Construction of Measures via Carathéodory's Theorem

25 Feb 2025

I prefer to approach Carathéodory’s theorem by dividing it into three subtheorems. They are as follows:

1. Carathéodory Construction Theorem: Any collection of sets and a positive function define an outer measure.
2. Carathéodory Restriction Theorem: Restricting the domain of an outer measure to measurable sets yields a measure.
3. Carathéodory Extension Theorem: There exists a unique extension of premeasure $\mu_0$ on an algebra $\mathcal{A}_0$ to a measure $\mu$ on the $\sigma$-algebra $\sigma(\mathcal{A}_0)$. Furthermore, $\mu$ is the measure obtained by applying the construction theorem with the collection of sets as $\mathcal{A}_0$ and the positive function as $\mu_0$, and then applying the restriction theorem.

Schematically, this can be understood as follows:

	(1)	(2)	(3)
Domain	Algebra $\mathcal{A}_0$	$\sigma$-algebra $\mathcal{A}$	Power set $\mathcal{P}(X)$
Function	Premeasure $\mu_0$	Measure $\mu$	Outer measure $\mu^\ast$

The construction theorem goes in the direction (1) → (3), the restriction theorem goes (3) → (2), and the extension theorem goes (1) → (2). Let us examine each one by one.

Definition. An outer measure $\mu^\ast: \mathcal{P}(X) \to [0, \infty]$ on $X$ is a function satisfying the following:

1. $\mu^\ast(\varnothing) = 0$
2. $A \subset B \implies \mu^\ast(A) \leq \mu^\ast(B)$
3. $\mu^\ast\left( \bigcup_{n \in \mathbb{N}} A_n \right) \leq \sum_{n \in \mathbb{N}} \mu^\ast(A_n)$

Carathéodory Construction Theorem. Given any collection $\mathcal{S}$ of subsets of $X$ and any function $l: \mathcal{S} \to [0, \infty]$ satisfying $l(\varnothing) = 0$, the function $\mu^\ast$ defined as follows is an outer measure:

\[\mu^*(E) = \inf \left\{ \sum_{n \in \mathbb{N}} l(A_n) : \{ A_n \} \subset \mathcal{S} \text{ covers }E \right\}\]

That is, $\mu^\ast$ is a function that defines something akin to a ‘measure’ of $E$ as an infimum over covers, and is conceptually similar to the upper sum_{upper sum} in Riemann integration.

Proof. Properties 1 and 2 are trivial. We prove property 3.

Let $A = \bigcup A_n$ and suppose an arbitrary $\epsilon > 0$ is given. By the definition of $\mu^\ast$, for each $n$, there exists a cover $\mathcal{C}_n = \lbrace A_{nm} \rbrace_{m \in \mathbb{N}} \subset \mathcal{S}$ of $A_n$ satisfying:

\[\sum l(A_{nm}) \leq \mu^*(A_n) + \epsilon/2^n\]

Therefore,

\[\sum_{n, m \in \mathbb{N}} l(A_{nm}) \leq \sum \mu^*(A_n) + \epsilon.\]

Since $\bigcup \mathcal{C}_n$ covers $A$, by the definition of $\mu^\ast$, we have $\mu^\ast(A) \leq \sum_{n, m \in \mathbb{N}} l(A_{nm})$. ■

Definition. Let $\mu^\ast$ be an outer measure. A set $A$ is said to be measurable with respect to $\mu^\ast$ if it satisfies the following for any $E$:

\[\mu^*(E) = \mu^*(E \cap A) + \mu^*(E \cap A^c)\]

Definition. Let $\mu^\ast$ be an outer measure. A set $N$ with $\mu^\ast(N) = 0$ is called a null set.

That is, a measurable set is one that ‘cleanly’ partitions any set with respect to $\mu^\ast$. Therefore, we can expect that $\mu^\ast$ will behave like an ordinary measure on the collection consisting only of measurable sets. The following restriction theorem validates this expectation.

Carathéodory Restriction Theorem. Let $\mu^\ast$ be an outer measure. Let $\mathcal{A}$ be the collection of sets that are measurable with respect to $\mu^\ast$. Then the following hold:

1. $\mathcal{A}$ is a $\sigma$-algebra.
2. $\mu^\ast |_\mathcal{A}$ is a measure.
3. $\mathcal{A}$ contains all null sets of $\mu^\ast$.

Proof.

1. $\mathcal{A}$ is an algebra.

Closure under complements is trivial. We show closure under finite intersections.

Suppose $A, B$ are measurable. To show that $A \cap B$ is measurable, it suffices to prove:

\[\mu^*(E \cap (A \cap B)) + \mu^*(E \cap (A \cap B)^c) \leq \mu^*(E) \quad \cdots \quad (*)\]

Since $A, B$ are measurable, the following holds:

\[\begin{aligned} \mu^*(E) &= \mu^*(E \cap A) + \mu^*(E \cap A^c) \\ &= \mu^*(E \cap A \cap B) + \mu^*(E \cap A \cap B^c) \\ &+ \mu^*(E \cap A^c \cap B) + \mu^*(E \cap A^c \cap B^c) \end{aligned}\]

Therefore, $(\ast)$ is equivalent to:

\[\begin{aligned} \mu^*(E \cap (A^c \cup B^c)) &\leq \mu^*(E \cap A \cap B^c) \\ &+ \mu^*(E \cap A^c \cap B) \quad \cdots \quad (**) \\ &+ \mu^*(E \cap A^c \cap B^c) \end{aligned}\]

From elementary set theory, we have:

\[A^c \cup B^c = (A \cap B^c) \cup (A^c \cap B) \cup (A^c \cap B^c)\]

Therefore, $(\ast\ast)$ holds. □

2. $\mathcal{A}$ is a $\sigma$-algebra.

It suffices to show that for $\lbrace A_n \rbrace \subset \mathcal{A}$ with $A_n \uparrow A$, we have $A \in \mathcal{A}$. (Why?)

By the definition of outer measure, $\mu^\ast(E \cap A_n) \leq \mu^\ast(E \cap A)$. Let $C_1 = A_1, C_n = A_n \setminus A_{n - 1}$. Then $\bigsqcup C_n = A$. Moreover,

\[\begin{aligned} \mu^*(E \cap A_n) &= \mu^*(E \cap A_n \cap C_n) + \mu^*(E \cap A_n \cap C_n^c) \\ &= \mu^*(E \cap C_n) + \mu^*(E \cap A_{n - 1}) \\ &= \cdots \\ &= \sum^n_{k = 1} \mu^*(E \cap C_k) \end{aligned}\]

Therefore, $\sum^\infty \mu^\ast(E \cap C_n) \leq \mu^\ast(E \cap A)$. Since $\lbrace E \cap C_n \rbrace $ covers $E \cap A$, we have $\sum^\infty \mu^\ast(E \cap C_n) \geq \mu^\ast(E \cap A)$. Therefore, $\sum^\infty \mu^\ast(E \cap C_n) = \mu^\ast(E \cap A)$.

Thus, for any $\epsilon > 0$, there exists a sufficiently large $N$ such that $\mu^\ast(E \cap A) - \epsilon \leq \mu^\ast(E \cap A_n)$. Therefore,

\[\begin{aligned} \mu^*(E \cap A_n^c) &= \mu^*(E) - \mu^*(E \cap A_n) \\ &\leq \mu^*(E) - \mu^*(E \cap A) + \epsilon \end{aligned}\]

whence we obtain:

\[\mu^*(E) \leq \mu^*(E \cap A_n) + \mu^*(E \cap A_n^c) \leq \mu^*(E) + \epsilon\]

Letting $n \to \infty$, we get $\mu^\ast(E \cap A) + \mu^\ast(E \cap A^c) = \mu^\ast(E)$. □

3. $\mu^\ast|_\mathcal{A}$ is a measure, 4. $\mathcal{A}$ contains all null sets.

Left as an exercise to the readers. (Not difficult) ■

Definition. Let $\mathcal{A}_0$ be an algebra. A function $\rho: \mathcal{A}_0 \to [0, \infty]$ is called a premeasure if it satisfies the following:

1. $\rho(\varnothing) = 0$
2. For any pairwise disjoint countable collection $\lbrace A_n \rbrace $, if $\bigcup A_n \in \mathcal{A}_0$, then $\rho\left( \bigcup A_n \right) = \sum \rho(A_n)$

Carathéodory Extension Theorem. For a premeasure $\rho$ on an algebra $\mathcal{A}_0$, define:

\[\mu^*(E) = \inf \left\{ \sum_{n \in \mathbb{N}} \mu_0(A_n) : \{ A_n \} \subset \mathcal{A}_0 \text{ covers }E \right\}\]

Then the following hold:

1. If $A \in \mathcal{A}_0$, then $\mu^\ast(A) = \rho(A)$.
2. $\sigma(\mathcal{A}_0)$ is measurable with respect to $\mu^\ast$.
3. If $\rho$ is $\sigma$-finite, then $\mu^\ast|_{\sigma(\mathcal{A}_0)}$ is the unique measure that extends $\rho$ to $\sigma(\mathcal{A}_0)$.

Proof.

1. If $A \in \mathcal{A}_0$, then $\mu^\ast(A) = \rho(A)$.

Since $A$ is a cover of $A$, we have $\mu^\ast(A) \leq \rho(A)$. If $\mu^\ast(A) < \rho(A)$, then there exists some cover $\lbrace A_n \rbrace $ of $A$ such that $\sum \rho(A_n) < \rho(A)$. However, since $\rho$ is a premeasure, this is a contradiction. □

2. $\sigma(\mathcal{A}_0)$ is measurable with respect to $\mu^\ast$.

First, we show that $\mathcal{A}_0$ is measurable. For any $A \in \mathcal{A}_0$, it suffices to show $\mu^\ast(E \cap A) + \mu^\ast(E \cap A^c) \leq \mu^\ast(E)$. By the definition of $\mu^\ast$, for any $\epsilon > 0$, there exists some cover $\mathcal{C}$ of $E$ such that:

\[\sum^\infty_{n = 1} \rho(C_n) \leq \mu^*(E) + \epsilon\]

Since $\mathcal{A}_0$ is an algebra, $A \cap C_n, A^c \cap C_n \in \mathcal{A}_0$. Therefore,

\[\begin{aligned} \mu^*(E \cap A) + \mu^*(E \cap A^c) &\leq \sum^\infty_{n = 1} \mu^*(A \cap C_n) + \sum^\infty_{n = 1} \mu^*(A^c \cap C_n) \\ &= \sum^\infty_{n = 1} \rho(A \cap C_n) + \sum^\infty_{n = 1} \rho(A^c \cap C_n) \\ &= \sum^\infty_{n = 1} \rho(C_n) \leq \mu^*(E) + \epsilon \end{aligned}\]

(Strictly speaking, one should consider $\sum^n_{k=1}$ and then take the limit $n \to \infty$) Therefore, $\mu^\ast(E \cap A) + \mu^\ast(E \cap A^c) \leq \mu^\ast(E)$, and $\mathcal{A}_0$ is measurable. By the Carathéodory restriction theorem, measurable sets form a $\sigma$-algebra, so $\sigma(\mathcal{A}_0)$ is also measurable. □

3. If $\rho$ is $\sigma$-finite, then $\mu^\ast|_{\sigma(\mathcal{A}_0)}$ is the unique measure that extends $\rho$ to $\sigma(\mathcal{A}_0)$.

First, assume $\rho < \infty$. Suppose a measure $\nu$ defined on $\sigma(\mathcal{A}_0)$ agrees with $\rho$ on $\mathcal{A}_0$. Let $\mu = \mu^\ast|_{\sigma(\mathcal{A}_0)}$. We show $\nu = \mu$.

Let $E \in \sigma(\mathcal{A}_0)$. There exists some cover $\lbrace A_n \rbrace \subset \mathcal{A}_0$ of $E$ such that:

\[\sum \rho(A_n) \leq \mu(E) + \epsilon\]

Since $\nu$ agrees with $\rho$ on $\mathcal{A}_0$, we have $\sum \rho(A_n) = \sum \nu(A_n) \geq \nu(E)$. Therefore, $\nu \leq \mu$.

Now define $B_n = \bigcup^n_{k=1}A_k$ and let $A = \bigcup^\infty_{n = 1} A_n = \lim_{n \to \infty} B_n$. Since $\mu(B) = \sum \rho(A_n) \leq \mu(E) + \epsilon$, we have $\mu(B \setminus E) \leq \epsilon$. Since $\nu$ is a measure,

\[\mu(A) = \lim \rho(B_n) = \lim \nu(B_n) = \nu(A)\]

Therefore,

\[\begin{aligned} \mu(E) \leq \mu(B) &= \mu(B \setminus E) + \mu(E) \\ &\leq \epsilon + \nu(E) \end{aligned}\]

That is, $\mu \leq \nu$. Therefore, $\mu = \nu$.

Now assume $\rho$ is $\sigma$-finite. There exists a collection $\lbrace C_n \rbrace $ such that $C_n \uparrow X$ and $\rho(C_n) < \infty$. By the preceding argument, $\mu$ and $\nu$ agree on $C_n$. Therefore,

\[\mu(A) = \lim \mu(A \cap C_i) = \lim \nu (A \cap C_i) = \nu(A)\]

so $\mu$ and $\nu$ agree on the entire space. ■

In the next article, we shall construct Lebesgue measure using Carathéodory’s theorem, and then show that whilst every Borel measurable set is Lebesgue measurable, the converse does not hold.