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Goal: one of the main goals of this course is to
construct a Borel measure $\mu$ on
$\mathcal{B}(\RR^d)$ such that such that
\[ \mu \left( \prod_{i = 1}^d (a_i, b_i) \right) =
\prod_{i = 1}^d (b_i - a_i) \qquad a_i < b_i \]
corresponding to the usual notion of volume of
rectangles. This measure will be called the
\emph{Lebesgue measure} after H. Lebesgue (1902).

We'll first look at $d = 1$.

\begin{flashcard}[borel-measure-1d]
\begin{theorem*}
  There exists a unique Borel measure $\mu$ on
  $\RR$ such that $\forall a, b \in \RR$ \cloze{with $a <
  b$, $\mu((a, b]) = b - a (\dag)$}. $\mu$ is
  called \cloze{the \fcemph{Lebesgue measure} on $\RR$.}
\end{theorem*}
\end{flashcard}

\begin{proof}
  \textbf{Existence:} Consider the ring
  $\mathcal{A}$ of finite unions of disjoint
  intervals of the form
  \[ A = (a_1, b_1] \cup (a_2, b_2] \cup \cdots
  \cup (a_n, b_n] \qquad a_1 \le b_1 \le a_2 \le
  b_2 \le \cdots \le a_n \le b_n \]
  Note that $\sigma(\mathcal{A}) = \mathcal{B}$
  (Example Sheet) (as all open intervals in
  $\sigma(\mathcal{A})$ and open intervals
  generate open sets).

  Define for such $A \in \mathcal{A}$,
  \[ \mu(A) = \sum_{i = 1}^n (b_i - a_i) \]
  This agrees with $(\dag)$ for $(a, b]$. This is
  additive and well-defined (check). (Note that
  this is important since representations aren't
  unique, for example $(0, 2] = (0, 1] \cup (1, 2]$).
  So, the existence of $\mu$ on $\mathcal{B}$
  follows from Caratheodory's theorem if we can
  show that $\mu$ is \fcemph{countably additive}.

  \fcscrap{
  \begin{remark*}
    (Example Sheet) $\mu$ a finitely additive set
    function on a ring $\mathcal{A}$. Then $\mu$
    is countably additive iff
    \begin{itemize}
      \item $A_n \uparrow A$, $A_n \in
        \mathcal{A} \implies \mu(A_n) \uparrow
        \mu(A)$.
      \item In addition, if $\mu$ is finite and
        $A_n \downarrow A$, $(A_n), A \in
        \mathcal{A}$, then $\mu(A_n) \downarrow
        \mu(A)$.
    \end{itemize}
    $A_n = [n, \infty)$, $A_n \downarrow
    \emptyset$.
  \end{remark*}
  }
  So, by example sheet, showing $\mu$ is countably
  additive on $\mathcal{A}$ is equivalent to
  showing the following:
  If $A_n \in \mathcal{A}$, $A_n \downarrow
  \emptyset$, then $\mu(A_n) \downarrow 0$, $A_1
  \supseteq A_2 \supseteq \cdots$. ($\mu$ finitely
  additive on a ring $\mathcal{A}$. THen $\mu$ is
  countably additive iff $A_n \downarrow
  \emptyset$, $\mu(A_1) < \infty$ implies
  $\mu(A_n) \downarrow 0$).

  We shall prove this by contradiction. Suppose
  $\exists B_n \in \mathcal{A}$, $B_n \downarrow
  \emptyset$, but for all $n$, $\mu(B_n) \ge 2\eps
  > 0$ for some $\eps > 0$. For each $n$, $B_n$
  can be approximated from within by $C_n \in
  \mathcal{A}$ such that $\ol{C_n} \subseteq B_n$
  and $\mu(B_n \setminus C_n) \le \eps / 2^n$. For
  example,
  \begin{align*}
    B_n
    &= (a_1, b_1] \cup (a_2, b_2] \\
    C_n
    &= \left(a_1 + \frac{\eps}{2 \cdot 2^n}, b_1
    \right] \cup \left( a_2 + \frac{\eps}{2 \cdot
    2^n}, b_2 \right]
  \end{align*}
  Then
  \begin{align*}
    \mu (b_n \setminus C_1 \cap C_2 \cap \cdots
    \cap C_n )
    &\le \mu \left( \bigcup_{i = 1}^n (B_i
    \setminus C_i \right) \\
    &\le \sum_{i = 1}^n \mu(B_i \setminus C_i) \\
    &\le \sum_{i = 1}^n \frac{\eps}{2^i} \\
    &= \eps
  \end{align*}
  Since $\mu(B_n) \ge 2\eps$ and $\mu(B_n
  \setminus (C_1 \cap \cdots \cap C_n)) \le \eps$,
  \[ \implies \mu(C_1 \cap \cdots \cap C_n) \ge
  \eps \implies C_1 \cap \cdots \cap C_n \neq
  \emptyset \]
  So $K_n = \ol{C_1} \cap \cdots \cap \ol{C_n}
  \neq \emptyset$. $(K_n)$ is a sequence of
  decreasing bounded closed sets, each non-empty, so
  $\bigcap_n K_n \neq \emptyset$ (by completeness
  of $\RR$). But then $\emptyset \neq \bigcap_n
  K_n \subseteq \bigcap_n B_n = \emptyset$, which
  is a contradiction.

  \textbf{Uniqueness:} For uniqueness, suppose
  $\mu$, $\lambda$ are measures on $\mathcal{B}$
  such that $(\dag)$ holds for sets of the form
  $(a, b]$. Define the truncated measures, for $A
  \in \mathcal{B}$,
  \[ \mu_n(A) = \mu(A \cap (n, n + 1]) \qquad
  \text{and} \qquad \lambda_n(A = \lambda(A \cap
  (n, n + 1]) \]
  Then $\mu_n$ and $\lambda_n$ are probability
  measures on $\mathcal{B}$ and $\mu_n =
  \lambda_n$ on the $\pi$-system of intervals of
  the form $(a, b]$, $a < b$, which generates
  $\mathcal{B}$. So by \nameref{uniqueness_thm},
  $\lambda_n = \mu_n$ on $\mathcal{B}$. Hence for
  all $A \in \mathcal{B}$,
  \begin{align*}
    \mu(A)
    &= \mu \left( \bigcup_n (A \cap (n, n + 1])
    \right) \\
    &= \sum_n \mu(A \cap (n, n + 1]) \\
    &= \sum_n \mu_n(A) \\
    &= \sum_n \lambda_n(A) \\
    &= \lambda(A) \qedhere
  \end{align*}
\end{proof}

\begin{remark*}
  \phantom{}
  \begin{enumerate}
    \item A set $B \in \mathcal{B}$ is called a
      Lebesgue ($\lambda$) null set if $\lambda(B)
      = 0$.
      Any singleton set
      \[ \{x\} = \bigcap_n \left( x - \frac{1}{n},
      x \right] \]
      is a Lebesgue-null set since $\lambda(\{x\})
      = \lim_{n \to \infty} \lambda((x - 1/n, x])
      = \lim_{n \to \infty} 1/n = 0$. In
      particular, $\lambda((a, b)) = \lambda([a,
      b]) = \lambda([a, b)) = b - a$. Any
      countable set $Q$ satisfies $\lambda(Q) =
      0$. There exist uncountable sets $C$ with
      $\lambda(C) = 0$ (for example the Cantor
      set).
    \item Lebesgue measure is translation
      invariant: for $x \in \RR$, and $B \in
      \mathcal{B}$, define $B + x = \{b + x : b
      \in B\}$, and define $\lambda_x(B) =
      \lambda(B + x)$. Then $\lambda_x((a, b]) =
      \lambda((a, b] + x) = \lambda((a + x, b +
      x]) = b - a = \lambda((a, b])$. So
      $\lambda_x = \lambda$ on the $\pi$-system of
      intervals, and hence $\lambda_x = \lambda$
      on $\mathcal{B}$. Question: Is the Lebesgue
      measure the only such translation invariant
      measure on $\mathcal{B}$?
    \item Lebesgue-measurable sets: Caratheodory
      extends $\lambda$ from $\mathcal{A}$ to not
      just $\sigma(A) = \mathcal{B}$, but actually
      to $\mathcal{M}$, the set of outer
      measurable sets. $\mathcal{M} \supseteq
      \mathcal{B}$, but how large is
      $\mathcal{M}$?
      \[ \mathcal{M} = \{A \cap N : A \in
      \mathcal{B}, N \subseteq B, B \in
      \mathcal{B}, \lambda(B) = 0\} \]
      (Lebesgue $\sigma$-algebra)
  \end{enumerate}
\end{remark*}
