Added the Singer representation theorem.

src/measure/radon/radon.tex

@@ -148,6 +148,27 @@
     Since $X$ is $\sigma$-compact, $\mu$ is $\sigma$-finite, so $\mu$ is regular by \autoref{proposition:radon-regular-sigma-finite}.
 \end{proof}
 
+\begin{lemma}
+\label{lemma:radon-compact-project}
+    Let $X$ be a LCH space, $Y$ be a compact Hausdorff space, $\mu$ be a finite Radon measure on $X \times Y$, and $\nu$ be a measure on $X$. If for each $A \in \cb_X$, $\nu(A) \le \mu(A \times Y)$, then $\nu$ is also a Radon measure. 
+\end{lemma}
+\begin{proof}
+    Let $A \in \cb_X$ and $\eps > 0$. By outer regularity of $\mu$, there exists $U \in \cn_{X \times Y}(A \times Y)$ such that $\mu(U \setminus (A \times Y)) < \eps$. By the \hyperref[Tube Lemma]{lemma:tube-lemma}, there exists $V \in \cn_X(A)$ such that $V \times Y \subset U$. In which case, 
+    \[
+    \nu(V \setminus A) \le \mu((V \setminus A) \times Y) \le \mu(U \setminus (A \times Y)) < \eps
+    \]
+
+    so $\nu$ is outer regular on $A$.
+
+    On the other hand, by \autoref{proposition:radon-regular-sigma-finite}, there exists $K \subset A \times Y$ compact such that $\mu((A \times Y) \setminus K) < \eps$. By \autoref{proposition:compact-extensions}, $\pi_1(K) \subset A$ is also compact. Thus
+    \[
+    \nu(A \setminus \pi_1(K)) \le \mu((A \setminus \pi_1(K)) \times Y) \le \mu((A \times Y) \setminus K)
+    \]
+
+    so $\nu$ is inner regular on $A$. 
+\end{proof}
+
+
 \begin{proposition}
 \label{proposition:radon-cc-dense}
     Let $X$ be a LCH space, $\mu: \cb_X \to [0, \infty]$ be a Radon measure, $E$ be a normed vector space, and $p \in [1, \infty)$, then $C_c(X; E)$ is dense in $L^p(X; E)$.