In our real analysis course, we are asked to prove Dynkin's $\pi$-$\lambda$ theorem. After searching on the Internet for a while, the proofs found are quite...verbose to me. Therefore, I write a "proof" on my own and want to ask if there is any flaw in it:
Notation.
- $X$: the underlying set.
- $\lambda(\mathcal C), \sigma(\mathcal C)$: the $\lambda$-system and $\sigma$-algebra generated by a collection $\mathcal C\subseteq \mathcal 2^X$, respectively.
- $\Pi, \Lambda$: the $\pi$-system and the $\lambda$-system given, respectively.
Proof. Since $\lambda(\Pi)\subseteq \Lambda$, if we can show that $\lambda(\Pi)$ is a $\pi$-system, then $\lambda(\Pi)$ is a $\sigma$-algebra and $\sigma(\Pi)\subseteq \lambda(\Pi)\subseteq \Lambda$ and we are done. Therefore, it remains to prove that $\lambda(\Pi)$ is closed under finite intersection as $\lambda(\Pi)$ is already nonempty by definition.
Set $\mathcal L_1=\{A\subseteq X: \forall B\in \Pi, A\cap B\in \lambda(\Pi)\}$. It is clear that $\Pi\subseteq \mathcal L_1$. Let $(A_j)_{j\geq 1}\subseteq \mathcal L_1$ be pairwise disjoint. Then for any $B\in \Pi$, $$\bigcup_{j\geq 1}\underbrace{A_j\cap B}_{\in \lambda(\Pi)}\in \lambda(\Pi)$$ as $\lambda(\Pi)$ is closed under countable pairwise disjoint union. Thus, $\bigcup_{j\geq 1}A_j\in \mathcal L_1$ and $\lambda(\Pi)\subseteq \mathcal L_1$ as $\mathcal L_1$ is a $\lambda$-system.
Set $\mathcal L_2=\{A\subseteq X: \forall B\in \lambda(\Pi), A\cap B\in \lambda(\Pi)\}$. Since $\lambda(\Pi)\subseteq \mathcal L_1$, $\Pi\subseteq \mathcal L_2$. Then a similar argument shows that $\mathcal L_2$ is a $\lambda$-system and $\lambda(\Pi)\subseteq \mathcal L_2$.
In conclusion, $\lambda(\Pi)$ is a $\pi$-system and we are done. $\square$
