From 48a0e63f6198c8d7c654bb970e4caa3791be0a61 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Sat, 27 Jun 2026 17:11:57 -0400 Subject: [PATCH] Added elementary facts about localisable measure spaces. --- src/measure/measure/index.tex | 1 + src/measure/measure/localisable.tex | 105 ++++++++++++++++++++++++++++ 2 files changed, 106 insertions(+) create mode 100644 src/measure/measure/localisable.tex diff --git a/src/measure/measure/index.tex b/src/measure/measure/index.tex index 97b6773..1069729 100644 --- a/src/measure/measure/index.tex +++ b/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} diff --git a/src/measure/measure/localisable.tex b/src/measure/measure/localisable.tex new file mode 100644 index 0000000..0c46ad3 --- /dev/null +++ b/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} + +