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%  Lecture 3
%  Primary source: handwritten notes (Mathpix mmd), lec1-4 lines 217-341
%  Fallback: kashlak.pdf (only for OCR/notation/curriculum clarity)
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\section[Lecture 3 -- Uniqueness; Dynkin's \texorpdfstring{$\pi$--$\lambda$}{pi-lambda}; Completeness]{Lecture 3 \textemdash{} Uniqueness; Dynkin's \texorpdfstring{$\pi$--$\lambda$}{pi-lambda} Theorem; Completeness}
\label{sec:lec03}

Lecture~2 produced an extension of a pre-measure to the generated
\(\sigma\)-field via Carath\'eodory. The natural follow-up is the
question that opens the handwritten page:
\begin{quote}
If \(\mu_{1}(A)=\mu_{2}(A)\) for every \(A\in\Acal\), does it follow
that \(\mu_{1}(B)=\mu_{2}(B)\) for every \(B\in\sigma(\Acal)\)?
\end{quote}
The answer is \emph{yes} provided \(\Acal\) is a \(\pi\)-system and the
measures are \(\sigma\)-finite. The vehicle is Dynkin's
\(\pi\)\textendash\(\lambda\) theorem, which we develop next, after
which we take a brief detour through completeness.

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

\begin{definition}{$\pi$-system}{pi-system}
A collection \(\Acal\) of subsets of \(\Omega\) is a \emph{\(\pi\)-system}
if it is closed under finite intersections: for every
\(A,B\in\Acal\), \(A\cap B\in\Acal\).
\end{definition}

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

\begin{remark}
A \(\lambda\)-system looks much like a \(\sigma\)-field, the difference
being that countable unions are required only for \emph{disjoint}
sequences. Every 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 lecture's central tool upgrades containment in a \(\lambda\)-system
to containment of the full generated \(\sigma\)-field, provided one
starts from a \(\pi\)-system.

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

\begin{remark}
The proof strategy from the handwriting: take \(\Lcal_{0}\) to be the
\emph{smallest} \(\lambda\)-system containing \(\Acal\). Show
\(\Lcal_{0}\) is also closed under intersections, by introducing
\[
  \Lcal' \;=\; \{B\in\Lcal_{0} : B\cap A\in\Lcal_{0}\text{ for all } A\in\Acal\}
\]
and verifying \(\Lcal'\) is a \(\lambda\)-system containing \(\Acal\),
hence \(\Lcal'=\Lcal_{0}\); then repeat the argument with
\(\Lcal''=\{B\in\Lcal_{0} : B\cap C\in\Lcal_{0}\text{ for all } C\in\Lcal_{0}\}\).
A \(\lambda\)-system that is also a \(\pi\)-system is a
\(\sigma\)-field, so \(\sigma(\Acal)\subseteq\Lcal_{0}\subseteq\Lcal\).
\end{remark}

\subsection{Uniqueness of extension}

The promised payoff:

\begin{theorem}{Uniqueness of extension}{uniqueness}
Let \(\Acal\) be a \(\pi\)-system on \(\Omega\) and let \(\mu_{1},\mu_{2}\)
be two \(\sigma\)-finite measures on \(\sigma(\Acal)\). 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}
The handwriting splits the proof in two:
\begin{itemize}
  \item \textbf{Finite case.} Assume
        \(\mu_{1}(\Omega)=\mu_{2}(\Omega)<\infty\) and set
        \(\Lcal=\{B\subset\Omega : \mu_{1}(B)=\mu_{2}(B)\}\). One
        verifies \(\Lcal\) is a \(\lambda\)-system: \(\Omega\in\Lcal\)
        by hypothesis; if \(A\subset B\) lie in \(\Lcal\) then
        \(\mu_{i}(B\setminus A)=\mu_{i}(B)-\mu_{i}(A)\) (here
        finiteness allows the subtraction), so
        \(B\setminus A\in\Lcal\); countable additivity handles disjoint
        unions. Since \(\Acal\subset\Lcal\), Dynkin gives
        \(\sigma(\Acal)\subset\Lcal\).
  \item \textbf{\(\sigma\)-finite case.} For each \(A\in\Acal\) with
        \(\mu_{1}(A)=\mu_{2}(A)<\infty\), define
        \(\Lcal_{A}=\{B\subseteq\Omega : \mu_{1}(A\cap B)=\mu_{2}(A\cap B)\}\),
        a \(\lambda\)-system; by Dynkin
        \(\sigma(\Acal)\subset\Lcal_{A}\). Decompose
        \(\Omega=\bigcup_{i\ge1}A_{i}\) with \(A_{i}\in\Acal\) and
        \(\mu_{1}(A_{i})=\mu_{2}(A_{i})<\infty\). Inclusion\textendash exclusion
        on the finitely many \(A_{1},\dots,A_{n}\) (using that \(\Acal\)
        is a \(\pi\)-system, so \(A_{i}\cap A_{j}\in\Acal\) and so on)
        gives \(\mu_{1}(B\cap\bigcup_{i\le n}A_{i})=\mu_{2}(B\cap\bigcup_{i\le n}A_{i})\)
        for every \(B\in\sigma(\Acal)\); let \(n\to\infty\).
\end{itemize}
\end{remark}

\begin{remark}
\(\pi\)-systems are very natural in probability theory: the joint event
\(A\cap B\) lives alongside \(A\) and \(B\), so any sensible event
class is closed under finite intersections.
\end{remark}

\begin{example}[$\sigma$-finiteness is needed]
Take \(\Omega=(0,1]\) and let \(\Acal\) be the collection of finite
unions of half-open intervals \((a,b]\). Two measures on
\(\sigma(\Acal)\) agreeing on \(\Acal\) but disagreeing on
\(\sigma(\Acal)\):
\begin{itemize}
  \item \(\mu\) sends \(\emptyset\mapsto 0\) and every non-empty
        element of \(\Acal\) to \(\infty\); the induced outer measure
        \(\mu^{*}\) then assigns \(\infty\) to every non-empty subset
        of \(\Omega\).
  \item Counting measure \(\nu\) also sends \(\emptyset\mapsto 0\) and
        each non-empty \((a,b]\) to \(\infty\), but
        \(\nu(\{\tfrac14,\tfrac12,\tfrac34\})=3\) while
        \(\mu^{*}(\{\tfrac14,\tfrac12,\tfrac34\})=\infty\).
\end{itemize}
The pre-measure on \(\Acal\) is not \(\sigma\)-finite, and uniqueness
breaks at the level of \(\sigma(\Acal)\).
\end{example}

\subsection{Completeness}

We now leave uniqueness behind and address a different deficiency: a
measure space may contain sets of measure zero whose subsets are not
themselves measurable. We patch this by enlarging the
\(\sigma\)-field to absorb every such ``negligible'' set. The geometry
behind the patch is the symmetric difference.

\begin{definition}{Symmetric difference}{symmetric-difference}
For sets \(A,B\subseteq\Omega\),
\[
  A\,\triangle\,B \;=\; (A\setminus B)\,\cup\,(B\setminus A).
\]
\end{definition}

\begin{figure}[h]
\centering
\begin{tikzpicture}[scale=1]
  \begin{scope}
    \clip (-1.1,0) circle (1.2);
    \fill[highlightred!35] (-2.4,-1.4) rectangle (2.4,1.4);
  \end{scope}
  \begin{scope}
    \clip (1.1,0) circle (1.2);
    \fill[highlightblue!30] (-2.4,-1.4) rectangle (2.4,1.4);
  \end{scope}
  \begin{scope}
    \clip (-1.1,0) circle (1.2);
    \clip (1.1,0) circle (1.2);
    \fill[white] (-2.4,-1.4) rectangle (2.4,1.4);
  \end{scope}
  \draw[thick, highlightred] (-1.1,0) circle (1.2);
  \draw[thick, highlightblue] (1.1,0) circle (1.2);
  \node[highlightred] at (-2.0,-0.95) {\(A\)};
  \node[highlightblue] at (2.0,-0.95) {\(B\)};
\end{tikzpicture}
\caption{The shaded region is \(A\,\triangle\,B\); the unshaded
overlap is \(A\cap B\).}
\label{fig:symmetric-difference}
\end{figure}

For a measure space \((X,\Fcal,\mu)\) the \emph{outer measure} extends
\(\mu\) to all of \(\Pcal(X)\) by
\[
  \mu^{*}(B) \;=\; \inf\{\,\mu(A) : B\subset A,\ A\in\Fcal\,\}.
\]
A subset \(N\subseteq X\) is \emph{\(\mu\)-null} if \(\mu^{*}(N)=0\);
write \(\Ncal_{\mu}\) for the collection of all such sets.

\begin{definition}{Complete measure space}{completeness}
The measure space \((X,\Fcal,\mu)\) is \emph{complete} if every
\(\mu\)-null set already lies in \(\Fcal\), i.e.\ \(\Ncal_{\mu}\subset\Fcal\).
\end{definition}

\begin{remark}[Exercise from the lecture]
Show that \(\Ncal_{\mu}\) is a ring of sets.
\end{remark}

When \((X,\Fcal,\mu)\) fails to be complete, we can replace \(\Fcal\)
by a slightly larger \(\sigma\)-field that contains all null sets.

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

\begin{remark}
The two descriptions of \(\Fcal\vee\Ncal_{\mu}\) match the picture
above: \(B\) is in the completion iff it differs from some
\(A\in\Fcal\) only on a set of outer measure zero, i.e.\
\(A\,\triangle\,B\) is null. Lebesgue measure on \(\R\) is the
completion of its restriction to the Borel \(\sigma\)-field; this
strict enlargement is the gap \(\Bcal\subsetneq\Mcal_{\lambda}\) used
in the next lecture.
\end{remark}