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/Part 4: Measure Theory and Integration/Chapter 14: Positive Measures

14.7 Carathéodory’s Extension Theorem

Definition 14.7.1 (Outer Measure). Let $X$ be a set and $\mu^{*}: 2^{X} \to [0, \infty]$, then $\mu^{*}$ is an outer measure if:

  1. $\mu^{*}(\emptyset) = 0$.

  2. For any $E, F \subset X$ with $E \subset F$, $\mu^{*}(E) \le \mu^{*}(F)$.

  3. For any $\seq{E_n}\subset X$,

    \[\mu^{*}\paren{\bigcup_{n \in \natp}E_n}\le \sum_{n \in \natp}\mu^{*}(E_{n})\]

Proposition 14.7.2. Let $X$ be a set, $\ce \subset 2^{X}$, and $\rho: \ce \to [0, \infty]$ such that $\emptyset \in \ce$ and $\rho(\emptyset) = 0$. Define

\[\mu^{*}: 2^{X} \to [0, \infty] \quad E \mapsto \inf\bracs{\sum_{n \in \natp}\rho(E_n) \bigg | \seq{E_n} \subset \ce, \bigcup_{n \in \natp}E_n \supset E}\]

then $\mu^{*}$ is an outer measure.

Proof. (OM1): $\emptyset \in \ce$ and $\rho(\emptyset) = 0$.

(OM2): Let $\seq{E_n}\subset 2^{X}$ and assume without loss of generality that $\sum_{n \in \natp}\mu^{*}(E_{n}) < \infty$. Let $\eps > 0$. For each $n \in \nat$, let $\bracs{F_{n, k}}_{k = 1}^{\infty} \subset \ce$ such that $\bigcup_{k \in \natp}F_{n, k}\supset E_{n}$ and $\sum_{n \in \natp}\rho(F_{n, k}) < \mu^{*}(E_{n}) + 2^{-n}\eps$, then

\begin{align*}\mu^{*}\paren{\bigcup_{n \in \natp}E_n}&\le \sum_{(n, k) \in \natp \times \natp}\rho(F_{n, k}) = \sum_{n \in \natp}\sum_{k \in \natp}\rho(F_{n, k}) \\&\le \sum_{n \in \natp}(\mu^{*}(E_{n}) + 2^{-n}\eps) \le \eps + \sum_{n \in \natp}\mu^{*}(E_{n})\end{align*}

Since $\eps$ is arbitrary, $\mu^{*}\paren{\bigcup_{n \in \natp}E_n}\le \sum_{n \in \natp}\mu^{*}(E_{n})$.$\square$

Definition 14.7.3 (Outer Measurable). Let $X$ be a set, $\mu^{*}: 2^{X} \to [0, \infty]$ be an outer measure, and $E \subset X$, then $E$ is $\mu^{*}$-measurable if for any $F \subset X$,

\[\mu^{*}(F) = \mu^{*}(E \cap F) + \mu^{*}(E \setminus F)\]

Theorem 14.7.4 (Carathéodory, [Theorem 1.11, Fol99]). Let $X$ be a set, $\mu^{*}: 2^{X} \to [0, \infty]$ be an outer measure, and $\cm \subset 2^{X}$ be the collection of all $\mu^{*}$-measurable sets, then:

  1. For any $F \subset X$ and $\seq{E_n}\subset \cm$ pairwise disjoint,

    \[\mu^{*}\paren{F \cap \bigsqcup_{n \in \natp}E_n}= \sum_{n \in \natp}\mu^{*}(F \cap E_{n})\]

  2. $\cm$ is a $\sigma$-algebra.

  3. $(X, \cm, \mu^{*}|_{\cm})$ is a complete measure space.

Proof. (1): Let $N \in \nat$, then

\[\mu^{*}\paren{F \cap \bigsqcup_{n \in \natp}E_n}\ge \mu^{*}\paren{F \cap \bigsqcup_{n =1}^N E_n}= \sum_{n = 1}^{N} \mu^{*}(F \cap E_{n})\]

As this holds for all $N \in \nat$,

\[\mu^{*}\paren{F \cap \bigsqcup_{n \in \natp}E_n}\ge \sum_{n \in \natp}\mu^{*}(F \cap E_{n})\]

(2): Firstly, $\emptyset, X \in \cm$ since $\mu^{*}(\emptyset) = 0$. For any $A \in \cm$, $A^{c} \in \cm$ as well because the definition of $\mu^{*}$-measurability is symmetric.

Let $A, B \in \cm$ and $F \subset X$, then

\begin{align*}\mu^{*}(F)&= \mu^{*}(F \cap A) + \mu^{*}(F \setminus A) \\&= \mu^{*}(F \cap A \cap B) + \mu^{*}(F \cap A \setminus B) + \mu^{*}(F \cap B \setminus A) + \mu^{*}(F \setminus (A \cup B)) \\&= \mu^{*}(F \cap B) + \mu^{*}(F \cap A \setminus B) + \mu^{*}(F \setminus (A \cup B)) \\&= \mu^{*}(F \cap (A \cup B)) + \mu^{*}(F \setminus (A \cup B))\end{align*}

so $\cm$ is an algebra. Since (1) implies that $\cm$ is closed under countable disjoint unions, $\cm$ is a $\sigma$-algebra by Lemma 13.1.5.

(3): By (1), $\mu|_{\cm}$ is a measure. Let $N \in \cm$ with $\mu^{*}(N) = 0$, then for any $E \subset N$ and $F \subset X$,

\begin{align*}\mu^{*}(E)&= \underbrace{\mu^*(E \cap N)}_{0} + \mu^{*}(E \setminus N) \\&= \underbrace{\mu^*(E \cap N \setminus F)}_{0} + \underbrace{\mu^*(E \cap F)}_{0} + \mu^{*}(E \setminus N) \\&= \mu^{*}((E \setminus F) \cap N) + \mu^{*}(E \cap F) + \mu^{*}((E \setminus F) \setminus N) \\&= \mu^{*}(E \cap F) + \mu^{*}(E \setminus F)\end{align*}
$\square$

Definition 14.7.5 (Premeasure). Let $X$ be a set, $\alg \subset 2^{X}$ be a ring, and $\mu_{0}: \alg \to [0, \infty]$, then $\mu_{0}$ is a premeasure if:

  1. $\mu_{0}(\emptyset) = 0$.

  2. For any $\seq{A_n}\subset \alg$ pairwise disjoint with $\bigsqcup_{n \in \natp}A_{n} \in \alg$,

    \[\mu_{0}\paren{\bigsqcup_{n \in \natp}A_n}= \sum_{n \in\natp}\mu_{0}(A_{n})\]

Theorem 14.7.6 (Carathéodory’s Extension Theorem, [Theorem 1.14, Fol99]). Let $X$ be a set, $\alg \subset 2^{X}$ be a ring, and $\mu_{0}: \alg \to [0, \infty]$ be a premeasure, then there exists a measure space $(X, \cm, \mu)$ such that

  1. $\cm \supset \alg$.

  2. $\mu|_{\alg} = \mu_{0}$.

  3. $\mu$ is complete.

  4. For any measure space $(X, \cn, \nu)$ satisfying (1) and (2),

    1. $\nu|_{\cm \cap \cn}\le \mu_{\cm \cap \cn}$.

    2. For any $E \in \cm \cap \cn$ with $\mu(E) < \infty$, $\mu(E) = \nu(E)$.

    3. If $\mu$ is $\sigma$-finite, then $\nu = \mu$.

Proof. Let

\[\mu^{*}: 2^{X} \to [0, \infty] \quad E \mapsto \inf\bracs{\sum_{n \in \natp}\mu_0(E_n) \bigg | \seq{E_n} \subset \alg, \bigcup_{n \in \natp}E_n \supset E}\]

then $\mu^{*}$ is an outer measure by Proposition 14.7.2. By Carathéodory’s theorem, there exists a measure space $(X, \cm, \mu)$ such that $\mu = \mu^{*}|_{\cm}$.

(1): Let $E \in \alg$ and $F \subset X$. If there exists $\seq{F_n}\subset \alg$ with $\bigcup_{n \in \natp}F_{n} \supset F$, then

\[\sum_{n \in \natp}\mu_{0}(F_{n}) = \sum_{n \in \natp}\mu_{0}(F_{n} \cap E) + \sum_{n \in \natp}\mu_{0}(F_{n} \setminus E) \ge \mu^{*}(F \cap E) + \mu^{*}(F \setminus E)\]

for all such $\seq{F_n}$, so $\mu^{*}(F) = \mu^{*}(F \cap E) + \mu^{*}(F \setminus E)$. Otherwise, $\mu^{*}(F) = \infty$ and the result holds directly.

(2): Let $E \in \alg$ and $\seq{E_n}\subset \alg$ such that $\bigcup_{n \in \natp}E_{n} \supset E$, then

\[\sum_{n \in \natp}\mu_{0}(E_{n}) = \sum_{n \in \natp}\mu_{0}(E_{n} \cap E) + \sum_{n \in \natp}\mu_{0}(E_{n} \setminus E) \ge \sum_{n \in \natp}\mu_{0}(E_{n} \cap E) = \mu_{0}(E)\]

As this holds for all such $\seq{E_n}$, $\mu^{*}(E) = \mu_{0}(E)$.

(U): Let $\seq{E_n}\subset \alg$, then by continuity from below (Proposition 14.1.5),

\[\nu\paren{\bigcup_{n \in \natp}E_n}= \limv{n}\nu\paren{\bigcup_{k = 1}^nE_k}= \limv{n}\mu\paren{\bigcup_{k = 1}^n E_k}= \mu\paren{\bigcup_{n \in \natp}E_n}\]

Thus if $E \in \cm \cap \cn$ with $\bigcup_{n \in \natp}E_{n} \supset E$, then

\[\nu(E) \le \nu\paren{\bigcup_{n \in \natp}E_n}= \mu\paren{\bigcup_{n \in \natp}E_n}\le \sum_{n \in \natp}\mu_{0}(E_{n})\]

As this holds for all such $\seq{E_n}$, $\nu(E) \le \mu^{*}(E) = \mu(E)$.

Suppose that $\mu(E) < \infty$, then there exists $\seq{E_n}\subset \alg$ such that $\bigcup_{n \in \natp}E_{n} \supset E$ and $\sum_{n \in \natp}\mu_{0}(E_{n}) < \infty$. In which case,

\begin{align*}\mu\paren{\bigcup_{n \in \natp}E_n}&= \nu\paren{\bigcup_{n \in \natp}E_n}= \nu(E) + \nu\paren{\bigcup_{n \in \natp}E_n \setminus E}\\&\le \nu(E) + \mu\paren{\bigcup_{n \in \natp}E_n \setminus E}\\ \mu(E)&\le \nu(E)\end{align*}

so $\nu(E) = \mu(E)$ whenever $\mu(E) < \infty$.

Finally, if $\mu$ is $\sigma$-finite, then there exists $\seq{E_n}\subset \cm$ such that $E_{n} \upto X$ and $\mu(E_{n}) < \infty$ for all $n \in \natp$. For each $n \in \natp$, there exists $\bracsn{E_{n, k}}_{k = 1}^{\infty} \subset \alg$ such that $E_{n} \subset \bigcup_{k \in \natp}E_{n, k}$ and $\mu\paren{\bigcup_{k \in \natp}E_{n, k}}< \infty$. Let $F_{n} = \bigcup_{j = 1}^{n} \bigcup_{k \in \natp}E_{j, k}$, then $\seq{F_n}\subset \sigma(\alg) \subset \cm \cap \cn$ with $F_{n} \upto X$ and $\mu(F_{n}) < \infty$ for all $n \in \natp$. For any $E \in \cm \cap \cn$, by continuity from below (Proposition 14.1.5),

\[\nu(E) = \limv{n}\nu(E \cap F_{n}) = \limv{n}\mu(E \cap F_{n}) = \mu(E)\]

so $\nu|_{\cm \cap \cn}= \mu|_{\cm \cap \cn}$.$\square$

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