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\begin{flashcard}[set-function-defn]
\begin{definition*}[Set-function]
  Let $\mathcal{A}$ be any collection of subsets
  of $E$\cloze{ such that $\emptyset \in \mathcal{A}$.
  A \emph{set-function} $\mu$ is a map $\mu :
  \mathcal{A} \to [0, \infty]$ such that
  $\mu(\emptyset) = 0$. We say}
  \begin{enumerate}[(1)]
    \item \cloze{$\mu$ is \emph{increasing} if $\mu(A)
      \le \mu(B)$ for all $A, B \in \mathcal{A}$
      such that $A \subset B$.}
    \item \cloze{$\mu$ is \emph{additive} if $A\mu(A \cup
      B) = \mu(A) + \mu(B)$ for all $A, B \in
      \mathcal{A}$ with $A$, $B$
      \fcemph{disjoint}.}
    \item \cloze{$\mu$ is \emph{countably additive} if
      $\mu \left( \bigcup_n A_n \right) = \sum_n
      \mu(A_n)$ for all $A_n$ disjoint such that
      $A_n, \bigcup_n A_n \in \mathcal{A}$.}
    \item \cloze{$\mu$ is \emph{countably subadditive} if
      $\mu \left( \bigcup_n A_n \right) \le \sum_n
      \mu(A_n)$ for all $A_n, \bigcup_n A_n \in
      \mathcal{A}$.}
  \end{enumerate}
\end{definition*}
\end{flashcard}

\begin{remark*}
  \hypertarget{countably-additive-remark}{If}
  $\mu$ is a \emph{countably additive} set
  function on $\mathcal{A}$, and $\mathcal{A}$
  \emph{is a ring}, then $\mu$ satisfies (Example
  Sheet 1) all of (1), (2), (3), (4).
\end{remark*}

\begin{flashcard}[caratheodory-thm]
\begin{theorem*}[Caratheodory]
  \cloze{
  Let $\mathcal{A}$ be a \emph{\fcemph{ring of
  subsets of}} $E$ and $\mu : \mathcal{A} \to [0,
  \infty]$ be a \fcemph{countably additive} set function on
  $\mathcal{A}$. Then $\mu$ \fcemph{extends to a
  measure} on $\sigma(\mathcal{A})$.
  }
\end{theorem*}
\end{flashcard}

\begin{proof}
  For any $B \subseteq E$, define
  \[ \mu^*(B) = \inf\left( \left\{ \sum \mu(A_i) : B
  \subseteq \bigcup_i A_i, A_i \in \mathcal{A}
  \right\} \cup \{\infty\} \right) \]

  Clearly $\mu^*(\phi) = 0$ and $\mu^*$ is
  increasing. So $\mu^*$ is an increasing set
  function on $\mathcal{P}(E)$.

  Call a set $A \subset E$ $\mu^*$ measurable if
  $\forall B \subseteq E$,
  \[ \mu^*(B) = \mu^*(B \cap A) + \mu^*(B \cap
  A^c) .\]
  Define $\mathcal{M} = \{A \subset E : \text{$A$
  is $\mu^*$ measurable}\}$. Shall show
  $\mathcal{M}$ is a $\sigma$-algebra that
  contains $\mathcal{A}$, $\mu^* |_\mathcal{M}$ is
  a measure on $\mathcal{M}$ that extends $\mu$
  (i.e. $\mu^* |_{\mathcal{A}} = \mu$).

  Step 1: $\mu^*$ is countably subadditive, i.e.
  if $B \subseteq \bigcup_n B_n$, will show
  $\mu^*(B) \le \sum_n \mu^*(B_n)$. Nothing to
  prove if $\mu^*(B_n) = \infty$ for some $n$. Sow
  assume $\mu^*(B_n) < \infty$ for all $n$. For
  all $\eps > 0$, there exists $(A_{n, m}) \in
  \mathcal{A}$ such that $B_n \subset \bigcup_m
  A_{n, m}$ and $\mu^*(B_n) + \frac{\eps}{2^n} \ge
  \sum_m \mu(A_{n, m})$. But then, $B \subseteq
  \bigcup_n B_n \subseteq \bigcup_{n, m} A_{n, m}$
  and $(A_{n, m}) \in \mathcal{A}$, so by
  definition of $\mu^*$,
  \[ \mu^*(B) \le \sum_n \sum_m \mu(A_{n, m}) \le
  \sum_n \left( \mu^*(B_n) + \frac{\eps}{2^n}
  \right) = \sum_n \mu^*(B_n) + \eps \]
  As $\eps > 0$ is arbitrary, we get the desired
  result.

  Step 2: $\mu^*$ extends $\mu$, i.e. for all
  $A \in \mathcal{A}$, $\mu^*(A) = \mu(A)$.
  ($\mu^*(A) \le \mu(A)$ by definition of $\mu^*$
  as $A \subseteq A$, $A \in \mathcal{A}$). Now we
  prove $\mu^*(A) \ge \mu(A)$. As $\mu$ is
  countably additive on $A\mathcal{A}$ and
  $\mathcal{A}$ is a ring, $\mu$ is countably
  sub-additive on $\mathcal{A}$ and increasing (by
  \hyperlink{countably-additive-remark}{earlier Remark}).
  Now, let $A \subset \bigcup_n (A \cap A_n)$, so
  by countable sub-additivity on $\mathcal{A}$,
  \[ \mu(A) \le \sum_n \mu(A \cap A_n) \le \sum_n
  \mu(A_n) \]
  (the second inequality is because $\mu$ is
  increasing).
  So by taking $\inf$ over all such $\{A_n\}$,
  $\mu(A) \le \mu^*(A)$.

  Step 3: $M \supseteq A$, i.e. $A \in
  \mathcal{A}$ and $B \subseteq E$, want to show
  \[ \mu^*(B) = \mu^*(B \cap A) + \mu^*(B \cap
  A^c) \]
  Since $B \subseteq (B \cap A) \cup (B \cap
  A^c)$, by Step 1 (countable sub additivity),
  $\mu^*(B) \le \mu^*(B \cap A) + \mu^*(B \cap
  A^c)$. To prove $\mu^*(B) \ge \mu^*(B \cap A) +
  \mu^*(B \cap A^c)$, without loss of generality
  assume $\mu^*(B) < \infty$. So again $\forall
  \eps > 0$, there exists $(A_n) \in \mathcal{A}$
  with $B \subseteq \bigcup_n A_n$ such that
  $\mu^*(B) + \eps \ge \sum_n \mu(A_n)$. Then
  $(B \cap A) \subseteq \bigcup_n (A_n \cap A)$
  and $(B \cap A^c) \subseteq \bigcup_n (A_n \cap
  A^c)$. So that
  \[ \mu^*(B \cap A) \le \sum \mu(A_n \cap A)
  \qquad \text{and} \qquad \mu^*(B \cap A^c) \le
  \sum_n \mu(A_n \cap A^c) ,\]
  so that
  \[ \mu^*)B \cap A) + \mu^*(B \cap A^c) \le
  \sum_n (\mu(A_n \cap A) + \mu(A_n \cap A^c)) =
  \sum \mu(A_n) \le \mu^*(B) + \eps \]
  As $\eps > 0$ is arbitrary, we get $\mu^*(B \cap
  A) + \mu^*(B \cap A^c) \le \mu^*(B)$.

  Step 4: $\mathcal{M}$ is an algebra: $\phi \in
  M$. Also if $A \in \mathcal{M}$, then $A^c \in
  \mathcal{M}$. Now let $A_1, A_2 \in \mathcal{M}$
  and $B \subset E$. Then
  \begin{align*}
    \mu^*(B)
    &= \mu^*(B \cap A_1) + \mu^*(B \cap A_1^c)
    &&\text{(as $A_1 \in \mathcal{M}$)} \\
    &= \mu^*(B \cap A_1 \cap A_2) = \mu^*(B \cap
    A_1 \cap A_2^c) + \mu^*(B \cap A_1^c)
    &&\text{(as $A_2 \in \mathcal{M}$)} \\
    &= \mu^*(B \cap (A_1 \cap A_2)) + \mu^*(B \cap
    (A_1 \cap A_2)^c \cap A_1) + \mu^*(B \cap (A_1
    \cap A_2)^c \cap A_1^c) \\
    &= \mu^*(B \cap (A_1 \cap A_2)) + \mu^*(B \cap
    (A_1 \cap A_2)^c)
    &&\text{(as $A_1 \in \mathcal{M}$, $B \cap (A_1
    \cap A_2)^c = \tilde{B}$)}
  \end{align*}
  So $A_1 \cap A_2 \in \mathcal{M}$.

  Step 5: $\mathcal{M}$ is a $\sigma$-algebra and
  $\mu^* |_{\mathcal{M}}$ is a measure (since
  $\mathcal{M}$ is an algebra, convince yourself), i.e. for
  any sequence $(A_n) \in \mathcal{M}$ with $A_n$
  pairwise disjoint, we want to prove $A \defeq
  \bigcup_n A_n \in \mathcal{M}$ and $\mu(A) =
  \sum_n \mu(A_n)$. So, as before for any $B
  \subseteq E$,
  \begin{align*}
    \mu^*(B)
    &= \mu^*(B \cap A_1) + \mu^*(B \cap A_1^c)
    &&\text{(as $A_1 \in \mathcal{M}$)} \\
    &= \mu^*(B \cap A_1) + \mu^*(B \cap A_1^c \cap
    A_2) + \mu^*(B \cap A_1^c \cap A_2^c)
    &&\text{(as $A_2 \in \mathcal{M}$)} \\
    &= \mu^*(B \cap A_1) + \mu^*(B \cap A_2) +
    \mu^*(B \cap A_1^c \cap A_2^c) \\
    &~~~\vdots \\
    &= \sum_{i = 1}^n \mu^*(B \cap A_i) + \mu^*(B
    \cap A_1^c \cap \cdots \cap A_n^c) \\
    &\ge \sum_{i = 1}^n \mu^*(B \cap A_1) +
    \mu^*(B \cap A^c)
  \end{align*}
  The last inequality comes from the fact that
  $\mu^*$ is increasing and since $A = \bigcup_i A_i$, we have $A^c = \cap
  A_i^c \subseteq A_1^c \cap \cdots \cap A_n^c$.

  So as $n \to \infty$,
  \begin{align*}
    \mu^*(B) 
    &\ge \sum_{i = 1}^\infty \mu^*(B \cap
    A_1) + \mu^*(B \cap A^c) \\
    &\ge \mu^*(B \cap A) + \mu^*(B \cap A^c) \\
  \end{align*}
  Also,
  \begin{align*}
    \mu^*(B)
    &\le \mu^*(B \cap A) + \mu^*(B \cap A^c)
  \end{align*}
  is obvious by sub additivity. So $\mu^*(B) =
  \mu^*(B \cap A) + \mu^*(B \cap A^c)$, i.e. $A
  \in \mathcal{M}$.
\end{proof}
