测度论：Measure Theory (3)

Definition. (Product $\sigma-$field)

Let $\Omega_1$ and $\Omega_2$ be two (universal) sets and $\Omega ~ = ~ \Omega_1 \times \Omega_2$ be the Cartesian product. Suppose ${\cal F}_1$ is a $\sigma$-field on $\Omega_1$ and ${\cal F}_2$ is a $\sigma$-field on $\Omega_2$.

For $\forall A_{1} \in \mathcal{F}_{1}, ~ \forall A_{2} \in \mathcal{F}_{2}$, the minimal $\sigma$-field $\cal F$ in $\Omega$ containing all "rectangles":

\[A_1 \times A_2 ~ = ~ \{(a_1,a_2)\in\Omega: ~ a_1\in A_1, ~ a_2\in A_2\}, \]

is called the product $\sigma$-field of ${\cal F}_1$ and ${\cal F}_2$, denoted by ${\cal F} ~ = ~ {\cal F}_1 \times {\cal F}_2$.

That is to say, the product $\sigma$-field ${\cal F}={\cal F}_1\times{\cal F}_2$ is the $\sigma-$field generated by:

\[\left\{ A_{1} \times A_{2}: ~ \forall A_{1} \in \mathcal{F}_{1}, ~ \forall A_{2} \in \mathcal{F}_{2} \right\} \]

Theorem 3.1

The product $\sigma$-field ${\cal F}_1\times{\cal F}_2$ is generated by the family of sets ("cylinders"):

\[{\cal C} ~ = ~ \{A_1\times\Omega_2:~A_1\in{\cal F}_1\} ~ \cup ~ \{\Omega_1\times A_2:~A_2\in{\cal F}_2\}. \]

$\textbf{在证明Theorem 3.1之前，先理解定理在说什么：}$

Notes 1. (Theorem 3.1)

Denote the product $\sigma$-field by $\cal F_{R} = F_{1} \times F_{2}$, where the rectangle $\cal R$ is defined as:

\[\mathcal{R} ~ := ~ \left\{ A_{1} \times A_{2}: ~ \forall A_{1} \in \mathcal{F}_{1}, ~ \forall A_{2} \in \mathcal{F}_{2} \right\} \]

$\cal F_R$ is the $\sigma-$field generated by $\mathcal{R}$. So the theorem is actually stating that, the $\sigma-$field generated by $\cal R$ is identical to the $\sigma-$field generated by $\cal C$ (i.e., $~\cal F_R = F_C$).

Notes 2. (Theorem 3.1)

设有全集 $\Omega_{1}, ~ \Omega_{2}$, 它们的笛卡尔积（Cartisian product）定义为：

\[\Omega ~ := ~ \Omega_{1} \times \Omega_{2} ~ = ~ \left\{ (w_{1}, w_{2}): ~ w_{1} \in \Omega_{1}, ~ w_{2} \in \Omega_{2} \right\} \]

设在 $\Omega_{1}, ~ \Omega_{2}$ 上分别有 $\cal F_{1}, F_{2}$ 两个 $\sigma-$field，则它们的笛卡尔积定义为包含所有“rectangle” $A_{1} \times A_{2}$ 的最小 $\sigma-$field。其中，对于每一个 $A_{1} \in \cal F_{1}$ 和 $A_{2} \in \cal F_{2}$，同理都有:

\[A_{1} \times A_{2} ~ := ~ \left\{ (a_{1}, a_{2}): ~ a_{1} \in A_{1}, ~ a_{2} \in A_{2} \right\} \]

也就是说两个 $\sigma$ 域的笛卡尔积可以由以下两种等价的定义方式给出：

$\cal F = F_{1} \times F_{2}$是包含所有"rectangle": $A_{1} \times A_{2}= \left\{ (a_{1}, a_{2}): ~ a_{1} \in A_{1}, ~ a_{2} \in A_{2} \right\} \mbox{ for } \forall A_{1} \in \mathcal{F_{1}}, \forall A_{2} \in \mathcal{F_{2}}$的最小$\sigma$域。定义包含所有rectangle的集类：

\[\mathcal{R} ~ := ~ \left\{ A_{1}\times A_{2}: ~ A_{1} \in \mathcal{F_{1}}, ~ A_{2} \in \mathcal{F_{2}} \right\} \]

$\cal F = F_{1} \times F_{2}$ 是由所有“cylinders”：${\cal C} = \{A_1\times\Omega_2:~A_1\in{\cal F}_1\}\cup\{\Omega_1\times A_2:~A_2\in{\cal F}_2\} ~ \mbox { for } \forall A_{1} \in \mathcal{F_{1}}, \forall A_{2} \in \mathcal{F_{2}}$ 构成的集类生成的 $\sigma$域。定义包含所有cylinder的集类：

\[{\cal C} ~ = ~ \{A_1\times\Omega_2: ~ A_1\in{\cal F}_1\} \cup \{\Omega_1\times A_2: ~ A_2\in{\cal F}_2\} \]

这里必须明确，犹如前文提到："由...生成的 $\sigma$域"，"..."是集类（集合的集合）而非集合(狭义的)。例如设全集 $= \left\{ 0, 1, 2, 3 \right\}$，由集类 $\left\{ \left\{ 1 \right\}\right\}$ 生成的 $\sigma$域：$\left\{ \emptyset, \Omega, \left\{ 1 \right\}, \left\{ 0, 2, 3 \right\} \right\}$。

此处，$A_{1} \in \mathcal{F_{1}}, ~ A_{2} \in \mathcal{F_{2}}$ 为两个集合，它们的笛卡尔积依然为集合————试想两个单点集合的笛卡尔积为一个二维坐标点集。但是，对于不同的 $A_{1} \in \mathcal{F_{1}}, A_{2} \in \mathcal{F_{2}}$， $\left\{ A_{1} \times A_{2}: A_{1} \in \mathcal{F_{1}}, A_{2} \in \mathcal{F_{2}} \right\}$ 却是一个集类，因为：

\[\left\{ A_{1} \times A_{2}: ~ A_{1} \in \mathcal{F_{1}}, A_{2} \in \mathcal{F_{2}} \right\} ~ := ~ \left\{ \left\{ (a_{1}, a_{2}): ~ a_{1} \in A_{1}, a_{2} \in A_{2} \right\}: ~ A_{1} \in \mathcal{F_{1}}, A_{2} \in \mathcal{F_{2}} \right\}。 \]

同理，对于特定的 $A_{1} \in \mathcal{F_{1}}, A_{2} \in \mathcal{F_{2}}$，$(A_{1} \times \Omega_{2}) \cup (A_{2} \times \Omega_{1})$ 是集合；但是对于 $\forall A_{1} \in \mathcal{F_{1}}, \forall A_{2} \in \mathcal{F_{2}}$，所有随之分别构造的 $(A_{1} \times \Omega_{2}) \cup (A_{2} \times \Omega_{1})$ 组成了集类 $\cal C$:

\[\mathcal{C} ~ := ~ \left\{ \left\{ (a_{1}, w_{2}): a_{1} \in A_{1}, w_{2} \in \Omega_{2} \right\} \cup \left\{ (w_{1}, a_{2}): w_{1} \in \Omega_{1}, a_{2} \in A_{2} \right\}: ~ \forall A_{1} \in \mathcal{F_{1}}, \forall A_{2} \in \mathcal{F_{2}} \right\} \]

总而言之，这里说的"rectangle"和"cylinder"都是集合，但对于所有的 $A_{1}, A_{2}$ 从而各自组成了集类。

$\textbf{正式证明之前再引入两条引理：}$

Lemma 3.1.1 (Theorem 3.1)

设 $A, B$ 为两个集合，则以下三个说法等价：

$(A \cup B) = (A \cap B)$
$A = B$
$A \Delta B = \emptyset$

Proof. (Lemma 3.1.1)

若$A = B$，则显然有$(A \cup B) = (A \cap B)$ 和 $A \Delta B = \emptyset$，则$2 \rightarrow 1, ~ 2 \rightarrow 3$自然成立。

现在证明：$1 \rightarrow 2$。若 $(A \cup B) = (A \cap B)$ 成立，假设 $A \not\subset B$，则：$\exists k \in A: ~ k \notin B$。而若 $k \in A$，则 $k \in (A \cup B)$，又由条件 $(A \cup B) = (A \cap B)$，则 $k \in (A \cap B) \implies k \in B$。故产生矛盾，所以假设不成立，有 $A \subset B$。同理可证 $B \subset A$，所以 $A = B$。

最后证明：$1 \rightarrow 3$。若 $(A \cup B) = (A \cap B)$ 成立，有：

\[(A \cup B) ~ = ~ (A \setminus B) \cup (B \setminus A) \cup (A \cap B) ~ = ~ (A \Delta B) \cup (A \cap B) ~ = ~ (A \cap B) \]

这说明：$(A \Delta B) \subset (A \cap B)$。若 $(A \Delta B) \neq \emptyset$，则有：

\[\forall k \in (A \Delta B): ~ k \in (A \cap B) \]

而 $k \in (A \Delta B)$ 意味着 $k \in A, ~ k \notin B$ 或 $k \in B, ~ k \notin A$，与结论矛盾，所以 $(A \Delta B) = \emptyset$。

至此我们完成证明 $1 \leftrightarrow 2$ 和 $1 \leftrightarrow 3$，得证：$1 \leftrightarrow 2 \leftrightarrow 3$。

Lemma 3.1.2 (Theorem 3.1)

若定义在同一全集上 $\Omega$ 的两个集类 $A, B$ 满足 $A \subset B$，则它们各自生成的 $\sigma$域满足：

\[\mathcal{F}_{A} ~ \subset ~ \mathcal{F}_{B} \]

Proof. (Lemma 3.1.2)

设 $A \subset B$。

当 $A = B$，显然 $\mathcal{F}_{A} = \mathcal{F}_{B}$，则 $\mathcal{F}_{A} \subset \mathcal{F}_{B}$ 自然成立。
当 $A = \emptyset$，则 $\mathcal{F}_{A} = \left\{ \emptyset, ~ \Omega \right\}$，而任意的 $\sigma$ 域都包含 $\emptyset, ~ \Omega$ 两个集合。故 $\mathcal{F}_{A} \subset \mathcal{F}_{B}$ 成立。
当 $A \subsetneqq B$ 时，则：

\[\forall a \in A: ~ a \in B \quad \mbox{ 且 } \quad \exists b \in B: ~ b \notin A \]

所以必存在非空集 $D$ 满足：

\[A \cap D = \emptyset \quad \mbox{ 且 } \quad A \cup D = B \]

因为 $\mathcal{F}_{A}$ 是由 $A$ 生成的 $\sigma$域，有：$A \subset \mathcal{F}_{A}$，同理可得 $B \subset \mathcal{F}_{B} \implies (A \cup D) \subset \mathcal{F}_{B}$，

$\textbf{个人认为这个定理比想象中难证明得多}$

Proof. (Theorem 3.1)

现在，所有cylinders构成的集类：$\mathcal{C} ~ = ~ \left\{ A_{1} \times \Omega_{2}: ~ A_{1} \in \mathcal{F}_{1} \right\} ~ \cup ~ \left\{ \Omega_{1} \times A_{2}: ~ A_{2} \in \mathcal{F}_{2} \right\}$。

注意 $\mathcal{C}$ 的构造方式，它与 $\left\{ (A_{1} \times \Omega_{2}) \cup (\Omega_{1} \times A_{2}): ~ A_{1} \in \mathcal{F}_{1}, A_{2} \in \mathcal{F}_{2} \right\}$ 完全不同！

所有rectangles构成的集类：$\mathcal{R} ~ = ~ \left\{ A_{1} \times A_{2}: ~ A_{1} \in \mathcal{F}_{1}, ~ A_{2} \in \mathcal{F}_{2} \right\}$，而它可以写作：

\[\begin{align*} \mathcal{R} & = \left\{ A_{1} \times A_{2}: ~ A_{1} \in \mathcal{F}_{1}, ~ A_{2} \in \mathcal{F}_{2} \right\}\\ & = \left\{ A_{1} \times \Omega_{2}: ~ A_{1} \in \mathcal{F}_{1} \right\} \cup \left\{ A_{1} \times A_{2}: ~ A_{1} \in \mathcal{F}_{1}, ~ A_{2} \in \mathcal{F}_{2} \setminus \left\{ \Omega_{2} \right\} \right\}\\ & \cup ~ \left\{ \Omega_{1} \times A_{2}: ~ A_{2} \in \mathcal{F}_{2} \right\} \cup \left\{ A_{1} \times A_{2}: ~ A_{1} \in \mathcal{F}_{1} \setminus \left\{ \Omega_{1} \right\}, ~ A_{2} \in \mathcal{F}_{2} \right\} \end{align*} \]

故可以发现：$\cal C \subset R$，由Lemma 3.1.2.，我们有：$\cal F_{C} \subset F_{R}$。

另一方面，对于确定的 $A_{1} \in \mathcal{F}_{1}, ~ A_{2} \in \mathcal{F}_{2}$，有：

\[A_{1} \times A_{2} = (A_{1} \times \Omega_{2}) \cap (\Omega_{1} \times A_{2}) \]

这很好理解，因为：

\[\begin{align*} (A_{1} \times \Omega_{2}) \cap (\Omega_{1} \times A_{2}) & = \left\{ (a_{1}, w_{2}): ~ a_{1} \in A_{1}, ~ w_{2} \in \Omega_{2} \right\} \cap \left\{ (w_{1}, a_{2}): ~ w_{1} \in \Omega_{1}, ~ a_{2} \in A_{2} \right\}\\ & = \left( \left\{ (a_{1}, a_{2}): ~ a_{1} \in A_{1}, ~ a_{2} \in A_{2} \right\} \cup \left\{ (a_{1}, w_{2}): ~ a_{1} \in A_{1}, ~ w_{2} \in \Omega_{2} \setminus A_{2} \right\} \right) \cap \\ & \qquad \quad \left( \left\{ (a_{1}, a_{2}): ~ a_{1} \in A_{1}, ~ a_{2} \in A_{2} \right\} \cup \left\{ (w_{1}, a_{2}): ~ w_{1} \in \Omega_{1} \setminus A_{1}, ~ a_{2} \in A_{2} \right\} \right)\\ & = \left( \left\{ (a_{1}, w_{2}): ~ a_{1} \in A_{1}, ~ w_{2} \in \Omega_{2} \setminus A_{2} \right\} \cap \left\{ (w_{1}, a_{2}): ~ w_{1} \in \Omega_{1} \setminus A_{1}, ~ a_{2} \in A_{2} \right\} \right) \cup \\ & \qquad \quad \left\{ (a_{1}, a_{2}): ~ a_{1} \in A_{1}, ~ a_{2} \in A_{2} \right\} \end{align*} \]

前半部分（即：$\left( \left\{ (a_{1}, w_{2}): ~ a_{1} \in A_{1}, ~ w_{2} \in \Omega_{2} \setminus A_{2} \right\} \cap \left\{ (w_{1}, a_{2}): ~ w_{1} \in \Omega_{1} \setminus A_{1}, ~ a_{2} \in A_{2} \right\} \right)$）为 $\emptyset$，则：

\[(A_{1} \times \Omega_{2}) \cap (\Omega_{1} \times A_{2}) ~ = ~ A_{1} \times A_{2} ~ = ~ \left\{ (a_{1}, a_{2}): ~ a_{1} \in A_{1}, ~ a_{2} \in A_{2} \right\} \]

所以对于 $\forall A_{1} \in \mathcal{F}_{1}, \forall A_{2} \in \mathcal{F}_{2}: ~ A_{1} \times A_{2} ~ = ~ (A_{1} \times \Omega_{2}) \cap (\Omega_{1} \times A_{2})$。

注意集类 $\mathcal{C} = \left\{ A_{1} \times \Omega_{2}: ~ A_{1} \in \mathcal{F}_{1} \right\} \cup \left\{ \Omega_{1} \times A_{2}: ~ A_{2} \in \mathcal{F}_{2} \right\}$，其中每一个 $A_{1} \times \Omega_{2}$ 与 $\Omega_{1} \times A_{2}$ 都为集合，分别装在两个集类中，再将两个集类进行并运算。因此，$\cal C$ 实际可以写为：

\[\mathcal{C} = \left\{ A_{1} \times \Omega_{2} ~ \mbox{ for } ~ A_{1} \in \mathcal{F}_{1}, ~~~ \Omega_{1} \times A_{2} ~ \mbox{ for } ~ A_{2} \in \mathcal{F}_{2} \right\} \]

那么，由 $\cal C$ 生成的 $\sigma$域 $\cal F_{C}$ 必定包含 $\cal C$，且满足 $\sigma$域的性质。这意味着：$\cal F_{C}$ 中必定包含以下元素：

\[A_{1} \times \Omega_{2} ~ \mbox{ for } ~ A_{1} \in \mathcal{F}_{1}, \quad \Omega_{1} \times A_{2} ~ \mbox{ for } ~ A_{2} \in \mathcal{F}_{2} \]

现在：$\mathcal{R} = \left\{ A_{1} \times A_{2}: ~ A_{1} \in \mathcal{F}_{1}, ~ A_{2} \in \mathcal{F}_{2} \right\}$，由于对于 $\cal R$ 中的每一个元素（集合） $A_{1} \times A_{2}$，都有 $(A_{1} \times \Omega_{2}) \cap (\Omega_{1} \times A_{2}) = A_{1} \times A_{2}$，而 $(A_{1} \times \Omega_{2})$ 和 $(\Omega_{1} \times A_{2})$ 都被包含在$\sigma$域 $\cal F_{C}$ 中。

由于 $\sigma$域有推论性质，对可数交运算封闭，则：

\[\left( (A_{1} \times \Omega_{2}) \cap (\Omega_{1} \times A_{2}) \right) ~ \in ~ \mathcal{F_{C}} \quad \implies \quad (A_{1} \times A_{2}) ~ \in ~ \cal F_{C} \]

于是：

\[\forall A_{1} \in \mathcal{F}_{1}, ~ \forall A_{2} \in \mathcal{F}_{2}: ~ (A_{1} \times A_{2}) \in \mathcal{F_{C}} \quad \implies \quad \forall r \in \mathcal{R}: ~ r \in \mathcal{F_{C}} \quad \implies \quad \mathcal{R} \subset \mathcal{F_{C}} \]

由于：$\mathcal{F_{R}}$ 是由 $\mathcal{R} = \left\{ A_{1} \times A_{2}: ~ A_{1} \in \mathcal{F}_{1}, ~ A_{2} \in \mathcal{F}_{2} \right\}$ 生成的最小$\sigma$域，这意味着，$\cal F_{R}$ 中包含着 $\cal R$ 中所有的元素（集合）$A_{1} \times A_{2}$，也仅包含着由这些集合交互进行补运算、可数并运算的结果集合。然而，对于 $\cal R$ 中的每一项，$ \cal F_{C}$ 都有：$\left( (A_{1} \times \Omega_{2}) \cap (\Omega_{1} \times A_{2}) \right)$ 与 $A_{1} \times A_{2}$ 对应，并且 $\cal F_{C}$ 也为 $\sigma$域，也必须要满足对补运算封闭、对可数并运算封闭的性质。所以，由 $\cal R$ 生成的 $\sigma$域 $\cal F_{R}$ 中的每一项，都在 $\cal F_{C}$ 中，i.e.：

\[\forall p \in \mathcal{F_{R}}: ~ p \in \mathcal{F_{C}} \quad \implies \quad \cal F_{R} ~ \subset ~ F_{C} \]

综上所述，已经证得 $\cal F_{C} \subset F_{R}, ~ \cal F_{R} \subset F_{C}$，因此：

\[\cal F_{R} ~ = ~ F_{C} ~ = ~ F ~ = ~ F_{1} \times F_{2} \]

直观理解

从直观的角度解释 $\cal R \subset F_{C}$ 的证明：

\[\begin{align*} & \mathcal{R} ~ := ~ \left\{ A_{1}\times A_{2}: ~ A_{1} \in \mathcal{F_{1}}, ~ A_{2} \in \mathcal{F_{2}} \right\}\\ & \mathcal{C} ~ := ~ \left\{ A_{1} \times \Omega_{2}: ~ A_{1} \in \mathcal{F_{1}} \right\} ~ \cup ~ \left\{ \Omega_{1} \times A_{2}: ~ A_{2} \in \mathcal{F_{2}} \right\} \end{align*} \]

设：

\[\begin{align*} & \mathcal{F_{1}} ~ := ~ \left\{ \Omega_{1}, ~ \emptyset, ~ A_{11}, ~ A_{12}, ~ A_{13}, \ldots \right\}\\ & \mathcal{F_{2}} ~ := ~ \left\{ \Omega_{2}, ~ \emptyset, ~ A_{21}, ~ A_{22}, ~ A_{23}, \ldots \right\} \end{align*} \]

设：

\[\begin{align*} & \vec{A_{1}} = \left( \Omega_{1}, ~ \emptyset, ~ A_{11}, ~ A_{12}, ~ A_{13}, \ldots \right)^{T}\\ & \vec{A_{2}} = \left( \Omega_{2}, ~ \emptyset, ~ A_{21}, ~ A_{22}, ~ A_{23}, \ldots \right)^{T}\\ \end{align*} \]

那么：

\[\begin{align*} \mathcal{R} ~ & = ~ \left\{ \vec{A_{1}} \cdot \vec{A_{2}}^{T} 矩阵中的每一个元素 \right\} = ~ \left\{ \Omega, ~ \emptyset, ~ A_{11} \cdot A_{21}, ~ A_{11} \cdot A_{22}, \ldots , ~ A_{12} \cdot A_{21}, \ldots \right\}\\ \ \; \\ \mathcal{C} ~ & = ~ \left\{ A_{11} \times \Omega_{2}, ~ A_{12} \times \Omega_{2}, ~ A_{13} \times \Omega_{2}, \ldots \right\} ~ \cup ~ \left\{ A_{21} \times \Omega_{1}, ~ A_{22} \times \Omega_{1}, ~ A_{23} \times \Omega_{1}, \ldots \right\}\\ & = \left\{ A_{11} \times \Omega_{2}, ~ A_{12} \times \Omega_{2}, \ldots, ~ A_{21} \times \Omega_{1}, ~ A_{22} \times \Omega_{1}, \ldots \right\} \end{align*} \]

由 $\cal C$ 生成的 $\sigma$域 $\cal F_{C}$ 自然也有以下这些元素：

\[A_{11} \times \Omega_{2}, \quad A_{12} \times \Omega_{2}, \quad \ldots \quad , \quad A_{21} \times \Omega_{1}, \quad A_{22} \times \Omega_{1}, \quad \ldots \quad \]

且对于 $\cal R$ 中的任意一项：$A_{1i} \cdot A_{2j}$，都能在 $\cal F_{C}$ 中找到 $(A_{1i} \times \Omega_{2}) ~ \cap ~ (\Omega_{1} \times A_{2j}) ~ = ~ A_{1i} \cdot A_{2j}$，所以：$\cal R \subset F_{C}$。

而后半段对于 $\cal F_{R} \subset F_{C}$ 的证明，实际上在告诉我们推论：

设 $\cal A \subset F$，则由 $\cal A$ 生成的 $\sigma$域 $\cal F_{A} \subset F$

posted @ 2022-05-27 15:05 车天健阅读(74) 评论(0) 编辑收藏举报

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