1 What is the Carathéodory Extension Theorem?
The Carathéodory extension theorem is a fundamental result in measure theory which gives the procedure by which any [[pre-measure|pre-measure]] defined on a given [[ring|ring]] (algebra) \(R\) of a given set \(X\) can be extended to a measure on a [[ring|σ-ring]] (σ-algebra) generated by \(R\), and this extension is unique if the pre-measure is \(\sigma\)-finite.
Given and outer measure \(\mu^*\) then 1. \(\mathcal{F}_{\mu^*}\) is a \(\sigma\)-algebra; and 2. \(\mu^*\) is a measure on \(\mathcal{F}_{\mu^*}\).
Note: We can show that this is in general not the largest \(\sigma\)-algebra on which \(\mu^*\) is a measure.
Proof: Since \(\mu^*(\emptyset)=0\implies \emptyset \in \mathcal{F}_{\mu^*}\) by the proposition above, hence \(\mathcal{F}_{\mu^*}\) is non-empty; and by definition of an outer-measurable set \(\mathcal{F}_{\mu^*}\) is closed under compliments. Next, if \(A,B \in \mathcal{F}_{\mu^*}\) and \(E\subset X\) we have that \[ \begin{align} \mu^*(E) & = \mu^*(E \cap A)+\mu^*(E \cap A^c) \\ & = \mu^*(E \cap A \cap B)+\mu^*(E \cap A \cap B^c)+\mu^*(E \cap A^c \cap B)+\mu^*(E \cap A^c \cap B^c) \\ & \geq \mu^*(E \cap(A\cup B))+\mu^*(E \cap(A \cup B)^c), \end{align} \] where the final inequality holds from subadditivity since \(A\cup B=(A \cap B)\cup(A \cap B^c)\cup (A^c\cap B)\). It follows that \(A \cup B\in \mathcal{F}_{\mu^*}\) and so it is an algebra. Moreover, if \(A,B \in \mathcal{F}_{\mu^*}\) and \(A \cap B=\emptyset\) then \[ \mu^*(A \cup B)=\mu^*((A\cup B)\cap A)+\mu^*((A \cup B)\cap A^c)=\mu^*(A)+\mu^*(B), \] hence \(\mu^*\) is finitely additive on \(\mathcal{F}_{\mu^*}\). To prove \(\mathcal{F}_{\mu^*}\) is a \(\sigma\)-algebra it suffices to show \(\mathcal{F}_{\mu^*}\) is closed under countable disjoint unions. If we consider disjoint \(\{ A_{j} \}_{j=1}^\infty \subseteq \mathcal{F}_{\mu^*}\) and define \(B_{n}:=\bigcup_{j=1}^nA_{{j}}\) and \(B :=\bigcup_{j=1}^\infty A_{j}\). Then, for any \(E \subset X\) we have \[ \mu^*(E \cap B_{n})=\mu^*(E \cap B_{n}\cap A_{n})+\mu^*(E \cap B_{n}\cap A_{n}^c)=\mu^*(E \cap A_{n})+\mu^*(E \cap B_{n-1}), \] hence, by induction we have that \(\mu^*(E \cap B_{n})=\sum_{j=1}^n\mu^*(E \cap A_{j})\). Therefore \[ \begin{align} \mu^*(E) & =\mu^*(E \cap B_{n})+\mu^*(E \cap B_{n}^c)\geq \sum_{j=1}^n\mu^*(E\cap A_{j})+\mu^*(E \cap B^c) \\ & \stackrel{n \to \infty}{\geq }\sum_{j=1}^\infty \mu^*(E \cap A_{j})+\mu^*(E \cap B^c)\geq \mu^*\left( \bigcup_{j=1}^\infty (E\cap A_{j}) \right)+\mu^*(E \cap B^c) \\ & = \mu^*(E \cap B)+\mu^*(E\cap B^c)\geq \mu^*(E), \end{align} \] and so all the inequalities become equalities thus \(B\in \mathcal{F}_{\mu^*}\) as required. Furthermore, taking \(E=B\) we have \(\mu^*(B)=\sum_{j=1}^\infty \mu^*(A_{j})\) and so \(\mu^*\) is countable additive on \(\mathcal{F}_{\mu^*}\). Finally, if \(\mu^*(A)=0\), for any \(E \subset X\) \[ \mu^*(E)\leq \mu^*(E \cap A)+\mu^*(E \cap A^c)=\mu^*(E \cap A^c)\leq \mu^*(E), \] so that \(A \in \mathcal{F}_{\mu^*}\), hence \(\mu^*\) is a complete measure on \(\mathcal{F}_{\mu^*}\). \(\square\)