\section{The Lebesgue-Radon-Nikodym Theorem}
\label{section:lebesgue-radon-nikodym}

\begin{theorem}[Lebesgue-Radon-Nikodym (Finite)]
\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 $\nu(dx) = \frac{d\nu_a}{d\mu}\mu(dx) = \frac{d\nu_a}{d\mu}\mu(dx)$. 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 Since $X = A \sqcup B$, $\nu = \nu_a + \nu_s$. 
        \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} 


\begin{theorem}[Lebesgue-Radon-Nikodym (Localisable)]
\label{theorem:lebesgue-radon-nikodym-localisable}
    Let $(X, \cm)$ be a measurable space, $\mu, \nu: \cm \to [0, \infty]$ be positive measures, and $\cf \subset \cm$ be an ideal such that:
    \begin{enumerate}[label=(\alph*)]
        \item $(X, \cm, \mu + \nu)$ and $(X, \cm, \mu)$ are localisable. 
        \item $\cf$ is a \hyperref[scaffold]{definition:measure-scaffold} for both $\mu$ and $\nu$. 
    \end{enumerate}
    
    then there exists a unique pair of positive measures $\nu_a, \nu_s: \cm \to [0, \infty]$ 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 $\mu$. 
        \item $\cf$ is a scaffold for $\nu_a$ and $\nu_s$. 
    \end{enumerate}

    The pair $(\nu_a, \nu_s)$ is the \textbf{Lebesgue decomposition} of $\nu$ with respect to $\mu$. Moreover, there exists a unique $\frac{d\nu_a}{d\mu} \in L^+(X, \cm, \mu)$ 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 $\nu(dx) = \frac{d\nu_a}{d\mu}\mu(dx) = \frac{d\nu_a}{d\mu}\mu(dx)$. 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}
    For each $A \in \cf$ and $E \in \cm$, let
    \[
    \mu^A(E) = \mu(A \cap E) \quad \nu^A(E) = \nu(A \cap E)
    \]

    then $\mu^A$ and $\nu^A$ are finite measures on $A$. By the \hyperref[finite case]{theorem:lebesgue-radon-nikodym}, there exists an a.e. unique $f_A \in L^+(A, \mu)$ and $\nu_s^A: \cm \to [0, \infty]$ such that:
    \begin{enumerate}
        \item $d\nu^A = f_A\mu^A + \nu_s^A = f_A\mu + \nu_s^A$. 
        \item $\nu_s^A$ is mutually singular with $\mu$. 
    \end{enumerate}

    The uniqueness given by the finite case implies that for any $A, B \in \cf$, $f_A|_{A \cap B} = f_B|_{A \cap B}$ $\mu$-almost everywhere. By the \hyperref[gluing lemma for measurable functions]{lemma:gluing-measurable}, there exists $f \in L^+(X, \mu)$ such that $f|_A = f_A$ $\mu$-almost everywhere for all $A \in \cf$. 

    By the \hyperref[gluing lemma for measures]{lemma:gluing-measure}, 
    \[
    \nu_s: \cm \to [0, \infty] \quad E \mapsto \sup_{A \in \cf}\nu_s^A(E \cap A)
    \]

    is a measure. 

    (4): By definition, $\cf$ is a scaffold for $\nu_s$. By \autoref{lemma:scaffolded-ac}, it is also a scaffold for $f d\mu$. 

    (1): Let $E \in \cm$, then since $\cf$ is a scaffold for $\nu$, 
    \begin{align*}
    \nu(E) &= \sup_{A \in \cf}\nu(A \cap E) = \sup_{A \in \cf}\nu^A(A \cap E) \\
    &= \sup_{A \in \cf}\braks{\int_{A \cap E}f_A d\mu + \nu_s^A(A \cap E)}
    \end{align*}

    As $\cf$ is an ideal, for any $A, B \in \cf$, $A \cup B \in \cf$, and
    \begin{align*}
        \int_{A \cap E}f_A d\mu \vee \int_{B \cap E}f_B d\mu &\le \int_{(A \cup B) \cap E}f_{A \cup B} d\mu \\
        \nu_s^A(A \cap E) \vee \nu_s^B(B \cap E) &\le \nu_s^{A \cup B}((A \cup B) \cap E)
    \end{align*}

    Thus the sum and the supremum may be interchanged, so
    \[
    \nu(E) = \sup_{A \in \cf}\int_{A \cap E}f_A d\mu + \sup_{A \in \cf}\nu_s^A(A \cap E)
    \]

    Now, since $\cf$ is a scaffold for $f\mu$, 
    \[
    \sup_{A \in \cf}\int_{A \cap E}f_A d\mu = \int_A f d\mu
    \]

    By definition of $\nu_s$, 
    \[
    \sup_{A \in \cf}\nu_s^A(A \cap E) = \nu_s(E)
    \]

    Therefore $\nu(E) = \int_E f d\mu + \nu_s(E)$, so $\nu(dx) = f(x)\mu(dx) + \nu_s(dx)$. 

    (2): $\nu_a(dx) = f(x)\mu(dx)$. 

    (3): For each $A \in \cf$, $f_A d\mu \perp \nu_s^A$, so there exists $E_A, F_A \in \cm$ such that:
    \begin{enumerate}[label=(\roman*)]
        \item $A = E_A \sqcup F_A$. 
        \item $\mu(F_A) = 0$ and $\nu_s^A(E_A) = 0$. 
    \end{enumerate}

    Let $F$ be an essential supremum of $\bracs{F_A}_{A \in \cf}$ with respect to $(\mu + \nu)$. By \autoref{lemma:gluing-measurable-sets}, for each $A \in \cf$,
    \[
    \mu(F \cap A) = \mu(F \cap E_A) \le (\mu + \nu)(F \cap E_A) = 0
    \]

    Since $\cf$ is a scaffold for $\mu$, $\mu(F) = 0$. On the other hand, for each $A \in \cf$, 
    \begin{align*}
    \nu_s(A \setminus F) &= \nu_s(E_A \setminus F) + \nu_s(F_A \setminus F) \\
    &\le \nu_s(E_A) + (\mu + \nu)(F_A \setminus F) = 0
    \end{align*}

    By definition of $\nu_s$, $\nu(F^c) = 0$. Therefore $\mu \perp \nu_s$. 

    (Uniqueness): By uniqueness of the finite case, the restriction of $(\nu_a, \nu_s)$ to each set in $\cf$ is unique. Since $\cf$ is a scaffold for both measures, they are uniquely determined by their restrictions to each set in $\cf$.  
    
\end{proof}