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# 数学代写|测度与积分代写Measure And Integration代考|BSMA2003 Set Operations

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## 数学代写|测度与积分代写Measure And Integration代考|Set Operations

2.1. Set Operations. Suppose that $X$ is a set. For $A \subset X$ let
$$A^c=X \backslash A={x \in X: x \notin A}$$
and more generally if $A, B \subset X$ let
$$B \backslash A={x \in B: x \notin A} .$$
We also define the symmetric difference of $A$ and $B$ by
$$A \triangle B=(B \backslash A) \cup(A \backslash B) .$$
As usual if $\left{A_\alpha\right}_{\alpha \in I}$ is an indexed collection of subsets of $X$ we define the union and the intersection of this collection by
\begin{aligned} & \cup_{\alpha \in I} A_\alpha:=\left{x \in X: \exists \alpha \in I \quad \ni x \in A_\alpha\right} \text { and } \ & \cap_{\alpha \in I} A_\alpha:=\left{x \in X: x \in A_\alpha \forall \alpha \in I\right} . \end{aligned}

Notice that $U$ is closely related to $\exists$ and $\cap$ is closely related to $\forall$. For example let $\left{A_n\right}_{n=1}^{\infty}$ be a sequence of subsets from $X$ and define
\begin{aligned} & \left{A_n \text { i.o. }\right}:=\left{x \in X: #\left{n: x \in A_n\right}=\infty\right} \text { and } \ & \left{A_n \text { a.a. }\right}:=\left{x \in X: x \in A_n \text { for all } n \text { sufficiently large }\right} \end{aligned}
Then $x \in\left{A_n\right.$ i.o. $}$ iff $\forall N \in \mathbb{N} \exists n \geq N \ni x \in A_n$ which may be written as
$$\left{A_n \text { i.o. }\right}=\cap_{N=1}^{\infty} \cup_{n \geq N} A_n$$
and similarly, $x \in\left{A_n\right.$ a.a. $}$ iff $\exists N \in \mathbb{N} \ni \forall n \geq N, x \in A_n$ which may be written as
$$\left{A_n \text { a.a. }\right}=\cup_{N=1}^{\infty} \cap_{n \geq N} A_n$$

## 数学代写|测度与积分代写Measure And Integration代考|Limits, Limsups, and Liminfs

Notation 2.1. The is the set $\overline{\mathbb{R}}:=\mathbb{R} \cup{\pm \infty}$, i.e. it is $\mathbb{R}$ with two new points called $\infty$ and $-\infty$. We use the following conventions, $\pm \infty \cdot 0=0, \pm \infty+a=\pm \infty$ for any $a \in \mathbb{R}, \infty+\infty=\infty$ and $-\infty-\infty=-\infty$ while $\infty-\infty$ is not defined.
If $\Lambda \subset \overline{\mathbb{R}}$ we will let $\sup \Lambda$ and $\inf \Lambda$ denote the least upper bound and greatest lower bound of $\Lambda$ respectively. We will also use the following convention, if $\Lambda=\emptyset$, then $\sup \emptyset=-\infty$ and $\inf \emptyset=+\infty$.
Notation 2.2. Suppose that $\left{x_n\right}_{n=1}^{\infty} \subset \overline{\mathbb{R}}$ is a sequence of numbers. Then
(2.1) $\quad \lim \inf {n \rightarrow \infty} x_n=\lim {n \rightarrow \infty} \inf \left{x_k: k \geq n\right}=\sup n \inf \left{x_k: k \geq n\right}$ and (2.2) $\quad \lim \sup {n \rightarrow \infty} x_n=\lim _{n \rightarrow \infty} \sup \left{x_k: k \geq n\right}=\inf _n \sup \left{x_k: k \geq n\right}$.
We will also write $\lim \inf$ for $\varliminf$ and $\lim \sup$ for $\varlimsup$.
Remark 2.3. Notice that if $a_k:=\inf \left{x_k: k \geq n\right}$ and $b_k:=\sup \left{x_k: k \geq n\right}$,then $\left{a_k\right}$ is an increasing sequence while $\left{b_k\right}$ is a decreasing sequence. Therefore the limits in Eq. (2.1) and Eq. (2.2) always exist.

The following proposition contains some basic properties of liminfs and limsups.
Proposition 2.4. Let $\left{a_n\right}_{n=1}^{\infty}$ and $\left{b_n\right}_{n=1}^{\infty}$ be two sequences of real numbers. Then
(1) Show $\liminf {n \rightarrow \infty} a_n \leq \limsup {n \rightarrow \infty} a_n$ and the $\lim {n \rightarrow \infty} a_n$ exists in $\overline{\mathbb{R}}$ iff $\liminf {n \rightarrow \infty} a_n=\limsup {n \rightarrow \infty} a_n \in \overline{\mathbb{R}}$. (2) Suppose that $\limsup {n \rightarrow \infty} a_n=M \in \overline{\mathbb{R}}$, show that there is a subsequence $\left{a_{n_k}\right}_{k=1}^{\infty}$ of $\left{a_n\right}_{n=1}^{\infty}$ such that $\lim {k \rightarrow \infty} a{n_k}=M$.
(3) Suppose that $\limsup {n \rightarrow \infty} a_n<\infty$ and $\lim _n \sup {n \rightarrow \infty} b_n>-\infty$, then prove that
$$\lim \sup {n \rightarrow \infty}\left(a_n+b_n\right) \leq \lim \sup {n \rightarrow \infty} a_n+\lim \sup {n \rightarrow \infty} b_n .$$ In other words, Eq. (2.3) holds provided the right side of the equation is well defined. (4) Suppose that $a_n \geq 0$ and $b_n \geq 0$ for all $n \in \mathbb{N}$. Show $$\lim \sup {n \rightarrow \infty}\left(a_n b_n\right) \leq \lim \sup {n \rightarrow \infty} a_n \cdot \lim \sup {n \rightarrow \infty} b_n,$$
provided the right hand side of (2.4) is not of the form $0 \cdot \infty$ or $\infty \cdot 0$.

## 数学代写|测度与积分代写 MEASURE AND INTEGRATION代 考|SET OPERATIONS

2.1. 设置操作。假设 $X$ 是一个集合。为了 $A \subset X$ 让
$$A^c=X \backslash A=x \in X: x \notin A$$

$$B \backslash A=x \in B: x \notin A .$$

$$A \triangle B=(B \backslash A) \cup(A \backslash B) .$$

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