Monotone Class

Theorem

If $M$ is a monotone class and $R \subset M$ is a ring, then $M$ contains the $\sigma$-ring $\sigma(R)$ generated by $R$ and use the following lemma.

Proof:

Consider the monotone class $M_0$ generated by $R$. By definition $M_0 \subset M$. The goal is show that $M_0$ is a ring that contains $R$.

Lemma

If $M_0$ is a ring, then it must be a $\sigma$-ring.

Pf: For each countable collection of sets $( E_n )$, $\cup_{n} E_n = \cup_{n} (\cup_{k = 1}^n E_k)$. Denote $F_n = \cup_{k = 1}^n E_n$, and $F = \cup_{n} E_n$. Clearly $F_n \uparrow F$, which implies that $F \subset M_0$.

$\blacksquare$

It immediately follows that $M_0 \supset \sigma(R)$. Indeed, as $\sigma(R)$ itself is a monotone class and $M_0$ is the smallest monotone class that contains $R$, we have $\sigma(R) \supset M_0$. It concludes that $M_0 = \sigma(R)$.

To show $M_0$ is a ring, we need to verify that $E, F \in M_0$, $E \cup F = M_0$ and $E \setminus F \in M_0$.

Define $S_E = \{ F \in M_0 : E \cup F \in M_0, E \setminus F \in M_0 \}$, the collection of sets that satisfies the conditions. Note that $S_E$ is a monotone class.

1. $(F_n) \subset S_E$, and $(F_n) \uparrow F$, then $E \cup F = E \cup [\cup_n F_n] = \cup_n [E \cup F_n]$. By definition of $S_E$, $E \cup F_n \in M_0$. As $F_n \uparrow$ and $M_0$ is monotone, $E \cup F \in M_0$. Hence $F \in M_0$.

2. $(F_n) \subset S_E$, and $(F_n) \uparrow F$, then $E \setminus F = E \setminus [\cup_n F_n] = \cap_n [E \cap F_n] \in M_0$.

We claim that for each $E \in M_0$, it holds that $F \in S_E$ for any $F \in M_0$, i.e., $M_0 \subset S_E$. But this is equivalent to show $E \in S_F$. Since $M_0$ is the smallest monotone class that contains $R$, it suffices to prove $R \subset S_F$. We need to utilize the following observation.

Let $O \subset R$, then $S_O \supset R$, which implies that $S_O \supset M_0$, as $S_O$ is a monotone class.

Therefore, $\forall F \in M_0$, $R \subset S_F$.

$\blacksquare$.