## 统计代写|随机过程代写stochastic process代考|STAT6540

2023年3月31日

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## 统计代写|随机过程代写stochastic process代考|Stability Properties of Measurable Functions

Measurability is a stable property, in the sense that all the usual operations on measurable functions preserve measurability. (This is not the case for continuity, which is in general not stable with respect to limits.) Also, the class of measurable functions is a rich one. In particular, “continuous functions are measurable”. More precisely:

Corollary 2.1.16 Let $X$ and $E$ be two topological spaces with respective Borel $\sigma$-fields $\mathcal{B}(X)$ and $\mathcal{B}(E)$. Any continuous function $f: X \rightarrow E$ is measurable with respect to $\mathcal{B}(X)$ and $\mathcal{B}(E)$

Proof. By definition of continuity, the inverse image of an open set of $E$ is an open set of $X$ and is therefore in $\mathcal{B}(X)$. By Theorem 2.1.15, since the open sets of $E$ generate $\mathcal{B}(E)$, the function $f$ is measurable with respect to $\mathcal{B}(X)$ and $\mathcal{B}(E)$.

Corollary 2.1.17 Let $(X, \mathcal{X})$ be a measurable space and let $n \geq 1$ be an integer. Then $f=\left(f_1, \ldots, f_n\right):(X, \mathcal{X}) \rightarrow\left(\mathbb{R}^n, \mathcal{B}\left(\mathbb{R}^n\right)\right)$ if and only if for all $1 \leq i \leq n,\left{f_i \leq a_i\right} \in \mathcal{X}$ for all $a_i \in \mathbb{Q}$ (the rational numbers).

Proof. Since by Theorem $2.1 .9, \mathcal{B}\left(\mathbb{R}^n\right)$ is generated by the sets $\prod_{i=1}^n\left(-\infty, a_i\right]$, where $a_i \in \mathbb{Q}$ for all $i \in{1, \ldots, n}$, it suffices by Theorem 2.1 .15 to show that for all $a \in \mathbb{Q}^n$, ${f \leq a} \in \mathcal{X}$. This is indeed the case since
$${f \leq a}=\cap_{i=1}^n\left{f_i \leq a_i\right}$$
and therefore ${f \leq a} \in \mathcal{X}$, being the intersection of a countable (actually: finite) number of sets in $\mathcal{X}$.

## 统计代写|随机过程代写stochastic process代考|Measure

The next most important notion of integration theory after that of measurable sets and measurable functions is that of measure.

Definition 2.1.28 Let $(X, \mathcal{X})$ be a measurable space. A set function $\mu: \mathcal{X} \rightarrow[0, \infty]$ is called a measure on $(X, \mathcal{X})$ if $\mu(\varnothing)=0$ and if for any countable sequence $\left{A_n\right}_{n \geq 0}$ of pairwise disjoint sets in $\mathcal{X}$, the following property ( $\sigma$-additivity) is satisfied
$$\mu\left(\sum_{n=0}^{\infty} A_n\right)=\sum_{n=0}^{\infty} \mu\left(A_n\right)$$
The triple $(X, \mathcal{X}, \mu)$ is then called a measure space.
The next two properties have been proved in the first chapter. We repeat the proofs for self-containedness. First, the monotonicity property:
$$A \subseteq B \text { and } A, B \in \mathcal{X} \Longrightarrow \mu(A) \leq \mu(B)$$
Indeed, $B=A+(B-A)$ and therefore, $\mu(B)=\mu(A)+\mu(B-A) \geq \mu(A)$. The sub- $\sigma$ additivity property:
$$A_n \in \mathcal{X} \text { for all } n \in \mathbb{N} \Longrightarrow \mu\left(\bigcup_{n=0}^{\infty} A_n\right) \leq \sum_{n=0}^{\infty} \mu\left(A_n\right),$$
is obtained by writing
$$\mu\left(\bigcup_{n=0}^{\infty} A_n\right)=\mu\left(\sum_{n=0}^{\infty} A_n^{\prime}\right)$$
where $A_0^{\prime}=A_0$ and for $n \geq 1$,
$$A_n^{\prime}=A_n \cap\left(\overline{\cup_{j=1}^{n-1} A_j}\right) \subseteq A_n$$
so that $\mu\left(A_n^{\prime}\right) \leq \mu\left(A_n\right)$ by the monotonicity property.

# 随机过程代考

## 统计代写|随机过程代写stochastic process代考|Stability Properties of Measurable Functions

$f=\left(f_1, \ldots, f_n\right):(X, \mathcal{X}) \rightarrow\left(\mathbb{R}^n, \mathcal{B}\left(\mathbb{R}^n\right)\right)$ 当且仅

1 lleq i lleq n, lleft{ffi i leq a_iliright $}$ lin Imathcal ${X}$ 对全部 $a_i \in \mathbb{Q}$ (有理数)。

$\prod_{i=1}^n\left(-\infty, a_i\right]$ ，在哪里 $a_i \in \mathbb{Q}$ 对全部 $i \in 1, \ldots, n$,

${f \backslash$ leq a $}=\backslash c a p _{i=1}^{\wedge} n \backslash e f t\left{f _i \backslash l e q\right.$ a_i $\backslash$ right $}$

## 统计代写|随机过程代写stochastic process代考|Measure

$$\mu\left(\sum_{n=0}^{\infty} A_n\right)=\sum_{n=0}^{\infty} \mu\left(A_n\right)$$

$$A \subseteq B \text { and } A, B \in \mathcal{X} \Longrightarrow \mu(A) \leq \mu(B)$$

$$A_n \in \mathcal{X} \text { for all } n \in \mathbb{N} \Longrightarrow \mu\left(\bigcup_{n=0}^{\infty} A_n\right) \leq \sum_{n=0}^{\infty} \mu\left(A_n\right)$$

$$\mu\left(\bigcup_{n=0}^{\infty} A_n\right)=\mu\left(\sum_{n=0}^{\infty} A_n^{\prime}\right)$$

$$A_n^{\prime}=A_n \cap\left(\overline{\cup_{j=1}^{n-1} A_j}\right) \subseteq A_n$$

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