测度论：Measure Theory (2) —— Monotone Class Theorem（单调类定理）

这是我的笔记以及一些个人理解，其中大多证明系本人完成。

单调类定理是我初次接触测度论时遇见的第一个难以理解其证明的定理，其中对三个集类的构造颇有“左脚踩右脚上天”的感觉。但在测度论中此类构造证明比比皆是，例如后面我会讲到Vitali对不可测集的构造。

Definition. (Monotone Class)

A monotone class $\cal G$ is a family of sets, which is closed under countable unions of increasing sets, and closed under countable intersections of decreasing sets, i.e.,

if $A_1\subset A_2\subset\ldots, \quad A_i\in{\cal G}$, then $\displaystyle\bigcup_{i=1}^\infty A_i\in{\cal G}$;

if $A_1\supset A_2\supset\ldots,\quad A_i\in{\cal G}$, then $\displaystyle\bigcap_{i=1}^\infty A_i\in{\cal G}$.

Lemma 2.1

Apparently, all $\sigma-$fields are monontone classes. However, there exists some monotone classes that are not closed under complement or finite unions (even if they are closed under countable unions), which means that:
Monotone class can neither be a field nor a $\sigma-$field.

Proof. (Lemma 2.1)

Firstly, let's prove that all $\sigma-$fields are monotone classes. Suppose we have a $\sigma-$field $\mathcal{F}$, and sets $E_{1}, E_{2}, \ldots \in \mathcal{F}$. By definition, we have $E_{1}^{c}, E_{2}^{c}, \ldots \in \mathcal{F}$.

Suppose $E_{1} \subset E_{2} \subset E_{3} \subset \cdots$, where $E_{i} \in \mathcal{F}$ for $\forall i \in \mathbb{N}^{+}$. Since $\sigma-$field is closed under countable unions, we have $\bigcup\limits^{\infty}_{i=1}E_{i} \in \mathcal{F}$.
Suppose $E_{1} \supset E_{2} \supset E_{3} \supset \cdots$, where $E_{i} \in \mathcal{F}$ for $\forall i \in \mathbb{N}^{+}$. Similarly, we have $\bigcap\limits^{\infty}_{i=1}E_{i} = \bigcup\limits^{\infty}_{i=1}E_{i}^{c} \in \mathcal{F}$.

Hence, we complete the proof that all $\sigma-$fields are monotone classes.

For the latter part of the statement, suppose that:

\[\mathcal{A}: = \left\{\mbox{all intervals in $\mathbb{R}$} \right\} \]

Suppose an arbitrary sequence of intervals $E_{1}, E_{2}, \ldots$, where $E_{i} \in \mathcal{F}$ for $\forall i \in \mathbb{N}^{+}$. We prove that $\mathcal{A}$ is a monotone class firstly.

W.l.o.g., assume all intervals are open, i.e., $E_{i}:= (a_{i}, b_{i})$.

When $E_{1} \subset E_{2} \subset \cdots$, we have $(a_{1}, b_{1}) \subset (a_{2}, b_{2}) \subset \cdots$, where $a_{1} \geq a_{2} \geq \cdots$, and $b_{1} \leq b_{2} \leq \cdots$, that is, $\left\{ a_{n} \right\}^{\infty}_{n=1}$ is deceasing, and $\left\{ b_{n} \right\}^{\infty}_{n=1}$ is increasing. Then $\lim\limits_{n \rightarrow \infty} a_{n} = a \in \mathbb{R}$ or $-\infty$, and $\lim\limits_{n \rightarrow \infty} b_{n} = b \in \mathbb{R}$ or $\infty$. However, any of $(a, b), (a, \infty), (-\infty, b), (-\infty, \infty)$ is interval. Therefore, $\bigcup\limits^{\infty}_{n=1}E_{n} = \lim\limits_{n \rightarrow \infty} (a_{n}, b_{n}) \in \mathcal{A}$.
When $E_{1} \supset E_{2} \supset \cdots$, similarly, we have $\bigcap\limits^{\infty}_{n=1}E_{n} \in \mathcal{A}$.

Therefore, we complete the proof that $\mathcal{A}$ is a monotone class.

However, suppose $E_{1} = (-2, -1) , E_{2} = (1, 2) $ and $E_{1}, E_{2} \in \mathcal{A}$.

$E_{1} \cup E_{2} = (-2, -1) \cup (1, 2)$ is not interval, hence $E_{1} \cup E_{2} \notin \mathcal{A}$, which means such monotone class is not closed under finite unions.
$E_{3} = \left(1, 3\right)$, then $E_{3}^{c} = \left(-\infty, 1\right] \cup \left[3, \infty \right)$ is not an interval, $E_{3}^{c} \notin \mathcal{A}$, which means such monotone class is not closed under complement either.

Theorem 2.1 (Monotone Class Theorem)

The smallest monotone class $\cal G_A$ containing a field $\cal A$ coincides with the $\sigma$-field $\cal F_A$ generated by $\cal A$.

Proof. (Monotone Class Theorem)

Firstly, be Lemma 2.1 above, $\cal F_A$ is a monotone class such that $\cal A \in F_A$, while $\cal G_A$ is the smallest monotone class which contains $\cal A$. Thus we have:

\[\cal G_A \subset F_A \]

Thus, it remains for us to prove that $\cal F_A \subset G_A$ as following. We now proving $\cal G_A$ is a field:

Let $\mathcal{G}:= \left\{A: A^{c} \in \cal G_A \right\}$.

Proof: $\cal A \subset G$.
For $\forall B \in \mathcal{A}$, since $\cal A$ is a field, we must have $B^{c} \in \cal A$, and because $\cal A \subset G_A$, we further have $B^{c} \in \cal G_A$. By the way that $\cal G$ was constructed, we have $B \in \cal G$. Thus,

\[\forall B \in \mathcal{A}: B \in \cal G_A \implies A \subset G_A \]

Proof: $\cal G$ is a monotone class.

Suppose $E_{1} \subset E_{2} \subset \cdots, E_{i} \in \cal G$. By the way that $\cal G$ was constructed, we have: $E_{1}^{c} \supset E_{2}^{c} \supset \cdots, E_{i}^{c} \in \cal G_A$. Since $\cal G_A$ is a monotone class, then $\bigcap\limits^{\infty}_{i=1} E_{i}^{c} = (\bigcup\limits^{\infty}_{i=1}E_{i} )^{c} \in \cal G_A$. By the way that $\cal G$ was constructed, we further have $\bigcup\limits^{\infty}_{i=1}E_{i} \in \cal G$

Suppose $E_{1} \supset E_{2} \supset \cdots, E_{i} \in \cal G$, we must have $E_{1}^{c} \subset E_{2}^{c} \subset \cdots, E_{i}^{c} \in \cal G_A$. Thus $\bigcup\limits^{\infty}_{i=1} E_{i}^{c} = (\bigcap\limits^{\infty}_{i=1} E_{i})^{c} \in \mathcal{G_A} \implies \bigcap\limits^{\infty}_{i=1}E_{i} \in \cal G$.

$\cal G_A$ is the smallest monotone class that contains $\cal A$, then $\cal G_A \subset G$, and thus for $\forall B \in \mathcal{G_A}: B \in \cal G$. By the way that $\cal G$ was constructed, we have $B^{c} \in \cal G_A$. Hence, for $\forall B \in \mathcal{G_A}: B^{c} \in \cal G_A$. Therefore, $\cal G_A$ is closed under complement.

Besides, since $\cal A$ is a field and $\cal A \subset G_A$, we have $\Omega \in \mathcal{A} \implies \Omega \in \cal G_A$.

Let $\mathcal{E}:= \left\{E: (A \cup E) \in \mathcal{G_A} \right\}$ for $\forall A \in \cal A$.

Proof: $ A \subset \mathcal{E}$.
For $\forall B \in \cal A$, since $\cal A$ is a field, then $A \cup B \subset \cal A$, and also $\cal A \subset G_A$, then $(A \cup B) \in \cal G_A$. By the way that $\cal E$ was constructed, we have: $B \in \cal E$. Thus, $\forall B \in \mathcal{A}: B \in \mathcal{E} \implies \cal A \subset E$.
Proof: $\mathcal{E}$ is a monotone class.
Suppose $E_{1} \subset E_{2} \subset \cdots, E_{i} \in \cal E$. By the way that $\cal E$ was constructed, we have $(A \cup E_{1}) \subset (A \cup E_{2}) \subset \cdots$, and $(A \cup E_{i}) \in \mathcal{G_A}$. Moreover, since $\cal G_A$ is a monotone class, we have $\bigcup\limits^{\infty}_{i=1}(A \cup E_{i}) = A \cup (\bigcup\limits^{\infty}_{i=1}E_{i}) \in \cal G_A$. By the way that $\cal E$ was constructed, we further have $\bigcup\limits^{\infty}_{i=1} E_{i} \in \cal E$.

Suppose $E_{1} \supset E_{2} \supset \cdots, E_{i} \in \cal E$, we must have $(A \cup E_{1}) \supset (A \cup E_{2}) \supset \cdots$, then $\bigcap\limits^{\infty}_{i=1}(A \cup E_{i}) = A \cup (\bigcap\limits^{\infty}_{i=1}E_{i}) \in \cal G_A$. By the way that $\cal E$ was constructed, we have: $\bigcap\limits^{\infty}_{i=1}E_{i} \in \cal E$.

Thus, $\cal E$ is a monotone class, with $\cal A \subset E$. Similar as the construction as in 1., since $\cal G_A$ is the smallest monotone class that contains $\cal A$, then: $\cal G_A \subset E$.

This means that, for $\forall B \in \mathcal{G_A}: B \in \cal E$. By the way that $\cal E$ was constructed, $(A \cup B) \in \cal G_A$.

Notice that, the first we have is $A \in \cal A$, and then construct $\mathcal{E}$ accordingly based on the fixed $A$.

Therefore, for $\forall A \in \mathcal{A}: (\exists \mathcal{E}): \forall B \in \mathcal{G_A}: A \cup B \in \cal G_A$. If we have omit the construction of $\cal E$, then:

\[\forall A \in \mathcal{A}: \forall B \in \mathcal{G_A}: A \cup B \in \cal G_A \]

Let $\mathcal{W}:= \left\{ E: (A \cup E) \in \cal G_A \right\}$ for $\forall A \in \cal G_A$.

Proof: $ A \subset \mathcal{W}$.
For $\forall B_{1} \in \cal A$, from the final conclusion of 2., we have: $\forall B_{2} \in \mathcal{G_A}: (B_{1} \cup B_{2})\in \cal G_A$. Besides, by the way that $\cal W$ was constructed, we have: $B_{1} \in \cal W$, hence $\forall B_{1} \in \mathcal{A}: B_{1} \in \mathcal{W} \implies \cal A \subset W$.
Proof: $\cal W$ is a monotone class.
Suppose $E_{1} \subset E_{2} \subset \cdots, E_{i} \in \cal W$, by the way that $\cal W$ was constructed, we have: $(A \cup E_{1}) \subset (A \cup E_{2}) \subset \cdots$, where $(A \cup E_{i}) \in \cal G_A$. Since $\cal G_A$ is a monotone class, then $\bigcup\limits^{\infty}_{i=1}(A \cup E_{i}) = A \cup (\bigcup\limits^{\infty}_{i=1}E_{i}) \in \cal G_A$. Furthermore, by the way that $\cal W$ was constructed, we have: $\bigcup\limits^{\infty}_{i=1}E_{i} \in \cal W$.

Suppose $E_{1} \supset E_{2} \supset \cdots$, where $E_{i} \in \cal W$, we must have $(A \cup E_{1}) \supset (A \cup E_{2}) \supset \cdots$, where $(A \cup E_{i}) \in \cal G_A$. Thus, $\bigcap\limits^{\infty}_{i=1} (A \cup E_{i}) = A \cup (\bigcap\limits^{\infty}_{i=1} E_{i}) \in \cal G_A$. By the way that $\cal W$ was constructed, we have: $\bigcap\limits^{\infty}_{i=1} E_{i} \in \cal W$.

Therefore, $\cal W$ is a monotone class, with $\cal A \subset W$.

Since $\cal G_A$ is the smallest monotone class that contains $\cal A$, then $\cal G_A \subset W$. Thus for $\forall B \in \mathcal{G_A}: B \in \cal W$, and by the way that $\cal W$ was constructed, we have $(A \cup B) \in \cal G_A$.

Notice that, similar as 2., here we firstly fix $A \in \cal G_A$, and then construct $\cal W$ based on the fixed $A$ accordingly.

Therefore, $\forall A \in \mathcal{G_A}: \forall B \in \mathcal{G_A}: (A \cup B) \in \cal G_A$.

According to what have proved above, $\cal G_A$ is closed under complement, therefore, we complete the proof that $\cal G_A$ is a field.

Lastly, we further prove that $\cal G_A$ is a $\sigma-$field, i.e., prove $\cal G_A$ is closed under countable unions.

Suppose $E_{1}, E_{2}, \ldots, E_{i} \in \cal G_A$. Since $\cal G_A$ is a monotone class, we have:

\[\begin{align*} \bigcup\limits^{\infty}_{i=1} E_{i} & = E_{1} \cup E_{2} \cup \cdots\\ & = E_{1} \cup (E_{1} \cup E_{2}) \cup (E_{1} \cup E_{2} \cup E_{3}) \cup \ldots\\ & = \bigcup\limits^{\infty}_{i=1}\left( (\bigcup\limits^{i-1}_{k=1} E_{k}) \cup E_{i} \right) \in \cal G_A\\ \end{align*} \]

Therefore, for $E_{1}, E_{2}, \ldots, E_{i} \in \mathcal{G_A}: \bigcup\limits^{\infty}_{i=1} E_{i} \in \cal G_A$. This shows that $\cal G_A$ is closed under countable unions, therefore, $\cal G_A$ is a $\sigma-$field.

$\cal F_A$ is the $\sigma-$field generated by $\cal A$ (i.e., the minimal $\sigma-$field containing $\cal A$). Hence $\cal F_A \subset G_A$.

Recall that, $\cal G_A \subset F_A$ is already proved at the beginning. Therefore, we complete the proof that $\cal G_A = F_A$.

posted @ 2022-05-23 10:39 车天健阅读(307) 评论(0) 编辑收藏举报

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