We fix $\Omega$ and let $G$ be an algebra on $\Omega$. I'm trying to prove below result. Could you please have a check on my attempt?
Theorem: Let $M(G)$ be the smallest monotone class containing $G$ and $\lambda(G)$ the smallest $\lambda$-system (on $\Omega$) that contains $G$. Then $M(G) = \lambda(G)$.
My attempt:
- $\lambda (G)$ is a monotone class.
Let $(A_n) \subset \lambda (G)$ such that $A_n \subset A_{n+1}$. Let $B_n := A_{n+1} \setminus A_n$. Because $\lambda (G)$ is a $\lambda$-system, we get $B_n \in \lambda (G)$. Clearly, $(B_n)$ is pairwise disjoint. Then $\bigcup B_n \in \lambda (G)$. This implies $\bigcup A_n = \bigcup B_n \in \lambda (G)$. By symmetry, if $(A_n) \subset \lambda (G)$ such that $A_n \supset A_{n+1}$ then $\bigcap A_n \in \lambda (G)$.
- $M(G)$ is a $\lambda$-system.
Because $G$ is an algebra on $\Omega$, we get $\Omega \in G$ and thus $\Omega \in M(G)$.
Let $\mathcal A := \{A \in M(G) \mid A^c \in M(G)\}$. Because $G$ is an algebra, we get that $G \subset \mathcal A$. Also, we can verify easily that $\mathcal A$ is a monotone class. Hence $\mathcal A = M(G)$.
Now we fix $B \in G$ and let $\mathcal A_B := \{A \in M(G) \mid A \cap B \in M(G)\}$. It's easy to see that $\mathcal A_B$ is a monotone class containing $G$. Hence $\mathcal A_B = M(G)$
Now we fix $B \in M(G)$ and let $\mathcal A_B := \{A \in M(G) \mid A \cap B \in M(G)\}$. If $A \in G$, then $\mathcal A_A = M(G)$ and thus $A \cap B \in M(G)$ and thus $A \in \mathcal A_B$. This means $\mathcal A_B$ contains $G$. It's easy to verify that $\mathcal A_B$ is a monotone class. Hence $\mathcal A_B = M(G)$.
This completes the proof.
