5. Monotone classes

(待修改! )

Definition 5.0.1. Fix a set $\Omega$, we say $\mathcal{M}\subset \Im (\Omega )$ is a monotone class if

•	${\left\{A_j\in \mathcal{M}\right\}}_{j⩾1}⟶A$ an increasing converges sequence $⟹$ $A\in \mathcal{M}$
•	${\left\{A_j\in \mathcal{M}\right\}}_{j⩾1}⟶A$ an decreasing converges sequence $⟹$ $A\in \mathcal{M}$

Observation $\mathcal{F}$ is a $\sigma$-field (i.e. a $\sigma$-algebra) $⟹$ $\mathcal{F}$ is a monotone class.

Observation $({\mathcal{M}}_{\alpha }{)}_{\alpha \in I}\subset \Im (\Omega )$ $⟹$ $\mathcal{M}={\bigcap }_{\alpha \in I}{\mathcal{M}}_{\alpha }$ is a monotone class.

Lemma 5.0.2. If $\mathcal{A}$ is an algebra over a set $\Omega$, then $$\mathcal{F}(\mathcal{A})=\mathcal{M}(\mathcal{A})$$

Proof.   $\mathcal{F}(\mathcal{A})$ is a monotone class and $\mathcal{A}\subset \mathcal{F}(\mathcal{A})$. By the definition of $\mathcal{M}(\mathcal{A})$, $$\mathcal{M}(\mathcal{A})\subset \mathcal{F}(\mathcal{A})$$The difficult part lies in its inverse $\mathcal{F}(\mathcal{A})\subset \mathcal{M}(\mathcal{A})$.

It sufficient to show that $\mathcal{M}(\mathcal{A})$ is a $\sigma$-algebra. We prove that $\mathcal{M}(\mathcal{A})$ is an algebra first.

Take $E\in \mathcal{M}(\mathcal{A})$, we define $$\mathcal{G}(E):=\left\{F\in \mathcal{M}(\mathcal{A})|E\backslash F,E\bigcap F,F\backslash E\in \mathcal{M}(\mathcal{A})\right\}$$

Claim 1

$E\in \mathcal{A}$ $⟹$ $\mathcal{M}(\mathcal{A})\subset \mathcal{G}(E)$.

•	$\mathcal{A}\subset \mathcal{G}(E)$
•	$\mathcal{G}(E)$ is a monotone class

1. Take $H\in \mathcal{A}$, then $E\backslash H$, $E\bigcap H$, $H\backslash E$ are all belongs to $\mathcal{A}$ and hence contain in $\mathcal{M}(\mathcal{A})$. So $\mathcal{A}\subset \mathcal{G}(E)$.
2. Assume ${\left\{H_k\in \mathcal{G}(E)\right\}}_{k⩾1}⟶H$ is an increasing converges sequence, then $E\backslash H_k\in \mathcal{M}(\mathcal{A})$. Since ${\left\{H_k\in \mathcal{G}(E)\right\}}_{k⩾1}⟶H$, it means that $\{E\backslash H_k\in$$\mathcal{M}$${(\mathcal{A})\}}_{k⩾1}$$⟶E\backslash H$ is a decreasing converges sequence. Then $E\backslash H\in \mathcal{M}(\mathcal{A})$. Similarly, $E\bigcap H_k\in \mathcal{M}(\mathcal{A})$, $\left\{E\bigcap H_k\in \mathcal{M}(\mathcal{A})\right\}⟶E\bigcap H$ is an increasing converges sequence. So $E\bigcap H\in \mathcal{M}(\mathcal{A})$. By the same argument, it is easy to see that $H\backslash F\in \mathcal{M}(\mathcal{A})$, hence $H\in \mathcal{G}(E)$. Then $\mathcal{G}(E)$ is closed by the increasing converges sequence, similarly argument shows that $\mathcal{G}(E)$ is closed by the decreasing converges sequence. Since $\mathcal{A}\subset \mathcal{G}(E)$ and $\mathcal{G}(E)$ is a monotone class, it implies that $\mathcal{M}(\mathcal{A})\subset \mathcal{G}(E)$.

Claim 2

$E\in \mathcal{M}(\mathcal{A})$ $⟹$ $\mathcal{M}(\mathcal{A})\subset \mathcal{G}(E)$.

•	$\mathcal{G}(E)$ is a monotone class
•	$\mathcal{A}\subset \mathcal{G}(E)$

1. Let $E\in \mathcal{M}(\mathcal{A})$ and ${\left\{H_k\in \mathcal{G}(E\right\}}_{k⩾1}⟶H$ is an increasing sequence, then $E\backslash H_k\in \mathcal{M}(\mathcal{A})$ and ${\left\{E\backslash H_k\in \mathcal{M}(\mathcal{A})\right\}}_{k⩾1}⟶E\backslash H$ is a decreasing converges sequence. So $E\backslash H\in \mathcal{M}(\mathcal{A})$. Similarly, $E\bigcap H$ and $H\backslash E$ are both contain in $\mathcal{M}(\mathcal{A})$. Therefore, $H\in \mathcal{G}(E)$. The same argument shows that $\mathcal{G}(E)$ is closed by the decreasing converges sequence.
2. Let $H\in \mathcal{A}$, then $H\in \mathcal{M}(\mathcal{A})$. So it sufficient to show that $E\backslash H$, $E\bigcap H$, $H\backslash E$ are all contain in $\mathcal{M}(\mathcal{A})$. However, $\mathcal{M}(\mathcal{A})\subset \mathcal{G}(H)$ since $H\in \mathcal{A}$. But $E\in$$\mathcal{M}(\mathcal{A})$, it belongs to $\mathcal{G}(H)$. So, $E\backslash H$, $E\bigcap H$, $H\backslash E$ are all contain in $\mathcal{M}(\mathcal{A})$ indeed.

Then we obtain that $\mathcal{M}(\mathcal{A})\subset \mathcal{G}(E)$.

•	$\Omega \in \mathcal{M}(\mathcal{A})$ It is clear that $\Omega \in \mathcal{A}\subset \mathcal{M}(\mathcal{A})$.
•	$E\in \mathcal{M}(\mathcal{A})$ $⟹$ $E^c\in \mathcal{M}(\mathcal{A})$ $E\in \mathcal{M}(\mathcal{A})$ $⟹$ $E\in \mathcal{G}(\Omega )$ (note that $\Omega \in \mathcal{A}$ $⟹$ $\mathcal{M}(\mathcal{A})\subset \mathcal{G}(\Omega )$), but $$\mathcal{G}(\Omega )=\left\{F\in \mathcal{M}(\mathcal{A})|F^c,F,\varnothing \in \mathcal{M}(\mathcal{A})\right\}$$So $E^c\in \mathcal{M}(\mathcal{A})$.
•	$E$, $F\in \mathcal{M}(\mathcal{A})$ $⟹$ $E\bigcap F\in \mathcal{M}(\mathcal{A})$ Since $E\in \mathcal{M}(\mathcal{A})$, $\mathcal{M}(\mathcal{A})\subset \mathcal{G}(E)$. Then $F\in \mathcal{G}(E)$, we have $E\bigcap F\in \mathcal{M}(\mathcal{A})$. So $\mathcal{M}(\mathcal{A})$ is an algebra. Now we claim that $\mathcal{M}(\mathcal{A})$ is closed by countable union.
•	$(A_j{)}_{j⩾1}\in \mathcal{M}(\mathcal{A})$ $⟹$ ${\bigcup }_{j⩾1}{A}_j\in \mathcal{M}(\mathcal{A})$

Since $\mathcal{M}(\mathcal{A})$ is an algebra, ${\bigcup }_{j=1}^n{A}_j\in \mathcal{M}(\mathcal{A})$. But ${\left\{{\bigcup }_{j=1}^n{A}_j\right\}}_{n\to +\infty }⟶{\bigcup }_{j⩾1}{A}_j$ is an increasing converges sequence. Therefore, ${\bigcup }_{j⩾1}{A}_j\in \mathcal{M}(\mathcal{A})$. Then we complete the proof that $\mathcal{M}(\mathcal{A})$ is a $\sigma$-algebra.

We now know that $\mathcal{A}\subset \mathcal{M}(\mathcal{A})$ and $\mathcal{M}(\mathcal{A})$ is a $\sigma$-algebra, then we obtain that $$\mathcal{F}(\mathcal{A})\subset \mathcal{M}(\mathcal{A})$$So $$\mathcal{M}(\mathcal{A})=\mathcal{F}(\mathcal{A})$$

$\square$