Caratheodory’s Extension Theorem
We already know given a \(\sigma\)-additive \(\mu: \mathscr{S} \to [0,\infty]\), there is an unique extension to an algebra with \(\sigma\)-additive \(\nu: \mathscr{A}(\mathscr{S}) \to [0,\infty]\), with \(\nu(A) = \mu(A) \forall A \in \mathscr{A}(\mathscr{S})\), in fact, the set function is
\[\nu(E) = \sum_{j=1}^{n}\mu(E_j)\]where \(E_j \in \mathscr{S}\) and they are disjoint.
Next we will prove, given a \(\sigma\)-additive \(\nu: \mathscr{A} \to [0,\infty]\), if \(\Omega\) is \(\nu\) \(\sigma\)-finite, then there is an unique extension to an \(\sigma\)-algebra \(\pi: \mathscr{F}(\mathscr{A}) \to [0,\infty]\), with \(\pi(A) = \nu(A) \forall A \in \mathscr{A}\).
We would
first define a set funtion on the whole \(\mathscr{P}(\Omega)\), called \(\pi^{*}: \mathscr{P}(\Omega) \to \mathbb{R}_+ \cup \{\infty\}\).
we can show \(\pi^{*}\) is an outer-measure, although it is not \(\sigma\)-additive.
then we will define a set \(\mathscr{M} \subseteq \mathscr{P}(\Omega)\), such that
it satisfies \(\mathscr{M}\) is an \(\sigma\)-algebra and \(\mathscr{M} \supseteq \mathscr{A}\)
it satisfies \(\pi^* \vert_{\mathscr{M}}\) is \(\sigma\)-additive
it satisfies \(\pi^* \vert_{\mathscr{A}(\mathscr{S})} = \nu\)
does \(\mathscr{M} = \mathscr{F}(\mathscr{A})\)? No, we can see example that \(\mathscr{M}\) is strictly bigger than \(\mathscr{F}(\mathscr{A})\)
Outer Measure
A set function \(\mu: \mathscr{C} \to \mathbb{R}_+ \cup \{\infty\}\) is an outer measure if and only if
(i) \(\mu(\emptyset) = 0\)
(ii) \(E \subseteq F, E, F \in \mathscr{C} \quad \implies \quad \mu(E) \le \mu(F)\)
(iii) \(\mu\) is countable monotone: if \(\{E_j\}_{j=1}^{\infty}\) is a countable covering of \(E\), then \(\mu(E) \le \sum_{j=1}^{\infty} \mu(E_j)\).
Step 1
We define a set function on the whole \(\mathscr{P}(\Omega)\), namely \(\pi^*: \mathscr{P}(\Omega) \to [0,\infty]\)
\[\pi^*(A) = \inf_{\{E_i\}} \sum_{i \ge 1} \nu(E_i)\]where \(\{E_i\}_{i \ge 1}\) is a countable covering of \(A\) in \(\mathscr{A}(\mathscr{S})\), i.e., \(E_i \in \mathscr{A}(\mathscr{S})\) and \(A \subseteq \cup_{i \ge 1} E_i\). And the function \(\pi^*\) is defined as the inf value for all possible covering of \(A\).
Proof: \(\pi^* : \mathscr{P}(\Omega) to [0,1]\) is an outer measure.
Step 2
Define a set \(\mathscr{M}\) (measurable subset). If \(A \in \mathscr{M}\), then \(\forall E \subseteq \Omega, \pi^*(E) = \pi^*(E \cap A) + \pi^*(E \cap A^c)\).
we can prove it satisfies \(\mathscr{M}\) is an \(\sigma\)-algebra and \(\mathscr{M} \supseteq \mathscr{A}(\mathscr{S})\)
Thus \(\mathscr{M} \supseteq \mathscr{F}(\mathscr{S})\).
Step 3
Consider \(\pi^* \vert_{\mathscr{M}}: \mathscr{M} \to [0,\infty]\).
it satisfies \(\pi^* \vert_{\mathscr{M}}\) is \(\sigma\)-additive
it satisfies \(\pi^* \vert_{\mathscr{A}(\mathscr{S})} = \nu\)
Step 4
For any two \(\mu_1, \mu_2: \mathscr{F}(\mathscr{A}) \to [0,\infty]\), if \(\mu_1 \vert_{\mathscr{A}} = \mu_2 \vert_{\mathscr{A}}\), and if \(\Omega\) is \(\sigma\)-finite on \(\mu_1 \vert_{\mathscr{A}} (\text{or } \mu_2 \vert_{\mathscr{A}})\), then
\[\mu_1 = \mu_2\]\(\sigma\)-Finite
If \(\Omega\) is \(\sigma\)-finite on \(\mu: \mathscr{C} \to [0,\infty]\), it means
\[\exists \text{ countable } \{E_j\}_{j=1}^{\infty}, E_j \in \mathscr{C}, E_j \uparrow \Omega, \text{ and } \mu(E_j) < \infty \forall j < \infty\]Monotone Class
Define, \(\mathscr{G} \subseteq \mathscr{P}(\Omega)\) is a monotone class if and only of
(i) \(A_j \in \mathscr{G} \forall j \ge 1\) and \(A_j \subseteq A_{j+1} \quad \implies \quad \cup_{j \ge 1}A_j \in \mathscr{G}\).
(ii) \(B_j \in \mathscr{G} \forall j \ge 1\) and \(B_j \supseteq B_{j+1} \quad \implies \quad \cap_{j \ge 1}B_j \in \mathscr{G}\).
Observations:
\(\mathscr{G}_{\alpha}\) is monotone class, where \(\alpha \in I\) and \(I\) is any index set. Then \(\cap_{\alpha \in I} \mathscr{G}_{\alpha}\) is a monotone class.
So we can talk about the smallest monotone class \(\mathscr{M}(\mathscr{C})\) generated by a class \(\mathscr{C}\) (contain \(\mathscr{C}\)).
Lemma
Given algebra \(\mathscr{A} \subseteq \mathscr{P}(\Omega)\), then
\[\mathscr{M}(\mathscr{A}) = \mathscr{F}(\mathscr{A})\]Summary
Given \(\nu: \mathscr{A} \to [0,\infty]\) and \(\Omega\) is \(\nu\) \(\sigma\)-finite. Then there is an unique extantion \(\pi: \mathscr{F}(\mathscr{A}): [0,\infty]\) with \(\pi(A) = \nu(A) \forall A \in \mathscr{A}\). We define
\[\begin{align} \pi^*&: \mathscr{P}(\Omega) \to [0,\infty]\\ \pi^*(A) &= \inf_{\{E_i\}} \sum_{i \ge 1} \nu(E_i) \end{align}\]where \(\{E_i\}\) is covering of \(A\). Then
\[\pi = \pi^* \vert_{\mathscr{F}(\mathscr{A})}\]