
\no\S
If $\SS$ is a $\lambda$-system and $A \subset \Omega$ then
\[
\SS_A := \{B \in 2^\Omega:~ BA \in \SS\}
\]
is also a $\lambda$-system.
Indeed, if $B_1 \subset B_2$ both belong to $\SS_A$ then
$B_1A, B_2A$ belong to $\SS$ and $(B_2-B_1)A =B_2A - B_1A \in \SS$,
because $\SS$ is closed under proper differences; 
so $B_2-B_1 \in \SS_A$.
Also, if $B_n \in \SS_A$, $B_n \subset B_{n+1}$ for all $n$, 
then $B_nA \in \SS$ for all $n$, and so 
$A\cup_n B_n = \cup_n AB_n \in \SS$ because $\SS$ 
is closed under increasing limits; so $\sup_n B_n \in \SS_A$.

\no\S
$\pi$ \& $\lambda$ \& $\Omega \equiv \sigma$. In other words,
if $\SS$ is both $\pi$ and $\lambda$ and contains $\Omega$
then it is a $\sigma$-field.
Indeed, if $A \in \SS$ then $A^c = \Omega-A \in \SS$ because $\SS$ is
$\lambda$. So $\SS$ is closed under complementations, intersections
and limits. So it is a $\sigma$-field.


\paragraph{Lemma [Sierpi\'nski-Dynkin]}
Let $\SS$ be a $\pi$-system containing $\Omega$.
Then $\lambda(\SS) = \sigma(\SS)$.

\no\proof
Since $\lambda(\SS)$ is the smallest $\lambda$-system containing $\SS$
and since $\sigma(\SS)$ is a $\lambda$-system, we have $\lambda(\SS)
\subset \sigma(\SS)$.
We need to show that $\sigma(\SS) \subset \lambda(\SS)$. If we show
that $\lambda(\SS)$ is also a $\pi$-system then it will be
a $\sigma$-field and we'll be done.
We know that
\begin{equation}
\label{nada}
A \in \SS,~ B \in \SS \Rightarrow AB \in \SS \subset \lambda(\SS).
\end{equation}
Hence
\[
\forall A \in \SS \quad \SS \subset \{B \in 2^\Omega: AB \in \lambda(\SS)\}
=: \lambda(\SS)_A.
\]
But $ \lambda(\SS)_A$ is $\lambda$ and so
\[
\forall A \in \SS \quad  \lambda(\SS)  \subset  \lambda(\SS)_A.
\]
Let's read this again:
\begin{equation}
\label{nada2}
 A \in \SS, ~ B \in  \lambda(\SS) \Rightarrow AB \in \lambda(\SS).
\end{equation}
Wonder of wonders: \eqref{nada} implied the seemingly stronger \eqref{nada2}.
Let's rewrite the \eqref{nada2} as:
\[
\forall B \in \lambda(\SS) \quad \SS \subset
\{A \in 2^\Omega:~ AB \in \lambda(\SS)\}=: \lambda(\SS)_B.
\]
But $ \lambda(\SS)_B$ is $\lambda$ and so
\[
\forall B \in \lambda(\SS) \quad \lambda(\SS)  \subset  \lambda(\SS)_B.
\]
Let's read this again:
\begin{equation}
\label{nada3}
B \in \lambda(\SS), ~ A \in \lambda(\SS) \Rightarrow AB \in \lambda(\SS).
\end{equation}
Wonder of wonders of wonders: \eqref{nada2} implied the seemingly stronger
\eqref{nada3} which is what we need.
\qed

