13.1 Algebras
Definition 13.1.1 (Algebra). Let $X$ be a set and $\alg \subset 2^{X}$, then $\alg$ is an algebra if:
$\emptyset, X \in \alg$.
For any $A \in \alg$, $A^{c} \in \alg$.
For any $A, B \in \alg$, $A \cup B \in \alg$.
Definition 13.1.2 (Ring). Let $X$ be a set and $\alg \subset 2^{X}$, then $\alg$ is a ring if:
$\emptyset \in \alg$.
For any $E, F \in \alg$ with $E \subset F$, $F \setminus E \in \alg$.
For any $A, B \in \alg$, $A \cup B \in \alg$.
Definition 13.1.3 ($\sigma$-Algebra). Let $X$ be a set and $\cm \subset 2^{X}$, then $\cm$ is a $\sigma$-algebra if:
$\emptyset, X \in \cm$.
For any $A \in \cm$, $A^{c} \in \cm$.
For any $\seq{A_n}\in \cm$, $\bigcup_{n \in \nat^+}A_{n} \in \cm$.
Proposition 13.1.4. Let $X$ be a set and $\alg \subset 2^{X}$ be an algebra, then:
For any $A, B \in \alg$, $A \cap B \in \alg$.
For any $A, B \in \alg$, $A \setminus B \in \alg$.
If $\alg$ is a $\sigma$-algebra, then:
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 13.1.5. Let $X$ be a set and $\alg \subset 2^{X}$ be an algebra, then the following are equivalent:
For any $\seq{A_n}\subset \alg$, $\bigcup_{n \in \nat^+}A_{n} \in \alg$.
For any $\seq{A_n}\subset \alg$ with $A_{n} \subset A_{n+1}$ for all $n \in \natp$, $\bigcup_{n \in \natp}A_{n} \in \alg$.
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 13.1.6 (Generated $\sigma$-Algebra). 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 13.1.7 (Borel $\sigma$-Algebra). Let $(X, \topo)$ be a topological space, then the Borel $\sigma$-algebra $\cb_{X}$ on $X$ is the $\sigma$-algebra generated by $\topo$.
Definition 13.1.8 (Borel $\sigma$-Algebra on $\ol{\real}$). The family
is the Borel $\sigma$-algebra on $\ol{\real}$.
Proposition 13.1.9. The following families of sets generate the Borel $\sigma$-algebra on $\real$:
$\bracs{(-\infty, a]| a \in \real}$.
$\bracs{(a, \infty)|a \in \real}$.
$\bracs{[a, \infty)| a \in \real}$.
$\bracs{(-\infty, a)| a \in \real}$.
$\bracs{[a, b)| -\infty < a < b < \infty}$.
$\bracs{[a, b]| -\infty < a < b < \infty}$.
$\bracs{(a, b]| -\infty < a < b < \infty}$.
$\bracs{(a, b)| -\infty < a < b < \infty}$.
Proof. It is sufficient to show that the $\sigma$-algebra generated by any of the above two families coincide, and that the resulting $\sigma$-algebra is the Borel $\sigma$-algebra on $\real$.
(1) $\to$ (2): For any $a \in \real$, $(a, \infty) = (-\infty, a)^{c}$.
(2) $\to$ (3): For any $a \in \real$, $[a, \infty) = \bigcap_{n \in \natp}(a - 1/n, \infty)$.
(3) $\to$ (4): For any $a \in \real$, $(-\infty, a) = [a, \infty)^{c}$.
(4) $\to$ (5): For any $a, b \in \real$, $[a, b) = (-\infty, b) \cap (-\infty, a)^{c}$.
(5) $\to$ (6): For any $a, b \in \real$, $[a, b]= \bigcap_{n \in \natp}(a - 1/n, b]$.
(6) $\to$ (7): For any $a, b \in \real$, $(a, b] = \bigcup_{n \in \natp}[a + 1/n, b]$.
(7) $\to$ (8): For any $a, b \in \real$, $(a, b) = \bigcup_{n \in \natp}(a, b - 1/n]$.
(8) $\to$ (1): For any $a \in \real$, $(-\infty, a] = \bigcup_{n \in \natp}\bigcap_{k \in \natp}(-n, a + 1/k]$.
For any $U \subset X$ open and $q \in U \cap \rational$, there exists $r_{q} > 0$ such that $(q - r_{q}, q + r_{q}) \subset U$. In which case,
so (8) generates all open sets in $\real$. Conversely, every element of (8) is open, so the $\sigma$-algebra generated by (8) is the Borel $\sigma$-algebra on $\real$.$\square$
Proposition 13.1.10. The following families of sets generate the Borel $\sigma$-algebra on $\ol \real$:
$\bracs{[-\infty, a]| a \in \real}$.
$\bracs{(a, \infty]|a \in \real}$.
$\bracs{[a, \infty]| a \in \real}$.
$\bracs{[-\infty, a)| a \in \real}$.
Proof. (1) $\to$ (2): For any $a \in \real$, $(a, \infty] = [-\infty, a]^{c}$.
(2) $\to$ (3): For any $a \in \real$, $[a, \infty] = \bigcap_{n \in \natp}(a - 1/n, \infty]$.
(3) $\to$ (4): For any $a \in \real$, $[-\infty, a) = [a, \infty]^{c}$.
(4) $\to$ (1): For any $a \in \real$, $[-\infty, a] = \bigcap_{n \in \natp}[-\infty, a + 1/n)$.
By definition, all elements of (1), (2), (3), and (4) belong to $\cb_{\ol{\real}}$. Let $\cm$ be the $\sigma$-algebra generated by (1), (2), (3), and (4), then
are elements of $\cm$. For any $a \in \real$, $(a, \infty) = (a, \infty] \setminus \bracs{\infty}$, so $\cm \supset \cb_{\real}$ by Proposition 13.1.9.
In addition, for any $E \in \cb_{\ol{\real}}$, $E = (E \cap \real) \cup (E \setminus \real)$, where $E \cap \real \in \cb_{\real}$. Since $\bracs{\infty}, \bracs{-\infty}\in \cm$ and $\cb_{\real} \subset \cm$, $E \in \cm$ and $\cm = \cb_{\ol \real}$.$\square$