Added universality of the zero-dimensional spaces.

src/measure/sets/borel.tex

@@ -8,6 +8,32 @@
 \end{definition}
 
 
+
+\begin{lemma}
+\label{lemma:monotone-borel-characterisation}
+    Let $X$ be a topological space, then $\cb_X$ is the smallest subset of $2^X$ that:
+    \begin{enumerate}
+        \item contains open and closed subsets of $X$. 
+        \item is closed under countable intersections.
+        \item is closed under countable disjoint unions. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma 8.2.4]{CohnMeasure}}}. ]
+    Let $\cf \subset 2^X$ be the smallest subset of $2^X$ satisfying the lemma. Since $\cb_X$ satisfies the lemma, $\cf \subset \cb_X$. On the other hand, let
+    \[
+    \cf_0 = \bracs{A \subset X| A \in \cf, A^c \in \cf}
+    \]
+
+    then by definition, $\cf_0$ is closed under complements. Let $\seq{A_n} \subset \cf_0$, then
+    \[
+    \bigcup_{n \in \natp}A_n = \bigsqcup_{n \in \natp}A_n \setminus \bigcup_{k = 1}^{n-1} A_k
+     = \bigsqcup_{n \in \natp}A_n \cap \bigcap_{k = 1}^{n - 1}A_k^c
+    \]
+
+    Since $\cf_0 \subset \cf$ is closed under complements, $\seq{A_n^c} \subset \cf$ as well. By (2) and (3), $\bigcup_{n \in \natp}A_n \in \cf$ and $\bigcap_{n \in \natp}A_n^c \in \cf$. Thus $\bigcup_{n \in \natp}A_n \in \cf$ as well. By (1), $\cf_0$ is a $\sigma$-algebra that contains all open subsets of $X$, so $\cf \supset \cf_0 \supset \cb_X$.  
+\end{proof}
+
+
 \begin{definition}[Borel $\sigma$-Algebra on $\ol{\real}$]
 \label{definition:borel-sigma-algebra-extended}
     The family