From 831acc66ccd0cc2602a6c7fde30c3b1a54aea159 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Mon, 29 Jun 2026 19:29:18 -0400 Subject: [PATCH] Added gluing for measures in terms of scaffoldings. --- src/measure/measure/index.tex | 1 + src/measure/measure/scaffold.tex | 93 ++++++++++++++++++++++++++++++ src/measure/measure/semifinite.tex | 18 ------ 3 files changed, 94 insertions(+), 18 deletions(-) create mode 100644 src/measure/measure/scaffold.tex diff --git a/src/measure/measure/index.tex b/src/measure/measure/index.tex index 1069729..a6a805e 100644 --- a/src/measure/measure/index.tex +++ b/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} diff --git a/src/measure/measure/scaffold.tex b/src/measure/measure/scaffold.tex new file mode 100644 index 0000000..6298f87 --- /dev/null +++ b/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} + + + diff --git a/src/measure/measure/semifinite.tex b/src/measure/measure/semifinite.tex index ba9972e..64fc10c 100644 --- a/src/measure/measure/semifinite.tex +++ b/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 -