Fixed typo.

src/cat/gluing/gluing.tex

@@ -0,0 +1,47 @@
+\section{Gluing Lemmas}
+\label{section:gluing}
+
+
+
+\begin{lemma}[Gluing for Functions]
+\label{lemma:glue-function}
+    Let $X, Y$ be sets, $\seq{U_i} \subset 2^X$, and $\seqi{f}$ with $f_i: U_i \to Y$ for all $i \in I$. If:
+    \begin{enumerate}
+        \item[(a)] $\bigcup_{i \in I}U_i = X$.
+        \item[(b)] For each $i, j \in I$, either $U_i \cap U_j = \emptyset$, or $f_i|_{U_i \cap U_j} = f_j|_{U_i \cap U_j}$.
+    \end{enumerate}
+
+    then there exists a unique $f: X \to Y$ such that $f|_{U_i} = f_i$ for all $i \in I$.
+\end{lemma}
+\begin{proof}
+    For each $i \in I$, let $\Gamma_i \subset U_i \times Y$ be the graph of $f_i$. Let $\Gamma = \bigcup_{i \in I}\Gamma_i$, then:
+    \begin{enumerate}
+        \item By assumption (a), $\bracs{x|(x, y) \in \Gamma} = \bigcup_{i \in I}U_i = X$.
+        \item For any $x \in X$, there exists $y \in Y$ with $(x, y) \in \Gamma$, and $i \in I$ such that $(x, y) \in \Gamma_i$. If $(x, y') \in \Gamma_j \subset \Gamma$, then $x \in U_i \cap U_j \ne \emptyset$. By assumption (b), $y = y'$.
+    \end{enumerate}
+
+    Thus $\Gamma$ is the graph of a function $f: X \to Y$ with $f|_{U_i} = f_i$ for all $i \in I$.
+\end{proof}
+
+
+\begin{lemma}[Gluing for Linear Functions]
+\label{lemma:glue-linear}
+    Let $E, F$ be vector spaces over a field $K$, $\fF$ be a family of subspaces of $E$, and $\bracs{T_V}_{V \in \fF}$ with $T_V \in \hom(V; F)$ for all $V \in \fF$. If:
+    \begin{enumerate}
+        \item[(a)] $\bigcup_{V \in \fF}V = E$.
+        \item[(b)] For each $V, W \in \fF$, $T_V|_{V \cap W} = T_W|_{V \cap W}$.
+        \item[(c)] $\fF$ is upward-directed with respect to includion.
+    \end{enumerate}
+
+    then there exists a unique $T \in \hom(E; F)$ such that $T|_{V} = T_V$ for all $V \in \fF$.
+\end{lemma}
+\begin{proof}
+    By (a), (b), and \autoref{lemma:glue-function}, there exists a unique $T: E \to F$ such that $T|_V = T_V$ for all $V \in \fF$.
+
+    Let $x, y \in E$ and $\lambda \in \fF$. By assumption (a), there exists $V_x, V_y \in \fF$ with $x \in V_x$ and $y \in V_y$. By assumption (3), there exists $V \in \fF$ with $V \supset V_x \cup V_y$. Hence
+    \[
+    T(\lambda x + y) = T_V(\lambda x + y) = \lambda T_Vx + T_Vy = \lambda Tx + Ty
+    \]
+
+    and $T \in \hom(E; F)$.
+\end{proof}

src/cat/gluing/index.tex

@@ -1,46 +1,8 @@
-\chapter{Gluing Lemmas}
+\chapter{Functions}
 \label{chap:gluing}
 
-
+The following chapter contains certain tricks in working with functions in an abstract setting. 
-\begin{lemma}[Gluing for Functions]
-\label{lemma:glue-function}
-    Let $X, Y$ be sets, $\seq{U_i} \subset 2^X$, and $\seqi{f}$ with $f_i: U_i \to Y$ for all $i \in I$. If:
-    \begin{enumerate}
-        \item[(a)] $\bigcup_{i \in I}U_i = X$.
-        \item[(b)] For each $i, j \in I$, either $U_i \cap U_j = \emptyset$, or $f_i|_{U_i \cap U_j} = f_j|_{U_i \cap U_j}$.
-    \end{enumerate}
-
-    then there exists a unique $f: X \to Y$ such that $f|_{U_i} = f_i$ for all $i \in I$.
-\end{lemma}
-\begin{proof}
-    For each $i \in I$, let $\Gamma_i \subset U_i \times Y$ be the graph of $f_i$. Let $\Gamma = \bigcup_{i \in I}\Gamma_i$, then:
-    \begin{enumerate}
-        \item By assumption (a), $\bracs{x|(x, y) \in \Gamma} = \bigcup_{i \in I}U_i = X$.
-        \item For any $x \in X$, there exists $y \in Y$ with $(x, y) \in \Gamma$, and $i \in I$ such that $(x, y) \in \Gamma_i$. If $(x, y') \in \Gamma_j \subset \Gamma$, then $x \in U_i \cap U_j \ne \emptyset$. By assumption (b), $y = y'$.
-    \end{enumerate}
-
-    Thus $\Gamma$ is the graph of a function $f: X \to Y$ with $f|_{U_i} = f_i$ for all $i \in I$.
-\end{proof}
 
 
-\begin{lemma}[Gluing for Linear Functions]
+\input{./gluing.tex}
-\label{lemma:glue-linear}
+\input{./level.tex}
-    Let $E, F$ be vector spaces over a field $K$, $\fF$ be a family of subspaces of $E$, and $\bracs{T_V}_{V \in \fF}$ with $T_V \in \hom(V; F)$ for all $V \in \fF$. If:
-    \begin{enumerate}
-        \item[(a)] $\bigcup_{V \in \fF}V = E$.
-        \item[(b)] For each $V, W \in \fF$, $T_V|_{V \cap W} = T_W|_{V \cap W}$.
-        \item[(c)] $\fF$ is upward-directed with respect to includion.
-    \end{enumerate}
-
-    then there exists a unique $T \in \hom(E; F)$ such that $T|_{V} = T_V$ for all $V \in \fF$.
-\end{lemma}
-\begin{proof}
-    By (a), (b), and \autoref{lemma:glue-function}, there exists a unique $T: E \to F$ such that $T|_V = T_V$ for all $V \in \fF$.
-
-    Let $x, y \in E$ and $\lambda \in \fF$. By assumption (a), there exists $V_x, V_y \in \fF$ with $x \in V_x$ and $y \in V_y$. By assumption (3), there exists $V \in \fF$ with $V \supset V_x \cup V_y$. Hence
-    \[
-    T(\lambda x + y) = T_V(\lambda x + y) = \lambda T_Vx + T_Vy = \lambda Tx + Ty
-    \]
-
-    and $T \in \hom(E; F)$.
-\end{proof}

src/cat/gluing/level.tex

@@ -0,0 +1,42 @@
+\section{Preimages}
+\label{section:preimage}
+
+\begin{definition}[Preimage Function]
+\label{definition:preimage-function}
+    Let $X, Y$ be sets and $P: 2^Y \to 2^X$, then $P$ is a \textbf{preimage function} if
+    \begin{enumerate}[label=(PF\arabic*)]
+        \item $P(\emptyset) = \emptyset$. 
+        \item For each $\mathcal{S} \subset 2^Y$, $\bigcup_{S \in \mathcal{S}}P(S) = P\paren{\bigcup_{S \in \mathcal{S}}S}$. 
+        \item For each $\mathcal{S} \subset 2^Y$, $\bigcap_{S \in \mathcal{S}}P(S) = P\paren{\bigcap_{S \in \mathcal{S}}}$. 
+    \end{enumerate}
+
+    and $P$ is \textbf{total} if
+    \begin{enumerate}
+        \item[(T)] $P(Y) = X$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{proposition}
+\label{proposition:preimage-gymnastics}
+    Let $X$ and $Y$ be sets, then:
+    \begin{enumerate}
+        \item For any $f: X \to Y$, the mapping $S \mapsto f^{-1}(S)$ is a total preimage function.
+        \item For any total preimage function $P: 2^Y \to 2^X$, there exists a unique $f: X \to Y$ such that $P(S) = f^{-1}(S)$ for all $S \in 2^Y$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2): Let $x, y \in Y$ with $x \ne y$, then by (PF1) and (PF3), 
+    \[
+    P(\bracs{x}) \cap P(\bracs{y}) = P(\bracs{x} \cap \bracs{y}) = P(\emptyset) = \emptyset
+    \]
+
+    By (T) and (PF2), $X = P(Y) = P\paren{\bigcup_{y \in Y}\bracs{y}} = \bigcup_{y \in Y}P(\bracs{y})$. Therefore $X = \bigsqcup_{y \in Y}P(\bracs{y})$. 
+
+    Therefore for each $x \in X$, there exists a unique $f(x) \in Y$ such that $x \in P(f(x))$. The association $x \mapsto f(x)$ then is the unique desired function. 
+\end{proof}
+
+
+
+
+
+

src/fa/duality/mackey.tex

@@ -66,3 +66,19 @@
     On the other hand, let $\mathcal{T} \subset 2^E$ be a locally convex topology consistent with $\dpn{E, F}{\lambda}$. By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, every $\mathcal{T}$-equicontinuous set is relatively $\sigma(F, E)$-compact. Therefore $\mathcal{T}$ is coarser than the topology of uniform convergence on relatively $\sigma(E, F)$-compact, convex, and circled sets. 
 \end{proof}
 
+\begin{definition}[Mackey Space]
+\label{definition:mackey-space}
+    Let $E$ be a separated locally convex space over $K \in \RC$, then $E$ is a \textbf{Mackey space} if $E$ is equipped with the Mackey topology of $\dpn{E, E^*}{E}$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:barreled-mackey}
+    Let $E$ be a separated barreled space over $K \in \RC$, then $E$ is a Mackey space. 
+\end{proposition}
+\begin{proof}
+    Let $\cf \subset E^*$ be a $\sigma(E^*, E)$-compact set and $U \in \cn_{K}(0)$ be a barrel, then $V = \bigcap_{\phi \in \cf}\phi^{-1}(U)$ is convex, circled, and closed. For each $x \in E$, $\cf(x) = \bracs{\dpn{x, \phi}{E}|\phi \in \cf}$ is bounded. Thus $V$ is absorbing and hence a barrel. Since $E$ is barreled, $V \in \cn_E(0)$. Therefore the Mackey topology is contained in the topology of $E$, and $E$ is a Mackey space. 
+\end{proof}
+
+
+
+

src/fa/lp/ui.tex

@@ -48,7 +48,7 @@
 
 \begin{theorem}[Vitali Convergence Theorem]
 \label{theorem:vitali-convergence}
-    Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, and $\fF \subset 2^{L^p(X; E)}$ be a filter, then $\fF$ is Cauchy in $L^p(X; E)$ if and only if:
+    Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a separable normed vector space over $K \in \RC$, and $\fF \subset 2^{L^p(X; E)}$ be a filter, then $\fF$ is Cauchy in $L^p(X; E)$ if and only if:
     \begin{enumerate}
         \item[(M)] $\fF$ is locally Cauchy in measure. 
         \item[(UI)] For each $\eps > 0$, there exists $M \ge 0$ and $F \in \fF$ such that
@@ -142,7 +142,7 @@
 
 \begin{corollary}[Dominated Convergence Theorem (In Measure)]
 \label{corollary:dct-filter}
-    Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, $\fF \subset 2^{L^p(X; E)}$ be a filter, and $g, h \in L^p(X; \real)$ such that:
+    Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a separable normed vector space over $K \in \RC$, $\fF \subset 2^{L^p(X; E)}$ be a filter, and $g, h \in L^p(X; \real)$ such that:
     \begin{enumerate}[label=(\alph*)]
         \item[(M)] $\fF \to g$ locally in measure.
         \item[(D)] There exists $F \in \fF$ such that $|f| \le h$ for all $f \in F$. 
@@ -159,7 +159,7 @@
 
 \begin{lemma}[Scheffé]
 \label{lemma:scheffe}
-    Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, $\fF \subset 2^{L^p(X; E)}$ be a filter, and $g \in L^p(X; E)$, then $\fF \to g$ in $L^p(X; E)$ if and only if:
+    Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a separable normed vector space over $K \in \RC$, $\fF \subset 2^{L^p(X; E)}$ be a filter, and $g \in L^p(X; E)$, then $\fF \to g$ in $L^p(X; E)$ if and only if:
     \begin{enumerate}
         \item[(M)] $\fF \to g$ locally in measure. 
         \item[(N)] $\lim_{f, \fF} \norm{f}_{L^p(X; E)} = \norm{g}_{L^p(X; E)}$. 

src/fa/tvs/equicontinuous.tex

@@ -43,7 +43,7 @@
 
     and that
     \begin{enumerate}
-        \item[(B2')] For each $x \in E$, $\alg(x) = \bracs{Tx|T \in \alg}$ is bounded in $F$. 
+        \item[(B2)] For each $x \in E$, $\alg(x) = \bracs{Tx|T \in \alg}$ is bounded in $F$. 
     \end{enumerate}
 
     then

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

@@ -3,7 +3,7 @@
 
 \begin{definition}[In Measure]
 \label{definition:in-measure}
-    Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a metric space. For each $\eps, \delta > 0$, let
+    Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a separable metric space. For each $\eps, \delta > 0$, let
     \[
     U(\delta, \eps) = \bracs{(f, g) \in \mathcal{L}^0(X; Y)| \mu\bracs{d(f, g) > \delta} < \eps}
     \]
@@ -33,7 +33,7 @@
 
 \begin{definition}[Ky Fan Metric]
 \label{definition:ky-fan}
-    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and
+    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a separable metric space, and
     \[
     \alpha: L^0(X; Y)^2 \to [0, \infty) \quad (f, g) \mapsto \inf\bracs{\eps > 0| \mu\bracs{d(f, g) > \eps} \le \eps} \wedge 1
     \]
@@ -76,7 +76,7 @@
 
 \begin{definition}[Locally In Measure]
 \label{definition:locally-in-measure}
-    Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a metric space. For each $\eps, \delta > 0$ and $A \in \cm$ with $\mu(A) < \infty$, let
+    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
     \[
     U(A, \delta, \eps) = \bracs{(f, g) \in \mathcal{L}^0(X; Y)| \mu(A \cap \bracs{d(f, g) > \delta}) < \eps}
     \]
@@ -106,7 +106,7 @@
 
 \begin{proposition}
 \label{proposition:convergence-in-measure}
-    Let $(X, \cm, \mu)$ be a semifinite measure space, $(Y, d)$ be a metric space, and $\fF$ be a filter of $(\cm, \cb_Y)$-measurable functions, then $\fF$ is Cauchy in measure if and only if:
+    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
@@ -135,7 +135,7 @@
 
 \begin{lemma}
 \label{lemma:ae-in-measure}
-    Let $(X, \cm, \mu)$ be a finite measure space, $(Y, d)$ be a 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.
+    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.
 \end{lemma}
 \begin{proof}
     Let $\eps > 0$, then for almost every $x \in X$, there exists $N \in \natp$ such that $d(f_n(x), f(x)) < \eps$ for all $n \ge N$, so
@@ -151,7 +151,7 @@
 
 \begin{theorem}
 \label{theorem:cauchy-in-measure-limit}
-    Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a complete metric space, then:
+    Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a Polish space, then:
     \begin{enumerate}
         \item For any $\seq{f_n} \subset L^0(X; Y)$ that is Cauchy in measure, there exists $f \in L^0(X; Y)$ and a subsequence $\seq{n_k}$ such that $f_{n_k} \to f$ almost everywhere and in measure.
         \item $L^0(X; Y)$ equipped with the uniform structure of convergence in measure is complete. 

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}
+
+

src/topology/main/index.tex

@@ -26,3 +26,4 @@
 \input{./baire.tex}
 \input{./cube.tex}
 \input{./compactify.tex}
+\input{./preimage.tex}

src/topology/main/preimage.tex

@@ -0,0 +1,120 @@
+\section{Open Preimage Functions}
+\label{section:preimage-function-topology}
+
+\begin{definition}[Open Preimage Function]
+\label{definition:open-preimage-function}
+    Let $X$ be a set, $(Y, \topo)$ be a topological space, and $P: \topo \to 2^X$, then $P$ is an \textbf{open preimage function} if
+    \begin{enumerate}[label=(PF\arabic*)]
+        \item $P(\emptyset) = \emptyset$. 
+        \item[(PF2')] For each $\mathcal{U} \subset \topo$, $\bigcup_{U \in \mathcal{U}}P(U) = P\paren{\bigcup_{U \in \mathcal{U}}U}$. 
+        \item[(PF3')] For each $U, V \in \topo$, $P(U \cap V) = P(U) \cap P(V)$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{proposition}
+\label{proposition:open-preimage-function-gymnastics}
+    Let $X$ be a set, $(Y, \topo)$ be a topological space, and $f: X \to Y$, then:
+    \begin{enumerate}
+        \item The mapping $U \mapsto f^{-1}(U)$ is an open preimage function. 
+        \item If $Y$ is T1, then for any $g: X \to Y$ with $g^{-1}(U) = f^{-1}(U)$ for all $U \in \topo$, $f = g$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2): Let $y \in Y$, then since $Y$ is T1, $\bracs{x} = \bigcap_{U \in \topo, y \in U}U$, so
+    \[
+    g^{-1}(\bracs{y}) = \bigcap_{\substack{U \in \topo \\ y \in U}}g^{-1}(U) = \bigcap_{\substack{U \in \topo \\ y \in U}}f^{-1}(U) = f^{-1}(\bracs{y})
+    \]
+    
+    Thus for any $x \in X$, $f(x) = y$ if and only if $g(x) = y$, so $f = g$. 
+\end{proof}
+
+
+\begin{definition}[Basic Preimage Function]
+\label{definition:basic-preimage-function}
+    Let $X$ be a set, $Y$ be a topological space, $\mathcal{B}$ be a base for $Y$ with $\emptyset \in \mathcal{B}$, and $p: \mathcal{B} \to 2^X$, then $p$ is a \textbf{basic preimage function} if:
+    \begin{enumerate}[label=(PF\arabic*)]
+        \item $P(\emptyset) = \emptyset$. 
+        \item[(PF2')] For each $\mathcal{U} \subset \mathcal{B}$ and $V \in \mathcal{B}$ with $V \subset \bigcup_{U \in \mathcal{U}}U$, $p(V) \subset \bigcup_{U \in \mathcal{U}}p(U)$.  
+        \item[(PF3')] For each $U, V \in \mathcal{B}$, $p(U) \cap p(V) \subset \bigcup_{W \in \mathcal{B}, W \subset U \cap V}p(W)$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{proposition}
+\label{proposition:basic-preimage-function}
+    Let $X$ be a set, $(Y, \topo)$ be a topological space, and $\mathcal{B}$ be a base for $Y$ with $\emptyset \in \mathcal{B}$, then:
+    \begin{enumerate}
+        \item For any open preimage function $P: \topo \to 2^X$, $P|_{\mathcal{B}}$ is a basic preimage function.
+        \item For any basic preimage function $p: \mathcal{B} \to 2^X$, there exists a unique open preimage function $P: \topo \to 2^X$ such that $p = P|_{\mathcal{B}}$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2): For each $U \in \topo$, let $\mathcal{B}(U) = \bracs{V \in \mathcal{B}|V \subset U}$, and
+    \[
+    P: \topo \to 2^X \quad U \mapsto \bigcup_{V \in \mathcal{B}(U)}p(V)
+    \]
+
+    then $P(\emptyset) = \emptyset$. For any $U \in\topo$, $U = \bigcup_{V \in \mathcal{B}(U)}V$, so if $P$ is an extension of $p$ to $\topo$ as an open preimage function, then $P$ must be unique by (PF2'). 
+
+    Let $\mathcal{U} \subset \topo$. Since $\bigcup_{U \in \mathcal{U}}\mathcal{B}(U) \subset \mathcal{B}\paren{\bigcup_{U \in \mathcal{U}}U}$, $\bigcup_{U \in \mathcal{U}}P(U) \subset P\paren{\bigcup_{U \in \mathcal{U}}U}$. On the other hand, for each $V \in \mathcal{B}\paren{\bigcup_{U \in \mathcal{U}}U}$, $V \subset \bigcup_{U \in \mathcal{U}}\bigcup_{W \in \mathcal{B}(U)}W$. By (PF2'),
+    \[
+    p(V) \subset \bigcup_{U \in \mathcal{U}}\bigcup_{W \in \mathcal{B}(U)}p(W) = \bigcup_{U \in \mathcal{U}}P(U)
+    \]
+
+    so
+    \[
+    P\paren{\bigcup_{U \in \mathcal{U}}U} = \bigcup_{V \in \mathcal{B}(\bigcup_{U \in \mathcal{U}}U)}p(V) \subset \bigcup_{U \in \mathcal{U}}P(U)
+    \]
+
+    Finally, let $U, V \in \topo$, then since $\mathcal{B}(U \cap V) = \mathcal{B}(U) \cap \mathcal{B}(V)$,
+    \begin{align*}
+    P(U \cap V) &= \bigcup_{W \in \mathcal{B}(U \cap V)}p(W) \\
+    &\subset \bigcup_{W \in \mathcal{B}(U)}p(W) \cap \bigcup_{W \in \mathcal{B}(V)}p(W) = P(U) \cap P(V)
+    \end{align*}
+    
+
+    On the other hand, by (PF3'), 
+    \begin{align*}
+        P(U) \cap P(V) &= \bigcup_{W \in \mathcal{B}(U)}\bigcup_{W' \in \mathcal{B}(V)}p(W) \cap p(W') \\
+        &\subset \bigcup_{W \in \mathcal{B}(U)}\bigcup_{W' \in \mathcal{B}(V)}\bigcup_{S \in \mathcal{B}(W \cap W')}p(S) \\
+        &\subset \bigcup_{W \in \mathcal{B}(U \cap V)} p(W) = P(U \cap V)
+    \end{align*}
+
+    so $P(U \cap V) = P(U) \cap P(V)$. Therefore $P$ is an open preimage function. 
+\end{proof}
+
+\begin{theorem}
+\label{theorem:open-preimage-function-existence}
+    Let $X$ be a set, $(Y, \fU)$ be a complete Hausdorff uniform space with topology $\topo$, and $P: \topo \to 2^X$ be an open preimage function such that:
+    \begin{enumerate}
+        \item[(S)] For each $x \in X$ and $U \in \fU$, there exists a $U$-small set $V \in \topo$ such that $x \in P(V)$. 
+    \end{enumerate}
+     
+    then there exists a unique $f: X \to Y$ such that $P(U) = f^{-1}(U)$ for all $U \in \topo$.
+\end{theorem}
+\begin{proof}
+    Since $P$ is an open preimage function and $Y$ is Hausdorff, \autoref{proposition:open-preimage-function-gymnastics} implies that such a function is unique if it exists, so it is sufficient to demonstrate existence. 
+
+    For each $x \in X$, let $\fF(x) = \bracs{U \in \topo|x \in P(U)}$, then by (S), $\fF(x)$ is non-empty. By (PF1) and (PF3'), $\fF(x)$ is a filter base in $(Y, \fU)$. Moreover, (S) implies that $\fF(x)$ is a Cauchy filter base. Since $Y$ is complete and Hausdorff, there exists a unique $y \in Y$ such that $\fF(x) \to y$. Thus the function in question must be
+    \[
+    f: X \to Y \quad x \mapsto \lim_{y, \fF(x)}y
+    \]
+
+    It remains to verify that $P$ is the open preimage function of $f$. To this end, let $U \in \topo$ and $x \in X$ such that $f(x) \in U$. Since $U \in \cn_Y(f(x))$, there exists a symmetric entourage $V \in \fU$ such that $(V \circ V)(f(x)) \subset U$. By (S), there exists a $V$-small set $W \in \fF(x)$. As $f(x) \in \ol W \subset V \circ W$ and $W$ is $V$-small, $W \subset (V \circ V)(f(x)) \subset U$. Thus $x \in P(W) \subset P(U)$, and $f^{-1}(U) \subset P(U)$. 
+
+    On the other hand, let $x \in P(U)$. Suppose for contradiction that $f(x) \not\in U$. For each $y \in U$, there exists $V_y \in \cn_Y^o(y)$ and $W_y \in \cn_Y^o(f(x))$ such that $V_y \subset U$ and $V_y \cap W_y = \emptyset$. Since $x \in f^{-1}(W_y) \subset P(W_y)$ and $P(V_y) \cap P(W_y) = P(V_y \cap W_y) = \emptyset$, $x \not\in P(V_y)$. By (PF2'), $\bigcup_{y \in U}P(V_y) = P(U)$, so $x \not\in P(U)$, which is a contradiction. Therefore $f^{-1}(U) = P(U)$ for all $U \in \topo$. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:basic-preimage-function-existence}
+    Let $X$ be a set, $(Y, \fU)$ be a complete Hausdorff uniform space, $\mathcal{B}$ be a base for $Y$ with $\emptyset \in \mathcal{B}$, and $p: \mathcal{B} \to 2^X$ be a basic preimage function such that:
+    \begin{enumerate}
+        \item[(S')] For each $x \in X$ and $U \in \fU$, there exists a $U$-small set $V \in \mathcal{B}$ such that $x \in P(V)$. 
+    \end{enumerate}
+     
+    then there exists a unique $f: X \to Y$ such that $p(U) = f^{-1}(U)$ for all $U \in \mathcal{B}$.
+\end{corollary}
+\begin{proof}
+    Let $\topo$ be the topology of $Y$. By \autoref{proposition:basic-preimage-function}, $p$ extends to a unique open preimage function $P: \topo \to 2^X$. Since $\mathcal{B} \subset \topo$, \autoref{theorem:open-preimage-function-existence} implies that there exists a unique $f: X \to Y$ such that $f^{-1}(U) = P^{-1}(U)$ for all $U \in \topo$, and in particular for all $U \in \mathcal{B}$. 
+\end{proof}
+
+

src/topology/metric/metric.tex

@@ -68,6 +68,12 @@
     (2) $\Rightarrow$ (3): Since $\fF$ is Cauchy, there exists $\seq{E_n} \subset \fF$ such that for each $n \in \natp$, $E_n \supset E_{n+1}$ and $\sup_{y, z \in E_n}d(y, z) \le 1/n$. For each $n \in \natp$, let $x_n \in E_n$, then there exists a subsequence $\seq{n_k}$ and $x \in X$ such that $x = \limv{n}x_n$. In which case, $x \in \bigcap_{n \in \natp}\overline{E_n}$. For each $n \in \natp$, $\sup_{y, z\in E_n}d(y, z) \le 1/n$, so $B_X(x, 2/n) \supset E_n$. Therefore $\fF \to x$. 
 \end{proof}
 
+\begin{definition}[Polish Space]
+\label{definition:polish-space}
+    Let $X$ be a topological space, then $X$ is \textbf{Polish} if it is completely metrisable and second countable. 
+\end{definition}
+
+
 
 \begin{theorem}[Banach's Fixed Point Theorem]
 \label{theorem:banach-fixed-point}
@@ -98,3 +104,6 @@
     \]
 \end{proof}
 
+
+
+