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%  Lecture 3
%  Source: handwritten notes (Mathpix-converted) + Kashlak STAT 571
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\section[Lecture 3 -- Uniqueness; Dynkin's \texorpdfstring{$\pi$--$\lambda$}{pi-lambda} Theorem; Completeness]{Lecture 3 \textemdash{} Uniqueness; Dynkin's \texorpdfstring{$\pi$--$\lambda$}{pi-lambda} Theorem; Completeness}
\label{sec:lec03}

The Carath\'eodory theorem of the previous lecture extends a pre-measure on a
ring $\Acal$ to a measure on $\sigma(\Acal)$. The natural follow-up question
is whether such an extension is \emph{unique}: if $\mu_{1}$ and $\mu_{2}$ are
two measures on $\sigma(\Acal)$ that coincide on $\Acal$, must they coincide
on all of $\sigma(\Acal)$? The Dynkin $\pi$-$\lambda$ theorem is the standard
machinery for answering this question.

\subsection{\texorpdfstring{$\pi$}{pi}-systems and \texorpdfstring{$\lambda$}{lambda}-systems}

\begin{definition}{\texorpdfstring{$\pi$}{pi}-system}{pisystem}
A collection $\Acal$ of subsets of $\Omega$ is a \emph{$\pi$-system} if
$\varnothing \in \Acal$ and, for any $A, B \in \Acal$, we have
$A \cap B \in \Acal$. That is, $\Acal$ is closed under finite intersections.
\end{definition}

\begin{definition}{\texorpdfstring{$\lambda$}{lambda}-system}{lambdasystem}
A collection $\Lcal$ of subsets of $\Omega$ is a \emph{$\lambda$-system} if
\begin{itemize}
  \item $\Omega \in \Lcal$;
  \item for $A, B \in \Lcal$ with $A \subset B$, we have $B \setminus A \in \Lcal$;
  \item for $\{A_{i}\}_{i=1}^{\infty} \subset \Lcal$ pairwise disjoint, we have
        $\bigcup_{i=1}^{\infty} A_{i} \in \Lcal$.
\end{itemize}
\end{definition}

\begin{remark}
A $\lambda$-system is very close to a $\sigma$-field; the difference is that
closure under countable unions is required only for \emph{disjoint} collections.
Note that any field is automatically a $\pi$-system, and a collection that is
\emph{both} a $\pi$-system and a $\lambda$-system is a $\sigma$-field.
\end{remark}

\subsection{Dynkin's \texorpdfstring{$\pi$--$\lambda$}{pi-lambda} theorem}

The next theorem is the workhorse: it lets us upgrade containment in a
$\lambda$-system to containment of the full generated $\sigma$-field, provided
we start from a $\pi$-system.

\begin{theorem}{Dynkin's \texorpdfstring{$\pi$--$\lambda$}{pi-lambda} Theorem}{dynkin}
Let $\Acal$ be a $\pi$-system and $\Lcal$ be a $\lambda$-system on $\Omega$
with $\Acal \subset \Lcal$. Then $\sigma(\Acal) \subset \Lcal$.
\end{theorem}

\subsection{Uniqueness of extension}

\begin{theorem}{Uniqueness of Extension}{uniqueness}
Let $\mu_{1}, \mu_{2}$ be $\sigma$-finite measures on $\sigma(\Acal)$, where
$\Acal$ is a $\pi$-system. If $\mu_{1}(A) = \mu_{2}(A)$ for every
$A \in \Acal$, then $\mu_{1}(B) = \mu_{2}(B)$ for every $B \in \sigma(\Acal)$.
\end{theorem}

\begin{remark}
For finite measures the proof is short: one checks that
$\Lcal := \{B \subset \Omega : \mu_{1}(B) = \mu_{2}(B)\}$ is a
$\lambda$-system containing $\Acal$, and then applies the
$\pi$-$\lambda$ theorem. The $\sigma$-finite case is handled by exhausting
$\Omega$ along a sequence $\Omega = \bigcup_{i} A_{i}$ with
$A_{i} \in \Acal$ and $\mu_{1}(A_{i}) = \mu_{2}(A_{i}) < \infty$ and using
inclusion--exclusion on the $\pi$-system.
\end{remark}

\begin{remark}[Probability spaces and \texorpdfstring{$\pi$}{pi}-systems]
$\pi$-systems arise naturally in probability: if we know two events, we can
also consider their intersection (the joint event). Rolling two fair dice,
for instance, the events $\{D_{1}+D_{2}=8\}$ and $\{D_{1} \equiv 0 \bmod 2\}$
both belong to a $\pi$-system, as does their intersection.
\end{remark}

\begin{example}[$\sigma$-finiteness is necessary]
Without $\sigma$-finiteness, uniqueness can fail. Take $\Omega = (0,1]$ and
$\Acal$ the collection of finite unions of half-open intervals $(a,b]$.
\begin{itemize}
  \item Let $\mu$ assign $0$ to $\varnothing$ and $\infty$ to every
        non-empty element of $\Acal$. Its outer measure $\mu^{*}$ assigns
        $\infty$ to every non-empty subset of $\Omega$.
  \item The counting measure also assigns $0$ to $\varnothing$ and
        $\infty$ to each non-empty $(a,b]$, but it assigns the finite
        value $3$ to $\{\tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}\}$.
\end{itemize}
The two measures agree on $\Acal$ but disagree on $\sigma(\Acal)$.
\end{example}

\subsection{Completeness}

We would like a measure space in which any set differing from a measurable
set by a null set is itself measurable, with the same measure. The original
$\sigma$-field need not have this property, but it can be enlarged so it does.

\begin{definition}{Symmetric Difference}{symdiff}
For two sets $A, B \subset \Omega$, the \emph{symmetric difference} is
\[
  A \triangle B := (A \setminus B) \cup (B \setminus A).
\]
\end{definition}

For a measure space $(X, \Fcal, \mu)$ with $\Fcal \subset \Pcal(X)$, recall the
outer measure
\[
  \mu^{*}(B) := \inf\{ \mu(A) : B \subset A,\ A \in \Fcal \}, \qquad B \in \Pcal(X).
\]

\begin{definition}{Null Sets and Completeness}{complete}
The collection of \emph{$\mu$-null sets} is
\[
  \Ncal_{\mu} := \{ N \subset X : \mu^{*}(N) = 0 \} \subset \Pcal(X).
\]
The measure space $(X, \Fcal, \mu)$ is \emph{complete} if
$\Ncal_{\mu} \subset \Fcal$.
\end{definition}

\begin{remark}[Exercise]
Show that $\Ncal_{\mu}$ is a ring.
\end{remark}

When the space is not complete, we may complete it by enlarging $\Fcal$ to
include all null sets and their unions with measurable sets.

\begin{proposition}{Completion of a Measure Space (Dudley 3.3.2)}{completion}
Let $(X, \Fcal, \mu)$ be a measure space with null sets $\Ncal_{\mu}$. Define
\[
  \Fcal \vee \Ncal_{\mu} := \{ A \cup N : A \in \Fcal,\ N \in \Ncal_{\mu} \}.
\]
Then
\[
  \Fcal \vee \Ncal_{\mu} = \{ B \subset X : \exists\, A \in \Fcal \text{ with } A \triangle B \in \Ncal_{\mu} \},
\]
and this is the smallest $\sigma$-field containing both $\Fcal$ and $\Ncal_{\mu}$.
The triple $(X, \Fcal \vee \Ncal_{\mu}, \bar{\mu})$ with
$\bar{\mu}(A \cup N) := \mu(A)$ is a complete measure space, called the
\emph{completion} of $(X, \Fcal, \mu)$.
\end{proposition}