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# 数学代写|Support of a measure数论代考

## 数论代考

Let $M$ be a topological space. If $M$ is either second countable (i.e., there is basis of open sets that is countable) or compact, then any Borel measure $\mu$ on M has a well-defined closed support, denoted $\operatorname{supp}(\mu)$, which is characterized by either of the following properties: (1) it is the complement of the largest open set U, with respect to inclusion, such that $\mu(\mathrm{U})=0 ;$ or $(2)$ it is $\mathrm{U}$ of $x$, we have $\mu(\mathrm{U})>0$.

If $\mathrm{X}$ is a random variable with values in $\mathrm{M}$, we will say that the support of $\mathrm{X}$ is the support of the law of X, which is a probability measure on M.
We need the following elementary property of the support of a measure:
LEMMA B.2.1. Let M and $\mathrm{N}$ be topological spaces that are each either second countable or compact. Let $\mu$ be a probability measure on $\mathrm{M}$, and let $f: \mathrm{M} \longrightarrow \mathrm{N}$ be a continuous map. The support of $f_{*}(\mu)$ is the closure of $f(\operatorname{supp}(\mu))$.

We recall that given a probability measure $\mu$ on M and a continuous map $f: \mathrm{M} \rightarrow \mathrm{N}$, the image measure $f_{}(\mu)$ is defined by $$f_{}(\mu)(\mathrm{A})=\mu\left(f^{-1}(\mathrm{~A})\right)$$
for a measurable set $\mathrm{A} \subset \mathrm{N}$, and it satisfies
$$\int_{\mathrm{N}} \varphi(x) d\left(f_{} \mu\right)(x)=\int_{\mathrm{M}} \varphi(f(y)) d \mu(y)$$ for $\varphi \geqslant 0$ and measurable, or $\varphi \circ f$ integrable with respect to $\mu$. Proof. First, if $y=f(x)$ for some $x \in \operatorname{supp}(\mu)$, and if $\mathrm{U}$ is an open neighborhood of $y$, then we can find an open neighborhood $\mathrm{V} \subset \mathrm{M}$ of $x$ such that $f(\mathrm{~V}) \subset \mathrm{U}$. Then $\left(f_{} \mu\right)(\mathrm{U}) \geqslant \mu(\mathrm{V})>0$. This shows that $y$ belongs to the support of $f_{} \mu$. Since the support is closed, we deduce that $\overline{f(\operatorname{supp}(\mu))} \subset \operatorname{supp}\left(f_{} \mu\right)$.

For the converse, let $y \in \mathrm{N}$ be in the support of $f_{} \mu$. For any open neighborhood $\mathrm{U}$ of $y$, we have $\mu\left(f^{-1}(\mathrm{U})\right)=\left(f_{} \mu\right)(\mathrm{U})>0$. This implies that $f^{-1}(\mathrm{U}) \cap \operatorname{supp}(\mu)$ is not empty, and since $\mathrm{U}$ is arbitrary, that $y$ belongs to the closure of $f(\operatorname{supp}(\mu))$.

Recall that a family $\left(\mathrm{X}{i}\right){i \in \mathrm{I}}$ of random variables, each taking possibly values in a different metric space $\mathrm{M}{i}$, is independent if, for any finite subset $\mathrm{J} \subset \mathrm{I}$, the joint distribution of $\left(\mathrm{X}{j}\right){j \in \mathrm{J}}$ is the measure on $\prod \mathrm{M}{j}$ which is the product measure of the laws of the $\mathrm{X}{j}$ ‘s. 132 LEMMA B.2.2. Let $\mathrm{X}=\left(\mathrm{X}{i}\right){i \in \mathrm{I}}$ be a finite family of random variables with values in a topological space $\mathrm{M}$ that is compact or second countable. Viewed as a random variable taking values in $\mathrm{M}^{\mathrm{I}}$, we have $$\operatorname{supp}(\mathrm{X})=\prod{i \in \mathrm{I}} \operatorname{supp}\left(\mathrm{X}{i}\right)$$ Proof. If $x=\left(x{i}\right) \in \mathrm{M}^{\mathrm{I}}$, then an open neighborhood $\mathrm{U}$ of $x$ contains a product set $\prod \mathrm{U}{i}$, where $\mathrm{U}{i}$ is an open neighborhood of $x_{i}$ in M. Then we have
$$\mathrm{P}(\mathrm{X} \in \mathrm{U}) \geqslant \mathrm{P}\left(\mathrm{X} \in \prod_{i} \mathrm{U}{i}\right)=\prod{i} \mathrm{P}\left(\mathrm{X}{i} \in \mathrm{U}{i}\right)$$
by independence. If $x_{i} \in \operatorname{supp}\left(\mathrm{X}{i}\right)$ for each $i$, then this is $>0$, and hence $x \in \operatorname{supp}(\mathrm{X})$. Conversely, if $x \in \operatorname{supp}(\mathrm{X})$, then for any $j \in \mathrm{I}$, and any open neighborhood U of $x{j}$, the set
$$\mathrm{V}=\left{y=\left(y_{i}\right){i \in \mathrm{I}} \in \mathrm{M}^{\mathrm{I}} \mid y{j} \in \mathrm{U}\right} \subset \mathrm{M}^{\mathrm{I}}$$
is an open neighborhood of $x$. Hence we have $\mathrm{P}(\mathrm{X} \in \mathrm{V})>0$, and since $\mathrm{P}(\mathrm{X} \in \mathrm{V})=$ $\mathbf{P}\left(\mathrm{X}{i} \in \mathrm{U}\right)$, it follows that $x{j}$ is in the support of $\mathrm{X}{j}$. B.3. Convergence in law Let M be a metric space. We view it as given with the Borel $\sigma$-algebra generated by open sets, and we denote by $C{b}(M)$ the Banach space of bounded complex-valued continuous functions on M, with the norm
$$|f|_{\infty}=\sup {x \in \mathrm{M}}|f(x)| .$$ Given a sequence $\left(\mu{n}\right)$ of probability measures on M, and a probability measure $\mu$ on M, one says that $\mu_{n}$ converges weakly to $\mu$ if and only if, for any bounded and continuous function $f: \mathrm{M} \longrightarrow \mathbf{R}$, we have
$$\int_{\mathrm{M}} f(x) d \mu_{n}(x) \longrightarrow \int_{\mathrm{M}} f(x) d \mu(x)$$
If $(\Omega, \Sigma, \mathbf{P})$ is a probability space and $\left(\mathrm{X}{n}\right){n \geqslant 1}$ is a sequence of M-valued random variables, and if $\mathrm{X}$ is an $\mathrm{M}$-valued random variable, then one says that $\left(\mathrm{X}_{n}\right)$ converges in

$$\int_{\mathrm{N}} \varphi(x) d\left(f_{} \mu\right)(x)=\int_{\mathrm{M}} \varphi(f(y)) d \亩(y)$$ 对于$\varphi \geqslant 0$ 和可测量的，或$\varphi \circ f$ 可与$\mu$ 积。 证明。首先，如果对于某个 $x \in \operatorname{supp}(\mu)$，$y=f(x)$，并且如果 $\mathrm{U}$ 是 $y$ 的一个开放邻域，那么我们可以找到$x$ 的开放邻域 $\mathrm{V} \subset \mathrm{M}$ 使得 $f(\mathrm{~V}) \subset \mathrm{U}$。那么$\left(f_{} \mu\right)(\mathrm{U}) \geqslant \mu(\mathrm{V})>0$。这说明$y$属于$f_{}\mu$的支持。由于支持是闭合的，我们推导出 $\overline{f(\operatorname{supp}(\mu))} \subset \operatorname{supp}\left(f_{} \mu\right)$。

$$\mathrm{P}(\mathrm{X} \in \mathrm{U}) \geqslant \mathrm{P}\left(\mathrm{X} \in \prod_{i} \mathrm{U}{i} \right)=\prod{i} \mathrm{P}\left(\mathrm{X}{i} \in \mathrm{U}{i}\right)$$

$$\mathrm{V}=\left{y=\left(y_{i}\right){i \in \mathrm{I}} \in \mathrm{M}^{\mathrm{I}} \mid y{j} \in \mathrm{U}\right} \subset \mathrm{M}^{\mathrm{I}}$$

$$|f|_{\infty}=\sup {x \in \mathrm{M}}|f(x)| .$$ 给定 M 上的概率测度序列 $\left(\mu{n}\right)$，以及 M 上的概率测度 $\mu$，可以说 $\mu_{n}$ 弱收敛到 $\mu$当且仅当，对于任何有界连续函数 $f: \mathrm{M} \longrightarrow \mathbf{R}$，我们有

\int_{\mathrm{M}} f(x) d \mu_{n}(x) \longrightarrow

## 数论代写

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## 编码理论代写

1. 数据压缩（或信源编码
2. 前向错误更正（或信道编码
3. 加密编码
4. 线路码

## 复分析代考

(1) 提到复变函数 ，首先需要了解复数的基本性左和四则运算规则。怎么样计算复数的平方根， 极坐标与 $x y$ 坐标的转换，复数的模之类的。这些在高中的时候囸本上都会学过。
(2) 复变函数自然是在复平面上来研究问题，此时数学分析里面的求导数之尖的运算就会很自然的 引入到复平面里面，从而引出解析函数的定义。那/研究解析函数的性贡就是关楗所在。最关键的 地方就是所谓的Cauchy一Riemann公式，这个是判断一个函数是否是解析函数的关键所在。
(3) 明白解析函数的定义以及性质之后，就会把数学分析里面的曲线积分 $a$ 的概念引入复分析中， 定义几乎是一致的。在引入了闭曲线和曲线积分之后，就会有出现复分析中的重要的定理: Cauchy 积分公式。 这个是易分析的第一个重要定理。