# 数学代写|实分析作业代写Real analysis代考|Measure of Open Sets: Compact Sets

## 数学代写|实分析作业代写Real analysis代考|Measure of Open Sets: Compact Sets

We begin our discussion of measure theory by first defining the measure of open and compact subsets of $\mathbb{R}$.

DEFINITION 10.2.1 If $J$ is an interval, we define the measure of $J$, denoted $m(J)$, to be the length of $J$.
Thus if $J$ is $(a, b),(a, b],[a, b)$, or $[a, b], a, b \in \mathbb{R}$, then
$$m(J)=b-a .$$
If $J$ is $\mathbb{R},(a, \infty),[a, \infty),(-\infty, b)$ or $(-\infty, b]$, we set $m(J)=\infty$. In dealing with the symbols $\infty$ and $-\infty$, it is customary to adopt the following conventions:

(a) If $x$ is real, then $x+\infty=\infty, \quad x-\infty=-\infty$.
(b) If $x>0$ then $x \cdot \infty=\infty, x \cdot(-\infty)=-\infty$.
(c) If $x<0$ then $x \cdot \infty=-\infty, x \cdot(-\infty)=\infty$.
(d) Also, $\infty+\infty=\infty,-\infty-\infty=-\infty, \infty \cdot( \pm \infty)= \pm \infty,-\infty \cdot( \pm \infty)=$ $\mp \infty$
The symbols $\infty-\infty$ and $-\infty+\infty$ are undefined, but we shall adopt the arbitrary convention that $0 \cdot \infty=0$.

## 数学代写|实分析作业代写Real analysis代考|Measure of Open Sets

Therefore equality holds in (1). The other two cases follow similarly.
Suppose $m\left(I_n\right)<\infty$ for all $n$. Since $U$ is unbounded, the collection $\left{I_n\right}$ must be infinite. If the collection were finite, then since each interval has finite length, each interval is bounded, and as a consequence $U$ must also be bounded. Let $\alpha \in \mathbb{R}$ with $\alpha\alpha $$there exists a positive integer N such that$$ \sum_{n=1}^N m\left(I_n\right)>\alpha $$Let V=\bigcup_{n=1}^N I_n. Then V is a bounded open set, and thus by the above,$$ m(V)=\lim {k \rightarrow \infty} m(V \cap(-k, k)) $$Since m(V)>\alpha, there exists k_o \in \mathbb{N} such that$$ m(V \cap(-k, k))>\alpha \quad \text { for all } k \geq k_o $$But V \cap(-k, k) \subset U_k for all k \in \mathbb{N}. Hence by Theorem 10.2.4, m(V \cap(-k, k)) \leq m\left(U_k\right) and as a consequence$$ m\left(U_k\right)>\alpha \quad \text { for all } k \geq k_o $$If m(U)=\infty, then since \alpha0, take \alpha=m(U)-\epsilon. By the above, there exists k_o \in \mathbb{N} such that$$ m(U)-\epsilon{k \rightarrow \infty} m\left(U_k\right)=m(U)$.
Remark. In proving results about open sets, the previous theorem allows us to first prove the result for the case where $U$ is a bounded open set, and then to use the limit process to extend the result to the unbounded case.

# 实分析代考

## 数学代写|实分析作业代写Real analysis代考|Measure of Open Sets: Compact Sets

$$m(J)=b-a$$

(a) 如果 $x$ 是真实的，那么 $x+\infty=\infty, \quad x-\infty=-\infty$.
(b) 如果 $x>0$ 然后 $x \cdot \infty=\infty, x \cdot(-\infty)=-\infty$.
(c) 如果 $x<0$ 然后 $x \cdot \infty=-\infty, x \cdot(-\infty)=\infty$.
(d) 此外， $\infty+\infty=\infty,-\infty-\infty=-\infty, \infty \cdot( \pm \infty)= \pm \infty,-\infty \cdot( \pm \infty)=$ $\mp \infty$

## 数学代写|实分析作业代写Real analysis代考|Measure of Open Sets

$U$ 也必须有界。让 $\alpha \in \mathbb{R}$ 和 $\alpha \alpha$ thereexistsapositiveinteger 否suchthat
$\sum_{n=1}^N m\left(I_n\right)>\alpha$ Let $\mathrm{V}=\backslash$ bigcup_ ${\mathrm{n}=1}^{\wedge} \mathrm{N}$ In. Then 在
isaboundedopenset, andthusbytheabove, $m(V)=\lim k \rightarrow \infty m(V \cap(-k, k))$
Sincem $(\mathrm{V})>\backslash$ alpha, thereexistsk_o $\backslash$ in $\backslash m a t h b b{N}$ suchthat
$m(V \cap(-k, k))>\alpha \quad$ for all $k \geq k_o$ But $\bigvee \backslash c a p(-\mathrm{k}, \mathrm{k}) \backslash$ 子集 $U _\mathrm{k}$ forall $\backslash \backslash$ in
$\backslash m a t h b b{N}$. HencebyTheorem $10.2 .4, \mathrm{~m}(\vee \backslash c a p(-k, k)) \backslash$ leqm $\backslash$ left(U_k $\backslash$ right)
andasaconsequencem $\left(U_k\right)>\alpha$ for all $k \geq k_o I f \mathrm{~m}(\mathrm{U})=$ linfty, thensince
$\backslash a l p h a 0$, take \alpha $=\mathrm{m}(\mathrm{U})$-\epsilon. Bytheabove, thereexistsk_o $\backslash$ in $\backslash m a t h b b{\mathrm{~N}}$
$\operatorname{suchthatm}(U)-\epsilon k \rightarrow \infty m\left(U_k\right)=m(U)$

myassignments-help数学代考价格说明

1、客户需提供物理代考的网址，相关账户，以及课程名称，Textbook等相关资料~客服会根据作业数量和持续时间给您定价~使收费透明，让您清楚的知道您的钱花在什么地方。

2、数学代写一般每篇报价约为600—1000rmb，费用根据持续时间、周作业量、成绩要求有所浮动(持续时间越长约便宜、周作业量越多约贵、成绩要求越高越贵)，报价后价格觉得合适，可以先付一周的款，我们帮你试做，满意后再继续，遇到Fail全额退款。

3、myassignments-help公司所有MATH作业代写服务支持付半款，全款，周付款，周付款一方面方便大家查阅自己的分数，一方面也方便大家资金周转，注意:每周固定周一时先预付下周的定金，不付定金不予继续做。物理代写一次性付清打9.5折。

Math作业代写、数学代写常见问题

myassignments-help擅长领域包含但不是全部: