Let $\Omega$ be a set. Let $a(C), \lambda(C), m(C)$ be the smallest algebra, the smallest $\lambda$-system, and the smallest monotone class that contain $C$ respectively.
Dynkin's $\pi-\lambda$ theorem: Let $C, D$ be a $\pi$-system and a $\lambda$-system on $\Omega$ respectively. If $C \subset D$ then $\sigma(C) \subset D$.
Monotone class theorem: Let $C$ be an algebra on $\Omega$. Then $m(C) = \sigma(C)$.
I'm trying to prove that the monotone class theorem is equivalent to Dynkin's $\pi-\lambda$ theorem. Could you please have a check on my attempt?
My attempt:
- Dynkin's $\pi-\lambda$ theorem $\implies$ Monotone class theorem:
Clearly, $\sigma(C)$ is a monotone class containing $C$, so $m(C) \subset \sigma(C)$. It remains to prove $\sigma(C) \subset m(C)$. Because $C$ is an algebra, it is a $\pi$-system. By Dynkin's $\pi-\lambda$ theorem, it suffices to prove $m(C)$ is a $\lambda$-system. It is indeed true by the following lemma 1, i.e.,
Lemma 1: Let $C$ be an algebra on $\Omega$. Then $m(C) = \lambda(C)$.
- Monotone class theorem $\implies$ Dynkin's $\pi-\lambda$ theorem:
We have $\sigma(C) = \sigma(a(C))$. By monotone class theorem, we get $\sigma(a(C)) = m(a(C))$. By lemma 1, $m(a(C)) = \lambda(a(C))$. It suffices to prove that $\lambda(a(C)) \subset \lambda(C)$. However, it is true by the following lemma 2, i.e.,
Lemma 2: The collection $\lambda(C)$ is a monotone class. If $C$ a $\pi$-system, then $a(C) \subset \lambda (C)$.
