14.2 The Borel $\sigma$-Algebra
Definition 14.2.1 (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 14.2.2 (Borel $\sigma$-Algebra on $\ol{\real}$). The family
is the Borel $\sigma$-algebra on $\ol{\real}$.
Proposition 14.2.3. 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 14.2.4. 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 14.2.3.
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 $\cm = \cb_{\ol \real}$.$\square$
Proposition 14.2.5. Let $X$ be a separable metric space, then the Borel $\sigma$-algebra on $X$ is generated by the following families of sets:
Open sets of $X$.
$\bracs{B(x, r)|x \in X, r > 0}$.
$\bracsn{\ol{B(x, r)}|x \in X, r > 0}$.
Proof. (1) $\subset$ (2): Let $U \subset X$ be open. By Definition 4.5.5, there exists a countable dense subset $S \subset U$. For each $x \in S$, let $r_{x} > 0$ such that $B(x, r) \subset U$, then $U = \bigcup_{x \in S}B(x, r_{x})$ is a countable union of open balls.
(2) $\subset$ (3): For any $x \in X$ and $r > 0$, $B(x, r) = \bigcup_{n \in \natp}\overline{B(x, r - 1/n)}$ is a countable union of closed balls.
(3) $\subset$ (1): For each $x \in X$ and $r > 0$, $\overline{B(x, r)}$ is closed.$\square$
Lemma 14.2.6. Let $X$ be a topological space and $Y \subset X$ be a subspace, then the Borel $\sigma$-algebra on $Y$ coincides with the $\sigma$-algebra on $Y$ induced by $\cb_{X}$.
Proof. Since $\bracsn{A \in \cb_X|A \cap Y \in \cb_Y}$ is a $\sigma$-algebra that contains all open sets in $X$, $\cb_{Y}$ contains the induced $\sigma$-algebra.
On the other hand, the induced $\sigma$-algebra contains all open sets in $Y$, so it contains $\cb_{Y}$.
Therefore the two $\sigma$-algebras coincide.$\square$