## 数学代写|几何测度论代写geometric measure theory代考|Lebesgue–Besicovitch differentiation theorem

Let $\mu$ and $v$ be Radon measures on $\mathbb{R}^n$. The upper $\mu$-density and the lower $\mu$ density of $v$ are the functions $D_\mu^{+} v: \operatorname{spt} \mu \rightarrow[0, \infty]$ and $D_\mu^{-} v: \operatorname{spt} \mu \rightarrow[0, \infty]$, defined as, respectively,
$$D_\mu^{+} v(x)=\underset{r \rightarrow 0^{+}}{\lim \sup } \frac{v(\bar{B}(x, r))}{\mu(\bar{B}(x, r))}, \quad D_\mu^{-} v(x)=\liminf {r \rightarrow 0^{+}} \frac{v(\bar{B}(x, r))}{\mu(\bar{B}(x, r))}, \quad x \in \operatorname{spt} \mu .$$ If the two limits exist and are finite, then we denote by $D\mu v(x)$ their common value, and call it the $\mu$-density of $v$ at $x$. We have thus defined a function
$$D_\mu v:\left{x \in \operatorname{spt} \mu: D_\mu^{+} v(x)=D_\mu^{-} v(x)\right} \rightarrow[0, \infty] .$$
Remark 5.7 By Exercise 4.27, and since spt $\mu$ is a closed set, $D_\mu^{+} v$ and $D_\mu^{-} v$ are Borel functions, which, by Remark $4.2$, we may consider as defined on the whole of $\mathbb{R}^n$. With the same caveat, $D_\mu v$ is a Borel function on $\mathbb{R}^n$. By Proposition 2.16, for every $x \in \mathbb{R}^n$ there exist at most countably many values of $r>0$ such that either $\mu(\partial B(x, r))>0$ or $v(\partial B(x, r))>0$. As a consequence, if $D_\mu v$ is defined at $x$, then it satisfies
$$D_\mu v(x)=\lim {r \rightarrow 0^{+}} \frac{v(B(x, r))}{\mu(B(x, r))} .$$ In other words, in evaluating $D\mu v$, we may indifferently use open or closed balls. The use of closed balls in the definition of $D_\mu^{+} v$ and $D_\mu^{-} v$ is instead necessary in order to apply Vitali’s property in the proof of the following theorem.
Theorem $5.8$ (Lebesgue-Besicovitch differentiation theorem) If $\mu$ and $v$ are Radon measures on $\mathbb{R}^n$, then $D_\mu v$ is defined $\mu$-a.e. on $\mathbb{R}^n, D_\mu v \in L_{\text {loc }}^1\left(\mathbb{R}^n, \mu\right)$, and, in fact, $D_\mu v$ is Borel measurable on $\mathbb{R}^n$. Furthermore,
$$v=\left(D_\mu v\right) \mu+v_\mu^s \quad \text { on } \mathcal{M}(\mu),$$
where the Radon measure $v_\mu^8$ is concentrated on the Borel set
\begin{aligned} Y &=\mathbb{R}^n \backslash\left{x \in \operatorname{spt} \mu: D_\mu^{+} v(x)<\infty\right} \ &=\left(\mathbb{R}^n \backslash \operatorname{spt} \mu\right) \cup\left{x \in \operatorname{spt} \mu: D_\mu^{+} v(x)=\infty\right} . \end{aligned}

## 数学代写|几何测度论代写geometric measure theory代考|Lebesgue points

By the mean value theorem, if $u \in C^0(\mathbb{R})$ and $\mu$ is a Radon measure on $\mathbb{R}^n$, then $\lim {r \rightarrow 0^{+}} \frac{1}{\mu(B(x, r))} \int{B(x, r)}|u(x)-u| \mathrm{d} \mu=0, \quad \forall x \in \mathbb{R}^n$.
If now $u \in L_{\mathrm{loc}}^1\left(\mathbb{R}^n, \mu\right)$, then this property still holds at $\mu$-a.e. $x \in \mathbb{R}^n$.
Theorem $5.16$ (Lebesgue points theorem) If $\mu$ is a Radon measure on $\mathbb{R}^n$, $p \in[1, \infty)$ and $u \in L_{\mathrm{loc}}^p\left(\mathbb{R}^n, \mu\right)$, then for $\mu$-a.e. $x \in \mathbb{R}^n$
$$\lim {r \rightarrow 0^{+}} \frac{1}{\mu(B(x, r))} \int{B(x, r)}|u(x)-u|^p \mathrm{~d} \mu=0 .$$
In this case, we say that $x$ is a Lebesgue point of $u$ with respect to $\mu$.
Example 5.17 Given $E \subset \mathbb{R}^n$ and $x \in \mathbb{R}^n$, if the limit
$$\theta_n(E)(x)=\lim {r \rightarrow 0^{+}} \frac{|E \cap B(x, r)|}{\omega_n r^n}$$ exists, it is called the $n$-dimensional density of $E$ at $x$. By the same argument as in Remark $5.7, \theta_n(E)$ defines a Borel function $\mathbb{R}^n$. If $E$ is a Lebesgue measurable set in $\mathbb{R}^n$, and we apply Theorem $5.16$ to the Radon measure $\mu=\mathcal{L}^n\llcorner E$, then we deduce immediately that $\theta_n(E)(x)$ exists for a.e. $x \in \mathbb{R}^n$. In particular, $$\theta_n(E)=1 \text { a.e. on } E, \quad \theta_n(E)=0 \text { a.e. on } \mathbb{R}^n \backslash E \text {. }$$ Given $t \in[0,1]$, the set of points of density $t$ of $E$ is defined as $$E^{(t)}=\left{x \in \mathbb{R}^n: \theta_n(E)(x)=t\right},$$ and it turns out to be a Borel set. Every Lebesgue measurable set is equivalent to the set of its points of density one, since, by (5.18), $$\left|E \Delta E^{(1)}\right|=0, \quad\left|\left(\mathbb{R}^n \backslash E\right) \Delta E^{(0)}\right|=0 .$$ Proof of Theorem $5.16$ We first note that for $\mu$-a.e. $x \in \mathbb{R}^n$ $$\lim {r \rightarrow 0^{+}} \frac{1}{\mu(B(x, r))} \int_{B(x, r)} u \mathrm{~d} \mu=u(x) .$$

## 数学代写|几何测度论代写geometric measure theory代考|Lebesgue–Besicovitch differentiation theorem

$$D_\mu^{+} v(x)=\limsup {r \rightarrow 0^{+}} \frac{v(\bar{B}(x, r))}{\mu(\bar{B}(x, r))}, \quad D\mu^{-} v(x)=\liminf r \rightarrow 0^{+} \frac{v(\bar{B}(x, r))}{\mu(\bar{B}(x, r))}, \quad x \in \operatorname{spt} \mu .$$

$$D_\mu v(x)=\lim r \rightarrow 0^{+} \frac{v(B(x, r))}{\mu(B(x, r))} .$$

$$v=\left(D_\mu v\right) \mu+v_\mu^s \quad \text { on } \mathcal{M}(\mu),$$

## 数学代写|几何测度论代写geometric measure theory代考|Lebesgue points

$$\lim r \rightarrow 0^{+} \frac{1}{\mu(B(x, r))} \int B(x, r)|u(x)-u|^p \mathrm{~d} \mu=0 .$$

$$\theta_n(E)(x)=\lim r \rightarrow 0^{+} \frac{|E \cap B(x, r)|}{\omega_n r^n}$$

$$\theta_n(E)=1 \text { a.e. on } E, \quad \theta_n(E)=0 \text { a.e. on } \mathbb{R}^n \backslash E .$$

$E^{\wedge}{(t)}=\backslash l_1 f t\left{x \backslash\right.$ in $\backslash$ mathbb ${R}^{\wedge} n: \backslash t$ theta_n$\left.(E)(x)=t \backslash r_{i g h t}\right}$,

$$\left|E \Delta E^{(1)}\right|=0, \quad\left|\left(\mathbb{R}^n \backslash E\right) \Delta E^{(0)}\right|=0 .$$

$$\lim r \rightarrow 0^{+} \frac{1}{\mu(B(x, r))} \int_{B(x, r)} u \mathrm{~d} \mu=u(x)$$

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