Classes of Subsets
Semi-Algebras
Definition: \(\mathscr{S} \subseteq \mathscr{P}(\Omega)\) is a semi-algebra, if and only if:
(i) \(\Omega \in \mathscr{S}\)
(ii) \(A, B \in \mathscr{S} \implies A \cap B \in \mathscr{S}\)
(iii) \(\forall A \in \mathscr{S}, \exists E_1, \dots, E_h \in \mathscr{S}\), such that \(A^c = \sum_{j=1}^h E_j\).
Example: \(\Omega = \mathbb{R}\). Then semi-algebra \(\mathscr{S}\) includes
\[\begin{align} &\mathbb{R}, \emptyset\\ & \{(a,b], a < b, a,b \in \mathbb{R}\}\\ & \{(-\infty, b], b \in \mathbb{R}\}\\ & \{(a, \infty), a \in \mathbb{R}\} \end{align}\]Example: \(\Omega = (0,1)\), the semi-algebra \(\mathscr{S}\) includes
\[\begin{align} &(0,1), \emptyset, \{(a,b], 0 \le a < b < 1\} \end{align}\]Algebras
Definition: \(\mathscr{A} \in \mathscr{P}(\Omega)\) is an algebra if and only if
(i) \(\Omega \in \mathscr{A}\)
(ii) \(A,B \in \mathscr{A} \implies A \cap B \in \mathscr{A}\)
(iii) \(A \in \mathscr{A} \implies A^c \in \mathscr{A}\)
Observation:
If \(\mathscr{A}\) is an algebra, then \(\mathscr{A}\) is semi-algebra. [assertion 01].
\(A, B \in \mathscr{A} \implies A \cup B \in \mathscr{A}\), since \(A \cup B = (A^c \cap B^c)^c\).
If \(\mathscr{A}_{\alpha} \subseteq \mathscr{P}(\Omega)\) is algebra for each \(\alpha \in I\), where \(I\) can be any index set (it can be uncountable). Then \(\cap_{\alpha \in I} \mathscr{A}_{\alpha}\) is an algebra. [assertion 02]
Algebra Generated By The Class \(\mathscr{C}\)
The algebra generabted by \(\mathscr{C} \subseteq \mathscr{P}(\Omega)\) is defined as \(\mathscr{A}(\mathscr{C})\), such that
(i) \(\mathscr{C} \subseteq \mathscr{A}(\mathscr{C})\)
(ii) For any algebra \(\mathscr{B}, \mathscr{B} \supseteq \mathscr{C} \implies \mathscr{B} \supseteq \mathscr{A}(\mathscr{C})\), i.e., \(\mathscr{A}(\mathscr{C})\) is the smallest algebra containing \(\mathscr{C}\).
\[\cap_{\alpha} \mathscr{A}_{\alpha}\]TODO: question, this is similar to infimum, but can we be sure the infimum exists like in the real number? For example, we take all the interval such that there exists element \(a < 0\) and \(b > 1\). Then does \(\cap_{\alpha} A_{\alpha}\) exists?
\(a \in A_{\alpha} \forall \alpha \quad \implies \quad a \in \cap_{\alpha} A_{\alpha}\)
\(\exists \alpha, a \notin A_{\alpha}, \quad \implies a \notin \cap_{\alpha} A_{\alpha}\).
Godel’s incompleteness theorems.
Lemma: \(\mathscr{S} \subseteq \mathscr{P}(\Omega)\) is a semi-algebra. Then
\[A \in \mathscr{A}(\mathscr{S}) \quad \iff \quad \exists \text{ disjoint } E_j \in \mathscr{S}, 1 \le j \le n, \text{ such that } A = \sum_{j=1}^{n} E_j\]or it is all the finite unions of disjoint sets.
\[\mathscr{A}(\mathscr{S}) = \{ \sum_{j=1}^n E_j, \forall \text{ disjoint } E_j \in \mathscr{S}, \forall n = 1, 2, 3, \dots\}\]\(\sigma\)-Algebras
Definition: \(\mathscr{F} \subseteq \mathscr{P}(\Omega)\) is \(\sigma\)-algebra if and only if
(i) \(\Omega \in \mathscr{F}\)
(ii) \(A_j \in \mathscr{F} \implies \cap_{j\ge 1} A_j \in \mathscr{F}\)
(iii) \(A \in \mathscr{F} \implies A^c \in \mathscr{F}\)
Observation:
If \(\mathscr{F}\) is a \(\sigma\)-algebra, then \(\mathscr{F}\) is algebra, \(\mathscr{F}\) is semi-algebra.
\(A_j \in \mathscr{F} \implies \cup_{j \ge 1} A_j \in \mathscr{F}\), since \(\cup_{j \ge 1} A_j = (\cap_{j \ge 1} A_j^c)^c\)
If \(\mathscr{F}_{\alpha} \subseteq \mathscr{P}(\Omega)\) is \(\sigma\)-algebra for each \(\alpha \in I\), where \(I\) can be any index set (it can be uncountable). Then \(\cap_{\alpha \in I} \mathscr{F}_{\alpha}\) is a \(\sigma\)-algebra.
\(\sigma\)-Algebra Generated By The Class \(\mathscr{C}\)
\(\mu\) Function
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 \(\mu(E) = \sum_{j=1}^n \mu(E_j)\).
\(\sigma\)-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 \(\mu(E) = \sum_{j\ge 1} \mu(E_j)\).
Example: Additive But Not \(\sigma\)-Additive
Let \(\Omega = (0,1), \mathscr{C} = {(a,b], 0 \le a < b < 1}\)
\[\mu((a,b]) = \begin{cases} \infty, & a = 0 \\ b-a, & a > 0 \end{cases}\]Proofs
- Assertion 01: If \(\mathscr{A}\) is an algebra, then \(\mathscr{A}\) is semi-algebra.
Proof: We can verify \(\mathscr{A}\) is semi-algebra.
(i) \(\Omega \in \mathscr{A}\)
(ii) \(A, B \in \mathscr{A} \implies A \cap B \in \mathscr{A}\)
(iii) \(\forall A \in \mathscr{A}, A^c = A^c \in \mathscr{A}\)
- Assertion 02: If \(\mathscr{A}_{\alpha} \subseteq \mathscr{P}(\Omega)\) is algebra for each \(\alpha \in I\), where \(I\) is any general index set. Then \(\cap_{\alpha \in I} \mathscr{A}_{\alpha}\) is an algebra.
Proof: We can verify \(\cap_{\alpha} \mathscr{A}_{\alpha}\) is an algebra
(i) \(\mathscr{A}_{\alpha}\) is an algebra \(\implies \Omega \in \mathscr{A}_{\alpha} \implies \Omega \in \cap_{\alpha} \mathscr{A}_{\alpha}\)
(ii) for any \(A, B \in \cap_{\alpha} \mathscr{A}_{\alpha}\), we know \(A, B \in \mathscr{A}_{\alpha} \forall \alpha\), then \(A \cap B \in \mathscr{A}_{\alpha} \forall \alpha\), then \(A \cap B \in \cap_{\alpha} \mathscr{A}_{\alpha}\)
(iii) similar to above \(A \in \cap_{\alpha} \mathscr{A}_{\alpha} \implies A^c \in \cap_{\alpha} \mathscr{A}_{\alpha}\)