Fleshed out the dual of measure spaces.

src/measure/vector/fin.tex

@@ -32,6 +32,15 @@
     by \hyperref[Fubini's theorem]{theorem:fubini-tonelli}.
 \end{proof}
 
+
+While the space of bounded Borel functions on $(X, \cm)$ forms a subspace of the dual of $M(X, \cm; \complex)$, it may not be immediately clear that they are insufficient. Before moving on to an explicit description of this dual, it is beneficial to consider a simple "example". 
+
+Let $X = [0, 1]$, equipped with its Borel $\sigma$-algebra, and $\mu$ be the Lebesgue measure on $X$, then for any $x \in [0, 1]$, $\mu$ is mutually singular with the delta mass at $x$. Therefore the closure of $\text{span}\bracs{\delta_x|x \in X}$ in $M_R(X, \cm; \complex)$ is a proper closed subspace of $M(X, \cm; \complex)$. By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, there exists $\phi \in M(X, \cm; \complex)^*$ such that $\dpn{\delta_x, \phi}{M(X, \cm; \complex)} = 0$ for all $x \in X$, but $\dpn{\mu, \phi}{M(X, \cm; \complex)} = 1$. If there exists a bounded Borel function $f: X \to \complex$ such that $\dpn{\nu, \phi}{M(X, \cm; \complex)} = \int f d\nu$ for all $\nu \in M(X, \cm; \complex)$, then $f(x) = \dpn{\delta_x, \phi}{M(X, \cm; \complex)} = 0$ for all $x \in [0, 1]$, which is impossible. Therefore $\phi$ cannot be represented as a bounded Borel function. 
+
+Aside from abusing the Hahn-Banach theorem, a more explicit construction of such a functional is via the atomic decomposition. However, it is not included here due to time constraints. 
+
+In any case, the above example shows that a linear functional on $M(X, \cm; \complex)$ may act as \textit{different} Borel functions on different families of measures that are mutually singular to each other. The following theorem will make this relation precise. 
+
 \begin{theorem}
 \label{theorem:hilbert-measures-dual}
     Let $(X, \cm)$ be a measurable space, $H$ be a Hilbert space over $K \in \RC$, and $\mathscr{M} \subset M(X, \cm; H)$ be a closed subspace such that:
@@ -60,7 +69,7 @@
         \item There exists a semifinite measure space $(\Omega, \cn, \mu)$ such that $\mathscr{M} \iso L^1(\Omega; H)$ and $\mathscr{M}^* \iso L^\infty(\Omega; H)$. 
     \end{enumerate}
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof, {{\cite{StackRadonDual}}}. ]
     (1): By (A), $\mu_f \in \mathscr{M}$ for all $f \in [l^1(I); L^1(\mu_i; H)]$. 
     
     On the other hand, let $\nu \in \mathscr{M}$. For each $i \in I$, let $\nu = \nu_a^{(i)} + \nu_s^{(i)}$ be the Lebesgue decomposition of $\nu$ with respect to $\mu_i$, then by the \hyperref[Radon-Nikodym theorem]{theorem:lebesgue-radon-nikodym}, there exists $f_i \in L^1(\mu_i; H)$ such that $\nu_a^{(i)}(dx) = f_i \mu_i(dx)$. 
@@ -84,5 +93,9 @@
     \]
 \end{proof}
 
+\begin{proposition}
+\label{proposition:measure-l-infinity-dominated-convergence}
+    Let $
+\end{proposition}