13.2 Lambda Systems
Definition 13.2.1 ($\pi$-System). Let $X$ be a set and $\mathcal{P}\subset 2^{X}$, then $\mathcal{P}$ is a $\pi$-system if:
$\emptyset \in \mathcal{P}$.
For any $A, B \in \mathcal{P}$, $A \cap B \in \mathcal{P}$.
Definition 13.2.2 ($\lambda$-System). Let $X$ be a set and $\alg \subset 2^{X}$, then $\alg$ is a $\lambda$-system/$d$-system if:
$\emptyset, X \in \alg$.
For any $E, F \in \alg$ with $E \subset F$, $F \setminus E \in \alg$.
For any $\seq{A_n}\subset \alg$ with $A_{n} \subset A_{n+1}$ for all $n \in \nat^{+}$, $\bigcup_{n \in \nat^+}A_{n} \in \alg$.
Definition 13.2.3 (Generated $\lambda$-System). Let $X$ be a set and $\ce \subset 2^{X}$, then the smallest $\lambda$-system $\lambda(\ce)$ over $X$ containing $\ce$ is the $\lambda$-system generated by $\ce$.
Lemma 13.2.4. Let $X$ be a set and $\alg \subset 2^{X}$, then the following are equivalent:
$\alg$ is a $\sigma$-algebra.
$\alg$ is a $\pi$-system and a $\lambda$-system.
Proof. $(2) \Rightarrow (1)$: Let $A, B \in \alg$, then $A \cup B = (A^{c} \cap B^{c})^{c} \in \alg$. Thus $\alg$ is an algebra. By Lemma 13.1.5, $\alg$ is a $\sigma$-algebra.$\square$
Theorem 13.2.5 (Dynkin’s $\pi$-$\lambda$ Theorem). Let $X$ be a set and $\mathcal{P}\subset 2^{X}$ be a $\pi$-system, then $\sigma(\mathcal{P}) = \lambda(\mathcal{P})$.
Proof. Let $\ce \subset \lambda(\mathcal{P})$ and
then
Let $E, F \in \cm$ with $E \subset F$, then for any $G \in \ce$,
\[(F \setminus E) \cap G = (F \cap G) \setminus (E \cap G) \in \lambda(\mathcal{P})\]Let $\seq{E_n}\in \cm$ with $E_{n} \subset E_{n+1}$ for all $n \in \natp$, and $F \in \ce$, then
\[\braks{\bigcup_{n \in \natp}E_n}\cap F = \bigcup_{n \in \natp}E_{n} \cap F \in \lambda(\mathcal{P})\]
so $\cm(\ce)$ is a $\lambda$-system.
Since $\mathcal{P}$ is a $\pi$-system, $\cm(\mathcal{P}) \supset \mathcal{P}$, so $\cm(\mathcal{P}) = \lambda(\mathcal{P})$. Thus for any $E \in \lambda(\mathcal{P})$ and $F \in \lambda(\mathcal{P})$, $E \cap F \in \lambda(\mathcal{P})$. Therefore $\cm(\lambda(\mathcal{P})) \supset \mathcal{P}$, $\cm(\lambda(\mathcal{P})) = \lambda(\mathcal{P})$, and $\lambda(\mathcal{P})$ satisfies (P2). By Lemma 13.2.4, $\lambda(\mathcal{P})$ is a $\sigma$-algebra.$\square$