17.1 Algebras

Definition 17.1.1 (Algebra).label Let $X$ be a set and $\alg \subset 2^{X}$, then $\alg$ is an algebra if:

  1. (A1)

    $\emptyset, X \in \alg$.

  2. (A2)

    For any $A \in \alg$, $A^{c} \in \alg$.

  3. (A3)

    For any $A, B \in \alg$, $A \cup B \in \alg$.

Definition 17.1.2 (Ring).label Let $X$ be a set and $\alg \subset 2^{X}$, then $\alg$ is a ring if:

  1. (A1)

    $\emptyset \in \alg$.

  2. (A2’)

    For any $E, F \in \alg$ with $E \subset F$, $F \setminus E \in \alg$.

  3. (A3)

    For any $A, B \in \alg$, $A \cup B \in \alg$.

Definition 17.1.3 ($\sigma$-Algebra).label Let $X$ be a set and $\cm \subset 2^{X}$, then $\cm$ is a $\sigma$-algebra if:

  1. (A1)

    $\emptyset, X \in \cm$.

  2. (A2)

    For any $A \in \cm$, $A^{c} \in \cm$.

  3. (A3’)

    For any $\seq{A_n}\in \cm$, $\bigcup_{n \in \nat^+}A_{n} \in \cm$.

Proposition 17.1.4.label Let $X$ be a set and $\alg \subset 2^{X}$ be an algebra, then:

  1. (1)

    For any $A, B \in \alg$, $A \cap B \in \alg$.

  2. (2)

    For any $A, B \in \alg$, $A \setminus B \in \alg$.

If $\alg$ is a $\sigma$-algebra, then:

  1. (1’)

    For any $\seq{A_n}\in \alg$, $\bigcap_{n \in \natp}A_{n} \in \alg$.

Proof. (1, 1’): $\bigcap_{i \in I}A_{i} = \braks{\bigcup_{i \in I}A_i^c}^{c}$.

(2): $A \setminus B = A \cap B^{c}$.$\square$

Lemma 17.1.5.label Let $X$ be a set and $\alg \subset 2^{X}$ be an algebra, then the following are equivalent:

  1. (1)

    For any $\seq{A_n}\subset \alg$, $\bigcup_{n \in \nat^+}A_{n} \in \alg$.

  2. (2)

    For any $\seq{A_n}\subset \alg$ with $A_{n} \subset A_{n+1}$ for all $n \in \natp$, $\bigcup_{n \in \natp}A_{n} \in \alg$.

  3. (3)

    For any $\seq{A_n}\subset \alg$ pairwise disjoint, $\bigsqcup_{n \in \natp}A_{n} \in \alg$.

Proof. (2) $\Rightarrow$ (1): For each $n \in \nat$, let $B_{n} = \bigcup_{k = 1}^{n} A_{n} \in \alg$, then $\bigcup_{n \in \natp}A_{n} = \bigcup_{n \in \nat}B_{n} \in \alg$.

(3) $\Rightarrow$ (2): Denote $A_{0} = \emptyset$ and $B_{n} = A_{n} \setminus A_{n-1}$, then $\seq{B_n}\subset \alg$ is pairwise disjoint and $\bigcup_{n \in \nat}A_{n} = \bigsqcup_{n \in \nat}B_{n} \in \alg$.$\square$

Definition 17.1.6 (Generated $\sigma$-Algebra).label Let $X$ be a set and $\ce \subset 2^{X}$, then the $\sigma$-algebra $\sigma(\ce)$ generated by $\ce$ is the smallest $\sigma$-algebra on $X$ containing $\ce$.

Definition 17.1.7 (Induced $\sigma$-Algebra).label Let $X$ be a set, $\cm \subset 2^{X}$ be a $\sigma$-algebra over $X$, and $E \subset X$, then the collection

\[\cm_{E} = \bracs{A \cap E|A \in \cm}\]

is the $\sigma$-algebra on $E$ induced by $\cm$.