Set Functions
Additive
Given \(\mathscr{C} \subseteq \mathscr{P}(\Omega)\) and \(\emptyset \in \mathscr{C}\). The function \(\mu: \mathscr{C} \to \mathbb{R}_+ \cup \{\infty\}\) is additive, if and only if
(i) \(\mu(\emptyset) = 0\)
(ii) if disjoint \(E_1, E_2, \dots, E_n \in \mathscr{C}\), and \(E = \sum_{j=1}^{n}E_j \in \mathscr{C}\), then \(\mu(E) = \sum_{j=1}^n \mu(E_j)\).
We can natually define a set function on the semi-algebra like \((a,b]\). We want to extend the set function to algebra and \(\sigma\)-algebra.
\(\sigma\)-Additive
Given \(\mathscr{C} \subseteq \mathscr{P}(\Omega)\) and \(\emptyset \in \mathscr{C}\). The function \(\mu: \mathscr{C} \to \mathbb{R}_+ \cup \{\infty\}\) is \(\sigma\)-additive, if and only if
(i) \(\mu(\emptyset) = 0\)
(ii) if disjoint \(E_1, E_2, \dots, E_j, \dots \in \mathscr{C}\), and \(E = \sum_{j\ge 1}E_j \in \mathscr{C}\), then \(\mu(E) = \sum_{j\ge 1} \mu(E_j)\).
Cover
Set \(E\) is covered by a collection \(\{E_k\}_{k=1}^{n}\) means
\[E \subseteq \cup_{k=1}^{n} E_k\]Finite Monotone
A set function \(\mu: \mathscr{C} \to [0,\infty]\) is said to be finite monotone provided that, whenever a set \(E \in \mathscr{C}\) is covered by a finite collection \(\{E_k\}_{k=1}^{n}\) of sets in \(\mathscr{C}\)
\[\mu(E) \le \sum_{k=1}^{n}\mu(E_k)\]Countable Monotone (subadditivity)
A set function \(\mu: \mathscr{C} \to [0,\infty]\) is said to be countable monotone provided that, whenever a set \(E \in \mathscr{C}\) is covered by a countable collection \(\{E_k\}_{k=1}^{\infty}\) of sets in \(\mathscr{C}\)
\[\mu(E) \le \sum_{k=1}^{\infty}\mu(E_k)\]Continuous Measure
Given \(\mathscr{C} \subseteq \mathscr{P}(\Omega)\), and \(\mu: \mathscr{C} \to \mathbb{R}_+ \cup \{\infty\}\).
for \(E \in \mathscr{C}\), \(\mu\) continuous from below at \(E\), if and only if \(\forall (E_n)_{n \ge 1}, E_n \in \mathscr{C}, E_n \uparrow E\), then \(\mu(E_n) \uparrow \mu(E)\).
for \(E \in \mathscr{C}\), \(\mu\) continuous from above at \(E\), if and only if \(\forall (E_n)_{n \ge 1}, E_n \in \mathscr{C}, E_n \downarrow E\) and \(\exists n_0, \mu(E_{n_0}) < \infty\), then \(\mu(E_n) \downarrow \mu(E)\).
- the need of \(\mu(E_{n_0})\), consider the example of \(E_n = (n, \infty)\), then \(E_n \downarrow \emptyset\).
Lemma: \(\mathscr{A} \subseteq \mathscr{P}(\Omega)\) is an algebra, then for an additive \(\mu: \mathscr{A} \to \mathbb{R}_+ \cup \{\infty\}\).
\(\mu\) is \(\sigma\)-additive \(\quad \implies \quad\) \(\mu\) continuous at \(E , \forall E \in \mathscr{A}\).
\(\mu\) is continuous from below \(\quad \implies \quad\) \(\mu\) is \(\sigma\)-additive.
\(\mu\) is continuous from above at \(\emptyset\) and \(\mu\) is finite (\(\mu(\Omega) < \infty\)) \(\quad \implies \quad\) \(\mu\) is \(\sigma\)-additive.
Measure Extantion
There exists only one unique extantion from \(\mathscr{S}\) to \(\mathscr{A}(\mathscr{S})\).
Theorem 1: \(\mathscr{S} \subseteq \mathscr{P}(\Omega)\) is a semi-algebra, \(\mu: \mathscr{S} \to \mathbb{R}_+ \cup \{\infty\}\) is additive. Then there exists \(\nu: \mathscr{A}(\mathscr{S}) \to \mathbb{R}_+ \cup \{\infty\}\) is additive, and
(a) \(\nu(A) = \mu(A) \forall A \in \mathscr{S}\)
(b) \(\nu_1(A) = \nu_2(A) \forall A \in \mathscr{S} \quad \implies \nu_1(E) = \nu_2(E) \forall E \in \mathscr{A}(\mathscr{\mathscr{S}})\)
Theorem 2: \(\mathscr{S} \subseteq \mathscr{P}(\Omega)\) is a semi-algebra, \(\mu: \mathscr{S} \to \mathbb{R}_+ \cup \{\infty\}\) is \(\sigma\)-additive. Then there exists \(\nu: \mathscr{A}(\mathscr{S}) \to \mathbb{R}_+ \cup \{\infty\}\) is \(\sigma\)-additive, and
(a) \(\nu(A) = \mu(A) \forall A \in \mathscr{S}\)
(b) \(\nu_1(A) = \nu_2(A) \forall A \in \mathscr{S} \quad \implies \nu_1(E) = \nu_2(E) \forall E \in \mathscr{A}(\mathscr{\mathscr{S}})\)
Using \(\mathscr{A}(\mathscr{S}) = \{ \sum_{j=1}^n E_j, \forall \text{ disjoint } E_j \in \mathscr{S}, \forall 1 \le n < \infty\}\), the set function extention is in fact
\[\nu(E) = \sum_{j=1}^n \mu(E_j)\]