Added measurability in separable metric spaces.

src/measure/sets/algebra.tex

@@ -72,7 +72,87 @@
     Let $X$ be a set and $\ce \subset 2^X$, then the $\sigma$-algebra $\sigma(\ce)$ \textbf{generated by} $\ce$ is the smallest $\sigma$-algebra on $X$ containing $\ce$.
 \end{definition}
 
+
 \begin{definition}[Borel $\sigma$-Algebra]
 \label{definition:borel-sigma-algebra}
     Let $(X, \topo)$ be a topological space, then the \textbf{Borel $\sigma$-algebra} $\cb_X$ on $X$ is the $\sigma$-algebra generated by $\topo$.
 \end{definition}
+
+
+\begin{definition}[Borel $\sigma$-Algebra on $\ol{\real}$]
+\label{definition:borel-sigma-algebra-extended}
+    The family
+    \[
+    \cb_{\ol{\real}} = \bracsn{E \subset \ol \real| E \cap \real \in \cb_\real}
+    \]
+    is the \textbf{Borel $\sigma$-algebra} on $\ol{\real}$.
+\end{definition}
+
+
+\begin{proposition}
+\label{proposition:borel-sigma-real-generators}
+    The following families of sets generate the Borel $\sigma$-algebra on $\real$:
+    \begin{enumerate}
+        \item $\bracs{(-\infty, a]| a \in \real}$.
+        \item $\bracs{(a, \infty)|a \in \real}$.
+        \item $\bracs{[a, \infty)| a \in \real}$.
+        \item $\bracs{(-\infty, a)| a \in \real}$.
+        \item $\bracs{[a, b)| -\infty < a < b < \infty}$.
+        \item $\bracs{[a, b]| -\infty < a < b < \infty}$.
+        \item $\bracs{(a, b]| -\infty < a < b < \infty}$.
+        \item $\bracs{(a, b)| -\infty < a < b < \infty}$.
+    \end{enumerate}
+\end{proposition}
+\begin{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,
+    \[
+    U = \bigcup_{q \in U \cap \rational}(q - r_q, q + r_q)
+    \]
+    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 $\real$.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:borel-sigma-extended-generators}
+    The following families of sets generate the Borel $\sigma$-algebra on $\ol \real$:
+    \begin{enumerate}
+        \item $\bracs{[-\infty, a]| a \in \real}$.
+        \item $\bracs{(a, \infty]|a \in \real}$.
+        \item $\bracs{[a, \infty]| a \in \real}$.
+        \item $\bracs{[-\infty, a)| a \in \real}$.
+    \end{enumerate}
+\end{proposition}
+\begin{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
+    \[
+    \bracs{\infty} = \bigcap_{n \in \nat}[n, \infty] \quad \bracs{-\infty} = \bigcap_{n \in \nat}[-\infty, n]
+    \]
+    are elements of $\cm$. For any $a \in \real$, $(a, \infty) = (a, \infty] \setminus \bracs{\infty}$, so $\cm \supset \cb_\real$ by \ref{proposition:borel-sigma-real-generators}.
+
+    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 $\cm = \cb_{\ol \real}$.
+\end{proof}