Updated the existing system to accomodate scaffolds.

.vscode/project.code-snippets

@@ -176,5 +176,10 @@
   "scope": "latex",
     "prefix": "cproof",
     "body": ["[Proof, {{\\cite[$1]{$2}}}. ]$0"]
+  },
+  "Scaffold": {
+    "scope": "latex",
+    "prefix": "scaf", 
+    "body": ["\\hyperref[scaffolded]{definition:measure-scaffold}$0"]
   }
 }

src/measure/measurable-maps/in-measure.tex

@@ -74,65 +74,6 @@
     On the other hand, if $\mu\bracs{d(f, g) > \eps} \le \eps$, then $d(f, g) \le \eps$. Therefore $\alpha$ induces the uniform structure of convergence in measure. 
 \end{proof}
 
-\begin{definition}[Locally In Measure]
-\label{definition:locally-in-measure}
-    Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a separable metric space. For each $\eps, \delta > 0$ and $A \in \cm$ with $\mu(A) < \infty$, let
-    \[
-    U(A, \delta, \eps) = \bracs{(f, g) \in \mathcal{L}^0(X; Y)| \mu(A \cap \bracs{d(f, g) > \delta}) < \eps}
-    \]
-
-    then
-    \[
-    \fB = \bracs{U(A, \delta, \eps)|\eps, \delta > 0, A \in \cm, \mu(A) < \infty}
-    \]
-
-    forms a fundamental system of entourages for a uniformity. The uniformity induced by $\fB$ is the \textbf{uniform structure of local convergence in measure} on $\mathcal{L}^0(X; Y)$.
-\end{definition}
-\begin{proof}
-    It is sufficient to check the conditions of \autoref{proposition:fundamental-entourage-criterion}:
-    \begin{enumerate}
-        \item[(FB1)] For each $\eps, \eps', \delta, \delta' > 0$ and $A, A' \in \cm$ with $\mu(A), \mu(A') < \infty$, 
-        \[
-        U(A \cup A', \delta \wedge \delta', \eps \wedge \eps') \subset U(A, \delta, \eps) \cap U(A', \delta', \eps')
-        \]
-        \item[(UB3)] For each $\eps, \delta > 0$, $A \in \cm$ with $\mu(A) < \infty$, and $f, g, h \in \mathcal{L}^0(X; Y)$, 
-        \[
-        \bracs{d(f, h) > \delta} \subset \bracs{d(f, g) > \delta} \cup \bracs{d(g, h) > \delta}
-        \]
-
-        so $U(A, \delta/2, \eps/2) \circ U(A, \delta/2, \eps/2) \subset U(A, \delta, \eps)$. 
-    \end{enumerate}
-\end{proof}
-
-\begin{proposition}
-\label{proposition:convergence-in-measure}
-    Let $(X, \cm, \mu)$ be a semifinite measure space, $(Y, d)$ be a separable metric space, and $\fF$ be a filter of $(\cm, \cb_Y)$-measurable functions, then $\fF$ is Cauchy in measure if and only if:
-    \begin{enumerate}
-        \item[(L)] $\fF$ is locally Cauchy in measure. 
-        \item[(T)] For each $\eps, \delta > 0$, there exists $F \in \fF$ and $A \in \cm$ with $\mu(A) < \infty$ such that
-        \[
-        \sup_{f, g \in F}\mu(A^c \cap \bracs{d(f, g) > \delta}) < \eps
-        \]
-    \end{enumerate}
-\end{proposition}
-\begin{proof}
-    (L) + (T) $\Rightarrow$ (In Measure): Let $\eps, \delta > 0$. By (T) then there exists $F_1 \in \fF$ and $A \in \cm$ with $\mu(A) < \infty$ such that 
-    \[
-    \sup_{f, g \in F_1}\mu(A^c \cap \bracs{d(f, g) > \delta}) < \eps
-    \]
-
-    By (L), there exists $F_2 \in \fF$ with $F_2 \subset F_1$ such that
-    \[
-    \sup_{f, g \in F_2}\mu(A \cap \bracs{d(f, g) > \delta}) < \eps
-    \]
-
-    Therefore
-    \[
-    \sup_{f, g \in F_2}\mu\bracs{d(f, g) > \delta} < 2\eps
-    \]
-\end{proof}
-
-
 \begin{lemma}
 \label{lemma:ae-in-measure}
     Let $(X, \cm, \mu)$ be a finite measure space, $(Y, d)$ be a separable metric space, and $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$ such that $f_n \to f$ almost everywhere, then $f_n \to f$ in measure.
@@ -157,7 +98,7 @@
         \item $L^0(X; Y)$ equipped with the uniform structure of convergence in measure is complete. 
     \end{enumerate}
 \end{theorem}
-\begin{proof}[Proof, [{{\cite[Theorem 2.30]{Folland}}}]. ]
+\begin{proof}[Proof, {{\cite[Theorem 2.30]{Folland}}}. ]
     (1): Since $\seq{f_n}$ is Cauchy in measure, there exists a subsequence $\seq{n_k}$ such that for each $k \in \natp$, $\mu(\bracsn{d(f_{n_k}, f_{n_{k+1}}) > 2^{-k}}) \le 2^{-k}$.
     
     In this case, for any $K \in \natp$ and $j \ge k \ge K$, 
@@ -196,48 +137,3 @@
     (2): Since the uniform structure of convergence in measure on $L^0(X; Y)$ is defined by the \hyperref[Ky Fan metric]{definition:ky-fan}, completeness follows from (1) and \autoref{proposition:complete-metric-space}. 
 \end{proof}
 
-\begin{theorem}[Monotone Convergence Theorem (in Measure)]
-\label{theorem:mct-measure}
-    Let $(X, \cm, \mu)$ be a semifinite measure space, $\net{f} \subset \mathcal{L}^+(X, \cm)$, and $f \in \mathcal{L}^+(X, \cm)$ such that
-    \begin{enumerate}[label=(\alph*)]
-        \item For each $x \in X$, $f_\alpha(x) \upto f(x)$. 
-        \item $f_\alpha \to f$ locally in measure. 
-    \end{enumerate}
-
-    then
-    \[
-    \lim_{\alpha \in A}\int f_\alpha d\mu = \int f d\mu
-    \]
-\end{theorem}
-\begin{proof}
-    By \autoref{definition:lebesgue-non-negative}, $\int f_\alpha d\mu \le \int f d\mu$ for each $\alpha \in A$. 
-
-    On the other hand, using \autoref{lemma:lebesgue-non-negative-strict}, it is sufficient to show that
-    \[
-    \lim_{\alpha \in A}\int f_\alpha d\mu = \sup_{\alpha \in A}\int f_\alpha d\mu \ge \int \phi d\mu
-    \]
-    
-    for any $\phi \in \Sigma^+(X, \cm)$ satisfying:
-    \begin{enumerate}[label=(\roman*)]
-        \item There exists $\delta > 0$ such that $\phi + \delta \le f$ on $\bracs{\phi > 0}$. 
-        \item $\phi \in L^1(X, \cm)$. 
-    \end{enumerate}
-
-    To this end, let $\eps > 0$. Since $\mu\bracs{\phi > 0} < \infty$, by (b), there exists $\alpha \in A$ such that
-    \[
-    \mu\bracs{\phi > 0, f_\alpha + \delta < \phi} \le \mu\bracs{\phi > 0, f_\alpha + \delta < f} < \frac{\eps}{\norm{\phi}_u}
-    \]
-
-    In which case, by \hyperref[linearity]{proposition:lebesgue-simple-properties},
-    \begin{align*}
-        \int \phi d\mu &= \int_{\bracs{\phi > 0}}\phi d\mu = \int_{\bracs{\phi > 0, f_\alpha + \delta \ge \phi}} \phi d\mu + \int_{\bracs{\phi > 0, f_\alpha + \delta < \phi}} \phi d\mu \\
-        &\le \int_{\bracs{\phi > 0, f_\alpha + \delta \ge \phi}} f_\alpha d\mu +\norm{\phi}_u\mu \bracs{\phi > 0, f_\alpha + \delta < \phi} \\
-        &\le \int f_\alpha d\mu + \norm{\phi}_u \frac{\eps}{\norm{\phi}_u} = \int f_\alpha d\mu + \eps
-    \end{align*}
-
-    As the above holds for all $\eps > 0$, $\int \phi d\mu \le \sup_{\alpha \in A}\int f_\alpha d\mu$. Therefore
-    \[
-    \int f d\mu = \sup_{\alpha \in A}\int f_\alpha d\mu = \lim_{\alpha \in A}\int f_\alpha d\mu
-    \]
-\end{proof}
-

src/measure/measurable-maps/index.tex

@@ -8,3 +8,4 @@
 \input{./metric.tex}
 \input{./approx.tex}
 \input{./in-measure.tex}
+\input{./locally-in-measure.tex}

src/measure/measurable-maps/local-in-measure.tex

@@ -0,0 +1,116 @@
+\section{Local Convergence in Measure}
+\label{section:locally-in-measure}
+
+\begin{definition}[Locally In Measure*]
+\label{definition:locally-in-measure}
+    Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} measure space and $(Y, d)$ be a separable metric space. For each $\eps, \delta > 0$ and $A \in \cf$, let
+    \[
+    U(A, \delta, \eps) = \bracs{(f, g) \in \mathcal{L}^0(X; Y)| \mu(A \cap \bracs{d(f, g) > \delta}) < \eps}
+    \]
+
+    then
+    \[
+    \fB = \bracs{U(A, \delta, \eps)|\eps, \delta > 0, A \in \cm, \mu(A) < \infty}
+    \]
+
+    forms a fundamental system of entourages for a uniformity. 
+    
+    The uniformity defined by $\fB$ is the \textbf{uniform structure of local convergence in measure}, and $\mathcal{L}_\cf^0(X; Y)$ denotes $\mathcal{L}^0(X; Y)$ equipped with this uniformity.
+\end{definition}
+\begin{proof}
+    It is sufficient to check the conditions of \autoref{proposition:fundamental-entourage-criterion}:
+    \begin{enumerate}
+        \item[(FB1)] For each $\eps, \eps', \delta, \delta' > 0$ and $A, A' \in \cm$ with $\mu(A), \mu(A') < \infty$, 
+        \[
+        U(A \cup A', \delta \wedge \delta', \eps \wedge \eps') \subset U(A, \delta, \eps) \cap U(A', \delta', \eps')
+        \]
+        \item[(UB3)] For each $\eps, \delta > 0$, $A \in \cf$, and $f, g, h \in \mathcal{L}^0(X; Y)$, 
+        \[
+        \bracs{d(f, h) > \delta} \subset \bracs{d(f, g) > \delta} \cup \bracs{d(g, h) > \delta}
+        \]
+
+        so $U(A, \delta/2, \eps/2) \circ U(A, \delta/2, \eps/2) \subset U(A, \delta, \eps)$. 
+    \end{enumerate}
+\end{proof}
+
+\begin{proposition}
+\label{proposition:convergence-in-measure}
+    Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} measure space, $(Y, d)$ be a separable metric space, and $\fF$ be a filter of $(\cm, \cb_Y)$-measurable functions, then $\fF$ is Cauchy in measure if and only if:
+    \begin{enumerate}
+        \item[(L)] $\fF$ is \hyperref[definition:locally-in-measure]{definition:locally-in-measure}. 
+        \item[(T)] For each $\eps, \delta > 0$, there exists $F \in \fF$ and $A \in \cf$ such that
+        \[
+        \sup_{f, g \in F}\mu(A^c \cap \bracs{d(f, g) > \delta}) < \eps
+        \]
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (L) + (T) $\Rightarrow$ (In Measure): Let $\eps, \delta > 0$. By (T) then there exists $F_1 \in \fF$ and $A \in \cf$ such that 
+    \[
+    \sup_{f, g \in F_1}\mu(A^c \cap \bracs{d(f, g) > \delta}) < \eps
+    \]
+
+    By (L), there exists $F_2 \in \fF$ with $F_2 \subset F_1$ such that
+    \[
+    \sup_{f, g \in F_2}\mu(A \cap \bracs{d(f, g) > \delta}) < \eps
+    \]
+
+    Therefore
+    \[
+    \sup_{f, g \in F_2}\mu\bracs{d(f, g) > \delta} < 2\eps
+    \]
+\end{proof}
+
+\begin{theorem}
+\label{theorem:locally-in-measure-complete}
+    Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} localisable measure space and $(Y, d)$ be a Polish space, then \hyperref[$\mathcal{L}^0_\cf(X; Y)$]{definition:locally-in-measure} is complete. 
+\end{theorem}
+\begin{proof}
+    Let $\fF \subset \mathcal{L}^0_\cf(X; Y)$ be a Cauchy filter. By \autoref{theorem:cauchy-in-measure-limit}, for each $A \in \cf$, there exists an almost everywhere unique $f_A \in \mathcal{L}^0(A; Y)$ such that $\fF$ converges to $f_A$ when restricted to $A$. Thus for any $A, B \in \cf$, $f_{A \cup B}|_{A \cap B} = f_A|_{A \cap B} = f_B|_{B \cap A}$ almost everywhere. By the \hyperref[gluing lemma for measurable functions]{lemma:gluing-measurable}, there exists $f \in \mathcal{L}^0(X; Y)$ such that $f|_A = f_A$ for all $A \in \cf$. Thus $\fF \to f$ locally in measure, and $\mathcal{L}^0(X; Y)$ is complete. 
+\end{proof}
+
+\begin{theorem}[Monotone Convergence Theorem (in Measure)]
+\label{theorem:mct-measure}
+    Let $(X, \cm, \mu)$ be a semifinite measure space, $\net{f} \subset \mathcal{L}^+(X, \cm)$, and $f \in \mathcal{L}^+(X, \cm)$ such that
+    \begin{enumerate}[label=(\alph*)]
+        \item For each $x \in X$, $f_\alpha(x) \upto f(x)$. 
+        \item $f_\alpha \to f$ locally in measure. 
+    \end{enumerate}
+
+    then
+    \[
+    \lim_{\alpha \in A}\int f_\alpha d\mu = \int f d\mu
+    \]
+\end{theorem}
+\begin{proof}
+    By \autoref{definition:lebesgue-non-negative}, $\int f_\alpha d\mu \le \int f d\mu$ for each $\alpha \in A$. 
+
+    On the other hand, using \autoref{lemma:lebesgue-non-negative-strict}, it is sufficient to show that
+    \[
+    \lim_{\alpha \in A}\int f_\alpha d\mu = \sup_{\alpha \in A}\int f_\alpha d\mu \ge \int \phi d\mu
+    \]
+    
+    for any $\phi \in \Sigma^+(X, \cm)$ satisfying:
+    \begin{enumerate}[label=(\roman*)]
+        \item There exists $\delta > 0$ such that $\phi + \delta \le f$ on $\bracs{\phi > 0}$. 
+        \item $\phi \in L^1(X, \cm)$. 
+    \end{enumerate}
+
+    To this end, let $\eps > 0$. Since $\mu\bracs{\phi > 0} < \infty$, by (b), there exists $\alpha \in A$ such that
+    \[
+    \mu\bracs{\phi > 0, f_\alpha + \delta < \phi} \le \mu\bracs{\phi > 0, f_\alpha + \delta < f} < \frac{\eps}{\norm{\phi}_u}
+    \]
+
+    In which case, by \hyperref[linearity]{proposition:lebesgue-simple-properties},
+    \begin{align*}
+        \int \phi d\mu &= \int_{\bracs{\phi > 0}}\phi d\mu = \int_{\bracs{\phi > 0, f_\alpha + \delta \ge \phi}} \phi d\mu + \int_{\bracs{\phi > 0, f_\alpha + \delta < \phi}} \phi d\mu \\
+        &\le \int_{\bracs{\phi > 0, f_\alpha + \delta \ge \phi}} f_\alpha d\mu +\norm{\phi}_u\mu \bracs{\phi > 0, f_\alpha + \delta < \phi} \\
+        &\le \int f_\alpha d\mu + \norm{\phi}_u \frac{\eps}{\norm{\phi}_u} = \int f_\alpha d\mu + \eps
+    \end{align*}
+
+    As the above holds for all $\eps > 0$, $\int \phi d\mu \le \sup_{\alpha \in A}\int f_\alpha d\mu$. Therefore
+    \[
+    \int f d\mu = \sup_{\alpha \in A}\int f_\alpha d\mu = \lim_{\alpha \in A}\int f_\alpha d\mu
+    \]
+\end{proof}
+

src/measure/measure/localisable.tex

@@ -184,9 +184,9 @@
         \item For each $A \in \cf$, $f_n|_A = f_{A, n}$ almost everywhere. 
     \end{enumerate}
     
-    Since $Y$ is Polish, \autoref{proposition:metric-measurable-limit} implies that $\bracsn{\limv{n}f_n \text{ exists}} \in \cm$. For each $A \in \cf$, $f_{A, n} \to f_n$ pointwise by (i). For each$n \in \natp$, $f_{A, n} = f_n|_A$ almost everywhere on $A$ by (2). Thus
+    Since $Y$ is Polish, \autoref{proposition:metric-measurable-limit} implies that $\bracsn{\limv{n}f_n \text{ exists}} \in \cm$. For each $A \in \cf$, $f_{A, n} \to f_A$ pointwise by (i). For each$n \in \natp$, $f_{A, n} = f_n|_A$ almost everywhere on $A$ by (2). Thus
     \begin{align*}
-        \bracs{\limv{n}f_n \text{ exists}} \cap A &\subset \bigcap_{n \in \natp}\bracs{f_{A, n} = f_n|_A} \\ 
+        \bracs{\limv{n}f_n \text{ exists}} \cap A &\supset \bigcap_{n \in \natp}\bracs{f_{A, n} = f_n|_A} \\ 
         \mu\paren{\bracs{\limv{n}f_n \text{ exists}} \cap A} &= \mu\paren{\bigcap_{n \in \natp}\bracs{f_{A, n} = f_n|_{A}}} = \mu(A) \\
         \mu\paren{\bracs{\limv{n}f_n \text{ does not exist}} \cap A} &= 0
     \end{align*}

src/measure/measure/semifinite.tex

@@ -63,23 +63,15 @@
 
 \begin{definition}[Scaffold*]
 \label{definition:measure-scaffold}
-    Let $(X, \cm, \mu)$ be a measure space and $\cf \subset \bracs{A \in \cm|\mu(A) < \infty}$, then $\cf$ is a \textbf{scaffold} for $\mu$ if for all $E \in \cm$, 
+    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}. 
-\end{definition}
 
-\begin{lemma}
-\label{lemma:measure-scaffold-semifinite}
-    Let $(X, \cm, \mu)$ be a measure space:
-    \begin{enumerate}
-        \item If $\cf \subset \cm$ is a \hyperref[scaffold]{definition:measure-scaffold} for $\mu$, then $\mu$ is semifinite.
-        \item If $\mu$ is semifinite, then $\bracs{A \in \cm|\mu(A) < \infty}$ is a scaffold for $\mu$. 
-    \end{enumerate}
-\end{lemma}
-% Omitted
+    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 $\mu$.