## 数学代写|实分析作业代写Real analysis代考|MATHS2100

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## 数学代写|实分析作业代写Real analysis代考|Measurable Functions

In this section we study measurable functions. In spite of the name, the concept of measurability of functions is defined in terms of $\sigma$-algebras and is not connected with measures. A connection with measures arises when for the basic $\sigma$-algebra we take the $\sigma$-algebra of all sets measurable with respect to a fixed measure. This important particular case is considered at the end of the section. Measurable functions are needed for constructing the integral.
3.1.1. Definition. Let $(X, \mathcal{A})$ be a space with a $\sigma$-algebra. A function $f: X \rightarrow \mathbb{R}^1$ is called measurable with respect to $\mathcal{A}$ (or $\mathcal{A}$-measurable) if we have ${x: f(x)<c} \in \mathcal{A}$ for every $c \in \mathbb{R}^1$.

The simplest example of an $\mathcal{A}$-measurable function is the indicator function $I_A$ of a set $A \in \mathcal{A}$ defined as follows: $I_A(x)=1$ if $x \in A, I_A(x)=0$ if $x \notin A$. The indicator function of a set $A$ is also called the characteristic function of $A$. The set $\left{x: I_A(x)1$. It is clear that the inclusion $A \in \mathcal{A}$ is also necessary for the $\mathcal{A}$-measurability of $I_A$.
3.1.2. Theorem. A function $f$ is measurable with respect to a $\sigma$-algebra $\mathcal{A}$ precisely when $f^{-1}(B) \in \mathcal{A}$ for all sets $B \in \mathcal{B}\left(\mathbb{R}^1\right)$.

Proof. Let $f$ be an $\mathcal{A}$-measurable function. Denote by $\mathcal{E}$ the collection of all sets $B \in \mathcal{B}\left(\mathbb{R}^1\right)$ such that $f^{-1}(B) \in \mathcal{A}$. We show that $\mathcal{E}$ is a $\sigma$-algebra. Indeed, if $B_n \in \mathcal{E}$, then $f^{-1}\left(\bigcup_{n=1}^{\infty} B_n\right)=\bigcup_{n=1}^{\infty} f^{-1}\left(B_n\right) \in \mathcal{A}$. In addition, for any $B \in \mathcal{E}$ we have $f^{-1}\left(\mathbb{R}^1 \backslash B\right)=X \backslash f^{-1}(B) \in \mathcal{A}$. Since $\mathcal{E}$ contains the rays $(-\infty, c)$, we obtain that $\mathcal{B}\left(\mathbb{R}^1\right) \subset \mathcal{E}$, i.e., $\mathcal{B}\left(\mathbb{R}^1\right)=\mathcal{E}$. The converse assertion is obvious, because all rays are Borel sets.
Let us write $f$ in the form $f=f^{+}-f^{-}$, where
$$f^{+}(x):=\max (f(x), 0) \text { and } f^{-}(x):=\max (-f(x), 0) .$$

## 数学代写|实分析作业代写Real analysis代考|Convergence in Measure and Almost Everywhere

Let $(X, \mathcal{A}, \mu)$ be a space with a nonnegative measure $\mu$. We shall say that a certain property is fulfilled almost everywhere (or $\mu$-almost everywhere) on $X$ if the set $Z$ of points in $X$ not possessing this property belongs to $\mathcal{A}\mu$ and has measure zero with respect to the measure $\mu$ extended to $\mathcal{A}\mu$. The complements of sets of measure zero are called sets of full measure. We shall use the following abbreviations for “almost everywhere”: a.e., $\mu$-a.e. It is clear from the definition of $\mathcal{A}\mu$ that there exists a set $Z_0 \in \mathcal{A}$ such that $Z \subset Z_0$ and $\mu\left(Z_0\right)=0$, i.e., the corresponding property is fulfilled outside some set from $\mathcal{A}$ of measure zero. This circumstance should be remembered when dealing with incomplete measures. For example, one can say that a sequence of functions $f_n$ converges a.e. or is Cauchy $a{.} e_2$, a function is nonnegative $a_a e_2$, etc. It is clear that convergence of $\left{f_n\right}$ a.e. follows from convergence of $\left{f_n(x)\right}$ for every $x$ (called the pointwise convergence), and the latter follows from the uniform convergence of $\left{f_n\right}$. A deeper connection between convergence almost everywhere and the uniform convergence is described by the following theorem due to the celebre Russian mathematician D. F. Egorov.

# 实分析代写

## 数学代写|实分析作业代写Real analysis代考|Measurable Functions

3.1.1. 定义。让 $(X, \mathcal{A})$ 是一个空间 $\sigma$-代数。一个功能 $f: X \rightarrow \mathbb{R}^1$ 被称为可测量的 $\mathcal{A}$ (或者 $\mathcal{A}$-可测量的) 如 果我们有 $x: f(x)<c \in \mathcal{A}$ 每一个 $c \in \mathbb{R}^1$.

$f^{+}(x):=\max (f(x), 0)$ and $f^{-}(x):=\max (-f(x)$

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