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数学代写|实分析代写real analysis代考| Lebesgue Measure

如果你也在 怎样代写实分析real analysis这个学科遇到相关的难题,请随时右上角联系我们的24/7代写客服。实分析real analysis是数学分析的一个分支,研究实数、实数序列和实数函数的行为。 实分析研究的实值序列和函数的一些特殊性质包括收敛性、极限、连续性、平稳性、可微分性和可整定性。实分析有别于复分析,后者涉及复数及其函数的研究。

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数学代写|实分析代写real analysis代考|Lebesgue Measure

数学代写|实分析代写real analysis代考|Definition and Basic Properties

To motivate the definition of measurability, suppose that $U$ is an open set that contains a set $E$. As we observed above, we do not know whether $|U|{e}$ and $|E|{e}+|U \backslash E|{e}$ will be equal. If it were the case that these quantities were equal, then we could combine this equality with equation (2.9) and infer that $|U \backslash E|{e} \leq \varepsilon$. The “measurable sets” are precisely the sets for which this inequality can be achieved. Here is the explicit definition.

Definition 2.2.1 (Lebesgue Measure). A set $E \subseteq \mathbb{R}^{d}$ is Lebesgue measurable, or simply measurable for short, if
$$
\forall \varepsilon>0, \quad \exists \text { open } U \supseteq E \text { such that }|U \backslash E|{e} \leq \varepsilon . $$ If $E$ is Lebesgue measurable, then its Lebesgue measure is its exterior Lebesgue measure, and in this case we denote this value by $|E|=|E|{e} . \diamond$
There is no difference between the numeric value of the Lebesgue measure and the exterior Lebesgue measure of a measurable set, but when we know that $E$ is measurable we write $|E|$ instead of $|E|_{e}$.

Notation 2.2.2. The collection of all Lebesgue measurable subsets of $\mathbb{R}^{d}$ will be denoted by
$$
\mathcal{L}=\mathcal{L}\left(\mathbb{R}^{d}\right)=\left{E \subseteq \mathbb{R}^{d}: E \text { is Lebesgue measurable }\right}
$$
We would like to know which types of subsets of $\mathbb{R}^{d}$ are measurable. A first observation is that $\mathcal{L}$ contains all of the open subsets of $\mathbb{R}^{d}$.

数学代写|实分析代写REAL ANALYSIS代考|Countable Additivity

(Countable Additivity). If $E_{1}, E_{2}, \ldots$ are disjoint, Lebesgue measurable subsets of $\mathbb{R}^{d}$, then
$$
\left|\bigcup_{k=1}^{\infty} E_{k}\right|=\sum_{k=1}^{\infty}\left|E_{k}\right| .
$$
60
2 Lebesgue Measure
Proof. Step 1. Assume first that each set $E_{k}$ is bounded. From subadditivity we obtain
$$
\left|\bigcup_{k=1}^{\infty} E_{k}\right| \leq \sum_{k=1}^{\infty}\left|E_{k}\right|,
$$
so our task is to prove the opposite inequality.
Fix $\varepsilon>0$. , there exists a closed set $F_{k} \subseteq E_{k}$ such that
$$
\left|E_{k} \backslash F_{k}\right|<\frac{\varepsilon}{2^{k}} .
$$
Since $E_{k}$ is bounded, $F_{k}$ is compact. Hence $\left{F_{k}\right}_{k \in \mathbb{N}}$ is a collection of disjoint compact sets. Let $N$ be any finite positive integer. Then, by using Corollary $2.2 .8$ and monotonicity, we see that
$$
\sum_{k=1}^{N}\left|F_{k}\right|=\left|\bigcup_{k=1}^{N} F_{k}\right| \leq\left|\bigcup_{k=1}^{N} E_{k}\right| \leq\left|\bigcup_{k=1}^{\infty} E_{k}\right|
$$
Taking the limit as $N \rightarrow \infty$,
$$
\sum_{k=1}^{\infty}\left|F_{k}\right|=\lim {N \rightarrow \infty} \sum{k=1}^{N}\left|F_{k}\right| \leq\left|\bigcup_{k=1}^{\infty} E_{k}\right|
$$
Therefore
$\begin{array}{rlr}\sum_{k=1}^{\infty}\left|E_{k}\right| & =\sum_{k=1}^{\infty}\left|F_{k} \cup\left(E_{k} \backslash F_{k}\right)\right| & \ & \leq \sum_{k=1}^{\infty}\left(\left|F_{k}\right|+\left|E_{k} \backslash F_{k}\right|\right) \quad \text { (by finite subadditivity) } \ & \leq \sum_{k=1}^{\infty}\left(\left|F_{k}\right|+\frac{\varepsilon}{2^{k}}\right) \quad \text { (by equation (2.15)) } \ & =\left(\sum_{k=1}^{\infty}\left|F_{k}\right|\right)+\varepsilon & \ & \leq\left|\bigcup_{k=1}^{\infty} E_{k}\right|+\varepsilon & \text { }\end{array}$

$\begin{aligned}\left|\bigcup_{k=1}^{\infty} E_{k}\right| &=\left|\bigcup_{k=1}^{\infty} \bigcup_{j=1}^{\infty} E_{k}^{j}\right| \ &=\sum_{k=1}^{\infty} \sum_{j=1}^{\infty}\left|E_{k}^{j}\right| \ &=\sum_{k=1}^{\infty}\left|\bigcup_{j=1}^{\infty} E_{k}^{j}\right| \ &=\sum_{k=1}^{\infty}\left|E_{k}\right| \end{aligned}$

It is worth noting that what makes Step 2 of the preceding proof possible is the fact that $\mathbb{R}^{d}$, whose measure is infinite, can be written as the union of countably many measurable sets that each have finite measure (in the language of abstract measure theory, this says that Lebesgue measure on $\mathbb{R}^{d}$ is $\sigma$-finite). While simple, this observation is extremely useful, as it often allows us to reduce issues about generic sets to sets that have finite measure. There are many ways to write $\mathbb{R}^{d}$ as a countable union of sets that have finite measures; here are a few typical examples.
(a) $\mathbb{R}^{d}=\cup_{n=1}^{\infty} B_{n}(0)$.
(b) $\mathbb{R}^{d}=\cup_{n=1}^{\infty}\left{x \in \mathbb{R}^{d}: n-1 \leq|x|<n\right}$.
(c) $\mathbb{R}^{d}=\bigcup_{k \in \mathbb{Z}^{d}}(Q+k)$ where $Q=[0,1]^{d}$.
The sets $B_{n}(0)$ in the union in (a) are not disjoint, whereas the sets in the union in (b) are disjoint. Although the sets in the union in (c) are not disjoint, they are nonoverlapping closed cubes.


数学代写|实分析代写real analysis代考|Equivalent Formulations of Measurability

As we have seen, the collection $\mathcal{L}$ of all Lebesgue measurable subsets of $\mathbb{R}^{d}$ is closed under countable unions and complements. Since $\mathcal{L}$ contains all of the open and closed subsets of $\mathbb{R}^{d}$, it must therefore also contain all of the following types of sets.
Definition 2.2.18 ( $G_{\delta}$-Sets and $F_{\sigma}$-Sets).
(a) A set $H \subseteq \mathbb{R}^{d}$ is a $G_{\delta}-$ set if there exist countably many open sets $U_{k}$ such that $H=\cap U_{k}$.
(b) A set $H \subseteq \mathbb{R}^{d}$ is an $F_{\sigma}$-set if there exist countably many closed sets $F_{k}$ such that $H=\bigcup F_{k}$.

The symbol $\sigma$ in this definition is reminiscent of the word “sums” and hence unions, while $\delta$ suggests the word “difference” and hence intersections. More precisely, $F_{\sigma}$ is derived from the French words fermé (closed) and somme (union), while $G_{\delta}$ is derived from the German Gebiet (area, neighborhood, open set) and Durchschnitt (average, intersection).

The half-open interval $[a, b)$ is neither an open nor a closed subset of $\mathbb{R}$, but it is both a $G_{\delta}$-set and an $F_{\sigma}$-set because we can write
$$
\bigcap_{k=1}^{\infty}\left(a-\frac{1}{k}, b\right)=[a, b)=\bigcup_{k=1}^{\infty}\left[a, b-\frac{1}{k}\right]
$$

数学代写|实分析代写real analysis代考|Lebesgue Measure

实分析代写

数学代写|实分析代写REAL ANALYSIS代考|DEFINITION AND BASIC PROPERTIES

为了激发可测量性的定义,假设ü是一个包含一个集合的开集和. 正如我们上面观察到的,我们不知道是否|ü|和和|和|和+|ü∖和|和将是平等的。如果这些量是相等的,那么我们可以将这个等式与方程结合起来2.9并推断|ü∖和|和≤e. “可测量的集合”正是可以实现这种不等式的集合。这是明确的定义。

定义 2.2.1大号和b和sG你和米和一种s你r和. 一套和⊆RdLebesgue 是可测量的,或者简称为可测量的,如果
∀e>0,∃ 打开 ü⊇和 这样 |ü∖和|和≤e.如果和是 Lebesgue 可测的,那么它的 Lebesgue 测度就是它的外部 Lebesgue 测度,在这种情况下,我们将这个值表示为|和|=|和|和.⋄
Lebesgue 测度的数值和可测集的外部 Lebesgue 测度之间没有区别,但是当我们知道和是可测量的,我们写|和|代替|和|和.

符号 2.2.2。的所有 Lebesgue 可测子集的集合Rd将表示为
\mathcal{L}=\mathcal{L}\left(\mathbb{R}^{d}\right)=\left{E \subseteq \mathbb{R}^{d}: E \text { 是勒贝格可测的}\对}\mathcal{L}=\mathcal{L}\left(\mathbb{R}^{d}\right)=\left{E \subseteq \mathbb{R}^{d}: E \text { 是勒贝格可测的}\对}
我们想知道哪些类型的子集Rd是可测量的。第一个观察结果是大号包含所有的开放子集Rd.

数学代写|实分析代写REAL ANALYSIS代考|COUNTABLE ADDITIVITY

C这你n吨一种b一世和一种dd一世吨一世v一世吨是. 如果和1,和2,…是不相交的,Lebesgue 可测子集Rd, 然后
|⋃到=1∞和到|=∑到=1∞|和到|.
60
2 Lebesgue 测量
证明。步骤 1. 首先假设每个集合和到是有界的。从次可加性我们得到
|⋃到=1∞和到|≤∑到=1∞|和到|,
所以我们的任务是证明相反的不等式。
使固定e>0. , 存在一个闭集F到⊆和到这样
|和到∖F到|<e2到.
自从和到有界,F到紧凑。因此\left{F_{k}\right}_{k \in \mathbb{N}}\left{F_{k}\right}_{k \in \mathbb{N}}是不相交的紧集的集合。让ñ是任何有限的正整数。然后,通过使用推论2.2.8和单调性,我们看到
∑到=1ñ|F到|=|⋃到=1ñF到|≤|⋃到=1ñ和到|≤|⋃到=1∞和到|
取极限为ñ→∞,
∑到=1∞|F到|=林ñ→∞∑到=1ñ|F到|≤|⋃到=1∞和到|
所以
∑到=1∞|和到|=∑到=1∞|F到∪(和到∖F到)| ≤∑到=1∞(|F到|+|和到∖F到|) (通过有限次可加性)  ≤∑到=1∞(|F到|+e2到) (由方程(2.15))  =(∑到=1∞|F到|)+e ≤|⋃到=1∞和到|+e 

|⋃到=1∞和到|=|⋃到=1∞⋃j=1∞和到j| =∑到=1∞∑j=1∞|和到j| =∑到=1∞|⋃j=1∞和到j| =∑到=1∞|和到|

值得注意的是,使上述证明的步骤 2 成为可能的原因是,Rd,其测度是无限的,可以写成可数多个可测集的并集,每个可测集都有有限测度一世n吨H和一世一种nG你一种G和这F一种bs吨r一种C吨米和一种s你r和吨H和这r是,吨H一世ss一种是s吨H一种吨大号和b和sG你和米和一种s你r和这n$Rd$一世s$σ$−F一世n一世吨和. 虽然简单,但这种观察非常有用,因为它通常允许我们将有关泛型集的问题减少到具有有限度量的集。有很多写法Rd作为具有有限测度的集合的可数并集;这里有几个典型的例子。
一种 Rd=∪n=1∞乙n(0).
b \mathbb{R}^{d}=\cup_{n=1}^{\infty}\left{x \in \mathbb{R}^{d}: n-1 \leq|x|<n\right }\mathbb{R}^{d}=\cup_{n=1}^{\infty}\left{x \in \mathbb{R}^{d}: n-1 \leq|x|<n\right }.
C Rd=⋃到∈从d(问+到)在哪里问=[0,1]d.
套装乙n(0)在工会中一种不是不相交的,而联合中的集合b是不相交的。虽然联合中的集合在C不是不相交的,它们是不重叠的封闭立方体。

数学代写|实分析代写REAL ANALYSIS代考|EQUIVALENT FORMULATIONS OF MEASURABILITY

正如我们所见,该系列大号的所有 Lebesgue 可测子集Rd在可数并集和补集下是封闭的。自从大号包含所有的开和闭子集Rd,因此它还必须包含以下所有类型的集合。
定义 2.2.18$Gd$−小号和吨s一种nd$Fσ$−小号和吨s.
一种一套H⊆Rd是一个Gd−如果存在可数个开集,则设置ü到这样H=∩ü到.
b一套H⊆Rd是一个Fσ-set 如果存在可数个闭集F到这样H=⋃F到.

符号σ在这个定义中让人想起“总和”这个词,因此联想到联合,而d暗示了“差异”一词,因此暗示了交叉点。更确切地说,Fσ源自法语单词ferméC一世这s和d和总和你n一世这n, 尽管Gd源自德国Gebiet一种r和一种,n和一世GHb这rH这这d,这p和ns和吨和平均一种v和r一种G和,一世n吨和rs和C吨一世这n.

半开区间[一种,b)既不是的开子集也不是闭子集R, 但它既是Gd-set 和一个Fσ-set 因为我们可以写
⋂到=1∞(一种−1到,b)=[一种,b)=⋃到=1∞[一种,b−1到]

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