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# 数学代写|测度与积分代写Measure And Integration代考|MX4083 Polar Coordinates and Surface Measure

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## 数学代写|测度与积分代写Measure And Integration代考|Polar Coordinates and Surface Measure

Polar Coordinates and Surface Measure. Let
$$S^{d-1}=\left{x \in \mathbb{R}^d:|x|^2:=\sum_{i=1}^d x_i^2=1\right}$$
be the unit sphere in $\mathbb{R}^d$. Let $\Phi: \mathbb{R}^d \backslash(0) \rightarrow(0, \infty) \times S^{d-1}$ and $\Phi^{-1}$ be the inverse map given by
$$\Phi(x):=\left(|x|, \frac{x}{|x|}\right) \text { and } \Phi^{-1}(r, \omega)=r \omega$$
respectively. Since $\Phi$ and $\Phi^{-1}$ are continuous, they are Borel measurable.
Consider the measure $\Phi_* m$ on $\mathcal{B}{(0, \infty)} \otimes \mathcal{B}{S^{d-1}}$ given by
$$\Phi_* m(A):=m\left(\Phi^{-1}(A)\right)$$
for all $A \in \mathcal{B}{(0, \infty)} \otimes \mathcal{B}{S^{d-1}}$. For $E \in \mathcal{B}{S^{d-1}}$ and $a>0$, let $$E_a:={r \omega: r \in(0, a] \text { and } \omega \in E}=\Phi^{-1}((0, a] \times E) \in \mathcal{B}{\mathbb{R}^d} \text {. }$$
Noting that $E_a=a E_1$, we have for $0<a<b, E \in \mathcal{B}_{S^{d-1}}, E$ and $A=(a, b] \times E$ that
\begin{aligned} \Phi^{-1}(A) & ={r \omega: r \in(a, b] \text { and } \omega \in E} \ & =b E_1 \backslash a E_1 . \end{aligned}

## 数学代写|测度与积分代写Measure And Integration代考|Hahn Decomposition Theorem

Definition 13.4. Let $\nu$ be a signed measure on $(X, \mathcal{M})$ and $E \in \mathcal{M}$, then
(1) $E$ is positive if for all $A \in \mathcal{M}$ such that $A \subset E, \nu(A) \geq 0$, i.e. $\left.\nu\right|{\mathcal{M}_E} \geq 0$. (2) $E$ is negative if for all $A \in \mathcal{M}$ such that $A \subset E, \nu(A) \leq 0$, i.e. $\left.\nu\right|{\mathcal{M}E} \leq 0$. (3) $E$ is null if for all $A \in \mathcal{M}$ such that $A \subset E$, i.e. $\left.\nu\right|{\mathcal{M}_E}=0$.
Here $\mathcal{M}_E \equiv{A \cap E: A \in \mathcal{M}}=$ trace of $M$ on $E$.
Lemma 13.5. Suppose that $\nu$ is a signed measure on $(X, \mathcal{M})$. Then
(1) Any subset of a positive set is positive.
(2) The countable union of positive (negative or null) sets is still positive (negative or null).
(3) Let us now further assume that $\nu(\mathcal{M}) \subset[-\infty, \infty)$ and $E \in \mathcal{M}$ is a set such that $\nu(E) \in(0, \infty)$. Then there exists a positive set $P \subseteq E$ such that $\nu(P) \geq \nu(E)$.

Proof. The first assertion is obvious. If $P_j \in \mathcal{M}$ are positive sets, let $P=$ $\bigcup_{n=1}^{\infty} P_n$. By replacing $P_n$ by the positive set $P_n \backslash\left(\bigcup_{j=1}^{n-1} P_j\right)$ we may assume that the $\left{P_n\right}_{n=1}^{\infty}$ are pairwise disjoint so that $P=\bigcup_{n=1}^{\infty} P_n$. Now if $E \subset P$ and $E \in \mathcal{M}$, $E=\coprod_{n=1}^{\infty}\left(E \cap P_n\right)$ so
$$\nu(E)=\sum_{n=1}^{\infty} \nu\left(E \cap P_n\right) \geq 0$$

## 数学代写测度与积分代写Measure And Integration代考|Polar Coordinates and Surface Measure

\left 缺分或无法识别的分隔符

$$\Phi(x):=\left(|x|, \frac{x}{|x|}\right) \text { and } \Phi^{-1}(r, \omega)=r \omega$$

$$\Phi_* m(A):=m\left(\Phi^{-1}(A)\right)$$

$$E_a:=r \omega: r \in(0, a] \text { and } \omega \in E=\Phi^{-1}((0, a] \times E) \in \mathcal{B}^d$$

## 数学代写|测度与积分代写Measure And Integration代考|Hahn Decomposition Theorem

left 缺少或无法识别的分隔符

$$\nu(E)=\sum_{n=1}^{\infty} \nu\left(E \cap P_n\right) \geq 0$$

## MATLAB代写

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