Polished A-A and added new lines for broken enumerates.

src/measure/radon/c0.tex

@@ -8,7 +8,8 @@
         \item $I = I^+ - I^-$.
         \item $I^+ \perp I^-$. 
         \item $\norm{I^+}_{C_0(X; \real)^*}, \norm{I^-}_{C_0(X; \real)^*} \le \norm{I}_{C_0(X; \real)^*}$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     
 \end{lemma}
 \begin{proof}

src/measure/radon/radon.tex

@@ -115,7 +115,8 @@
     \begin{enumerate}
         \item[(a)] For any $U \subset X$ open, $U$ is $\sigma$-compact.
         \item[(b)] For any $K \subset X$ compact, $\mu(K) < \infty$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     then $\mu$ is a regular measure on $X$.
 \end{proposition}
 \begin{proof}

src/measure/radon/riesz.tex

@@ -71,7 +71,8 @@
         \]
 
         As this holds for all $\eps > 0$, $\mu^*(E) \le \sum_{n \in \natp}\mu^*(E_n)$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     Therefore $\mu^*: 2^E \to [0, \infty]$ is an outer measure.
 
     To see that all Borel sets are $\mu^*$-measurable, let $E \subset X$ and $U \in \topo$. First suppose that $E$ is open. Let $f \prec E \cap U$, then for any $g \prec E \setminus \supp{f}$, $\supp{f} \cap \supp{g} = \emptyset$ and $f + g \prec E$, so