> [!tldr] > If $\mathcal{A} \subseteq \mathcal{M}$, and $\mathcal{A}$ is a $\pi$-system and $\mathcal{M}$ a $\lambda$-system, then $\sigma(\mathcal{A}) \subseteq \mathcal{M}$. ## Applications The main strategy to prove that "$P_{A}$ is true for all $A \in \mathcal{F}quot; (where $P_{A}$ is some statement about $A$) is to show that (1) the set $\mathcal{T}_{P}$ on which $P_{A}$ is true is a $\lambda$-system (or even a $\sigma$-algebra), and then (2) a generating $\pi$-system of $\mathcal{F}$ is contained in $\mathcal{T}_{A}$. Then $\mathcal{T}_{A}=\mathcal{F}$ by double inclusion. > [!proof] Suppose $(\Omega_{1},\mathcal{F}_{1}),(\Omega_{2}, \mathcal{F}_{2})$, $\mathcal{F}=\mathcal{F}_{1} \times \mathcal{F}_{2}$, and for any $D \in \mathcal{F}$ define function $D(\cdot): \omega_{1} \mapsto \{ \omega_{2}:(\omega_{1},\omega_{2}) \in D \}$. Prove that $D(\omega_{1}) \in \mathcal{F}_{2}$ for any $D, \omega_{1}$. > > Consider the family of sets $D$ for which $D(\omega_{1}) \in \mathcal{F}_{2}$ is true for all $\omega_{1}$; call it $\mathcal{D}$. We show that it is a $\lambda$-system. > - Obviously $\Omega=\Omega_{1}\times \Omega_{2}$ is in the set. > - If $A,B \in \mathcal{D}$, and $A \subseteq B$, then $\begin{align*} (B-A)(\omega_{1})&= \{ \omega_{2}:(\omega_{1},\omega_{2}) \in (B-A) \}\\ &= \{ \omega_{2}:(\omega_{1},\omega_{2}) \in B \}-\{ \omega_{2}:(\omega_{1},\omega_{2}) \in A\}\\ &= B(\omega_{1})-A(\omega_{1}) \in \mathcal{F}_{2}. \end{align*}$ > - If $(A_{n}) \subseteq \mathcal{D}$ is a rising collection of sets, then $\begin{align*} (\cup_{n}A_{n})(\omega_{1})&= \{ \omega_{2}:(\omega_{1},\omega_{2}) \in \cup_{n}A_{n} \}\\ &= \cup_{n}\{ \omega_{2}:(\omega_{1},\omega_{2}) \in A_{n} \}\\ &= \cup_{n}A_{n}(\omega_{1}) \in \mathcal{F}_{2} \end{align*}$ > > Now we only need to find a generating $\pi$-system is contained in it: namely the rectangles $\{ D:= A_{1}\times A_{2} ~|~ A_{i} \in \mathcal{F}_{i},~ i=1,2 \}$. Clearly $D(\omega_{1})=A_{2}$ if $\omega_{1} \in A_{1}$, and $\emptyset$ if $w_{1}\ne A_{1}$, so indeed this is true.