Chapter 6 Construction of a Probability Measure

南京审计大学统计学研究生第一学期课程，《高等概率论》。

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概率测度的构造 adveprobab 2

对于 B\mathcal{B}B 是 {ω}ω∈Ω\{\omega\}_{\omega \in \Omega}{ω}ω∈Ω ，即 B\mathcal{B}B 是单点集组成的事件类

如果概率 PPP 建立在 B\mathcal{B}B 上，这样构造的概率比较好计算，现在的想法是：想通过有限集合或者可数集合，来进行拓展，将概率测度扩展开

即将 PPP 扩展到 A=2Ω\mathcal{A}=2^{\Omega}A=2Ω

核心：

finite⇒uncountablefinite \qquad \Rightarrow \qquad uncountable finite⇒uncountable

Assume given Ω\OmegaΩ (countable or uncountable) and a σ\sigmaσ-algebra A∈2Ω\mathcal{A} \in 2^{\Omega}A∈2Ω
(Ω,A)(\Omega,\mathcal{A})(Ω,A) is called a measurable space. 构造PPP
Want to construct probability measure on A\mathcal{A}A
- When Ω\OmegaΩ is finite or countable, we have already seen this is simple to do.
- When Ω\OmegaΩ is uncountable, the same technique does not work. 失效
  - Indeed, a “typical” probability PPP will have P({ω})=0P(\{\omega\})=0P({ω})=0 for all ω\omegaω, and thus the family of all numbers P({ω})P(\{\omega\})P({ω}) for ω∈Ω\omega \in \Omegaω∈Ω does not characterize probability PPP in general.
  - 怎么理解上一点？对于不可数集合。。。

In many “concrete” situations — that it is often relatively simple to construct a “probability” on an algebra which generates the σ\sigmaσ-algebra A\mathcal{A}A , and the problem at hand is then to extend this probability to the σ\sigmaσ-algebra itself.

在由 A\mathcal{A}A 生成的 sigma 代数里面构建一个代数上的概率是相对比较容易的，现在要将其拓展到 sigma 代数里面

Suppose: 在一个代数上建立概率 PPP
- A0\mathcal{A}_0A0 is an algebra and, A=σ(A0)\mathcal{A}=\sigma(\mathcal{A}_0)A=σ(A0).
- Given a probability PPP on the algebra A0\mathcal{A}_0A0 : that is, a set function P:A0→[0,1]P: \mathcal{A}_0 \to [0,1]P:A0→[0,1] satisfying
  - P(Ω)=1P(\Omega)=1P(Ω)=1
  - (Countable Additivity) for any sequence {An}n≥1⊂A0\{A_n\}_{n\ge 1} \subset \mathcal{A}_0{An}n≥1⊂A0, pairwise disjoint, and such that ∪n≥1An∈A0\cup_{n\ge 1}A_n \in \mathcal{A}_0∪n≥1An∈A0, we have

P(∪n≥1An)=∑n=1∞P(An)P \left( \cup_{n\ge 1} A_n \right) = \sum_{n=1}^{\infty} P(A_n) P(∪n≥1An)=n=1∑∞P(An)

It might seem natural to use for A\mathcal{A}A, the set of all subsets of Ω\OmegaΩ, as we did in the case where Ω\OmegaΩ was countable.

We do not do so for the following reason, illustrated by an example: 不可数的不满足

Suppose Ω=[0,1]\Omega=[0,1]Ω=[0,1], and define a set function PPP on intervals of the form P((a,b])=b−aP((a,b])=b-aP((a,b])=b−a, where 0≤a≤b≤10\le a\le b \le 10≤a≤b≤1
It is a natural “probability measure” that assigns the usual length of an interval as its probability.

概率表示为长度
Suppose we want to extend PPP in a unique way to 2Ω=2[0,1]=2^{\Omega}=2^{[0,1]}=2Ω=2[0,1]= all subsets of [0,1][0,1][0,1] such that
- P(Ω)=1P(\Omega)=1P(Ω)=1
- P(∪n=1∞An)=∑n=1∞P(An)P( \cup_{n=1}^{ \infty}A_n)= \sum_{n=1}^{\infty} P(A_n)P(∪n=1∞An)=∑n=1∞P(An) for any sequence of subsets {An}n≥1\{A_n\}_{n\ge 1}{An}n≥1 with An∩Am=∅A_n \cap A_m = \emptysetAn∩Am=∅ for n≠mn \neq mn=m

One can prove that no such PPP exists! 怎么证明其不存在性？
The collection of sets 2[0,1]2^{[0,1]}2[0,1] is simply too big for this to work.

Borel realized that we can however do this on a smaller collection of sets, namely the smallest σ\sigmaσ-algebra containing intervals of the form (a,b](a,b](a,b].

Borel set :

the sigma-algebra generated by the open sets \qquad所有开集生成的 sigma 代数
the sigma-algebra generated by the open intervals \quad所有开区间生成的 sigma 代数
the sigma-algebra generated by the (−∞,a],a∈Q(- \infty , a], a \in \mathbb{Q}(−∞,a],a∈Q \quad所有 (−∞,a](- \infty,a](−∞,a] 生成的 sigma 代数

这三个定义构造的 Borel set 越来越小，事件类越小，构建概率 PPP 越简单

由定理2.1：在R\mathbb{R}R上的博雷尔集是由(−∞,a](-\infty,a](−∞,a]这种形式的区间所生成的sigma代数

Proof.

Let C\mathcal{C}C denote all open intervals. Since every open set in R\mathcal{R}R is the countable union of open intervals, we have σ(C)=\sigma(\mathcal{C})=σ(C)= the Borel σ−algebra\sigma-algebraσ−algebra of R\mathbb{R}R

Let D\mathcal{D}D denote all intervals of the form (−∞,a](- \infty,a](−∞,a], where a∈Qa \in \mathbb{Q}a∈Q.

Let (a,b)∈C(a,b) \in \mathcal{C}(a,b)∈C

Let (an)n≥1(a_n)_{n \ge 1}(an)n≥1 be a sequence of rationals decreasing to a

Let (bn)n≥1(b_n)_{n \ge_1}(bn)n≥1 be a sequence of rationals increasing strictly to b.

Then

(a,b)=∪n=1∞(an,bn]=∪n=1∞((−∞,bn]∩(−∞,an]c)\begin{aligned} (a,b) &= \cup_{n=1}^{ \infty} (a_n, b_n] \\ &= \cup_{n=1}^{ \infty} \left( (- \infty,b_n] \cap (- \infty,a_n]^c \right) \\ \end{aligned} (a,b)=∪n=1∞(an,bn]=∪n=1∞((−∞,bn]∩(−∞,an]c)

Therefore , C⊂σ(D)\mathcal{C} \subset \sigma(\mathcal{D})C⊂σ(D), where σ(C)⊂σ(D)\sigma(\mathcal{C}) \subset \sigma(\mathcal{D})σ(C)⊂σ(D)

However, since each element of D\mathcal{D}D is a closed set, it is also a Borel set, and therefore D\mathcal{D}D is contained in the Borel sets B\mathcal{B}B, Thus we have
B=σ(C)⊂σ(D)⊂B\mathcal{B} = \sigma(\mathcal{C}) \subset \sigma(\mathcal{D}) \subset \mathcal{B} B=σ(C)⊂σ(D)⊂B

and hence σ(D)=B\sigma(\mathcal{D}) = \mathcal{B}σ(D)=B

这个定理给出了更容易验证 Borelσ−algebraBorel \sigma-algebraBorelσ−algebra 的一种方法，给出了验证博雷尔集的一个充要条件

这个定理的证明核心在于 $(a,b) =\cup_{n=1}^{ \infty} \left( (- \infty,b_n] \cap (- \infty,a_n]^c \right) \$

需要证明左边的开区间(a,b)(a,b)(a,b)所构成的 sigma 代数B\mathcal{B}B和右边的 (−∞,a](-\infty,a](−∞,a]这种形式构成的sigma代数σ(D)\sigma(\mathcal{D})σ(D)是相等的，则需要证明两个包含关系

Theorem 6.1 概率延拓定理（唯一延拓）

Each probability PPP defined on the algebra A0\mathcal{A}_0A0 has a unique extension (also call PPP) on A\mathcal{A}A.

每一个定义在代数 A0\mathcal{A_0}A0 上的概率，在 A\mathcal{A}A 上都有唯一的拓展

We will show only the uniqueness. For the existence one can consult any standard text on measure theory.

Definition 6.1

A class C\mathcal{C}C of subsets of Ω\OmegaΩ is closed under finite intersections C\qquad \mathcal{C}C

对有限并封闭 （代数满足对有限并封闭）
- If for when A1,A2,...,An∈CA_1,A_2,...,A_n \in \mathcal{C}A1,A2,...,An∈C ,
  then A1∩A2∩⋯∩An∈CA_1 \cap A_2 \cap \cdots \cap A_n \in \mathcal{C}A1∩A2∩⋯∩An∈C as well
  
  (n arbitrary but finite)
A class C\mathcal{C}C is closed under increasing limits C\qquad \mathcal{C}C

对单増并封闭 （ σ\sigmaσ-代数满足，且 σ\sigmaσ-代数更小）
- If wherever A1⊂A2⊂⋯⊂An⊂⋯A_1 \subset A_2 \subset \cdots \subset A_n \subset \cdotsA1⊂A2⊂⋯⊂An⊂⋯ is a sequence of events in C\mathcal{C}C, then ∪n=1∞An∈C\cup_{n=1}^{ \infty} A_n \in \mathcal{C}∪n=1∞An∈C as well.
A class C\mathcal{C}C is closed under differences C\qquad \mathcal{C}C

对差封闭 （代数满足对差封闭，但满足对差封闭的不一定是代数）
- If whenever A,B∈CA,B \in \mathcal{C}A,B∈C with A⊂BA \subset BA⊂B, then B−A∈CB-A \in \mathcal{C}B−A∈C.

⋆⋆⋆⋆⋆\star\star\star\star\star⋆⋆⋆⋆⋆Monotone Class Theorem 单调类定理

Let C\mathcal{C}C be a class of subsets of Ω\OmegaΩ, closed under finite intersections and containing Ω\OmegaΩ.

C\mathcal{C}C 满足两个条件：包含 Ω\OmegaΩ、对有限交封闭

Let B\mathcal{B}B be the smallest class containing C\mathcal{C}C which is closed under increasing limits and by difference.

B\mathcal{B}B 满足两个条件：包含C\mathcal{C}C 的最小类、对单増并、差封闭

Then B=σ(C)\mathcal{B} = \sigma(C)B=σ(C)

Proof.

Note that

The intersection of classed of sets closed under increasing limits and differences is again a class of that type.

一列事件，对单増并封闭，则把之交起来也对单増并封闭

Proof.

题设：

有若干个事件类 Aα,α∈I\mathcal{A}_{\alpha},\alpha \in \mathcal{I}Aα,α∈I

对每个 α∈I,Aα\alpha \in \mathcal{I},\quad \mathcal{A}_{\alpha}α∈I,Aα 对单増并封闭

推导：

若 A1⊂A2⊂⋯A_1 \subset A_2 \subset \cdotsA1⊂A2⊂⋯ 且 Ai∈∩α∈IAαA_i \in \cap_{\alpha \in I} \mathcal{A}_{\alpha}Ai∈∩α∈IAα

对任意固定的 α∈I,Ai∈Aα,,i=1,2,...\alpha \in \mathcal{I},\quad A_i \in \mathcal{A}_{\alpha}, \quad ,i=1,2,...α∈I,Ai∈Aα,,i=1,2,...

则 ⋃i=1∞Ai∈Aα\bigcup_{i=1}^{ \infty}A_i \in \mathcal{A}_{\alpha}⋃i=1∞Ai∈Aα ，于是 ⋃i=1∞Ai∈⋂α∈IAα\bigcup_{i=1}^{ \infty} A_i \in \bigcap_{\alpha \in I} \mathcal{A}_{\alpha}⋃i=1∞Ai∈⋂α∈IAα （每个都属于 Aα\mathcal{A}_{\alpha}Aα ，则属于 Aα\mathcal{A}_{\alpha}Aα 的交）

一列事件，对差封闭，则把之交起来也对差封闭

自己证明一下

So, by taking the intersection of all such classed,

there always exists a smallest class containing C\mathcal{C}C which is closed under increasing limits and by differences.

总存在一个最小的类，满足：包含 C\mathcal{C}C ，且对单増并封闭、差封闭

所有满足（对单増并封闭、差封闭、包含 C\mathcal{C}C ）的都交起来，最小

For each set BBB, denote BB\mathcal{B}_{B}BB to be the collection of sets AAA such that A∈BA \in \mathcal{B}A∈B and A∩BA \cap BA∩B, i.e.i.e.i.e.

BB={A:A∈B,A∩B∈B}\mathcal{B}_{B} = \left\{ A: A \in \mathcal{B}, A \cap B \in \mathcal{B} \right\} BB={A:A∈B,A∩B∈B}

看上去 BB\mathcal{B}_{B}BB 跟 BBB 的选择有关，但实际上无关

Given the properties of B\mathcal{B}B, one easily checks that BB\mathcal{B}_{B}BB is closed under increasing limits and by differences.

证明对单调并封闭，即证明：若A1⊂A2⊂...A_1 \subset A_2 \subset ...A1⊂A2⊂... 且 Ai∈BBA_i \in \mathcal{B}_{B}Ai∈BB ，证明 ∪i=1∞Ai∈BB\cup_{i=1}^{ \infty} A_i \in \mathcal{B}_{B}∪i=1∞Ai∈BB

则证明其满足两条：

∪i=1∞Ai∈B\cup_{i=1}^{ \infty} A_i \in \mathcal{B}∪i=1∞Ai∈B
(∪i=1∞Ai)B∈B(\cup_{i=1}^{ \infty}A_i) B \in \mathcal{B}(∪i=1∞Ai)B∈B

证明第一条：∪i=1∞Ai∈B\cup_{i=1}^{ \infty} A_i \in \mathcal{B}∪i=1∞Ai∈B

∵B\because \quad \mathcal{B}∵B 对单増并封闭（由定义）

∴∪i=1∞Ai∈B\therefore \quad \cup_{i=1}^{ \infty}A_i \in \mathcal{B}∴∪i=1∞Ai∈B

证明第二条：(∪i=1∞Ai)B∈B(\cup_{i=1}^{ \infty}A_i) B \in \mathcal{B}(∪i=1∞Ai)B∈B

(∪i=1∞Ai)B=∪i=1∞(AiB)(\cup_{i=1}^{ \infty}A_i) B = \cup_{i=1}^{\infty} (A_i B)(∪i=1∞Ai)B=∪i=1∞(AiB) 且 A1B⊂A2B⊂...A_1 B \subset A_2 B \subset ...A1B⊂A2B⊂...

则 ∪i=1∞(AiB)⊂B\cup_{i=1}^{ \infty}(A_i B) \subset \mathcal{B}∪i=1∞(AiB)⊂B （定义）

则说明 ∪i=1∞Ai∈BB\cup_{i=1}^{ \infty} A_i \in \mathcal{B}_{B}∪i=1∞Ai∈BB ，即 BB\mathcal{B}_{B}BB 对单増并封闭

仿照这个证明，对差封闭的自己证明一下

Let B∈CB \in \mathcal{C}B∈C; 给定 B∈CB \in \mathcal{C}B∈C;

For each C∈CC \in \mathcal{C}C∈C one has B∩C⊂C⊂BB \cap C \subset \mathcal{C} \subset \mathcal{B}B∩C⊂C⊂B and C∈BC \in \mathcal{B}C∈B, thus C∈BBC \in \mathcal{B}_{B}C∈BB

Hence C⊂BB⊂B\mathcal{C} \subset \mathcal{B}_{B} \subset \mathcal{B}C⊂BB⊂B. 后一个是定义

∵BB\because \mathcal{B}_{B}∵BB 是包含了 C\mathcal{C}C 的对有限并、差封闭的

∴B=BB\therefore \mathcal{B}=\mathcal{B}_{B}∴B=BB (by the properties of B\mathcal{B}B and of BB\mathcal{B}_{B}BB )

说明在B∈CB \in \mathcal{C}B∈C时是可以证明B=BB\mathcal{B}=\mathcal{B}_{B}B=BB的

下面将其拓展到 B\mathcal{B}B 上

Now let B∈BB \in \mathcal{B}B∈B.

For each C∈C⊂BC \in \mathcal{C} \subset \mathcal{B}C∈C⊂B

由上一条可知，在B∈CB \in \mathcal{C}B∈C时是可以证明B=BB\mathcal{B}=\mathcal{B}_{B}B=BB的，则在C∈CC \in \mathcal{C}C∈C时是可以证明B=BC\mathcal{B}=\mathcal{B}_{C}B=BC的，则 B∈B=BCB \in \mathcal{B}=\mathcal{B}_{C}B∈B=BC

we have B∈BCB \in \mathcal{B}_{C}B∈BC, and because of the preceding, B∩C∈BB \cap C \in \mathcal{B}B∩C∈B, hence C∈BBC \in \mathcal{B}_{B}C∈BB, whence C⊂BB⊂B\mathcal{C} \subset \mathcal{B}_{B} \subset \mathcal{B}C⊂BB⊂B, hence B=BB\mathcal{B}=\mathcal{B}_{B}B=BB

B∈BC⇓定义1.B∈B2.BC∈B⇒C∈BCB∈B⇓定义C∈BB⇓C⊂BB⊂BB=BB\begin{array}{lll} \begin{array}{lll} B \in \mathcal{B}_{C} \\ \Downarrow \text{定义}\\ 1. B \in \mathcal{B} \\ 2. BC \in \mathcal{B} \\ \end{array} & \Rightarrow & \begin{array}{lll} C \in \mathcal{B} \\ CB \in \mathcal{B} \\ \Downarrow \text{定义}\\ C \in \mathcal{B}_{B} \end{array} & \Downarrow & \\ \mathcal{C} \subset \mathcal{B}_{B} \subset \mathcal{B} \\ \mathcal{B} = \mathcal{B}_{B} \end{array} B∈BC⇓定义1.B∈B2.BC∈BC⊂BB⊂BB=BB⇒C∈BCB∈B⇓定义C∈BB⇓
即BB\mathcal{B}_{B}BB 跟 BBB 无关，但是BBB 必须在 B\mathcal{B}B 里面

Since B=BB\mathcal{B}=\mathcal{B}_{B}B=BB for all B∈BB \in \mathcal{B}B∈B, we conclude B\mathcal{B}B is closed by finite intersections.

∵BB\because \mathcal{B}_{B}∵BB 对有限并封闭

∀A,B∈B\forall A,B \in \mathcal{B}∀A,B∈B

∵B∈BA∈B=BB\because B \in \mathcal{B} \qquad A \in \mathcal{B}=\mathcal{B}_{B}∵B∈BA∈B=BB

⇒AB∈B\Rightarrow AB \in \mathcal{B}⇒AB∈B

即 B\mathcal{B}B 对有限并封闭

可以由定义推导

Furthermore Ω∈B\Omega \in \mathcal{B}Ω∈B, and B\mathcal{B}B is closed by difference, hence also under complementation.

Since B\mathcal{B}B is closed by increasing limits as well, we conclude B\mathcal{B}B is a σ\sigmaσ-algebra, and it is clearly the smallest such containing C\mathcal{C}C.

证明：B\mathcal{B}B 是 σ\sigmaσ-代数

缺一个可列并封闭：

证明：

若 Ai∈BA_i \in \mathcal{B}Ai∈B

令 Bn=∪i=1nAiB_n = \cup_{i=1}^{n}A_iBn=∪i=1nAi，则 ∪i=1nAi=∪i=1nBi\cup_{i=1}^{n}A_i = \cup_{i=1}^{n}B_i∪i=1nAi=∪i=1nBi ，且 B1⊂B2⊂...B_1 \subset B_2 \subset ...B1⊂B2⊂...

Bnc=∩i=1nAic∈BB_n^c = \cap_{i=1}^{n}A_i^c \in \mathcal{B}Bnc=∩i=1nAic∈B

Ai∈BA_i \in \mathcal{B}Ai∈B
Aic∈BA_i^c \in \mathcal{B}Aic∈B
AicA_i^cAic 有限并 ∈B\in \mathcal{B}∈B

⇒Bn∈B\Rightarrow \quad B_n \in \mathcal{B}⇒Bn∈B

则 B\mathcal{B}B 对可列并封闭

B\mathcal{B}B 是 σ\sigmaσ-代数，B⊃C⇒B⊃σ(C)\mathcal{B} \supset \mathcal{C} \qquad \Rightarrow \quad \mathcal{B} \supset \sigma(\mathcal{C})B⊃C⇒B⊃σ(C)

由 σ(C)⊃C\sigma(\mathcal{C}) \supset \mathcal{C}σ(C)⊃C 且 σ(C)\sigma(\mathcal{C})σ(C) 对单増并、差封闭

由 B\mathcal{B}B 的最小性，可得 σ(C)⊃B\sigma(\mathcal{C}) \supset \mathcal{B}σ(C)⊃B

则 B=σ(C)\mathcal{B}=\sigma(\mathcal{C})B=σ(C)

The proof of the uniqueness in Theorem 6.1 is an immediate consequence of the following Corollary 6.1, itself a consequence of the Monotone Class Theorem

Corollary 6.1 推论：概率延拓定理

Let PPP and QQQ be two probabilities defined on A\mathcal{A}A

Suppose PPP and QQQ agree on a class C⊂A\mathcal{C} \subset \mathcal{A}C⊂A which is closed under finite intersections.

If σ(C)=A\sigma(\mathcal{C}) = \mathcal{A}σ(C)=A, we have P=QP=QP=Q

这个定理说明：两个定义在对交封闭的 C\mathcal{C}C 上相等的概率测度 P,QP,QP,Q ，可以将其延拓到 σ(C)\sigma(\mathcal{C})σ(C) 上

Proof. We can assume w.l.o.g.w.l.o.g.w.l.o.g. that Ω∈C\Omega \in \mathcal{C}Ω∈C , since

Ω∈A\Omega \in \mathcal{A}Ω∈A, because A\mathcal{A}A is a σ\sigmaσ-algebra

P(Ω)=Q(Ω)=1P( \Omega ) = Q( \Omega ) =1P(Ω)=Q(Ω)=1, because they are both probabilities.

Let

B={A∈A:P(A)=Q(A)}\mathcal{B} = \left\{ A \in \mathcal{A}:P(A)=Q(A) \right\} B={A∈A:P(A)=Q(A)}
By the definition of a Probability measure and Theorem 2.3, B\mathcal{B}B is closed by difference and by increasing limits.

Also B\mathcal{B}B contains C\mathcal{C}C by hypothesis.

Therefore since σ(C)=A\sigma(\mathcal{C})=\mathcal{A}σ(C)=A, we have B=A\mathcal{B}=\mathcal{A}B=A by the Monotone Class Theorem.

不妨设 Ω∈C\Omega \in \mathcal{C}Ω∈C （若不在里面，定义 C′=C∪{Ω}\mathcal{C}'=\mathcal{C} \cup \left\{ \Omega \right\}C′=C∪{Ω} ）

B={A∈A:P(A)=Q(A)}\mathcal{B}= \left\{ A \in \mathcal{A}:P(A)=Q(A) \right\}B={A∈A:P(A)=Q(A)}

证明： B=A\mathcal{B}=\mathcal{A}B=A ，则在 A\mathcal{A}A 上 P(A)=Q(A)P(A)=Q(A)P(A)=Q(A)

∵\because∵ C\mathcal{C}C 上 P(A)=Q(A)P(A)=Q(A)P(A)=Q(A) ∴C⊂B\quad \therefore \mathcal{C} \subset \mathcal{B}∴C⊂B

若 A1⊂A2⊂...Ai∈BA_1 \subset A_2 \subset ...\qquad A_i \in \mathcal{B}A1⊂A2⊂...Ai∈B

则 ∪i=1∞Ai\cup_{i=1}^{ \infty}A_i∪i=1∞Ai

P(∪i=1∞Ai)=lim⁡n→∞P(An)=lim⁡n→∞Q(An)=Q(∪i=1∞Ai)P \left( \cup_{i=1}^{ \infty} A_i \right) = \lim_{n \to \infty} P(A_n) = \lim_{n \to \infty} Q(A_n) = Q \left( \cup_{i=1}^{ \infty} A_i \right) P(∪i=1∞Ai)=n→∞limP(An)=n→∞limQ(An)=Q(∪i=1∞Ai)

∴∪i=1∞Ai∈B\therefore \cup_{i=1}^{ \infty} A_i \in \mathcal{B}∴∪i=1∞Ai∈B

对差也一样

又 B⊃C\mathcal{B} \supset \mathcal{C}B⊃C 且 B\mathcal{B}B 对单増并、差封闭

∴σ(C)=A⊂B⊂A\therefore \quad \sigma(\mathcal{C})=\mathcal{A} \subset \mathcal{B} \subset \mathcal{A}∴σ(C)=A⊂B⊂A

∴A=B\therefore \quad \mathcal{A} = \mathcal{B}∴A=B

Definition 6.2

Let PPP be a probability on A\mathcal{A}A.

A null set (or negligible set) for PPP is a subset AAA of Ω\OmegaΩ such that there exists a B∈AB \in \mathcal{A}B∈A satisfying A⊂BA \subset BA⊂B and P(B)=0P(B)=0P(B)=0

可忽略的集合

A\mathcal{A}A 是一个概率测度

因为概率一定满足规范性和可列可加性

AAA 不一定在 A\mathcal{A}A 上

Remark

We say that a property holds almost surely (a.s.a.s.a.s. in short) if it holds outside a negligible set.
This notion clearly depends on the probability, so we say sometimes P-almost surely , or P-a.s.

几乎处处收敛

ξn→a.s.ξ⇔∃A⊂Bs.t.B⊂Aand P(B)=0limn→∞ξn(ω)=ξ\begin{aligned} \xi_n \to_{a.s.} \xi & \Leftrightarrow \exists A \subset B \\ & \qquad s.t. \ B \subset \mathcal{A} \text{ and } P(B)=0 \\ & \qquad lim_{n \to \infty} \xi_n(\omega) = \xi \end{aligned} ξn→a.s.ξ⇔∃A⊂Bs.t. B⊂A and P(B)=0limn→∞ξn(ω)=ξ

rrr 收敛

ξn→rξ⇔E∣ξn−ξ∣r→0n→∞\xi_n \to_{r} \xi \Leftrightarrow E |\xi_n - \xi|^{r} \to 0 \quad n \to \infty ξn→rξ⇔E∣ξn−ξ∣r→0n→∞

Remark

The negligible set are not necessarily in A\mathcal{A}A
Nevertheless it is natural to say that they have probability zero.
In the following theorem, we extend the probability to the σ\sigmaσ-algebra which is generated by A\mathcal{A}A and all PPP-negligible sets.

目标：将概率 PPP 延拓到 A∪{allP-negligible}\mathcal{A} \cup \left\{ \text{all} P \text{-negligible} \right\}A∪{allP-negligible}

Theorem 6.4

Let PPP be a probability on A\mathcal{A}A and let N\mathcal{N}N be the class of all PPP-negligible sets.

Then

A′={A∪N:A∈A,N∈N}\mathcal{A}' = \left\{ A \cup N : A \in \mathcal{A}, N \in \mathcal{N} \right\} A′={A∪N:A∈A,N∈N}

is a σ\sigmaσ-algebra, called the PPP-completion of A\mathcal{A}A

A′\mathcal{A}'A′ is the smallest σ\sigmaσ-algebra containing A\mathcal{A}A and N\mathcal{N}N
PPP extends uniquely as a probability (still denoted by PPP) on A′\mathcal{A}'A′,

by setting P(A∪N)=P(A)P(A \cup N)=P(A)P(A∪N)=P(A) for A∈AA \in \mathcal{A}A∈A and N∈NN \in \mathcal{N}N∈N

Proof.

由 Corollary 6.1 可知，存在 A∪N\mathcal{A} \cup\mathcal{N}A∪N

∀A∈A∪N即 A∈AorA∈N\forall A \in \mathcal{A} \cup \mathcal{N} \text{ 即 } A \in \mathcal{A} \ or \ A \in \mathcal{N}∀A∈A∪N 即 A∈A or A∈N

下面证明 A∪N\mathcal{A} \cup \mathcal{N}A∪N 对有限交封闭

若 A∈A,B∈A⇒AB∈A⊂A∪NA \in \mathcal{A}, B \in \mathcal{A} \qquad \Rightarrow AB \in \mathcal{A} \subset \mathcal{A} \cup \mathcal{N}A∈A,B∈A⇒AB∈A⊂A∪N

若 A∈A,B∈NAB∈N⊂A∪NA \in \mathcal{A}, B \in \mathcal{N} \qquad AB \in \mathcal{N} \subset \mathcal{A} \cup \mathcal{N}A∈A,B∈NAB∈N⊂A∪N

∵B∈N→∃C∈As.t.B⊂CandP(C)=0∴AB⊂B⊂C⇒AB∈N\begin{aligned} \because & B \in \mathcal{N} \to \exists C \in \mathcal{A} \quad s.t. \quad B \subset C \ and \ P(C)=0 \\ \therefore & AB \subset B \subset C \\ \Rightarrow & AB \in \mathcal{N} \\ \end{aligned} ∵∴⇒B∈N→∃C∈As.t.B⊂C and P(C)=0AB⊂B⊂CAB∈N

若 A∈N,B∈N⇒AB∈N⊂A∪NA \in \mathcal{N}, B \in \mathcal{N} \qquad \Rightarrow AB \in \mathcal{N} \subset \mathcal{A} \cup\mathcal{N}A∈N,B∈N⇒AB∈N⊂A∪N

由 123 说明 A∪N\mathcal{A} \cup \mathcal{N}A∪N 对有限交封闭

证明 A∪N⊂A′\mathcal{A} \cup \mathcal{N} \subset \mathcal{A}'A∪N⊂A′

∀A∈A∪N\forall A \in \mathcal{A} \cup \mathcal{N}∀A∈A∪N ，即 A∈AorA∈NA \in \mathcal{A} \ or \ A \in \mathcal{N}A∈A or A∈N

∵∅∈Aand∅∈N∀B∈A,B=B∪∅∈A⇒A⊂A′∀N∈N,N=N∪∅∈A′⇒N⊂A′}⇒A∪N⊂A′\left. \begin{array}{lll} \because \emptyset \in \mathcal{A} \ and \ \emptyset \in \mathcal{N} \\ \forall B \in \mathcal{A} , \quad B = B \cup \emptyset \in \mathcal{A} \Rightarrow \mathcal{A} \subset \mathcal{A}' \\ \forall N \in \mathcal{N},\quad N = N \cup \emptyset \in \mathcal{A}' \Rightarrow \mathcal{N} \subset \mathcal{A}' \end{array} \right\} \Rightarrow \mathcal{A} \cup \mathcal{N} \subset \mathcal{A}' ∵∅∈A and ∅∈N∀B∈A,B=B∪∅∈A⇒A⊂A′∀N∈N,N=N∪∅∈A′⇒N⊂A′⎭⎬⎫⇒A∪N⊂A′

证明A′=σ(A∪N)\mathcal{A}'=\sigma(\mathcal{A} \cup \mathcal{N})A′=σ(A∪N)

Denote σ(A∪N)=A′′\sigma(\mathcal{A} \cup \mathcal{N})= \mathcal{A}''σ(A∪N)=A′′

1.A⊂A′andN⊂A′已证⇕A∪N⊂A′⇒A′′⊂A′2.∀B∈A′B=A∪N,A∈A,N∈N∵A′′is σ-algebraand A∈A⊂A′′,N∈A⊂A′′B=A∪N∈A′′⇒A′⊂A′′⇓A′=A′′=σ(A∪N)\begin{array}{c} \hline \begin{array}{lll} \begin{array}{lll} 1. \\ \mathcal{A} \subset \mathcal{A}' \ and \ \mathcal{N} \subset \mathcal{A}' \text{已证}\\ \Updownarrow \\ \mathcal{A} \cup \mathcal{N} \subset \mathcal{A}' \\ \Rightarrow \mathcal{A}'' \subset \mathcal{A}' \end{array} && \begin{array}{lll} 2. \\ \forall B \in \mathcal{A}' \quad B = A \cup N , A \in \mathcal{A}, N \in \mathcal{N} \\ \because \mathcal{A}'' \text{ is $\sigma$-algebra} \\ \text{ and } A \in \mathcal{A} \subset \mathcal{A}'', N \in \mathcal{A} \subset \mathcal{A}'' \\ B = A \cup N \in \mathcal{A}'' \Rightarrow \mathcal{A}' \subset \mathcal{A}'' \end{array} \\ & & \\ \end{array} \\ \hline \Downarrow \\ \mathcal{A}' = \mathcal{A}'' = \sigma(\mathcal{A} \cup\mathcal{N}) \end{array} 1.A⊂A′ and N⊂A′已证⇕A∪N⊂A′⇒A′′⊂A′2.∀B∈A′B=A∪N,A∈A,N∈N∵A′′ is σ-algebra and A∈A⊂A′′,N∈A⊂A′′B=A∪N∈A′′⇒A′⊂A′′⇓A′=A′′=σ(A∪N)

证明 A′\mathcal{A}'A′ 是一个 σ\sigmaσ-algebra

Ω∈A′Ω=Ω∪∅,Ω∈A,∅∈N\Omega \in \mathcal{A}' \qquad \Omega=\Omega \cup \emptyset,\quad \Omega \in \mathcal{A}, \emptyset \in \mathcal{N}Ω∈A′Ω=Ω∪∅,Ω∈A,∅∈N

若 B∈A′B \in \mathcal{A}'B∈A′ ，则 ∃A∈A,N∈N\exists A \in \mathcal{A}, N \in \mathcal{N}∃A∈A,N∈N

s.t.B=A∪NBC=AC∩NC=(AC∩CC)∪(AC∩NC∩C)(AC∩CC)∈A(AC∩NC∩C)⊂C∈N⇓BC∈A′\begin{aligned} s.t. \\ B &= A \cup N \\ B^C &= A^C \cap N^C = (A^C \cap C^C) \cup (A^C \cap N^C \cap C) \\ & (A^C \cap C^C) \in \mathcal{A} \\ & (A^C \cap N^C \cap C) \subset C \in \mathcal{N} \\ \Downarrow \\ B^C & \in \mathcal{A}' \\ \end{aligned} s.t.BBC⇓BC=A∪N=AC∩NC=(AC∩CC)∪(AC∩NC∩C)(AC∩CC)∈A(AC∩NC∩C)⊂C∈N∈A′

可列可加

Bn∈A′,n=1,2,...B_n \in \mathcal{A}',n=1,2,...Bn∈A′,n=1,2,...

Nn⊂Cn∈A,P(Cn)=0N_n \subset C_n \in \mathcal{A},P(C_n)=0Nn⊂Cn∈A,P(Cn)=0

P(∪n=1∞Cn)≤∑n=1∞P(Cn)=0P( \cup_{n=1}^{ \infty} C_n) \le \sum_{n=1}^{\infty} P(C_n) =0P(∪n=1∞Cn)≤∑n=1∞P(Cn)=0

Bn=An∪NnAn∈A,Nn∈N∪n=1∞Bn=(∪n=1∞An)∪(∪n=1∞Nn)∈A′∪n=1∞An∈A∪n=1∞Nn∈N\begin{aligned} B_n &= A_n \cup N_n \qquad A_n \in \mathcal{A}, N_n \in \mathcal{N} \\ \cup_{n=1}^{ \infty} B_n &= \left( \cup_{n=1}^{ \infty}A_n \right) \cup \left( \cup_{n=1}^{ \infty} N_n \right) \in \mathcal{A}' \\ & \cup_{n=1}^{ \infty} A_n \in \mathcal{A} \\ & \cup_{n=1}^{ \infty} N_n \in \mathcal{N} \\ \end{aligned} Bn∪n=1∞Bn=An∪NnAn∈A,Nn∈N=(∪n=1∞An)∪(∪n=1∞Nn)∈A′∪n=1∞An∈A∪n=1∞Nn∈N

则 A′\mathcal{A}'A′ 是一个 σ\sigmaσ-algebra ， A′=σ(A∪N)\mathcal{A}'=\sigma(\mathcal{A} \cup \mathcal{N})A′=σ(A∪N)

由 corollary 6.1 知, The uniqueness of the extension is straightforward.

现在想知道，若 A\mathcal{A}A 上有不一样的分解，其概率是否会于分解有关？

Suppose now that A1∪N1=A2∪N2A_1 \cup N_1 = A_2 \cup N_2A1∪N1=A2∪N2 with Ai∈AA_i \in \mathcal{A}Ai∈A and Ni∈NN_i \in \mathcal{N}Ni∈N

The symmetrical difference A1△A2=(A1∩A2c)∪(A1c∩A2)A_1 \triangle A_2 = (A_1 \cap A_2^c ) \cup (A_1^c \cap A_2 )A1△A2=(A1∩A2c)∪(A1c∩A2) is contained in N1∪N2N_1 \cup N_2N1∪N2

Proof.

A1A2c⊂(A1∪N1)A2c=(A2∪N2)A2c=A2A2c∪N2A2c=N2A2c⊂N2A1cA2⊂A1c(A2∪N2)=A1c(A1∪N1)=A1cA1∪A1cN1=A1cN1⊂N1∴A1A2c∪A1cA2⊂N1∪N2\begin{aligned} A_1A_2^c & \subset (A_1 \cup N_1)A_2^c = (A_2 \cup N_2 ) A_2^c = A_2 A_2^c \cup N_2A_2^c = N_2 A_2^c \subset N_2 \\ A_1^c A_2 & \subset A_1^c (A_2 \cup N_2) = A_1^c (A_1 \cup N_1 ) = A_1^c A_1 \cup A_1^c N_1 = A_1^c N_1 \subset N_1 \\ \therefore &\ A_1 A_2^c \cup A_1^c A_2 \quad \subset \quad N_1 \cup N_2 \end{aligned} A1A2cA1cA2∴⊂(A1∪N1)A2c=(A2∪N2)A2c=A2A2c∪N2A2c=N2A2c⊂N2⊂A1c(A2∪N2)=A1c(A1∪N1)=A1cA1∪A1cN1=A1cN1⊂N1 A1A2c∪A1cA2⊂N1∪N2

∃Ci∈AP(Ci)=0,Ni⊂Ci,i=1,2,...Ai∈A\exists C_i \in \mathcal{A} \quad P(C_i)=0 ,\quad N_i \subset C_i, \quad i=1,2,... \quad A_i \in \mathcal{A}∃Ci∈AP(Ci)=0,Ni⊂Ci,i=1,2,...Ai∈A

A1△A2⊂N1∪N2⊂C1∪C2⇒A1△A2∈N,P(C1∪C2)≤P(C1)+P(C2)=0A_1 \triangle A_2 \subset N_1 \cup N_2 \subset C_1 \cup C_2 \quad \Rightarrow \quad A_1 \triangle A_2 \in \mathcal{N}, \quad P(C_1 \cup C_2 ) \le P(C_1)+P(C_2)=0A1△A2⊂N1∪N2⊂C1∪C2⇒A1△A2∈N,P(C1∪C2)≤P(C1)+P(C2)=0

P(A1)=P(A1A2+A1A2c)=P(A1A2)+P(A1A2c)=P(A1A2)A1A2c⊂A1△A2⊂N1∪N2⊂C1∪C2⇒P(A1A2c)≤P(C1∪C2)=0P(A1)=P(A1A2)P(A2)=P(A1A2)⇒P(A1)=P(A2)⇒Q(A1∪N1)=Q(A2∪N2)\begin{aligned} P(A_1) &= P(A_1A_2 +A_1A_2^c) \\ &= P(A_1A_2)+P(A_1A_2^c) \\ &= P(A_1A_2) \\ & A_1A_2^c \subset A_1 \triangle A_2 \subset N_1 \cup N_2 \subset C_1 \cup C_2 \quad\Rightarrow P(A_1A_2^c) \le P(C_1 \cup C_2)=0 \\ P(A_1) &= P(A_1A_2) \\ P(A_2)&=P(A_1A_2) \\ &\Rightarrow P(A_1)=P(A_2) \\ &\Rightarrow Q(A_1 \cup N_1) = Q(A_2 \cup N_2) \end{aligned} P(A1)P(A1)P(A2)=P(A1A2+A1A2c)=P(A1A2)+P(A1A2c)=P(A1A2)A1A2c⊂A1△A2⊂N1∪N2⊂C1∪C2⇒P(A1A2c)≤P(C1∪C2)=0=P(A1A2)=P(A1A2)⇒P(A1)=P(A2)⇒Q(A1∪N1)=Q(A2∪N2)

则不管怎么分解，延拓的概率都相等

现在证明 QQQ 是概率，概率需要验证两条：

Q(Ω)=Q(Ω∪∅)=P(Ω)=1Q(\Omega) = Q (\Omega \cup \emptyset) = P(\Omega)=1Q(Ω)=Q(Ω∪∅)=P(Ω)=1

假设 Bn∈A′B_n \in \mathcal{A}'Bn∈A′ 且 BnB_nBn 两两互斥 n=1,2,...n=1,2,...n=1,2,...

Q(∪i=1∞Bn)=Q{∪i=1∞(An∪Nn}=Q{∪i=1∞An∪∪i=1∞Nn}=P{∪i=1∞An}An是Bn的一部分，则 Bn两两互斥得到 An两两互斥=∑n=1∞P{An}Q(Bn)=Q(An∪Nn)=P(An)∴Q(∪i=1∞Bn)=∑n=1∞Q(Bn)\begin{aligned} Q( \cup_{i=1}^{ \infty} B_n) &= Q \left\{ \cup_{i=1}^{ \infty} (A_n \cup N_n \right\} \\ &= Q \left\{ \cup_{i=1}^{ \infty} A_n \cup \cup_{i=1}^{ \infty} N_n \right\} \\ &= P \left\{ \cup_{i=1}^{ \infty} A_n \right\} \\ & \qquad \text{ $A_n$ 是$B_n$ 的一部分，则 $B_n$ 两两互斥得到 $A_n$ 两两互斥}\\ &= \sum_{n=1}^{\infty} P \left\{ A_n \right\} \\ \\ Q(B_n) &= Q(A_n \cup N_n) = P(A_n) \\ \therefore & \ Q( \cup_{i=1}^{ \infty} B_n ) = \sum_{n=1}^{\infty} Q(B_n) \\ \end{aligned} Q(∪i=1∞Bn)Q(Bn)∴=Q{∪i=1∞(An∪Nn}=Q{∪i=1∞An∪∪i=1∞Nn}=P{∪i=1∞An} An 是Bn 的一部分，则 Bn 两两互斥得到 An 两两互斥=n=1∑∞P{An}=Q(An∪Nn)=P(An) Q(∪i=1∞Bn)=n=1∑∞Q(Bn)

则 QQQ 是 A′\mathcal{A}'A′ 上的概率测度，仍然记 QQQ 为 PPP

定理6.4证明完毕

回顾一下定理6.1以及其证明过程：

定理6.1：概率 PPP 是 A\mathcal{A}A 上的概率， N\mathcal{N}N 是所有的可忽略集合类

定义 A′={A∪N:A∈A,N∈N}\mathcal{A}'= \left\{ A \cup N: A \in \mathcal{A}, N \in \mathcal{N} \right\}A′={A∪N:A∈A,N∈N}

则可以得到：

A′\mathcal{A}'A′ 是一个 σ\sigmaσ-代数
A′=σ(A∪N)\mathcal{A}'=\sigma(\mathcal{A} \cup \mathcal{N})A′=σ(A∪N)
概率 PPP （原来定义在 A\mathcal{A}A 上）可在 A′\mathcal{A}'A′ 上唯一延拓

定理6.1把概率又往前延拓，现在是可以延拓到一个 σ\sigmaσ-代数和一个可忽略集的并集构成的 σ\sigmaσ-代数上了。

定理的证明需要证明几步：

A′\mathcal{A}'A′ 是一个 σ\sigmaσ-代数 {∅Ω在里面A在A′里面，则 Ac也在A′里面对可列并封闭\left\{ \begin{array}{lll} \emptyset \ \Omega \text{在里面} \\ A \text{在} \mathcal{A}' \text{里面，则 } A^c \text{也在} \mathcal{A}' \text{里面} \\ 对可列并封闭 \end{array}\right.⎩⎨⎧∅ Ω在里面A在A′里面，则 Ac也在A′里面对可列并封闭
证明 A′=σ(A∪N)\mathcal{A}'=\sigma(\mathcal{A } \cup \mathcal{N})A′=σ(A∪N) ，由两个包含关系可以证明

其中，证明 A∈N⊂A′\mathcal{A} \in \mathcal{N} \subset \mathcal{A}'A∈N⊂A′ 就是为了证明这一步
证明概率 PPP 延拓到 A′\mathcal{A}'A′ 上之后（记为 QQQ )仍然是一个概率（由概率的定义来证明）
证明唯一性，延拓之后的概率是唯一的，则仍然可以记为 PPP.

唯一性的证明由性质6.1可以得到，需要证明
- A∪N\mathcal{A} \cup \mathcal{N}A∪N 对有限交封闭
- A′=σ(A∪N)\mathcal{A}'=\sigma(\mathcal{A} \cup \mathcal{N})A′=σ(A∪N)
- 概率与分解无关
  
  性质6.1：概率 PPP 和概率 QQQ 是定义在A\mathcal{A}A 上的两个概率，若 P,QP,QP,Q 在 C⊂A\mathcal{C} \subset \mathcal{A}C⊂A 上相等，并且对有限交封闭，如果 σ(C)=A\sigma(\mathcal{C})=\mathcal{A}σ(C)=A ，则 P=QP=QP=Q

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