Added gluing for measures in terms of scaffoldings.

src/measure/measure/index.tex

@@ -4,6 +4,7 @@
 \input{./measure.tex}
 \input{./complete.tex}
 \input{./semifinite.tex}
+\input{./scaffold.tex}
 \input{./sigma-finite.tex}
 \input{./localisable.tex}
 \input{./regular.tex}

src/measure/measure/scaffold.tex

@@ -0,0 +1,93 @@
+\section{Scaffolds}
+\label{section:scaffold}
+
+\begin{definition}[Scaffold*]
+\label{definition:measure-scaffold}
+    Let $(X, \cm, \mu)$ be a measure space and $\cf \subset \bracs{A \in \cm|\mu(A) < \infty}$ be an ideal, then $\cf$ is a \textbf{scaffold} for $\mu$ if for all $E \in \cm$, 
+    \[
+    \mu(E) = \sup\bracs{\mu(E \cap A)|A \in \cf}
+    \]
+
+    and the quadruple $(X, \cm, \cf, \mu)$ is a \textbf{scaffolded measure space}. 
+
+    For any semifinite measure space $(X, \cm, \mu)$, $\cf = \bracs{A \in \cm|\mu(A) < \infty}$ is the \textbf{canonical scaffold} for $\mu$, and $(X, \cm, \mu)$ will be equipped with this scaffold unless specified otherwise. 
+\end{definition}
+
+\begin{example}
+    Let $X$ be a LCH space, $\mu$ be a Radon measure, and $\mathcal{K}$ be the collection of compact subsets of $X$, then $\mathcal{K}$ is a scaffold for $\mu$. 
+\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:
+    \begin{enumerate}[label=(\alph*)]
+        \item For each $A \in \cf$, $\mu_A$ is a finite measure on $A$.
+        \item For each $A, B \in \cf$ and $E \in \cm$, $\mu_A(E \cap A \cap B) = \mu_B(E \cap A \cap B)$. 
+    \end{enumerate}
+
+    Let
+    \[
+    \mu: \cm \to [0, \infty] \quad E \mapsto \sup\bracsn{\mu_A(E \cap A)|A \in \cf}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item $\mu$ is a measure.
+        \item $\cf$ is a \hyperref[scaffold]{definition:measure-scaffold} for $(X, \cm, \mu)$.
+        \item For each $E \in \cm$, $\mu(A \cap E) = \mu_A(A \cap E)$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (3): Let $E \in \cm$, then $\mu_A(A \cap E) \le \mu(A \cap E)$ by definition. On the other hand, for any $B \in \cf$, 
+    \[
+    \mu_B(A \cap B \cap E) = \mu_A(A \cap B \cap E) \le \mu_A(A \cap E)
+    \]
+
+    Since the above holds for all $B \in \cf$, $\mu(A \cap E) \le \mu_A(A \cap E)$. 
+
+    (1): Let $\seq{E_n} \subset \cm$ be pairwise disjoint, then for each $A \in \cm$, 
+    \[
+    \mu\paren{A \cap \bigsqcup_{n \in \natp}E_n} = \sum_{n \in \natp}\mu_A(A \cap E_n) \le \sum_{n \in \natp}\mu(E_n)
+    \]
+
+    so $\mu\paren{\bigsqcup_{n \in \natp}E_n} \le \sum_{n \in \natp}\mu(E_n)$. 
+
+    On the other hand, let $n \in \natp$, $\seqf{A_k} \subset \cf$, and $A = \bigcup_{k = 1}^n A_k \in \cf$, then
+    \begin{align*}
+        \sum_{k = 1}^n \mu(A_k \cap E_k) &= \sum_{k = 1}^n \mu_{A_k}(A_k \cap E_k)  \\
+        &\le \mu_A\paren{A \cap \bigsqcup_{k = 1}^n E_k} \\
+        &\le \mu\paren{\bigsqcup_{k \in \natp}E_k}
+    \end{align*}
+
+    As this holds for all choice of $\seq{A_k} \subset \cf$, 
+    \[
+    \sum_{k = 1}^n \mu(E_k) \le \mu\paren{\bigsqcup_{k \in \natp}E_k}
+    \]
+
+    and as the above holds for all $n \in \natp$, $\sum_{n \in \natp}\mu(E_n) \le \mu\paren{\bigsqcup_{n \in \natp}E_n}$. 
+    
+    (2): By definition of $\mu$. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:scaffolded-part}
+    Let $(X, \cm, \mu)$ be a measure space, $\cf \subset \bracs{A \in \cm|\mu(A) < \infty}$ be an ideal, and
+    \[
+    \mu_\cf: \cm \to [0, \infty] \quad E \mapsto \sup\bracs{\mu(A \cap E)|A \in \cf}
+    \]
+
+    then
+    \begin{enumerate}
+        \item $\mu_\cf$ is a measure on $(X, \cm)$. 
+        \item $\cf$ is a \hyperref[scaffold]{definition:measure-scaffold} for $\mu_\cf$/
+    \end{enumerate}
+    
+    and $\mu_\cf$ is the \textbf{$\cf$-scaffolded part} of $\mu$. 
+\end{corollary}
+\begin{proof}
+    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}
+
+
+

src/measure/measure/semifinite.tex

@@ -60,21 +60,3 @@
 
 \end{proof}
 
-
-\begin{definition}[Scaffold*]
-\label{definition:measure-scaffold}
-    Let $(X, \cm, \mu)$ be a measure space and $\cf \subset \bracs{A \in \cm|\mu(A) < \infty}$ be an ideal, then $\cf$ is a \textbf{scaffold} for $\mu$ if for all $E \in \cm$, 
-    \[
-    \mu(E) = \sup\bracs{\mu(E \cap A)|A \in \cf}
-    \]
-
-    and the quadruple $(X, \cm, \cf, \mu)$ is a \textbf{scaffolded measure space}. 
-
-    For any semifinite measure space $(X, \cm, \mu)$, $\cf = \bracs{A \in \cm|\mu(A) < \infty}$ is the \textbf{canonical scaffold} for $\mu$, and $(X, \cm, \mu)$ will be equipped with this scaffold unless specified otherwise. 
-\end{definition}
-
-\begin{example}
-    Let $X$ be a LCH space, $\mu$ be a Radon measure, and $\mathcal{K}$ be the collection of compact subsets of $X$, then $\mathcal{K}$ is a scaffold for $\mu$. 
-\end{example}
-% Omitted
-