Added elementary facts about localisable measure spaces.

src/measure/measure/index.tex

@@ -5,6 +5,7 @@
 \input{./complete.tex}
 \input{./semifinite.tex}
 \input{./sigma-finite.tex}
+\input{./localisable.tex}
 \input{./regular.tex}
 \input{./outer.tex}
 \input{./lebesgue-stieltjes.tex}

src/measure/measure/localisable.tex

@@ -0,0 +1,105 @@
+\section{Localisable Measures}
+\label{section:localisable-measure}
+
+\begin{definition}[Essential Supremum]
+\label{definition:esssup-measure-space}
+    Let $(X, \cm, \mu)$ be a measure space, $\ce \subset \cm$, and $S \in \cm$, then $S$ is an \textbf{essential upper bound} of $\ce$ if for any $E \in \ce$, $\mu(E \setminus S) = 0$. If in addition, for any essential upper bound $T$ of $\ce$, $\mu(S \setminus T) = 0$, then $S$ is an \textbf{essential supremum} of $\ce$.  
+\end{definition}
+
+
+\begin{definition}[Localisable]
+\label{definition:localisable-measure}
+    Let $(X, \cm, \mu)$ be a measure space, then $X$ is \textbf{localisable} if:
+    \begin{enumerate}
+        \item $\mu$ is semifinite. 
+        \item For every $\ce \subset \cm$, there exists an essential supremum $A$ of $\ce$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Decomposable]
+\label{definition:decomposable-measure}
+    Let $(X, \cm, \mu)$ be a measure space and $\seqi{A} \subset \cm$, then $\seqi{A}$ is a \textbf{decomposition} of $(X, \cm, \mu)$ if:
+    \begin{enumerate}
+        \item For each $i \in I$, $\mu(A_i) < \infty$. 
+        \item $X = \bigsqcup_{i \in I}X_i$. 
+        \item $\cm = \bracs{E \subset X|E \cap A_i \in \cm \forall i \in I}$. 
+        \item For each $E \in \cm$, $\mu(E) = \sum_{i \in I}\mu(E \cap A_i)$. 
+    \end{enumerate}
+
+    If $(X, \cm, \mu)$ admits a decomposition, then it is \textbf{decomposable}. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:essential-upperbound-finite}
+    Let $(X, \cm, \mu)$ be a finite measure space, $\ce \subset \cm$, and $\mathcal{S}$ be the set of all essential upper bounds of $\ce$, then:
+    \begin{enumerate}
+        \item $\mathcal{S} \ne \emptyset$. 
+        \item For any $S \in \cm$, $S \in \mathcal{S}$ if and only if $\mu(S \cap E) = \mu(E)$ for all $E \in \ce$. 
+        \item For any $\seq{S_n} \subset \mathcal{S}$, $\bigcap_{n \in \natp}S_n \in \mathcal{S}$. 
+        \item There exists $S \in \mathcal{S}$ such that $\mu(S) = \inf\bracs{\mu(T)|T \in \mathcal{S}}$. 
+        \item For any $S \in \mathcal{S}$ with $\mu(S) = \inf\bracs{\mu(T)|T \in \mathcal{S}}$, $S$ is an essential supremum of $\ce$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): $X \in \mathcal{S}$. 
+
+    (2): Since $\mu$ is finite, for any $E \in \ce$, $\mu(S \cap E) = \mu(E)$ if and only if $\mu(E \setminus S) = 0$. 
+
+    (3): Firstly, for any $S, T \in \mathcal{S}$ and $E \in \ce$, 
+    \[
+    \mu(E \setminus (S \cap T)) \le \mu(E \setminus S) + \mu(E \setminus T) = 0
+    \]
+    
+    so $S \cap T \in \mathcal{S}$. Now let $\seq{S_n} \subset X$ and $E \in \ce$, then since $\mu$ is finite, 
+    \[
+    \mu(E) = \limv{N}\mu\paren{E \cap \bigcap_{n = 1}^N S_n} = \mu\paren{E \cap \bigcap_{n \in \natp}S_n}
+    \]
+    
+    by \hyperref[continuity from above]{proposition:measure-properties}. By (2), $\bigcap_{n \in \natp}S_n \in \mathcal{S}$. 
+
+    (4): Let $M = \inf\bracs{\mu(T)|T \in \mathcal{S}}$ and $\seq{S_n} \subset \mathcal{S}$ such that $\limv{n}\mu(S_n) = M$, then by continuity from above,
+    \[
+    M \le \mu\paren{\bigcap_{n \in \natp}S_n} \le \limv{n}\mu(S_n) = M
+    \]
+
+    By (3), $\bigcap_{n \in \natp}S_n \in \mathcal{S}$, therefore the minimum is achieved. 
+
+    (5): Let $R \in \mathcal{S}$. By (3), $S \cap R \in \mathcal{S}$ with $\mu(S \cap R) = \inf\bracs{\mu(T)|T \in \mathcal{S}} = \mu(S)$, so
+    \[
+    \mu(S \setminus R) = \mu(S \setminus (S \cap R)) = \mu(S) - \mu(S \cap R) = 0
+    \]
+
+    and $S$ is the essential supremum of $\ce$. 
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:decomposable-localisable}
+    Let $(X, \cm, \mu)$ be a decomposable measure space, then $X$ is localisable. 
+\end{proposition}
+\begin{proof}
+    Let $\ce \subset \cm$ and $\seqi{A} \subset \cm$ be a decomposition of $X$. 
+    
+    For each $i \in I$, let $\ce_i = \bracs{E \cap A_i|E \in \ce}$. By \autoref{lemma:essential-upperbound-finite}, there exists an essential upper bound $S_i \in \cm$ of $\ce_i$ contained in $A_i$ with respect to the restricted measure $\mu|_{A_i}$. In other words,  
+    \begin{enumerate}[label=(\roman*)]
+        \item $S_i$ is an essential upper bound of $\ce_i$. 
+        \item For any essential upper bound $T \in \cm$ of $\ce_i$ with $T \subset A_i$, $\mu(S_i \setminus T) = 0$. 
+    \end{enumerate}
+
+    Now, let $S = \bigsqcup_{i \in I}S_i$, then since $X = \bigsqcup_{i \in I}A_i$ and $S_i \subset A_i$ for all $i \in I$, $S \cap A_i = S_i \in \cm$ for all $i \in I$, so $S \in \cm$. For any $E \in \ce$, $E \cap A_i \in \ce_i$. Passing through the decomposition, (i) implies that, 
+    \begin{align*}
+        \mu(E \setminus S) &= \sum_{i \in I}\mu((E \setminus S) \cap A_i) = \sum_{i \in I}\mu((E \cap A_i) \setminus (S \cap A_i)) \\
+        &= \sum_{i \in I}\mu((E \cap A_i) \setminus S_i) = \sum_{i \in I}0 = 0
+    \end{align*}
+
+    and $S$ is an essential upper bound of $\ce$. 
+    
+    Finally, let $T \in \cm$ be an essential upper bound of $\ce$, then $T \cap A_i$ is an essential upper bound of $\ce_i$ for all $i \in I$. The decomposition and (ii) then shows that
+    \[
+    \mu(S \setminus T) = \sum_{i \in I}\mu((S \cap A_i) \setminus (T \cap A_i)) = \sum_{i \in I}\mu(S_i \setminus (T \cap A_i)) = 0
+    \]
+
+    therefore $S$ is an essential supremum of $\ce$. 
+\end{proof}
+
+