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% 10/10/2023 10AM

To address the uniqueness of extension, we
introduce further subclasses of $\mathcal{P}(E)$.
Let $\mathcal{A}$ be a collection of subsets of
$E$.

\begin{flashcard}[pi-system-defn]
\begin{definition*}[$\pi$-system]
  \cloze{$\mathcal{A}$ is called a
  \emph{$\pi$-system} if $\emptyset \in
  \mathcal{A}$ and $\forall A, B \in \mathcal{A}$,
  $A \cap B \in \mathcal{A}$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[d-system-defn]
\begin{definition*}[$d$-system]
  \cloze{$\mathcal{A}$ is called a \emph{$d$-system}
  (for Dynkin, alternatively called a
  \emph{$\lambda$-system}) if
  \begin{itemize}
    \item $E \in \mathcal{A}$
    \item $A, B \in \mathcal{A}$ and $A \subseteq
      B$, then $B \setminus A \in \mathcal{A}$
    \item $\{A_n\} \in \mathcal{A}$ and $A_n
      \uparrow A = \bigcup_n A_n$, then $A \in
      \mathcal{A}$ ($A_1 \subseteq A_2 \subseteq
      A_3 \subseteq \cdots$)
  \end{itemize}
  }
\end{definition*}
\end{flashcard}
\vspace{-1em}

Equivalently, $\mathcal{A}$ is a $d$-system if
\begin{itemize}
  \item $\emptyset \in \mathcal{A}$
  \item $A \in \mathcal{A} \implies A^c \in
    \mathcal{A}$
  \item $\{A_n\} \in \mathcal{A}$ and $A_n$'s
    \emph{disjoint}, then $\bigcup_n A_n \in
    \mathcal{A}$
\end{itemize}
Proof is on Example Sheet 1.

\textbf{Fact (Example Sheet 1):} A $\pi$-system
$\mathcal{A}$ which is also a $d$-system is
actually a $\sigma$-algebra.

\begin{flashcard}[dynkins-lemma]
\begin{lemma*}[Dynkin's lemma]
  \refstepcounter{customlemma}
  \label{dynkins_lemma}
  \cloze{Let $\mathcal{A}$ be a $\pi$-system. Then
  any $d$-system that contains $\mathcal{A}$,
  contains also $\sigma(\mathcal{A})$.}
\end{lemma*}
\end{flashcard}

\begin{flashcard}[dynkin-proof]
\begin{proof}
  \cloze{
  Define
  \[ \mathcal{D} =
  \bigcap_{\substack{
    \text{$\ol{\mathcal{D}}$ a $d$-system} \\
    \text{$\ol{\mathcal{D}} \supseteq \mathcal{A}$}
  }} \ol{\mathcal{D}} \]
  Then $\mathcal{D}$ is itself a $d$-system (proof
  same as in $\sigma(\mathcal{A})$). We will show
  that $\mathcal{D}$ is also a $\pi$-system, then
  we are done.

  To achieve this, define
  \[ \mathcal{D}' = \{B \in \mathcal{D} : B \cap
  A \in \mathcal{D} ~\forall A \in \mathcal{A}\} \]
  Then $\mathcal{A} \subseteq \mathcal{D}'$ (as
  $\mathcal{A}$ is a $\pi$-system). $\mathcal{D}'$
  is in fact a $d$-system ($B \in \mathcal{A}$,
  then for any $A \in \mathcal{A}$, $B \cap A \in
  \mathcal{A}$ as $\mathcal{A}$ is a $\pi$-system,
  hence $B \in \mathcal{D}$). Fix $A \in
  \mathcal{A}$.
  \begin{itemize}
    \item Then $E \cap A = A \in \mathcal{A}
      \subseteq \mathcal{D}$. So $E \in
      \mathcal{D}'$.
    \item If $B_1 \subseteq B_2$, and $B_1, B_2
      \in \mathcal{D}'$, then $(B_2 \setminus B_1)
      \cap A = (B_2 \cap A) \setminus (B_1 \cap A)
      \in \mathcal{D}$. But by definition, as
      $B_1, B_2 \in \mathcal{D}'$, we get $B_1
      \cap A, B_2 \cap A \in \mathcal{D}$. Also,
      $B_1 \cap A \subseteq B_2 \cap A$ and
      $\mathcal{D}$ is a $d$-system, so $(B_2 \cap
      A) \setminus (B_1 \cap A) \in \mathcal{D}$.
      So $(B_2 \setminus B_1) \cap A \in \mathcal{D}$,
      i.e. $B_2 \setminus B_1 \in \mathcal{D}'$.
    \item Finally, if $\{B_n\} \in \mathcal{D}'$
      and $B_n \uparrow B = \bigcup_n B_n$, then
      $B_n \cap A \in \mathcal{D}$ and $(B_n \cap
      A) \uparrow (B \cap A)$ and $\mathcal{D}$ is
      a $d$-system, so $B \cap A \in \mathcal{D}$.
      So $B \in \mathcal{D}'$.
  \end{itemize}
  So $\mathcal{D}'$ is a $d$-system. Also,
  $\mathcal{D}' \subseteq \mathcal{D}$. But also,
  $\mathcal{A} \subseteq \mathcal{D}'$ and
  $\mathcal{D}'$ is a $d$-system, so $\mathcal{D}
  \subseteq \mathcal{D}'$ (by minimality of
  $\mathcal{D}$). So $\mathcal{D} = \mathcal{D}'$,
  i.e. for all $B \in \mathcal{D}$ and $A \in
  \mathcal{A}$, $B \cap A \in \mathcal{D}$ ($*$). Now we
  can repeat this argument one level higher.
  Define
  \[ \mathcal{D}'' = \{B \in \mathcal{D} : B \cap
  A \in \mathcal{D} ~\forall A \in \mathcal{D}\} \]
  Then $\mathcal{A} \subseteq \mathcal{D}''$ (by
  ($*$)). Then check, as before, that
  $\mathcal{D}''$ is a $d$-system. So
  $\mathcal{D}'' = \mathcal{D}$. But then (by the
  definition of $\mathcal{D}''$), $\forall B \in
  \mathcal{D}$, $A \in \mathcal{D} \implies B \cap
  A \in \mathcal{D}$, i.e. $\mathcal{D}$ is a
  $\pi$-system. So $\mathcal{D}$ is a
  $\sigma$-algebra containing $\mathcal{A}$, hence
  $\mathcal{D} \supseteq \sigma(\mathcal{A})$
  (check that $\emptyset \in \mathcal{D}$).
}
\end{proof}
\end{flashcard}

Now we get the \emph{uniqueness theorem}.

\begin{flashcard}[uniqueness-of-extension-thm]
\begin{theorem*}[Uniqueness of Extension]
  \refstepcounter{customtheorem}
  \label{uniqueness_thm}
  \cloze{
  Let $\mu_1$, $\mu_2$ be some measures on $(E,
  \mathcal{E})$ with $\mu_1(E) = \mu_2(E) <
  \infty$. Suppose that $\mu_1 = \mu_2$ on
  $\mathcal{A}$ for some $\pi$-system
  $\mathcal{A}$ that generates $\mathcal{E}$ (i.e.
  $\sigma(\mathcal{A}) = \mathcal{E}$). Then
  $\mu_1 = \mu_2$ on $\mathcal{E}$.
  }
\end{theorem*}
\end{flashcard}

\begin{proof}
  Consider $\mathcal{D} = \{A \in \mathcal{E} :
  \mu_1(A) = \mu_2(A)\}$. Then, by hypothesis,
  $\mathcal{A} \subseteq \mathcal{D}$. Also,
  $\mathcal{A}$ is a $\pi$-system. So enough to
  show that $\mathcal{D}$ is a $d$-system (by
  \nameref{dynkins_lemma}, $\sigma(\mathcal{A} =
  \mathcal{E} \subseteq \mathcal{D} \subseteq
  \mathcal{E}$, so $\mathcal{D} = \mathcal{E}$).
  \begin{itemize}
    \item $\emptyset \in \mathcal{D}$ since
      $\mu_1(\emptyset) = \mu_2(\emptyset) = 0$.
    \item $A \in \mathcal{D} \implies \mu_1(A) =
      \mu_2(A)$. So $\mu_1(A^c) = \mu_1(E) -
      \mu_1(A) = \mu_2(E) - \mu_2(A) =
      \mu_2(A^c)$. So $A^c \in \mathcal{D}$.
    \item $\{A_n\} \in \mathcal{D}$ disjoint, then
      $\mu_1 \left( \bigcup_n A_n \right) = \sum
      \mu_1(A_n) = \sum \mu_2(A_n) = \mu_2 \left(
      \bigcup_n A_n \right)$. So $\bigcup_n A_n
      \in \mathcal{D}$.
  \end{itemize}
  So $\mathcal{D}$ is a $d$-system.
\end{proof}

% \begin{remark*}
%   If $A_n \uparrow A$, then $\mu(A) = \lim_{n \to
%   \infty} \mu(A_n)$. Use this to show
%   $\mathcal{D}$ is a $d$-system satisfying
%   ($+$).
% \end{remark*}

\textbf{Question:} How to show all sets of a
$\sigma$-algebra $\mathcal{E}$ generated by
$\mathcal{E}$ has a certain property
$\mathcal{P}$?
\[ \mathcal{G} = \{A \subseteq E : \text{$A$ has
the property $\mathcal{P}$}\} \]
and all elements of $\mathcal{A}$ has the property
$\mathcal{P}$.
Possible methods:
\begin{enumerate}[(1)]
  \item Show that $\mathcal{G}$ is a
    $\sigma$-algebra.
  \item Show that $\mathcal{G}$ is a $d$-system
    and $\mathcal{A}$ is a $\pi$-system, and use
    \nameref{dynkins_lemma}.
  \item MCT (monotone convergence theorem?)
\end{enumerate}

\begin{flashcard}[Borel-set-defn]
\begin{definition*}[Borel Sets]
  \cloze{
  Let $E$ be a Hausdorff topological space. The
  $\sigma$-algebra generated by the set of open
  sets, i.e. $\sigma(\mathcal{A})$ where
  $\mathcal{A} = \{A \subseteq E :
  \text{$\mathcal{A}$ open}\}$ is called the
  \emph{Borel $\sigma$-algebra} $\mathcal{B}(E)$
  of $E$. $\mathcal{B}(\RR)$ is written as
  $\mathcal{B}$. A measure $\mu$ on $(E,
  \mathcal{B}(E))$ is called a \emph{Borel
  measure} on $E$. If $\mu(K) < \infty$ for all
  $K$ compact, then $\mu$ is called a \emph{Radon}
  measure on $E$. If $\mu(E) = 1$, $\mu$ is called
  a \emph{probability measure} on $E$, and $(E,
  \mathcal{E}, \mu)$ is called a probability space
  (usually use $(\Omega, \mathcal{F}, \PP)$). If
  $\mu(E) < \infty$, $\mu$ is a finite measure on
  $E$. If $\exists (E_n)$ in $\mathcal{E}$ such
  that $\mu(E_n) < \infty$ for all $n$ and $E =
  \bigcup_n E_n$, then $\mu$ is called a
  \emph{$\sigma$-finite measure} (the
  \nameref{uniqueness_thm} holds for
  $\sigma$-finite measures).
  }
\end{definition*}
\end{flashcard}