From Ring to $\sigma$-Algebra: Carathéodory's Theorem

As mentioned in the previous section, the question is whether a $\sigma$-additive pre-measure on a ring $\mathcal{R}$ can be extended to the $\sigma$-algebra it generates. In case such an extension exists, a further relevant question is whether such an extension is unique.

Before assessing the extension, we first address some central properties of the $\sigma$-additivity in terms of continuity.

Lemma

Let $\mathcal{R}$ be a ring and $\mu :\mathcal{R}\to [0,\mathrm{\infty }]$ a finite content, that is $\mu [A]<\mathrm{\infty }$ for every $A\in \mathcal{R}$. Then the following properties of $\mu$ hold:

1. $\sigma$-additivity
2. $\sigma$-sub-additivity
3. Lower semi-continuity: $\underset{n}{sup}\mu [A_n]=\mu [\cup A_n]$ for every countable family $(A_n)$ of increasing elements in $\mathcal{R}$ such that $\cup A_n$ is in $\mathcal{R}$.
4. Upper semi-continuity: $\underset{n}{inf}\mu [A_n]=\mu [\cap A_n]$ for every countable family $(A_n)$ of decreasing elements in $\mathcal{R}$ such that $\cap A_n$ is in $\mathcal{R}$.
5. Continuous at $\mathrm{\varnothing }$: $\underset{n}{inf}\mu [A_n]=0$ for every countable family $(A_n)$ of decreasing elements in $\mathcal{R}$ such that $\cap A_n=\mathrm{\varnothing }$.

Proof

• The equivalence between $\sigma$-additivity and $\sigma$-sub-additivity is already shown in a previous Lemma in the context of a semi-ring.

• $\sigma$-additivity implies lower semi-continuity: Let $(A_n)$ be an increasing sequence of elements in $\mathcal{R}$ such that $A=\cup A_n$ is in $\mathcal{R}$.
  Defining $B_n=A_n\setminus {\cup }_{k<n}{A}_k=A_n\setminus B_{n-1}$ for $n>1$ and $B_1=A_1$ provides a disjoint sequence of elements in $\mathcal{R}$.
  Since $A_n={\cup }_{1\le k\le n}{B}_k$ and $A=\cup B_n$, it follows from $\sigma$-additivity that

  $$\mu [A]=\sum \mu [B_n]=sup\sum _{1\le k\le n}\mu [B_k]=sup\mu [{\cup }_{1\le k\le n}{B}_k]=sup\mu [A_n]$$

• Lower semi-continuity implies upper semi-continuity:
  Let $(A_n)$ be a decreasing sequence of elements in $\mathcal{R}$ such that $A=\cap A_n$ is in $\mathcal{R}$. It follows that $B_n=A_1\setminus A_n$ defines an increasing sequence such that $B=\cup B_n=A_1\setminus \cap A_n=A_1\setminus A\in \mathcal{R}$. Lower semi-continuity, additivity, and the properties of Lemma [lem-propcontentsemiring] imply that

  $$\mu [A_1]-inf\mu [A_n]=sup(\mu [A_1]-\mu [A_n])=sup\mu [A_1\setminus A_n]=\mu [\cup A_1\setminus A_n]=\mu [A_1]-\mu [A]$$

• Upper semi-continuity implies continuity at $\mathrm{\varnothing }$ (immediate).

• Continuity at $\mathrm{\varnothing }$ implies $\sigma$-additivity: Let $(A_n)$ be a sequence of mutually disjoint elements of $\mathcal{R}$ such that $A=\cup A_n$ is in $\mathcal{R}$. By means of Lemma, it follows that

  $$\sum \mu [A_n]\le \mu [A]$$

  It follows that $B_n=A\setminus ({\cup }_{1\le k\le n}{A}_k)$ is a decreasing family of elements of $\mathcal{R}$ such that $\cap B_n=\mathrm{\varnothing }$. By additivity of $\mu$, we have

  $$\mu [A]=\mu [B_n\cup ({\cup }_{1\le k\le n}{A}_k)]=\mu [B_n]+\sum _{1\le k\le n}\mu [A_k]$$

  for every $n$. Since $\sum _{q\le k\le n}\mu [A_k]\to \sum \mu [A_n]$ and by continuity at $\mathrm{\varnothing }$, $\mu [B_n]\to 0$, it follows that $\mu [A]=\sum \mu [A_n]$, hence the $\sigma$-additivity.

We are now in position to formulate the central extension theorem of Carathéodory.

Theorem: Carathéodory's Extension Theorem

Let $X$ be a non-empty set, $\mathcal{S}$ a semi-ring such that $X=\cup A_n$ for some countable family $(A_n)$ of elements in $\mathcal{S}$. Suppose that $\mu :\mathcal{R}\to [0,\mathrm{\infty }]$ is a pre-measure such that:

1. $\mu [A_n]<\mathrm{\infty }$ for every $n$.
2. $\mu$ is $\sigma$-sub-additive or equivalently $\sigma$-additive.

Then $\mu$ can be uniquely extended to a measure $\nu$ on $\mathcal{F}=\sigma (\mathcal{S})$.

Remark

We already know from the assumption of this theorem that according to a previous proposition there exists a unique extension $\nu$ to the ring $\mathcal{R}$ generated by $\mathcal{S}$ such that $\nu$ is a $\sigma$-additive content. We can therefore assume for the proof of Carathéodory's theorem that $\mu$ is a $\sigma$-additive content on a ring.

The existence is the complex and lengthy part of the proof. However uniqueness can be adressed through the following proposition.

Proposition

Let $(X,\mathcal{F})$ be a measurable space and $\mathcal{P}$ a $\pi$-system on $X$ that generates $\mathcal{F}$ and such that there exists a sequence $(A_n)$ of elements of $\mathcal{P}$ with $\cup A_n=X$. Let $\mu$ and $\nu$ be two measures on $\mathcal{F}$ that coincide on $\mathcal{P}$ and such that $\mu (A_n)=\nu (A_n)<\mathrm{\infty }$ for every $n$. Then $\mu =\nu$.

Proof

Let $n$ be an integer and consider the collection of sets ${\mathcal{C}}^n=\{A\in \mathcal{F}:\mu [A\cap A_n]=\nu [A\cap A_n]\}$. Then ${\mathcal{C}}^n$ is a $\lambda$-system. Indeed

• $X$ is in ${\mathcal{C}}^n$ since $\mu [X\cap A_n]=\mu [A_n]=\nu [A_n]=\nu [A_n\cap X]$.
• Stability under relative complements: For $A\subseteq B$ with $A$ and $B$ in ${\mathcal{C}}^n$, it holds that $\mu [(B\setminus A)\cap A_n]=\mu [B\cap A_n]-\mu [A\cap A_n]=\nu [B\cap A_n]-\nu [A\cap A_n]=\nu [(B\setminus A)\cap A_n]$, showing that $B\setminus A$ belongs to ${\mathcal{C}}^n$.

• Stability under countable disjoint unions: If $(B_m)$ is a disjoint sequence in ${\mathcal{C}}^n$, then

  $$\mu [(\cup B_m)\cap A_n]=\sum _m\mu [B_m\cap A_n]=\sum _m\nu [B_m\cap A_n]=\nu [(\cup B_m)\cap A_n]$$

  showing that $\cup B_m\in {\mathcal{C}}^n$.

Since $\mathcal{P}\subseteq {\mathcal{C}}^n$, and $\mathcal{P}$ is stable under intersection, by means of the Dynkin Lemma, it follows that $\mathcal{F}=\sigma (\mathcal{P})\subseteq {\mathcal{C}}^n\subseteq \mathcal{F}$. Thus, $\mu =\nu$.

Example: Lebesgue/Stieljes Measure

The conditions of Carathéodory's Theorem are fulfilled in the case of the Lebesgue/Stieltjes measure. Therefore, it defines a unique measure on the Borel $\sigma$-algebra of $\mathbb{R}$.

We are left to show Carathéodory's extension theorem. The proof relies on the concept of outer measure. So far we tried to built measures from the bottom up, from simple class of sets and trying to reach $\sigma$-algebra. The idea of outer measure is a top down approach hopping that both concepts would meet at the right place, that is, the $\sigma$-algebra.

Outer Measure

Definition: Outer Measure

A set function ${\mu }^{\ast }:2^X\to [0,\mathrm{\infty }]$ is called an outer measure if

• ${\mu }^{\ast }[\mathrm{\varnothing }]=0$;

• $\sigma$-sub-additivity: For any countable family $(B_n)$ of subsets of $X$ and $A\subseteq \cup B_n$ it holds

  $${\mu }^{\ast }[A]\le \sum {\mu }^{\ast }[B_n]$$

A set $A\subseteq X$ is called ${\mu }^{\ast }$-measurable if

$${\mu }^{\ast }[E]={\mu }^{\ast }[E\cap A]+{\mu }^{\ast }[E\cap A^c]$$

for every $E\subseteq X$.

Note that the second condition implies that an outer measure is monotone, that is $\mu [A]\le \mu [B]$ for any $A\subseteq B$. The notion of ${\mu }^{\ast }$-measurability of a set depends naturally of the outer measure. Clearly, due to subadditivity, for any subsets $E$ and $A$ of $X$ it holds

$${\mu }^{\ast }[E]\le {\mu }^{\ast }[E\cap A]+{\mu }^{\ast }[E\cap A^c]$$

Hence, $A$ is ${\mu }^{\ast }$-measurable if the reverse inequality holds for any set $E\subseteq X$. The denomination measurable is justified by the following theorem.

Theorem

Let ${\mu }^{\ast }$ be an outer measure on $X$. Then, the collection $\mathcal{M}$ of all ${\mu }^{\ast }$-measurable is a $\sigma$-algebra. Furthermore, the restriction of ${\mu }^{\ast }$ on this collection is a measure.

Proof

We show that $\mathcal{M}$ is a $\sigma$-algebra.

Step 1 - $\mathrm{\varnothing }$ belongs to $\mathcal{M}$: Clearly, $\mathrm{\varnothing }$ is obviously ${\mu }^{\ast }$ measurable.

Step 2 - Stability under complementation: For $A$ in $\mathcal{M}$, then for any $E\subseteq X$ it holds that

$${\mu }^{\ast }(E)={\mu }^{\ast }(E\cap A)+{\mu }^{\ast }(E\cap A^c).$$

Interchanging $A$ and $A^c$ yields $A^c$ belongs to $\mathcal{M}$.

Step 3 - Stability under union: Let $A$ and $B$ be two ${\mu }^{\ast }$-measurable sets. For any set $E$ it holds

$${\mu }^{\ast }[E]={\mu }^{\ast }[E\cap A]+{\mu }^{\ast }[E\cap A^c]$$

In particular for $B$ it holds

$${\mu }^{\ast }[E\cap A^c]={\mu }^{\ast }[(E\cap A^c)\cap B]+{\mu }^{\ast }[(E\cap A^c)\cap B^c]$$

Since by sub-additivity it holds that ${\mu }^{\ast }[E\cap (A\cup B)]\le {\mu }^{\ast }[E\cap A]+{\mu }^{\ast }[E\cap (A^c\cap B)]$. Hence, it holds that

$$\begin{array}{rl}{\mu }^{\ast }[E]& ={\mu }^{\ast }(E\cap A)+{\mu }^{\ast }[(E\cap A^c)\cap B]+{\mu }^{\ast }[(E\cap A^c)\cap B^c]\\ & ={\mu }^{\ast }(E\cap A)+{\mu }^{\ast }[(E\cap A^c)\cap B]+{\mu }^{\ast }[(E\cap (A\cup B{)}^c]\\ & \ge {\mu }^{\ast }[E\cap (A\cup B)]+{\mu }^{\ast }[E\cap (A\cup B{)}^c]\end{array}$$

showing that $A\cup B$ is ${\mu }^{\ast }$-measurable.

In particular $\mathcal{M}$ is a ring.

Step 4 - stability under countable union: Since $\mathcal{M}$ is a ring, it is enough to show that $\cup A_n$ is ${\mu }^{\ast }$-measurable for any countable family $(A_n)$ of pairwise disjoints ${\mu }^{\ast }$-measurable sets. Let $B_n={\cup }_{k\le n}{A}_k$ which is measurable for each $n$. By induction, it holds that ${\mu }^{\ast }[E\cap B_n]=\sum _{k\le n}{\mu }^{\ast }[E\cap A_k]$ for any set $E\subseteq X$. Indeed $B_n$ is ${\mu }^{\ast }$-measurable and ${\mu }^{\ast }[E\cap B_{n+1}]={\mu }^{\ast }[E\cap B_{n+1}\cap B_n]+{\mu }^{\ast }[E\cap B_{n+1}\cap B_n^c]={\mu }^{\ast }[E\cap B_n]+{\mu }^{\ast }[E\cap A_{n+1}]$. By monotonicity, it also holds that

$${\mu }^{\ast }[E\cap A]\ge \underset{n}{sup}{\mu }^{\ast }[E\cap B_n]=\sum {\mu }^{\ast }[E\cap A_k]$$

which together with sub-additivity yields ${\mu }^{\ast }[E\cap A]=\sum {\mu }^{\ast }[E\cap A_n]$. Hence, for any $E$ we get

$${\mu }^{\ast }[E]={\mu }^{\ast }[E\cap B_n]+{\mu }^{\ast }[E\cap B_n^c]\ge \sum _{k\le n}{\mu }^{\ast }[E\cap B_k]+{\mu }^{\ast }[E\cap A^c]$$

from which follows that ${\mu }^{\ast }[E]\ge {\mu }^{\ast }[E\cap A]+{\mu }^{\ast }[E\cap A^c]$ showing that $A=\cup A_n$ is ${\mu }^{\ast }$-measurable. Thus $\mathcal{M}$ is a $\sigma$-Algebra.

The fact that ${\mu }^{\ast }$ restricted to $\mathcal{M}$ is immediate since it is $\sigma$-sub-additive.

Carathéodory's Extension Theorem Proof

We are now in position to show the proof of Carathéodory's extension theorem.

Proof: Carathéodory's Extension Theorem

From the previous results, we know that we can extend uniquely $\mu$ to a $\sigma$-additive and finite measure to the ring $\mathcal{R}$ generated by $\mathcal{S}$.

We define the set function

$${\mu }^{\ast }[E]:=inf\{\sum {\mu }^{\ast }[A_n]:E\subseteq \cup A_n:(A_n)\text{ is a countable family in }\mathcal{R}\}$$

By definition, ${\mu }^{\ast }$ coincides with $\mu$ on $\mathcal{R}$. Let us show that ${\mu }^{\ast }$ is an outer measure. Clearly, ${\mu }^{\ast }[\mathrm{\varnothing }]=0$. Let us show that ${\mu }^{\ast }$ is $\sigma$-sub-additive by taking a countable family $(E_n)$ and $E$ of sets such that $E\subseteq \cup E_n$. Since if ${\mu }^{\ast }[E_n]=\mathrm{\infty }$ for some $n$, then trivialy it holds that ${\mu }^{\ast }[E]\le \sum {\mu }^{\ast }[E_n]$, we can then assume that ${\mu }^{\ast }[E_n]<\mathrm{\infty }$ for any $n$. Choose families $(A_{nk})$ in $\mathcal{R}$ such that $E_n{\subseteq }_k{A}_{nk}$ and $\sum _k{\mu }^{\ast }[A_{nk}]\le {\mu }^{\ast }[E_n]+\epsilon 2^{-k}$. It follows that $E\subseteq {\cup }_{nk}{A}_{nk}$ and

$${\mu }^{\ast }[E]\le \sum _{nk}{\mu }^{\ast }[A_{nk}]\le \sum {\mu }^{\ast }[E_n]+\epsilon$$

which by arbitrariness of $\epsilon$ show $\sigma$-subadditivity.

From the previous Theorem on outer measure, it follows that the set $\mathcal{M}$ of ${\mu }^{\ast }$-measurable sets is a $\sigma$-algebra on which ${\mu }^{\ast }$ is a measure. Since any element of $\mathcal{R}$ is ${\mu }^{\ast }$-measurable, it follows that $\sigma (\mathcal{R})\subseteq \mathcal{M}$.

Hence ${\mu }^{\ast }$ defines a measure on $\sigma (\mathcal{R})$ coinciding with $\mu$ on $\mathcal{R}$ which ends the proof.

Example: Lebesgue/Stieljes measure

From the previous derivation together with this extension theorem, we conclude that the Lebesgue/Stieljes measure is indeed a measure.