-theoretic notion of dimension. Precisely, given $E \subset \mathbb{R}^n$ we define the Hausdorff dimension of $E$ as
$$\operatorname{dim}(E)=\inf \left{s \in[0, \infty): \mathcal{H}^s(E)=0\right} .$$
Its use as a notion of dimension is justified by the following statements.
(i) If $E \subset \mathbb{R}^n$ then $\operatorname{dim}(E) \in[0, n]$. Moreover $\mathcal{H}^s(E)=\infty$ for every $s<$ $\operatorname{dim}(E)$ and $\mathcal{H}^s(E) \in(0, \infty)$ implies $s=\operatorname{dim}(E)$ (the converse is not necessarily true: it may happen that $\mathcal{H}^s(E) \in{0,+\infty}$ for $s=\operatorname{dim}(E)$ ). (ii) $\mathcal{H}^0$ is the counting measure. (iii) If $E$ is a curve, then $\mathcal{H}^1(E)$ coincides with the classical length of $E$. (iv) If $1 \leq k \leq n-1, k \in \mathbb{N}$, and $E$ is a $k$-dimensional $C^1$-surface, then $\mathcal{H}^k(E)$ coincides with the classical $k$-dimensional area of $E$. (v) If $E \subset \mathbb{R}^n$, then $\mathcal{H}^n(E)=\mathcal{L}^n(E)$. (vi) If $s>n$, then $\mathcal{H}^s=0$.
(vii) If $A$ is an open set in $\mathbb{R}^n$, then $\operatorname{dim}(A)=n$.
(viii) For every $s \in[0, n]$ there exists a compact set $K$ such that $\operatorname{dim}(K)=s$.

We now prove properties (i), (ii), and (vi). Properties (iii) and (v) are proved in Sections $3.2$ and 3.3, respectively. Property (vii) follows from (i) and (v), since $\left|(0,1)^n\right|=1$. Property (iv) is a consequence of the area formula; see Chapter 8. For property (viii), see [Hut81] and [Fal86].
Proposition 3.1 If $s>n$, then $\mathcal{H}^s=0$.
Proof Let $Q=(0,1)^n$. Since $\lambda^s \mathcal{H}^s(Q)=\mathcal{H}^s(\lambda Q) \rightarrow \mathcal{H}^s\left(\mathbb{R}^n\right)$ as $\lambda \rightarrow \infty$, it suffices to prove $\mathcal{H}^s(Q)=0$. This follows by letting $k \rightarrow \infty$ in the following inequalities, which are obtained by considering a partition of $Q$ by $k^n$ cubes of diameter $k^{-1} \sqrt{n}$ :
$$\mathcal{H}_{\sqrt{n} / k}^s(Q) \leq \omega_s k^n\left(\frac{\sqrt{n}}{2 k}\right)^s=\frac{\omega_s n^{s / 2}}{2^s} k^{n-s} .$$
Proposition $3.2$ If $E \subset \mathbb{R}^n$, then $\operatorname{dim}(E) \in[0, n]$, and $\mathcal{H}^s(E)=\infty$ for every $s<\operatorname{dim}(E)$.

# 几何测度论代考

## 数学代写|几何测度论代写geometric measure theory代考|Radon measures. Restriction, support

$$\mu(E)=\inf \mu(A): E \subset A, A \text { is open } \quad=\sup \mu(K): K \subset E, K \text { is compact }$$

$$\mu(E)=\mu(F)=\inf \mu(A): F \subset A, A \text { is open } \quad \geq \inf \mu(A): E \subset A, A \text { is open } \geq \mu(E) .$$

$$\mu(E)=\sup \mu(C): C \subset E, C \text { is closed . }$$

## 数学代写|几何测度论代写geometric measure theory代考|Hausdorff measures and the notion of dimension

《left 的分隔符缺失或无法识别

$\left(\right.$ (一) 如果 $E \subset \mathbb{R}^n$ 然后 $\operatorname{dim}(E) \in[0, n]$. 而且 $\mathcal{H}^s(E)=\infty$ 对于每个 $s<\operatorname{dim}(E)$ 和 $\mathcal{H}^s(E) \in(0, \infty)$ 暗示 $s=\operatorname{dim}(E)$ (反过来不一定正确: 可能会发生 $\mathcal{H}^s(E) \in 0,+\infty$ 为了 $\left.s=\operatorname{dim}(E)\right)$ 。(二) $\mathcal{H}^0$ 是计数度量。(iii) 如果 $E$ 是一条曲线，那么 $\mathcal{H}^1(E)$ 与经典长度一致 $E$. (iv) 如果 $1 \leq k \leq n-1, k \in \mathbb{N}$ ，和 $E$ 是一个 $k$ 维 $C^1$-表面，然后 $\mathcal{H}^k(E)$ 与经典不谋而合 $k$-维面积 $E$. (v) 如果 $E \subset \mathbb{R}^n$ ，然后 $\mathcal{H}^n(E)=\mathcal{L}^n(E)$. (vi) 如果 $s>n$ ，然后 $\mathcal{H}^s=0$.
(vii) 如果 $A$ 是一个开集 $\mathbb{R}^n$ ，然后 $\operatorname{dim}(A)=n$.
(viii) 对于每个 $s \in[0, n]$ 存在一个紧集 $K$ 这样 $\operatorname{dim}(K)=s$.

$$\mathcal{H}_{\sqrt{n} / k}^s(Q) \leq \omega_s k^n\left(\frac{\sqrt{n}}{2 k}\right)^s=\frac{\omega_s n^{s / 2}}{2^s} k^{n-s}$$

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assignmentutor™您的专属作业导师
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