Added the Radon-Nikodym theorem for finite measures.

src/measure/vector/rn.tex

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+\section{The Lebesgue-Radon-Nikodym Theorem}
+\label{section:lebesgue-radon-nikodym}
+
+\begin{theorem}[Lebesgue-Radon-Nikodym]
+\label{theorem:lebesgue-radon-nikodym}
+    Let $(X, \cm)$ be a measurable space, $\mu$ be a $\sigma$-finite positive measure on $(X, \cm)$, $H$ be a Hilbert space over $K \in \RC$ and $\nu: \cm \to H$ be a finite vector measure, then there exists a unique pair of finite vector measures $\nu_a, \nu_s: \cm \to H$ such that:
+    \begin{enumerate}
+        \item $\nu = \nu_a + \nu_s$. 
+        \item $\nu_a$ is absolutely continuous with respect to $\mu$. 
+        \item $\nu_s$ is mutually singular with $\mu$. 
+    \end{enumerate}
+
+    The pair $(\nu_a, \nu_s)$ is the \textbf{Lebesgue decomposition} of $\nu$ with respect to $\mu$. Moreover, there exists a unique $\frac{d\nu_a}{d\mu} \in L^1(X, \cm, \mu; H)$ such that for every $A \in \cm$, 
+    \[
+    \nu_a(A) = \int_{A} \frac{d\nu_a}{d\mu} d\mu
+    \]
+
+    If $\nu \ll \mu$, then $\nu_s = 0$ and $\nu(dx) = \frac{d\nu}{d\mu}\mu(dx)$. In which case, the function $\frac{d\nu}{d\mu}$ is the \textbf{Radon-Nikodym derivative} of $\nu$ with respect to $\mu$.
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Exercise 6.18]{Folland}}}\footnote{I decided to abuse Hilbert spaces for this theorem because it is more fun, and because I will use the Riesz representation theorem twice.}. ]
+    (Finite + Positive): First suppose that $\mu$ is finite and $\nu$ is positive. Let $\lambda = \mu + \nu$, then the mapping
+    \[
+    I_\nu: L^2(X, \cm, \lambda; K) \to K \quad f \mapsto \int f d\nu
+    \]
+
+    is a continuous linear functional on $L^2(X, \cm, \lambda; K)$. By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}, there exists a unique $g \in L^2(X, \cm, \lambda; K)$ such that for every $f \in L^2(X, \cm, \mu; K)$, $\int f d\nu = \int f g d\lambda$. By uniqueness of the Riesz representation, $\int f d\mu = \int f(1 - g)d\lambda$ for all $f \in L^2(X, \cm, \lambda; K)$. 
+
+    Since $0 \le \int f d\nu \le \norm{f}_{L^1(X, \cm, \lambda; K)}$ for all $f \in L^2(X, \cm, \lambda; K) \cap L^+(X, \cm)$, $0 \le g \le 1$ $\lambda$-almost everywhere. Let $A = \bracs{g < 1}$, $B = \bracs{g = 1}$, 
+    \[
+    \nu_a: \cm \to H \quad E \mapsto \nu(E \cap A)
+    \]
+    
+    and
+    \[
+    \nu_s: \cm \to H \quad E \mapsto \nu(E \cap B)
+    \]
+
+    then
+    \begin{enumerate}
+        \item By definition, $\nu = \nu_a + \nu_s$. 
+        \item Let $E \in \cm$ with $\mu(E) = 0$, then $(1 - g)|_E = 0$ $\lambda$-almost everywhere. Thus $E \subset B$ modulo a $\lambda$-null set, and
+        \[
+            \nu_a(E) = \nu(E \cap A) \le \lambda(A \cap B) = 0
+        \]
+
+        so $\nu_a \ll \mu$. 
+        
+        \item Since $\mu(B) = 0$ and $\nu_s(A) = 0$, $\mu \perp \nu_s$. 
+    \end{enumerate}
+    
+    Let
+    \[
+    h: X \to [0, \infty) \quad x \mapsto \frac{\one_A(x)g(x)}{1 - g(x)}
+    \]
+
+    then for each $E \in \cm$, 
+    \begin{align*}
+        \nu_a(E) &= \int_{E \cap A} g d\lambda = \int_E \one_A g d\lambda \\
+        &= \int_{E \cap A} \frac{\one_A g}{1 - g} \cdot (1 - g)d\lambda = \int_E \frac{\one_A g}{1 - g}d\mu
+    \end{align*}
+    
+    so $\one_A g/(1 - g)$ is the desired Radon-Nikodym derivative. 
+
+    ($\sigma$-Finite + Positive): Now suppose that $\mu$ is $\sigma$-finite and $\nu$ is still positive. Let $\seq{E_n} \subset \cm$ be a pairwise disjoint sequence such that $\bigsqcup_{n \in \natp}E_n = X$ and $\mu(E_n) < \infty$ for all $n \in \natp$. For each $n \in \natp$ and $E \in \cm$, let $\mu_n(E) = \mu(E \cap E_n)$ and $\nu_n(E) = \nu(E \cap E_n)$, then there exists $g_n \in L^1(X, \cm, \mu)$ and a positive measure $\nu_s^{(n)}$ such that
+    \begin{enumerate}
+        \item $\nu_n(dx) = g_n\mu(dx) + \nu_s^{(n)}(dx)$. 
+        \item[(3)] $\nu_s^{(n)}$ is mutually singular with $\mu$. 
+    \end{enumerate}
+
+    Let $g = \sum_{n \in \natp}g_n$ and $\nu_s = \sum_{n \in \natp}\nu_s^{(n)}$, then by the \hyperref[Monotone Convergence Theorem]{theorem:mct}, completeness of $L^1(X, \cm, \mu)$, and \hyperref[completeness of $M(X, \cm; \real)$]{definition:vector-measure-finite-space}, $g \in L^1(X, \cm, \mu)$ and $\nu_s$ is a measure with
+    \begin{enumerate}
+        \item $\nu(dx) = g\mu(dx) + \nu_s$. 
+        \item[(3)] $\nu_s$ is mutually singular with $\mu$. 
+    \end{enumerate}
+
+    (Hilbert): Finally, suppose that $\nu$ is an $H$-valued finite vector measure, then the mapping
+    \[
+    I_\nu: L^2(X, \cm, |\nu|; H) \to K \quad f \mapsto \int f d\nu
+    \]
+
+    is a continuous linear functional. By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}, there exists a unique $g \in L^2(X, \cm, |\nu|; H)$ such that for every $f \in L^2(X, \cm, |\nu|; H)$, $\int f d\nu = \int fg d|\nu|$. For each $E \in \cm$, let $\nu_a(E) = \int_E g d|\nu|_a$ and $\nu_s(E) = \int_E g d|\nu|_s$, then
+    \begin{enumerate}
+        \item $\nu(E) = \int_E g d|\nu| = \int_E g d|\nu|_a + \int_E g d|\nu|_s = \nu_a(E) + \nu_s(E)$. 
+        \item Since $|\nu|_a$ is absolutely continuous with respect to $\mu$, so is $\nu_a$. 
+        \item Since $|\nu|_s$ is mutually singular with $\mu$, so is $\nu_s$. 
+    \end{enumerate}
+
+    and
+    \[
+    \nu_a(E) = \int_E g d|\nu|_a = \int_E g \cdot \frac{d|\nu|_a}{d\mu} d\mu
+    \]
+
+    so $\frac{d\nu_a}{d\mu} = g \cdot \frac{d|\nu|_a}{d\mu}$ is the desired Radon-Nikodym derivative. 
+    
+    
+    (Uniqueness): For any decomposition $\nu = \rho_a + \rho_s$ satisfying the above, then $\rho_a - \nu_a = \nu_s - \rho_s$ with $\rho_a - \nu_a \perp \nu_s - \rho_s$. Therefore $\rho_a = \nu_a$, $\rho_s = \nu_s$, and the decomposition is unique. 
+\end{proof} 
+