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：

  1. \(\cal F = F_{1} \times F_{2}\)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：

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

  1. \(\cal F = F_{1} \times F_{2}\) 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：

  1. \((A \cup B) = (A \cap B)\)
  2. \(A = B\)
  3. \(A \Delta B = \emptyset\)


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 \(A \subset B\)。

  1. 当 \(A = B\)，显然 \(\mathcal{F}_{A} = \mathcal{F}_{B}\)，则 \(\mathcal{F}_{A} \subset \mathcal{F}_{B}\) 自然成立。

  2. 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。

  3. 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(100)  评论(0)  编辑  收藏  举报