Added elements of localisable measures.

document.tex

@@ -6,7 +6,7 @@
 
 Hi, welcome to my digital garden, where I collect math results that I learn. 
 
-Occasionally, I make up some definitions to play with. These definition blocks will always have a * at the end of its tital to indicate that it lives mostly in my head. These terms will always be referenced with a link to their definition block. 
+Occasionally, I make up some definitions to play with. These definition blocks will always have a * at the end of its title to indicate that it lives mostly in my head. These terms will always be referenced with a link to their definition block. 
 
 \input{./src/cat/index}
 \input{./src/topology/index}

src/measure/measure/scaffold.tex

@@ -18,6 +18,7 @@
 \end{example}
 % Omitted
 
+
 \begin{lemma}[Gluing Lemma for Measures]
 \label{lemma:gluing-measure}
     Let $(X, \cm)$ be a measurable space, $\cf \subset \cm$ be an ideal, and $\bracsn{\mu_A}_{A \in \cf}$ such that:
@@ -89,5 +90,30 @@
     For each $A \in \cf$ and $E \in \cm$, let $\mu_A(E) = \mu(E \cap A)$, then $\bracsn{\mu_A}_{A \in \cf}$ is a family of measures satisfying \autoref{lemma:gluing-measure}. Therefore $\mu_\cf$ as defined is a measure, and $\cf$ is a scaffold for $\mu_\cf$. 
 \end{proof}
 
+\begin{lemma}
+\label{lemma:scaffolded-ac}
+    Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} measure space and $f \in L^+(X)$, then for any $E \in \cm$, 
+    \[
+    \int_E f d\mu = \sup_{A \in \cf}\int_{E \cap A} f d\mu
+    \]
+\end{lemma}
+\begin{proof}
+    For any $F \in \cm$, 
+    \[
+        \int_{E} \one_F d\mu = \mu(E \cap F) = \sup_{A \in \cf}\mu(A \cap E \cap F) = \sup_{A \in \cf}\int_{E \cap A} \one_F d\mu
+    \]
 
+    so by linearity, the above holds for all simple functions in $L^+(X)$. 
+
+    Now, for each simple function $\phi \in \Sigma^+(X)$ with $\phi \le f$, 
+    \[
+    \int_E \phi d\mu = \sup_{A \in \cf}\int_{E \cap A}\phi d\mu \le \sup_{A \in \cf} \int_{E \cap A} f d\mu
+    \]
+
+    As the above holds for all $\phi \in \Sigma^+(X)$ with $\phi \le f$, 
+    \[
+    \int_E f d\mu = \sup_{A \in \cf}\int_{E \cap A} f d\mu
+    \]
+    
+\end{proof}
 

src/measure/vector/rn-draft.tex

@@ -0,0 +1,95 @@
+\begin{theorem}[Lebesgue-Radon-Nikodym (Localisable)]
+\label{theorem:lebesgue-radon-nikodym-localisable}
+    Let $(X, \cm)$ be a measurable space, $\mu: \cm \to [0, \infty]$ be a localisable measure, $\nu: \cm \to [0, \infty]$ be a positive measure, and $\cf \subset \cm$ be a \hyperref[scaffold]{definition:measure-scaffold} for both $\mu$ and $\nu$, 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. 
+
+    (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 $\mu$ as well, \autoref{lemma:scaffolded-ac} implies that
+    \[
+    \sup_{A \in \cf}\int_{A \cap E}f_A d\mu = \int_A f d\mu
+    \]
+
+    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 $\bracs{F_A}_{A \in \cf}$. 
+
+    Let $E$ and $F$ be essential suprema of $\bracs{E_A}_{A \in \cf}$ and $\bracs{F_A}_{A \in \cf}$ in $(X, \cm, \mu)$, respectively. By \autoref{lemma:gluing-measurable-sets}, $\mu(E \cap F) = 0$. After modification on a null set, assume without loss of generality that $X = E \sqcup F$. 
+    
+
+
+    
+
+    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 $\mu$.
+
+    Let $A \in \cf$, then
+    \[
+    \mu^A: \cm \to [0, \infty] \quad 
+    \]
+\end{proof}

src/measure/vector/rn.tex

@@ -1,7 +1,7 @@
 \section{The Lebesgue-Radon-Nikodym Theorem}
 \label{section:lebesgue-radon-nikodym}
 
-\begin{theorem}[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}
@@ -96,3 +96,4 @@
     (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} 
 
+