Progress over the past week.

src/cat/cat/cat-func.tex

@@ -1,7 +1,5 @@
-\chapter{Categories and Functors}
-\label{chap:categories}
-
-
+\section{Categories and Functors}
+\label{section:category}
 
 \begin{definition}[Category]
 \label{definition:category}

src/cat/cat/index.tex

@@ -0,0 +1,7 @@
+\chapter{Category Theory}
+\label{chap:category}
+
+\textit{I do not know much about categories, however some concepts from it are useful in phrasing certain properties.}
+
+\input{./src/cat/cat/cat-func.tex}
+\input{./src/cat/cat/universal.tex}

src/cat/cat/universal.tex

@@ -1,5 +1,5 @@
-\chapter{Universal Construction}
-\label{chap:universal}
+\section{Universal Construction}
+\label{section:universal}
 
 \begin{definition}[Universal Object]
 \label{definition:universal}
@@ -69,6 +69,12 @@
     The directed set is \textbf{upward-directed} if it satisfies (3U), and \textbf{downward-directed} if it satisfies (3D).
 \end{definition}
 
+\begin{definition}[Cofinal]
+\label{definition:cofinal}
+    Let $(I, \lesssim)$ be a upward/downward directed set, then $J \subset I$ is \textbf{cofinal} if for every $\alpha \in I$, there exists $\beta \in J$ with $\beta \gtrsim \alpha$/$\beta \lesssim \alpha$.
+\end{definition}
+
+
 \begin{definition}[Directed System]
 \label{definition:directed-system}
     Let $\catc$ be a category and $(I, \lesssim)$ be a directed set. A \textbf{directed system} is a pair $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim j})$ such that:

src/cat/gluing/index.tex

@@ -0,0 +1,41 @@
+\chapter{Gluing Lemmas}
+\label{chap: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/index.tex

@@ -1,7 +1,5 @@
-\part{Categories}
+\part{General Tools}
 \label{part:part-categories}
 
-\textit{I do not know much about categories, however some concepts from it are useful in phrasing certain properties.}
-
-\input{./src/cat/cat-func.tex}
-\input{./src/cat/universal.tex}
+\input{./src/cat/cat/index.tex}
+\input{./src/cat/gluing/index.tex}

src/conventions/index.tex

@@ -0,0 +1,8 @@
+
+definitions: use ([acronym]1), ([acronym]2), etc. If the axioms resemble an existing structure, use their naming instead.
+
+Set builders must use bar.
+
+(U) indicates some kind of universal property, or some kind of property that makes this object unique.
+
+$\natp = \bracs{1, \cdots}$, $\nat_0 = \bracs{0, 1, \cdots}$. Nobody knows what $\nat$ is.

src/fa/index.tex

@@ -1,5 +1,5 @@
 \part{Functional Analysis}
-\label{part:part-fa}
+\label{part:fa}
 
 
 \input{./src/fa/tvs/index.tex}

src/fa/lc/convex.tex

@@ -1,6 +1,7 @@
 \section{Seminorms}
 \label{section:seminorms}
 
+
 \begin{definition}[Convex]
 \label{definition:convex}
     Let $E$ be a vector space over $K \in \RC$, then $A \subset E$ is \textbf{convex} if for any $x, y \in A$, $\bracs{\lambda x + (1 - \lambda) y| \lambda \in [0, 1]} \subset A$.

src/fa/lc/hahn-banach.tex

@@ -0,0 +1,66 @@
+\section{The Hahn-Banach Theorem}
+\label{section:hahn-banach}
+
+\begin{lemma}
+\label{lemma:hahn-banach}
+    Let $E$ be a vector space over $\real$, $\rho: E \to \real$ be a sublinear functional, $F \subset E$ be a subspace, $\phi \in \hom(F; \real)$ with $\phi(x) \le \rho(x)$ for all $x \in F$.
+
+    Let $x_0 \in E \setminus F$, $\lambda \in \real$, and define
+    \[
+    \phi_{x_0, \lambda}: F + \real x_0 \quad (y + tx_0) \mapsto \phi(y) + \lambda t
+    \]
+    then there exists $\lambda \in \real$ such that $\phi_{x_0, \lambda}(x) \le \rho(x)$ for all $x \in F + \real x_0$.
+\end{lemma}
+\begin{proof}
+    Let $x, y \in F$, then
+    \begin{align*}
+        \phi(x) + \phi(y) = \phi(x + y) &\le \rho(x + y) \le \rho(x - x_0) + \rho(y + x_0) \\
+        \phi(x) - \rho(x - x_0) &\le \rho(y + x_0) - \phi(y) \\
+        \sup_{x \in F}[\phi(x) - \rho(x - x_0)] &\le \inf_{x \in F}[\rho(x + x_0) - \phi(x)]
+    \end{align*}
+    Let
+    \[
+    \lambda \in \braks{\sup_{x \in F}[\phi(x) - \rho(x - x_0)], \inf_{x \in F}[\rho(x + x_0) - \phi(x)]}
+    \]
+    then for any $x \in F$ and $t > 0$,
+    \begin{align*}
+        \phi(x + tx_0) &= \phi(x) + t\lambda = t\braks{\phi(t^{-1}x) + \lambda} \\
+        &\le t\braks{\rho(t^{-1}x + x_0) - \phi(t^{-1}x) + \phi(t^{-1}x)} \\
+        &= t\rho(t^{-1}x + x_0) = \rho(x + tx_0) \\
+        \phi(x - tx_0) &= \phi(x) - t\lambda = t\braks{\phi(t^{-1}x) - \lambda} \\
+        &\ge t\braks{\rho(t^{-1}x - x_0) + \phi(t^{-1}x) - \phi(t^{-1}x)} \\
+        &= t\rho(t^{-1}x - x_0) = \rho(x - tx_0)
+    \end{align*}
+\end{proof}
+
+\begin{theorem}[Hahn-Banach, Analytic Form {{\cite[Theorem 5.6, 5.7]{Folland}}}]
+\label{theorem:hahn-banach}
+    Let $E$ be a vector space over $K \in \RC$, $\rho: E \to \real$ be a sublinear functional, $F \subsetneq E$ be a subspace,
+    \begin{enumerate}
+        \item For any $\phi \in \hom(F; \real)$ with $\phi \le \rho|_F$, there exists $\Phi \in \hom(E; \real)$ such that $\Phi \le \rho$ and $\Phi|_F = \phi$.
+        \item If $\rho$ is a seminorm, then for any $\phi \in \hom(F; \complex)$ with $\abs{\phi} \le \rho|_F$, there exists $\Phi \in \hom(E; \complex)$ such that $\abs{\Phi} \le \rho$ and $\Phi|_F = \phi$.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): Let $x_0 \in E \setminus F$, then by \ref{lemma:hahn-banach}, there exists $\lambda \in \real$ such that if
+    \[
+    \phi_{x_0, \lambda}: F + \real x_0 \to \real \quad (x + tx_0) \mapsto x + \lambda t
+    \]
+    then $\phi_{x_0, \lambda}(x) \le \rho(x)$ for all $x \in F + \real x_0$. Thus for any $F \subsetneq E$, $\phi$ admits an extension to a subspace that strictly contains $F$.
+
+    Let
+    \[
+    \mathbf{F} = \bracs{\Phi \in \hom(F'; \real)|F' \supset F, \Phi|_F = \phi, \Phi \le \rho|_{F'}}
+    \]
+    For any $\Phi, \Phi' \in \mathbf{F}$, define $\Phi \le \Phi'$ if $\Phi'$ is an extension of $\Phi$. Let $\cf \subset \mathbf{F}$ be a chain. For any $\Phi \in \cf$, let $\cd(\Phi)$ be its domain. Since $\bigcup_{\phi \in \cf}\cd(\Phi)$ is a subspace and $\cf$ is a chain, by \ref{lemma:glue-linear}, there exists $\Phi \in \hom\paren{\bigcup_{\Phi' \in \cf}\cd(\Phi'); \real}$ such that $\Phi|_{\cd(\Phi')} = \Phi'$ for all $\Phi' \in \cf$. Thus $\Phi \in \mathbf{F}$ is an upper bound of $\cf$.
+
+    By Zorn's lemma, $\mathbf{F}$ admits a maximal element $\Phi$. If $\cd(\Phi) \subsetneq E$, then $\Phi$ is not maximal by the preceding discussion. Therefore $\cd(\Phi) = E$ and $\Phi$ is a desired extension.
+
+    (1): Given that $\rho$ is a seminorm, for any $u \in \hom(E; \real)$, $u \le \rho$ if and only if $\abs{u} \le \rho$.
+
+    Let $u = \re{\phi}$, then $u \in \hom(E; \real)$ by \ref{proposition:polarisation-linear}. By (1), there exists $U \in \hom(E; \real)$ such that $\abs{U} \le \rho$ and $U|_F = u$. For each $x \in E$, let $\Phi(x) = U(x) - iU(ix)$, then $\Phi \in \hom(E; \complex)$ and $\Phi|_F = \phi$ by \ref{proposition:polarisation-linear}. In addition, for any $x \in E$, if $\alpha = \overline{\sgn(\Phi(x))}$, then
+    \[
+    |\Phi(x)| = \alpha\Phi(x) = \Phi(\alpha x) = U(\alpha x) \le \rho(\alpha x) = \rho(x)
+    \]
+    so $\abs{\Phi} \le \rho$.
+\end{proof}

src/fa/lc/index.tex

@@ -4,3 +4,4 @@
 
 \input{./src/fa/lc/convex.tex}
 \input{./src/fa/lc/continuous.tex}
+\input{./src/fa/lc/hahn-banach.tex}

src/fa/rs/bv.tex

@@ -47,13 +47,13 @@
     \]
     By assumption (b), $[0, M_\rho]$ is in the filter generated by $V_{\rho, P}(\fF)$. Thus $V_{\rho, P}(f) \le M_\rho$. As this holds for all $P \in \scp([a, b])$, $V_{\rho, P}(f) \le M_\rho$, and $f \in BV([a, b]; E)$.
 
-    (5): For each $n \in \nat$, let
+    (5): For each $n \in \nat^+$, let
     \[
     D_n = \bracs{x \in [a, b]|\forall \eps > 0, \exists y \in (x - \eps, x + \eps): \norm{f(x) - f(y)}_E \ge 1/n}
     \]
-    then $D = \bigcup_{n \in \nat}D_n$ is the set of discontinuity points of $f$. If $D$ is uncountable, then there exists $N \in \nat$ such that $D_n$ is infinite.
+    then $D = \bigcup_{n \in \nat^+}D_n$ is the set of discontinuity points of $f$. If $D$ is uncountable, then there exists $N \in \nat^+$ such that $D_n$ is infinite.
 
-    Fix $N \in \nat$. Let $E_1 = D_n \cap (a, b)$ and $I_1 = (a, b)$, then
+    Fix $N \in \nat^+$. Let $E_1 = D_n \cap (a, b)$ and $I_1 = (a, b)$, then
     \begin{enumerate}
         \item[(a)] $|E_k| \ge N - k$.
         \item[(b)] $E_k \subset I_k^o$.
@@ -62,5 +62,5 @@
 
     Let $k \le N$ and suppose inductively that $E_k, I_k$ have been constructed. Let $x_k \in E_k$, then by (b), there exists $\eps > 0$ such that $[x_k - \eps, x_k + \eps] \subset I_k$ and $|E_k \setminus [x_k - \eps, x_k + \eps]| \ge N - k$. Let $y_k \in [x_k - \eps, x_k + \eps]$ such that $\norm{f(x_k) - f(y_k)} \ge 1/n$, $I_{k + 1} = I_k \setminus [x_k - \eps, x_k + \eps]$, and $E_{k+1} = E_k \setminus [x_k - \eps, x_k + \eps]$, then $I_k$ and $E_k$ satisfies (a) and (b).
 
-    Therefore there exists pairs $\bracs{(x_k, y_k)|1 \le k \le N}$ such that $\norm{f(x_k) - f(y_k)}_E \ge 1/n$ for all $n$, and the smallest interval containing each $(x_k, y_k)$ are pairwise disjoint. Thus $[f]_{\var} \ge N/n$ for all $N \in \nat$, so $[f]_{\var} = \infty$.
+    Therefore there exists pairs $\bracs{(x_k, y_k)|1 \le k \le N}$ such that $\norm{f(x_k) - f(y_k)}_E \ge 1/n$ for all $n$, and the smallest interval containing each $(x_k, y_k)$ are pairwise disjoint. Thus $[f]_{\var} \ge N/n$ for all $N \in \nat^+$, so $[f]_{\var} = \infty$.
 \end{proof}

src/fa/rs/rs-bv.tex

@@ -1,4 +1,4 @@
-\section{Riemann-Stieltjes Integrals and Functions of Bounded Variationo}
+\section{Riemann-Stieltjes Integrals and Functions of Bounded Variation}
 \label{section:rs-bv}
 
 \begin{proposition}
@@ -36,7 +36,7 @@
     then $f \in RS([a, b], G)$ and $\int_a^b f dG = \lim_{\alpha \in A}\int_a^b f_\alpha dG$. In particular,
     \begin{enumerate}
         \item If $H$ is complete, then condition (a) may be omitted.
-        \item If $H$ is sequentially complete and $A = \nat$, then condition (b) may be omitted.
+        \item If $H$ is sequentially complete and $A = \nat^+$, then condition (b) may be omitted.
     \end{enumerate}
 \end{proposition}
 \begin{proof}

src/fa/tvs/continuous.tex

@@ -84,3 +84,41 @@
     \end{enumerate}
     The uniformity $\fU$ and its induced topology are the \textbf{product uniformity/topology}, and $E$ equipped with $\fU$ is the \textbf{product TVS} of $\seqi{E}$.
 \end{definition}
+
+\begin{definition}[Uniform/Bounded Operator Topology]
+\label{definition:uniform-op-topo}
+    Let $E, F$ be TVSs over $K \in \RC$. For each bounded set $B \subset E$ and entourage $V \subset F \times F$, let
+    \[
+    U(B, V) = \bracs{(S, T) \in L(E; F) \times L(E; F)|(Sx, Tx) \in V \forall x \in B}
+    \]
+    then
+    \[
+    \fB = \bracs{U(B, V)| B \subset E \text{ bounded}, U \subset F \times F \text{ entourage}}
+    \]
+    is a fundamental system of entourages for a vector space uniformity on $L(E; F)$, which induces the \textbf{uniform/bounded operator topology} on $L(E; F)$.
+\end{definition}
+\begin{proof}
+    Firstly,
+    \begin{enumerate}
+        \item[(FB1)] Let $U(B, V), U(B', V') \in \fB$, then $U(B, V) \cap U(B', V') \supset U(B \cup B', V \cap V') \in \fB$.
+        \item[(UB1)] For any $U(B, V) \in \fB$, the diagonal is contained in $V$, so the diagonal is also contained in $U(B, V)$.
+        \item[(UB2)] For any $U(B, V) \in \fB$, there exists an entourage $W \subset F \times F$ such that $W \circ W \subset V$. So $U(B, W) \in \fB$ with
+        \[
+        U(B, W) \circ U(B, W) \subset U(B, W \circ W) \subset U(B, V)
+        \]
+    \end{enumerate}
+    By \ref{proposition:fundamental-entourage-criterion}, $\fB$ forms a fundamental system of entourages for a uniformity on $L(E; F)$.
+
+    TODO SCHAEFER WOLFF PAGE 79
+\end{proof}
+
+
+\begin{definition}[Strong Operator Topology]
+\label{definition:strong-op-topo}
+    Let $E, F$ be TVSs over $K \in \RC$, then the \textbf{strong operator topology} on $L(E; F)$ is the initial topology generated by the maps $\bracs{T \mapsto Tx| x \in E}$, where $F$ is equipped with the strong topology.
+\end{definition}
+
+\begin{definition}[Weak Operator Topology]
+\label{definition:weak-op-topo}
+    Let $E, F$ be TVSs over $K \in \RC$, then the \textbf{weak operator topology} on $L(E; F)$ is the initial topology generated by the maps $\bracs{T \mapsto Tx| x \in E}$, where $F$ is equipped with the weak topology.
+\end{definition}

src/fa/tvs/dual.tex

@@ -0,0 +1,34 @@
+\section{The Dual Space}
+\label{section:tvs-dual}
+
+\begin{proposition}[Polarisation, {{\cite[Proposition 5.5]{Folland}}}]
+\label{proposition:polarisation-linear}
+    Let $E$ be a vector space over $\complex$, $\phi \in \hom(E; \complex)$, and $u = \re{\phi}$, then
+    \begin{enumerate}
+        \item $u \in \hom(E; \real)$ when $E$ is viewed as a vector space over $\real$.
+        \item For any $x \in E$, $\dpb{x, \phi}{E} = \dpb{x, u}{E} - i \dpb{ix, u}{E}$.
+    \end{enumerate}
+    Conversely, if $u \in \hom(E; \real)$ and $\phi \in \hom(E; \complex)$ is defined by $\dpb{x, \phi}{E} = \dpb{x, u}{E} - i \dpb{ix, u}{E}$ for all $x \in E$, then $f \in \hom(E; \complex)$.
+\end{proposition}
+\begin{proof}
+    (1): Given that $\phi \in \hom(E; \real)$, $u \in \hom(E; \real)$. For any $x \in E$,
+    \[
+    \im{\dpb{x, \phi}{E}} = \re{-i \dpb{x, \phi}{E}} = -\dpb{ix, u}{E}
+    \]
+    so (2) holds.
+
+    (2): Since $u \in \hom(E; \real)$, so is $\phi$. On the other hand, for any $x \in E$,
+    \[
+    \dpb{ix, \phi}{E} = \dpb{ix, u}{E} - i\dpb{-x, u}{E} = i(\dpb{x, u}{E} - i \dpb{ix, u}{E}) = i \dpb{x, \phi}{E}
+    \]
+\end{proof}
+
+\begin{definition}[Topological Dual]
+\label{definition:topological-dual}
+    Let $E$ be a TVS over $K \in \RC$, then the \textbf{topological dual} $E^*$ of $E$ is the set of all \textit{continous} linear functionals on $E$.
+\end{definition}
+
+\begin{definition}[Weak Topology]
+\label{definition:weak-topology}
+    Let $E$ be a TVS over $K \in \RC$, then the \textbf{weak topology} on $E$ is the initial topology generated by $E^*$. The space $E$ equipped with its weak topology is denoted $E_w$.
+\end{definition}

src/fa/tvs/index.tex

@@ -3,4 +3,6 @@
 
 \input{./src/fa/tvs/definition.tex}
 \input{./src/fa/tvs/bounded.tex}
+\input{./src/fa/tvs/dual.tex}
 \input{./src/fa/tvs/continuous.tex}
+\input{./src/fa/tvs/completion.tex}

src/measure/index.tex

@@ -0,0 +1,5 @@
+\part{Measure Theory and Integration}
+\label{part:measure}
+
+\input{./src/measure/sets/index.tex}
+\input{./src/measure/measure/index.tex}

src/measure/measure/complete.tex

@@ -0,0 +1,69 @@
+\section{Complete Measures}
+\label{section:complete-measure}
+
+\begin{definition}[Complete Measure]
+\label{definition:complete-measure}
+    Let $(X, \cm, \mu)$ be a measure space, then $\mu$ is \textbf{complete} if for any null set $E \in \cm$ and $F \subset E$, $F \in \cm$.
+\end{definition}
+
+\begin{definition}[Completion]
+\label{definition:measure-completion}
+    Let $(X, \cm, \mu)$ be a measure space, $\cn = \bracs{N \in \cm| \mu(N) = 0}$, and
+    \[
+    \ol{\cm} = \bracs{E \cup F| E \in \cm, F \subset N, N \in \cn}
+    \]
+    then
+    \begin{enumerate}
+        \item For any $A \in \ol{\cm}$, there exists $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $E \cap N = \emptyset$ and $A = E \sqcup F$.
+        \item $\ol{\cm} \supset \cm$ is a $\sigma$-algebra.
+        \item There exists an extension $\ol{\mu}$ of $\mu$ as a measure on $\cm$.
+        \item $(X, \ol{\cm}, \ol{\mu})$ is a complete measure space.
+        \item[(U)] For any complete measure space $(X, \cf, \nu)$ where $\cf \supset \cm$ and $\nu|_\cm = \mu$, $\cf \supset \ol{\cm}$ and $\nu|_{\ol{\cm}} = \ol{\mu}$.
+    \end{enumerate}
+    and space $(X, \ol{\cm}, \mu)$ is the \textbf{completion} of $(X, \cm, \mu)$.
+\end{definition}
+\begin{proof}
+    (1): Let $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $A = E \cup F$. Define $E' = E \setminus N$ and $F' = F \cup (E \cap N)$, then $E' \cap N = \emptyset$ and $A = E \cup F = E' \sqcup F'$.
+
+    (2): Let $A \in \ol{\cm}$, $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $A = E \sqcup F$, then
+    \begin{align*}
+        A^c &= (A^c \cap N^c) \cup (A^c \cap N) = (N^c \setminus A) \cup (N \setminus A) \\
+        &= \underbrace{(N^c \setminus E)}_{\in \cm} \cup \underbrace{(N \setminus F)}_{\subset N} \in \cm
+    \end{align*}
+    Let $\seq{A_n} \subset \cm$. For each $n \in \natp$, let $E_n \in \cm$, $N_n \in \cn$, and $F_n \subset N$ such that $A_n = E_n \cup F_n$, then
+    \[
+    \bigcup_{n \in \natp}A_n = \underbrace{\braks{\bigcup_{n \in \natp}E_n}}_{\in \cm} \cup \underbrace{\braks{\bigcup_{n \in \natp}F_n}}_{\subset \bigcup_{n \in \natp}N_n} \in \ol{\cm}
+    \]
+
+    (3): Let $A \in \ol{\cm}$, $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $A = E \cup F$. Since $A \setminus N = E \setminus N \in \cm$ and $A \cup N = E \cup N \in \cm$,
+    \[
+    \mu(A \setminus N) \le \mu(E) \le \mu(A \cup N)
+    \]
+    Let $E' \in \cm$, $N' \in \cn$, and $F' \subset N'$ such that $A = E' \cup F'$, then
+    \[
+    \mu(A \setminus (N \cup N')) \le \mu(A \setminus N') \le \mu(E') \le \mu(A \cup N') \le \mu(A \cup (N \cup N'))
+    \]
+    Now, since $N$ and $N'$ are null,
+    \[
+    \mu(A \cup (N \cup N')) = \mu(A \setminus (N \cup N')) + \mu(N \cup N') = \mu(A \setminus (N \cup N'))
+    \]
+    so
+    \[
+    \mu(A \setminus (N \cup N')) = \mu(E) = \mu(E') = \mu(A \cup (N \cup N'))
+    \]
+    Thus for any decomposition $A = E \cup F$ where $E \in \cm$ and $F$ is contained in a null set, $\ol{\mu}(A) = \mu(E)$ is well-defined.
+
+    To see that $\ol{\mu}$ is a measure, let $\seq{A_n} \subset \ol{\cm}$ be pairwise disjoint. For each $n \in \nat$, let $E_n \in \cm$, $N_n \in \cn$, and $F_n \subset N_n$ such that $A_n = E_n \cup F_n$, then
+    \[
+    \ol{\mu}\paren{\bigsqcup_{n \in \natp}A_n} = \mu\paren{\bigsqcup_{n \in \natp}E_n} = \sum_{n \in \natp}\mu(E_n) = \sum_{n \in \natp}\mu(A_n)
+    \]
+
+    (4): Let $A \in \ol{\cm}$ with $\ol{\mu}(A) = 0$, $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $A = E \cup F$, then $\mu(E) = \mu(E \cup N) = 0$. By definition of $\ol{\cm}$, any subset of $E \cup N$ is also in $\ol{\cm}$. Thus $\ol{\mu}$ is complete.
+
+    (U): Let $N \in \cn \subset \cm \subset \cf$, then for any $F \subset N$, $\nu(F) \le \nu(N) = \mu(N)$, and $F \in \cf$ by completeness of $(X, \cf, \nu)$. Thus $\cf \supset \ol{\cm}$.
+
+    For any $A \in \ol{\cm}$, let $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $A = E \cup N$, then
+    \[
+    \nu(A) = \nu(E \cup F) = \nu(E) = \mu(E) = \ol{\mu}(A)
+    \]
+\end{proof}

src/measure/measure/index.tex

@@ -0,0 +1,8 @@
+\chapter{Positive Measures}
+\label{chap:measures}
+
+\input{./src/measure/measure/measure.tex}
+\input{./src/measure/measure/complete.tex}
+\input{./src/measure/measure/semifinite.tex}
+\input{./src/measure/measure/sigma-finite.tex}
+\input{./src/measure/measure/outer.tex}

src/measure/measure/measure.tex

@@ -0,0 +1,121 @@
+\section{Measures}
+\label{section:measures}
+
+\begin{definition}[Measurable Space]
+\label{definition:measurable-space}
+    Let $X$ be a set and $\cm \subset 2^X$ be a $\sigma$-algebra, then the pair $(X, \cm)$ is a \textbf{measurable space}.
+\end{definition}
+
+
+
+\begin{definition}[Measure]
+\label{definition:measure}
+    Let $(X, \cm)$ be a measurable space and $\mu: \cm \to [0, \infty]$, then $\mu$ is a \textbf{(countably-additive) measure} if
+    \begin{enumerate}
+        \item[(M1)] $\mu(\emptyset) = 0$.
+        \item[(M2)] For any $\seq{E_n} \subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{n \in \natp}E_n} = \sum_{n \in \natp} \mu(E_n)$.
+    \end{enumerate}
+    In which case, $(X, \cm, \mu)$ is a \textbf{measure space}.
+
+    If $\mu: \cm \to [0, \infty]$ instead satisfies (M1) and
+    \begin{enumerate}
+        \item[(M2')] For any $\seqf{E_j} \subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{j = 1}^n E_j} = \sum_{j = 1}^n \mu(E_j)$.
+    \end{enumerate}
+    then $\mu$ is a \textbf{finitely-additive measure}.
+\end{definition}
+
+\begin{definition}[Null Set]
+\label{definition:null}
+    Let $(X, \cm, \mu)$ be a measure space, then $E \in \cm$ is \textbf{$\mu$-null/null} if $\mu(E) = 0$.
+\end{definition}
+
+\begin{definition}[Almost Everywhere]
+\label{definition:almost-everywhere}
+    Let $(X, \cm, \mu)$ be a measure space. A statement holds \textbf{$\mu$-almost everywhere/$\mu$-a.e./a.e.} if it is false on a subset of a $\mu$-null set.
+\end{definition}
+
+\begin{proposition}[{{\cite[Theorem 1.8]{Folland}}}]
+\label{proposition:measure-properties}
+    Let $(X, \cm, \mu)$ be a measure space, then:
+    \begin{enumerate}
+        \item For any $E, F \in \cm$ with $E \subset F$, $\mu(E) \le \mu(F)$.
+        \item For any $\seq{E_n} \subset \cm$, $\mu\paren{\bigcup_{n \in \natp}E_n} \le \sum_{n \in \natp}\mu(E_n)$.
+        \item For any $\seq{E_n} \subset \cm$ with $E_n \subset E_{n+1}$ for all $n \in \nat$, $\mu\paren{\bigcup_{n \in \nat}E_n} = \limv{n}\mu(E_n)$.
+        \item For any $\seq{E_n} \subset \cm$ with $E_n \supset E_{n+1}$ for all $n \in \nat$ and $\mu(E_1) < \infty$, $\mu(\bigcap_{n \in \natp}E_n) = \limv{n}\mu(E_n)$.
+        \item For any $\seq{E_n} \subset \cm$,
+        \[
+        \mu\paren{\liminf_{n \to \infty}E_n} \le \liminf_{n \to \infty}\mu(E_n)
+        \]
+        \item For any $\seq{E_n} \subset \cm$ with $\mu\paren{\bigcup_{n \in \nat}E_n} < \infty$,
+        \[
+        \mu\paren{\limsup_{n \to \infty}E_n} \ge \limsup_{n \to \infty}\mu(E_n)
+        \]
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): $\mu(F) = \mu(E) + \mu(F \setminus E) \ge \mu(E)$.
+
+    (2): For each $n \in \natp$, let $F_n = E_n \setminus \bigcup_{k = 1}^{n-1}E_k$, then $\seq{F_n}$ is pairwise disjoint with $\bigsqcup_{n \in \natp}F_n = \bigcup_{n \in \natp}E_n$. In which case,
+    \[
+    \mu\paren{\bigcup_{n \in \nat}E_n} = \mu\paren{\bigsqcup_{n \in \nat}F_n} = \sum_{n \in \natp}\mu(F_n) \le \sum_{n \in \natp}\mu(E_n)
+    \]
+
+    (3): Denote $E_0 = \emptyset$. For each $n \in \natp$, let $F_n = E_n \setminus E_{n - 1}$, then $\seq{F_n}$ is pairwise disjoint with $\bigsqcup_{n \in \natp}F_n = \bigcup_{n \in \natp}E_n$. In which case,
+    \[
+    \mu\paren{\bigcup_{n \in \natp}E_n} = \mu\paren{\bigsqcup_{n \in \natp}F_n} = \limv{n}\sum_{k = 1}^n \mu(F_n) = \limv{n}\mu(E_n)
+    \]
+
+    (4): Since $\mu(E_1) < \infty$,
+    \[
+    \limv{n}\mu(E_n) = \mu(E_1) - \limv{n}\mu(E_1 \setminus E_n) = \mu(E_1) - \mu\paren{E_1 \setminus \bigcap_{n \in \natp}E_n} = \mu\paren{\bigcap_{n \in \natp}E_n}
+    \]
+    by (3).
+
+    (5): Using (1) and (3),
+    \begin{align*}
+        \mu\paren{\liminf_{n \to \infty}E_n} &= \mu\paren{\bigcup_{n \in \natp}\bigcap_{m \ge n}E_m} = \limv{n}\mu\paren{\bigcap_{m \ge n}E_m} \\
+        &\le \limv{n} \inf_{m \ge n}\mu(E_m) = \liminf_{n \to \infty}\mu(E_n)
+    \end{align*}
+
+    (6): Using (1) and (4),
+    \begin{align*}
+        \mu\paren{\limsup_{n \to \infty}E_n} &= \mu\paren{\bigcap_{n \in \natp}\bigcup_{m \ge n}E_m} = \limv{n}\mu\paren{\bigcup_{m \ge n}E_m} \\
+        &\ge \limv{n}\sup_{m \ge n}\mu(E_m) = \limsup_{n \to \infty}\mu(E_n)
+    \end{align*}
+\end{proof}
+
+\begin{theorem}[Dynkin's Uniqueness Theorem]
+\label{theorem:dynkin-uniqueness}
+    Let $(X, \cm)$ be a measurable space, $\mathcal{P} \subset \cm$ be a $\pi$-syste, and $\mu, \nu: \cm \to [0, \infty]$ be measures. If
+    \begin{enumerate}
+        \item[(a)] $\sigma(\mathcal{P}) = \cm$.
+        \item[(b)] $\mu(E) = \nu(E)$ for all $E \in \mathcal{P}$.
+        \item[(c)] There exists $\seq{E_n} \subset \mathcal{P}$ such that $E_n \upto X$ and $\mu(E_n) < \infty$ for all $n \in \natp$.
+    \end{enumerate}
+    then $\mu = \nu$.
+\end{theorem}
+\begin{proof}
+    Let $F \in \mathcal{P}$ with $\mu(F) < \infty$ and
+    \[
+    \alg(F) = \bracs{E \in \cm|\mu(E \cap F) = \nu(E \cap F)}
+    \]
+    then $\alg(F) \supset \mathcal{P}$ by (b), and
+    \begin{enumerate}
+        \item[(L2)] For any $E, E' \in \alg(F)$ with $E \subset E'$ and $F \in \mathcal{P}$,
+        \begin{align*}
+        \mu((E' \setminus E) \cap F) &= \mu(E' \cap F) - \mu(E \cap F) \\
+        &= \nu(E' \cap F) - \nu(E \cap F) = \nu((E' \setminus E) \cap F)
+        \end{align*}
+        \item[(L3)] For any $\seq{E_n} \subset \alg$ with $E_n \subset E_{n+1}$ for all $n \in \natp$ and $F \in \mathcal{P}$,
+        \[
+        \mu\paren{\bigcup_{n \in \nat}E_n \cap F} = \limv{n}\mu(E_n \cap F) = \limv{n}\nu(E_n \cap F) = \nu\paren{\bigcup_{n \in \nat}E_n \cap F}
+        \]
+        by continuity from below (\ref{proposition:measure-properties}).
+    \end{enumerate}
+    so $\alg(F)$ is a $\lambda$-system. By (a) and Dynkin's $\pi$-$\lambda$ theorem (\ref{theorem:pi-lambda}), $\alg(F) = \cm$.
+
+    Let $\seq{E_n}$ as in assumption (c), then $\mu(E_n \cap F) = \mu(E_n \cap F)$ for all $n \in \natp$ and $F \in \cm$. Thus by continuity from below (\ref{proposition:measure-properties}),
+    \[
+    \mu(F) = \limv{n}\mu(E_n \cap F) = \limv{n}\nu(E_n \cap F) = \nu(F)
+    \]
+\end{proof}

src/measure/measure/outer.tex

@@ -0,0 +1,154 @@
+\section{Carathéodory's Extension Theorem}
+\label{section:caratheodory}
+
+\begin{definition}[Outer Measure]
+\label{definition:outer-measure}
+    Let $X$ be a set and $\mu^*: 2^X \to [0, \infty]$, then $\mu^*$ is an \textbf{outer measure} if:
+    \begin{enumerate}
+        \item[(M1)] $\mu^*(\emptyset) = 0$.
+        \item[(OM1)] For any $E, F \subset X$ with $E \subset F$, $\mu^*(E) \le \mu^*(F)$.
+        \item[(OM2)] For any $\seq{E_n} \subset X$,
+        \[
+        \mu^*\paren{\bigcup_{n \in \natp}E_n} \le \sum_{n \in \natp}\mu^*(E_n)
+        \]
+    \end{enumerate}
+\end{definition}
+
+\begin{proposition}[{{\cite[Proposition 1.10]{Folland}}}]
+\label{proposition:outer-measure-inf}
+    Let $X$ be a set, $\ce \subset 2^X$, and $\rho: \ce \to [0, \infty]$ such that $\emptyset, X \in \ce$ and $\rho(\emptyset) = 0$. Define
+    \[
+    \mu^*: 2^X \to [0, \infty] \quad E \mapsto \inf\bracs{\sum_{n \in \natp}\rho(E_n) \bigg | \seq{E_n} \subset \ce, \bigcup_{n \in \natp}E_n \supset E}
+    \]
+    then $\mu^*$ is an outer measure.
+\end{proposition}
+\begin{proof}
+    (OM2): Let $\seq{E_n} \subset 2^X$ and assume without loss of generality that $\sum_{n \in \natp}\mu^*(E_n) < \infty$. Let $\eps > 0$. For each $n \in \nat$, let $\bracs{F_{n, k}}_{k = 1}^\infty \subset \ce$ such that $\bigcup_{k \in \natp}F_{n, k} \supset E_n$ and $\sum_{n \in \natp}\rho(F_{n, k}) < \mu^*(E_n) + 2^{-n}\eps$, then
+    \begin{align*}
+    \mu^*\paren{\bigcup_{n \in \natp}E_n} &\le \sum_{(n, k) \in \natp \times \natp}\rho(F_{n, k}) = \sum_{n \in \natp}\sum_{k \in \natp}\rho(F_{n, k}) \\
+    &\le \sum_{n \in \natp}(\mu^*(E_n) + 2^{-n}\eps) \le \eps + \sum_{n \in \natp}\mu^*(E_n)
+    \end{align*}
+    Since $\eps$ is arbitrary, $\mu^*\paren{\bigcup_{n \in \natp}E_n} \le \sum_{n \in \natp}\mu^*(E_n)$.
+\end{proof}
+
+\begin{definition}[Outer Measurable]
+\label{definition:outer-measurable}
+    Let $X$ be a set, $\mu^*: 2^X \to [0, \infty]$ be an outer measure, and $E \subset X$, then $E$ is \textbf{$\mu^*$-measurable} if for any $F \subset X$,
+    \[
+    \mu^*(F) = \mu^*(E \cap F) + \mu^*(E \setminus F)
+    \]
+\end{definition}
+
+\begin{theorem}[Carathéodory, {{\cite[Theorem 1.11]{Folland}}}]
+\label{theorem:caratheodory}
+    Let $X$ be a set, $\mu^*: 2^X \to [0, \infty]$ be an outer measure, and $\cm \subset 2^X$ be the collection of all $\mu^*$-measurable sets, then:
+    \begin{enumerate}
+        \item For any $F \subset X$ and $\seq{E_n} \subset \cm$ pairwise disjoint,
+        \[
+        \mu^*\paren{F \cap \bigsqcup_{n \in \natp}E_n} = \sum_{n \in \natp}\mu^*(F \cap E_n)
+        \]
+        \item $\cm$ is a $\sigma$-algebra.
+        \item $(X, \cm, \mu^*|_\cm)$ is a complete measure space.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): Let $N \in \nat$, then
+    \[
+    \mu^*\paren{F \cap \bigsqcup_{n \in \natp}E_n} \ge \mu^*\paren{F \cap \bigsqcup_{n =1}^N E_n} = \sum_{n = 1}^N \mu^*(F \cap E_n)
+    \]
+    As this holds for all $N \in \nat$,
+    \[
+    \mu^*\paren{F \cap \bigsqcup_{n \in \natp}E_n} \ge \sum_{n \in \natp}\mu^*(F \cap E_n)
+    \]
+
+    (2): Firstly, $\emptyset, X \in \cm$ since $\mu^*(\emptyset) = 0$. For any $A \in \cm$, $A^c \in \cm$ as well because the definition of $\mu^*$-measurability is symmetric.
+
+    Let $A, B \in \cm$ and $F \subset X$, then
+    \begin{align*}
+        \mu^*(F) &= \mu^*(F \cap A) + \mu^*(F \setminus A) \\
+        &= \mu^*(F \cap A \cap B) + \mu^*(F \cap A \setminus B) + \mu^*(F \cap B \setminus A) + \mu^*(F \setminus (A \cup B)) \\
+        &= \mu^*(F \cap B) + \mu^*(F \cap A \setminus B) + \mu^*(F \setminus (A \cup B)) \\
+        &= \mu^*(F \cap (A \cup B)) + \mu^*(F \setminus (A \cup B))
+    \end{align*}
+    so $\cm$ is an algebra. Since (1) implies that $\cm$ is closed under countable disjoint unions, $\cm$ is a $\sigma$-algebra by \ref{lemma:sigma-algebra-condition}.
+
+    (3): By (1), $\mu|_\cm$ is a measure. Let $N \in \cm$ with $\mu^*(N) = 0$, then for any $E \subset N$ and $F \subset X$,
+    \begin{align*}
+        \mu^*(E) &= \underbrace{\mu^*(E \cap N)}_0 + \mu^*(E \setminus N) \\
+        &= \underbrace{\mu^*(E \cap N \setminus F)}_0 + \underbrace{\mu^*(E \cap F)}_0 + \mu^*(E \setminus N) \\
+        &= \mu^*((E \setminus F) \cap N) + \mu^*(E \cap F) + \mu^*((E \setminus F) \setminus N) \\
+        &= \mu^*(E \cap F) + \mu^*(E \setminus F)
+    \end{align*}
+\end{proof}
+
+
+\begin{definition}[Premeasure]
+\label{definition:premeasure}
+    Let $X$ be a set, $\alg \subset 2^X$ be an algebra, and $\mu_0: \alg \to [0, \infty]$, then $\mu_0$ is a \textbf{premeasure} if:
+    \begin{enumerate}
+        \item[(M1)] $\mu_0(\emptyset) = 0$.
+        \item[(PM)] For any $\seq{A_n} \subset \alg$ pairwise disjoint with $\bigsqcup_{n \in \natp}A_n \in \alg$,
+        \[
+        \mu_0\paren{\bigsqcup_{n \in \natp}A_n} = \sum_{n \in\natp}\mu_0(A_n)
+        \]
+    \end{enumerate}
+\end{definition}
+
+\begin{theorem}[Carathéodory's Extension Theorem, {{\cite[Theorem 1.14]{Folland}}}]
+\label{theorem:caratheodory-extension}
+    Let $X$ be a set, $\alg \subset 2^X$ be an algebra, and $\mu_0: \alg \to [0, \infty]$ be a premeasure, then there exists a measure space $(X, \cm, \mu)$ such that
+    \begin{enumerate}
+        \item $\cm \supset \alg$.
+        \item $\mu|_\alg = \mu_0$.
+        \item $\mu$ is complete.
+        \item[(U)] For any measure space $(X, \cn, \nu)$ satisfying (1) and (2),
+        \begin{enumerate}
+            \item[(a)] $\nu|_{\cm \cap \cn} \le \mu_{\cm \cap \cn}$.
+            \item[(b)] For any $E \in \cm \cap \cn$ with $\mu(E) < \infty$, $\mu(E) = \nu(E)$.
+            \item[(c)] If $\mu$ is $\sigma$-finite, then $\nu = \mu$.
+        \end{enumerate}
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    Let
+    \[
+    \mu^*: 2^X \to [0, \infty] \quad E \mapsto \inf\bracs{\sum_{n \in \natp}\mu_0(E_n) \bigg | \seq{E_n} \subset \alg, \bigcup_{n \in \natp}E_n \supset E}
+    \]
+    then $\mu^*$ is an outer measure by \ref{proposition:outer-measure-inf}. By Carathéodory's theorem (\ref{theorem:caratheodory}), there exists a measure space $(X, \cm, \mu)$ such that $\mu = \mu^*|_{\cm}$.
+
+    (1): Let $E \in \alg$ and $F \subset X$. For any $\seq{F_n} \subset \alg$ with $\bigcup_{n \in \natp}F_n \supset F$,
+    \[
+    \sum_{n \in \natp}\mu_0(F_n) = \sum_{n \in \natp}\mu_0(F_n \cap E) + \sum_{n \in \natp}\mu_0(F_n \setminus E) \ge \mu^*(F \cap E) + \mu^*(F \setminus E)
+    \]
+    As this holds for all such $\seq{F_n}$, $\mu_0^*(F) = \mu^*(F \cap E) + \mu^*(F \setminus E)$.
+
+    (2): Let $E \in \alg$ and $\seq{E_n} \subset \alg$ such that $\bigcup_{n \in \natp}E_n \supset E$, then
+    \[
+    \sum_{n \in \natp}\mu_0(E_n) = \sum_{n \in \natp}\mu_0(E_n \cap E) + \sum_{n \in \natp}\mu_0(E_n \setminus E) \ge \sum_{n \in \natp}\mu_0(E_n \cap E) = \mu_0(E)
+    \]
+    As this holds for all such $\seq{E_n}$, $\mu^*(E) = \mu_0(E)$.
+
+    (U): Let $\seq{E_n} \subset \alg$, then by continuity from below (\ref{proposition:measure-properties}),
+    \[
+    \nu\paren{\bigcup_{n \in \natp}E_n} = \limv{n}\nu\paren{\bigcup_{k = 1}^nE_k} = \limv{n}\mu\paren{\bigcup_{k = 1}^n E_k} = \mu\paren{\bigcup_{n \in \natp}E_n}
+    \]
+    Thus if $E \in \cm \cap \cn$ with $\bigcup_{n \in \natp}E_n \supset E$, then
+    \[
+    \nu(E) \le \nu\paren{\bigcup_{n \in \natp}E_n} = \mu\paren{\bigcup_{n \in \natp}E_n} \le \sum_{n \in \natp}\mu_0(E_n)
+    \]
+    As this holds for all such $\seq{E_n}$, $\nu(E) \le \mu^*(E) = \mu(E)$.
+
+    Suppose that $\mu(E) < \infty$, then there exists $\seq{E_n} \subset \alg$ such that $\bigcup_{n \in \natp}E_n \supset E$ and $\sum_{n \in \natp}\mu_0(E_n) < \infty$. In which case,
+    \begin{align*}
+    \mu\paren{\bigcup_{n \in \natp}E_n} &= \nu\paren{\bigcup_{n \in \natp}E_n} = \nu(E) + \nu\paren{\bigcup_{n \in \natp}E_n \setminus E} \\
+    &\le \nu(E) + \mu\paren{\bigcup_{n \in \natp}E_n \setminus E}  \\
+    \mu(E) &\le \nu(E)
+    \end{align*}
+    so $\nu(E) = \mu(E)$ whenever $\mu(E) < \infty$.
+
+    Finally, if $\mu$ is $\sigma$-finite, then there exists $\seq{E_n} \subset \cm$ such that $E_n \upto X$ and $\mu(E_n) < \infty$ for all $n \in \natp$. For each $n \in \natp$, there exists $\bracsn{E_{n, k}}_{k = 1}^\infty \subset \alg$ such that $E_n \subset \bigcup_{k \in \natp}E_{n, k}$ and $\mu\paren{\bigcup_{k \in \natp}E_{n, k}} < \infty$. Let $F_n = \bigcup_{j = 1}^n \bigcup_{k \in \natp}E_{n, k}$, then $\seq{F_n} \subset \sigma(\alg) \subset \cm \cap \cn$ with $F_n \upto X$ and $\mu(F_n) < \infty$ for all $n \in \nat$. For any $E \in \cm \cap \cn$, by continuity from below (\ref{proposition:measure-properties}),
+    \[
+    \nu(E) = \limv{n}\nu(E \cap F_n) = \limv{n}\mu(E \cap F_n) = \mu(E)
+    \]
+    so $\nu_{\cm \cap \cn} = \mu_{\cm \cap \cn}$.
+\end{proof}

src/measure/measure/semifinite.tex

@@ -0,0 +1,53 @@
+\section{Semifinite Measures}
+\label{section:semifinite}
+
+\begin{definition}[Semifinite]
+\label{definition:semifinite-measure}
+    Let $(X, \cm, \mu)$ be a measure space, then the following are equivalent:
+    \begin{enumerate}
+        \item For any $E \in \cm$ with $\mu(E) = \infty$, there exists $F \subset E$ with $0 < \mu(F) < \infty$.
+        \item For any $E \in \cm$,
+        \[
+        \mu(E) = \sup\bracs{\mu(F)| F \in \cm, F \subset E, \mu(F) < \infty}
+        \]
+    \end{enumerate}
+    If the above holds, then $\mu$ is a \textbf{semifinite measure}.
+\end{definition}
+\begin{proof}
+    $(1) \Rightarrow (2)$: If $\mu(E) < \infty$, then the result holds directly. Assume that $\mu(E) = \infty$, and suppose for contradiction that
+    \[
+    M = \sup\bracs{\mu(F)| F \in \cm, F \subset E, \mu(F) < \infty} < \infty
+    \]
+    Let $\seq{F_n} \subset \cm$ such that $F \subset E$, $\mu(F) < \infty$, and $\mu(F_n) \upto M$. Let $F = \bigcup_{n \in \nat}F_n$, then $\mu(F) = M$, and $\mu(E \setminus F) = \infty$. By (1), there exists $G \in \cm$ with $G \subset E \setminus F$ and $0 < \mu(G) < \infty$. Thus $F \cup G \subset E$ with $M < \mu(F) + \mu(G) = \mu(F \cup G) < \infty$. This contradicts the fact that $M$ is the supremum.
+\end{proof}
+
+\begin{definition}[Semifinite Part]
+\label{definition:semifinite-part}
+    Let $(X, \cm, \mu)$ be a measure space and
+    \[
+    \mu_0: \cm \to [0, \infty] \quad E \mapsto \sup\bracs{\mu(F)| F \in \cm, F \subset E, \mu(F) < \infty}
+    \]
+    then $\mu_0$ is a semifinite measure, and the \textbf{semifinite part} of $\mu$.
+\end{definition}
+\begin{proof}
+    Firstly, for any $F \in \cm$ with $\mu(F) < \infty$, $\mu(F) = \mu_0(F)$.
+
+    Let $\seq{E_n} \subset \cm$ be pairwise disjoint. For any $F \in \cm$ with $F \subset \bigsqcup_{n \in \natp}E_n$ and $\mu(F) < \infty$,
+    \[
+    \mu(F) = \sum_{n \in \natp}\mu(F \cap E_n) \le \sum_{n \in \natp}\mu_0(E_n)
+    \]
+    Thus $\mu_0\paren{\bigcup_{n \in \natp}E_n} \le \sum_{n \in \natp}\mu_0(E_n)$.
+
+    On the other hand, let $n \in \natp$ and $\seqf{F_k} \subset \cm$ such that $F_k \subset E_k$ for each $1 \le k \le n$ and $\mu(F_k) < \infty$, then $F = \bigsqcup_{k = 1}^n F_k \subset \bigsqcup_{k \in \natp}E_k$. Thus
+    \[
+    \sum_{k = 1}^n \mu(F_k) = \mu(F) \le \mu_0\paren{\bigsqcup_{k \in \natp}E_k}
+    \]
+    As this holds for all such $\seqf{F_k}$,
+    \[
+    \sum_{k = 1}^n \mu_0(E_k) \le \mu_0\paren{\bigsqcup_{k \in \natp}E_k}
+    \]
+    Since this holds for all $n \in \natp$,
+    \[
+    \sum_{n \in \natp}\mu_0(E_n) \le \mu_0\paren{\bigsqcup_{k \in \natp}E_k}
+    \]
+\end{proof}

src/measure/measure/sigma-finite.tex

@@ -0,0 +1,11 @@
+\section{$\sigma$-Finite Measures}
+\label{section:sigma-finite}
+
+\begin{definition}[$\sigma$-Finite Measure]
+\label{definition:sigma-finite}
+    Let $(X, \cm, \mu)$ be a measure space, then the following are equivalent:
+    \begin{enumerate}
+        \item There exists $\seq{E_n} \subset \cm$ pairwise disjoint such that $\bigsqcup_{n \in \nat}E_n = X$ and $\mu(E_n) < \infty$ for all $n \in \nat$.
+        \item There exists $\seq{E_n} \subset \cm$ such that $E_n \upto X$ and $\mu(E_n) < \infty$ for all $n \in \nat$.
+    \end{enumerate}
+\end{definition}

src/measure/sets/algebra.tex

@@ -0,0 +1,67 @@
+\section{Algebras}
+\label{section:set-algebra}
+
+\begin{definition}[Algebra]
+\label{definition:algebra}
+    Let $X$ be a set and $\alg \subset 2^X$, then $\alg$ is an \textbf{algebra} if:
+    \begin{enumerate}
+        \item[(A1)] $\emptyset, X \in \alg$.
+        \item[(A2)] For any $A \in \alg$, $A^c \in \alg$.
+        \item[(A3)] For any $A, B \in \alg$, $A \cup B \in \alg$.
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[$\sigma$-Algebra]
+\label{definition:sigma-algebra}
+    Let $X$ be a set and $\cm \subset 2^X$, then $\cm$ is a \textbf{$\sigma$-algebra} if:
+    \begin{enumerate}
+        \item[(A1)] $\emptyset, X \in \cm$.
+        \item[(A2)] For any $A \in \cm$, $A^c \in \cm$.
+        \item[(A3')] For any $\seq{A_n} \in \cm$, $\bigcup_{n \in \nat^+}A_n \in \cm$.
+    \end{enumerate}
+\end{definition}
+
+\begin{proposition}
+\label{proposition:algebra-properties}
+    Let $X$ be a set and $\alg \subset 2^X$ be an algebra, then:
+    \begin{enumerate}
+        \item For any $A, B \in \alg$, $A \cap B \in \alg$.
+        \item For any $A, B \in \alg$, $A \setminus B \in \alg$.
+    \end{enumerate}
+    If $\alg$ is a $\sigma$-algebra, then:
+    \begin{enumerate}
+        \item[(1')] For any $\seq{A_n} \in \alg$, $\bigcap_{n \in \natp}A_n \in \alg$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1, 1'): $\bigcap_{i \in I}A_i = \braks{\bigcup_{i \in I}A_i^c}^c$.
+
+    (2): $A \setminus B = A \cap B^c$.
+\end{proof}
+
+
+\begin{lemma}
+\label{lemma:sigma-algebra-condition}
+    Let $X$ be a set and $\alg \subset 2^X$ be an algebra, then the following are equivalent:
+    \begin{enumerate}
+        \item For any $\seq{A_n} \subset \alg$, $\bigcup_{n \in \nat^+}A_n \in \alg$.
+        \item For any $\seq{A_n} \subset \alg$ with $A_n \subset A_{n+1}$ for all $n \in \natp$, $\bigcup_{n \in \natp}A_n \in \alg$.
+        \item For any $\seq{A_n} \subset \alg$ pairwise disjoint, $\bigsqcup_{n \in \natp}A_n \in \alg$.
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (2) $\Rightarrow$ (1): For each $n \in \nat$, let $B_n = \bigcup_{k = 1}^n A_n \in \alg$, then $\bigcup_{n \in \natp}A_n = \bigcup_{n \in \nat}B_n \in \alg$.
+
+    (3) $\Rightarrow$ (2): Denote $A_0 = \emptyset$ and $B_n = A_n \setminus A_{n-1}$, then $\seq{B_n} \subset \alg$ is pairwise disjoint and $\bigcup_{n \in \nat}A_n = \bigsqcup_{n \in \nat}B_n \in \alg$.
+\end{proof}
+
+
+\begin{definition}[Generated $\sigma$-Algebra]
+\label{definition:generated-sigma-algebra}
+    Let $X$ be a set and $\ce \subset 2^X$, then the $\sigma$-algebra $\sigma(\ce)$ \textbf{generated by} $\ce$ is the smallest $\sigma$-algebra on $X$ containing $\ce$.
+\end{definition}
+
+\begin{definition}[Borel $\sigma$-Algebra]
+\label{definition:borel-sigma-algebra}
+    Let $(X, \topo)$ be a topological space, then the \textbf{Borel $\sigma$-algebra} $\cb_X$ on $X$ is the $\sigma$-algebra generated by $\topo$.
+\end{definition}

src/measure/sets/elementary.tex

@@ -0,0 +1,35 @@
+\section{Elementary Families}
+\label{section:elementary-families}
+
+\begin{definition}[Elementary Family]
+\label{definition:elementary-family}
+    Let $X$ be a set and $\ce \subset 2^X$, then $\ce$ is an \textbf{elementary family} if:
+    \begin{enumerate}
+        \item[(P1)] $\emptyset \in \mathcal{P}$.
+        \item[(P2)] For any $A, B \in \mathcal{P}$, $A \cap B \in \mathcal{P}$.
+        \item[(E1)] For any $E \in \ce$, there exists $\seqf{E_j} \subset \ce$ such that $E^c = \bigsqcup_{j = 1}^n E_j$.
+    \end{enumerate}
+\end{definition}
+
+\begin{proposition}[{{\cite[Proposition 1.7]{Folland}}}]
+\label{proposition:elementary-family-algebra}
+    Let $X$ be a set and $\ce \subset 2^X$ be an elementary family, then
+    \[
+    \alg = \bracs{\bigsqcup_{i = 1}^n E_j \bigg | \seqf{E_j} \subset \text{ pairwise disjoint}}
+    \]
+    is an algebra.
+\end{proposition}
+\begin{proof}
+    (A1): By (P1) $\emptyset \in \ce$. By (E1), there exists $\seqf{E_j} \subset \ce$ such that $X = \emptyset^c = \bigsqcup_{j = 1}^nE_j \in \alg$.
+
+    (A2): Let $A = \bigsqcup_{j = 1}^n A_j \in \alg$. By including additional empty sets, assume without loss of generality that there exists $\bracs{E_{i, j}| 1 \le i \le n, 1 \le j \le m}$ such that $A_i^c = \bigsqcup_{j = 1}^m E_{i, j}$ for each $1 \le j \le n$. In which case,
+    \[
+    A^c = \bigcap_{i = 1}^n A_i^c = \bigcap_{i = 1}^n \bigsqcup_{j = 1}^m E_{i, j} = \bigsqcup_{\alpha \in [1, m]^n} \underbrace{\bigcap_{i = 1}^n E_{i, \alpha_i}}_{\in \ce} \in \ce
+    \]
+
+    (A3): Let $A = \bigsqcup_{j = 1}^n A_j, B = \bigsqcup_{j = 1}^m B_j \in \alg$, then
+    \[
+    A \cap B = \braks{\bigsqcup_{j = 1}^n A_j} \cap \braks{\bigsqcup_{j = 1}^m B_j} = \bigsqcup_{i = 1}^n \bigcup_{j = 1}^m \underbrace{A_i \cap B_j}_{\in \ce} \in \alg
+    \]
+    so $\alg$ is closed under intersections. Thus using (A2), $A \cup B = (A^c \cap B^c)^c \in \alg$.
+\end{proof}

src/measure/sets/index.tex

@@ -0,0 +1,6 @@
+\chapter{Set Systems}
+\label{chap:set-system}
+
+\input{./src/measure/sets/algebra.tex}
+\input{./src/measure/sets/lambda.tex}
+\input{./src/measure/sets/elementary.tex}

src/measure/sets/lambda.tex

@@ -0,0 +1,64 @@
+\section{Lambda Systems}
+\label{section:lambda-system}
+
+\begin{definition}[$\pi$-System]
+\label{definition:pi-system}
+    Let $X$ be a set and $\mathcal{P} \subset 2^X$, then $\mathcal{P}$ is a \textbf{$\pi$-system} if:
+    \begin{enumerate}
+        \item[(P1)] $\emptyset \in \mathcal{P}$.
+        \item[(P2)] For any $A, B \in \mathcal{P}$, $A \cap B \in \mathcal{P}$.
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[$\lambda$-System]
+\label{definition:lambda-system}
+    Let $X$ be a set and $\alg \subset 2^X$, then $\alg$ is a \textbf{$\lambda$-system/$d$-system} if:
+    \begin{enumerate}
+        \item[(L1)] $\emptyset, X \in \alg$.
+        \item[(L2)] For any $E, F \in \alg$ with $E \subset F$, $F \setminus E \in \alg$.
+        \item[(L3)] For any $\seq{A_n} \subset \alg$ with $A_n \subset A_{n+1}$ for all $n \in \nat^+$, $\bigcup_{n \in \nat^+}A_n \in \alg$.
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Generated $\lambda$-System]
+\label{definition:generated-lambda-system}
+    Let $X$ be a set and $\ce \subset 2^X$, then the smallest $\lambda$-system $\lambda(\ce)$ over $X$ containing $\ce$ is the \textbf{$\lambda$-system generated by $\ce$}.
+\end{definition}
+
+\begin{lemma}
+\label{lemma:pi-lambda}
+    Let $X$ be a set and $\alg \subset 2^X$, then the following are equivalent:
+    \begin{enumerate}
+        \item $\alg$ is a $\sigma$-algebra.
+        \item $\alg$ is a $\pi$-system and a $\lambda$-system.
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    $(2) \Rightarrow (1)$: Let $A, B \in \alg$, then $A \cup B = (A^c \cap B^c)^c \in \alg$. Thus $\alg$ is an algebra. By \ref{lemma:sigma-algebra-condition}, $\alg$ is a $\sigma$-algebra.
+\end{proof}
+
+
+\begin{theorem}[Dynkin's $\pi$-$\lambda$ Theorem]
+\label{theorem:pi-lambda}
+    Let $X$ be a set and $\mathcal{P} \subset 2^X$ be a $\pi$-system, then $\sigma(\mathcal{P}) = \lambda(\mathcal{P})$.
+\end{theorem}
+\begin{proof}
+    Let $\ce \subset \lambda(\mathcal{P})$ and
+    \[
+    \cm(\ce) = \bracs{E \in \lambda(\mathcal{P})| E \cap F \in \lambda(\mathcal{P}) \forall F \in \ce}
+    \]
+    then
+    \begin{enumerate}
+        \item[(L2)] Let $E, F \in \cm$ with $E \subset F$, then for any $G \in \ce$,
+        \[
+        (F \setminus E) \cap G = (F \cap G) \setminus (E \cap G) \in \lambda(\mathcal{P})
+        \]
+        \item[(L3)] Let $\seq{E_n} \in \cm$ with $E_n \subset E_{n+1}$ for all $n \in \natp$, and $F \in \ce$, then
+        \[
+        \braks{\bigcup_{n \in \natp}E_n} \cap F = \bigcup_{n \in \natp}E_n \cap F \in \lambda(\mathcal{P})
+        \]
+    \end{enumerate}
+    so $\cm(\ce)$ is a $\lambda$-system.
+
+    Since $\mathcal{P}$ is a $\pi$-system, $\cm(\mathcal{P}) \supset \mathcal{P}$, so $\cm(\mathcal{P}) = \lambda(\mathcal{P})$. Thus for any $E \in \lambda(\mathcal{P})$ and $F \in \lambda(\mathcal{P})$, $E \cap F \in \lambda(\mathcal{P})$. Therefore $\cm(\lambda(\mathcal{P})) \supset \mathcal{P}$, $\cm(\lambda(\mathcal{P})) = \lambda(\mathcal{P})$, and $\lambda(\mathcal{P})$ satisfies (P2). By \ref{lemma:pi-lambda}, $\lambda(\mathcal{P})$ is a $\sigma$-algebra.
+\end{proof}

src/topology/functions/index.tex

@@ -0,0 +1,4 @@
+\chapter{Function Spaces}
+\label{chap:function-spaces}
+
+\input{./src/topology/functions/set-systems.tex}

src/topology/functions/set-systems.tex

@@ -0,0 +1,55 @@
+\section{Topology With Respect to Families of Sets}
+\label{section:pointwise}
+
+\begin{definition}[Set-Open Topology]
+\label{definition:set-open}
+    Let $T$ be a set, $\mathfrak{S} \subset 2^T$ be a non-empty family of sets, and $(X, \topo)$ be a topological space. For each $S \in \mathfrak{S}$ and $U \subset X$ open, let
+    \[
+    M(S, U) = \bracs{f \in X^T| f(S) \subset U}
+    \]
+    and
+    \[
+    \ce(\mathfrak{S}, \topo) = \bracs{M(S, U)| S \in \mathfrak{S}, U \in \topo}
+    \]
+    then the topology generated by $\ce$ is the \textbf{$\mathfrak{S}$-open topology} on $T^X$.
+
+    If $\cb \subset \topo$ generates $\topo$, then $\ce(\mathfrak{S}, \cb)$ generates the $\mathfrak{S}$-open topology.
+\end{definition}
+
+
+\begin{definition}[Set Uniformity]
+\label{definition:set-uniform}
+    Let $T$ be a set, $\mathfrak{S} \subset 2^T$ be a non-empty family of sets, and $(X, \fU)$ be a uniform space. For each $S \in \mathfrak{S}$ and $U \in \fU$, let
+    \[
+    E(S, U) = \bracs{(f, g) \in X^T|(f(x), g(x)) \in U \forall x \in S}
+    \]
+    and
+    \[
+    \mathfrak{E}(\mathfrak{S}, \fU) = \bracs{E(S, U)| S \in \mathfrak{S}, U \in \fU}
+    \]
+    then
+    \begin{enumerate}
+        \item $\mathfrak{E}(\mathfrak{S}, \fU)$ generates a uniformity $\fV$ on $X^T$.
+        \item The topology induced by $\fV$ is finer than the $\mathfrak{S}$-topology on $T^X$.
+        \item If $\mathfrak{S}$ is upward-directed with respect to inclusion, then $\mathfrak{E}(\mathfrak{S}, \fU)$ is forms a fundamental system of entourages for $\fV$.
+    \end{enumerate}
+    and the topology induced by $\fV$ is the \textbf{topology of uniform convergence on the sets $\mathfrak{S}$}, or the $\mathfrak{S}$-topology.
+\end{definition}
+\begin{proof}
+    (1): Since $\Delta \subset E(S, U)$ for all $S \in \mathfrak{S}$ and $U \in \fU$, $\mathfrak{E}(\mathfrak{S}, \fU)$ generates a uniformity on $X^T$.
+
+    (2): Let $U \subset X$ be open, then for each $x \in U$, there exists $V_x \in \fU$ such that $x \in V_x(x) \subset U$. In which case, $U = \bigcup_{x \in U}V_x(x)$ and $M(S, U) = \bigcup_{x \in U}M(S, V_x)(x)$.
+
+    (3): It is sufficient to verify
+    \begin{enumerate}
+        \item[(FB1)] For any $S, S'  \in \mathfrak{S}$, there exists $T \in \mathfrak{S}$ with $T \supset S, S'$. In which case, for any $U, U' \in \fU$,
+        \[
+        E(T, U \cap U') \subset E(S \cup S', U \cap U') \subset E(S, U) \cap E(S', U')
+        \]
+        \item[(FB3)] For any $U \in \fU$, there exists $V \in \fV$ with $V \circ V \subset U$. Thus for any $S \in \mathfrak{S}$,
+        \[
+        E(S, V) \circ E(S, V) \subset E(S, V \circ V) \subset E(S, U)
+        \]
+    \end{enumerate}
+    By \ref{proposition:fundamental-entourage-criterion}, $\mathfrak{E}$ is a fundamental system of entourages for the uniformity that it generates.
+\end{proof}

src/topology/index.tex

@@ -1,5 +1,6 @@
 \part{General Topology}
-\label{part:-part-topology}
+\label{part:topology}
 
 \input{./src/topology/main/index.tex}
 \input{./src/topology/uniform/index.tex}
+\input{./src/topology/functions/index.tex}

src/topology/main/baire.tex

@@ -6,10 +6,10 @@
 \label{definition:baire}
     Let $X$ be a topological space, then the following are equivalent:
     \begin{enumerate}
-        \item For any $\seq{A_n}$ nowhere dense, $\bigcup_{n \in \nat}A_n \subsetneq X$.
-        \item For any $\seq{A_n}$ closed with empty interior, $\bigcup_{n \in \nat}A_n \subsetneq X$.
-        \item For any $\seq{A_n}$ closed with $\bigcup_{n \in \nat}A_n = X$, there exists $N \in \nat$ such that $\bigcup_{n \le N}A_n$ has non-empty interior.
-        \item For any $\seq{U_n}$ open and dense, $\bigcap_{n \in \nat}U_n$ is dense.
+        \item For any $\seq{A_n}$ nowhere dense, $\bigcup_{n \in \nat^+}A_n \subsetneq X$.
+        \item For any $\seq{A_n}$ closed with empty interior, $\bigcup_{n \in \nat^+}A_n \subsetneq X$.
+        \item For any $\seq{A_n}$ closed with $\bigcup_{n \in \nat^+}A_n = X$, there exists $N \in \nat^+$ such that $\bigcup_{n \le N}A_n$ has non-empty interior.
+        \item For any $\seq{U_n}$ open and dense, $\bigcap_{n \in \nat^+}U_n$ is dense.
     \end{enumerate}
     If the above holds, then $X$ is a \textbf{Baire space}.
 \end{definition}

src/topology/main/definition.tex

@@ -54,7 +54,7 @@
     Conversely, if $\cb \subset 2^X$ is a family such that:
     \begin{enumerate}
         \item[(TB1)] For every $x \in X$, there exists $U \in \cb$ such that $x \in U$.
-        \item[(TB2)] For every $x \in X$ and $U, V \in \cb$ such that $x \in U \cap$, there exists $W \in \cb$ such that $x \in W \subset U \cap V$.
+        \item[(TB2)] For every $x \in X$ and $U, V \in \cb$ such that $x \in U \cap V$, there exists $W \in \cb$ such that $x \in W \subset U \cap V$.
     \end{enumerate}
     then $\topo(\cb)$ is a topology on $X$, and $\cb$ is a base for $\topo(\cb)$.
 \end{definition}

src/topology/main/metric.tex

@@ -29,7 +29,7 @@
 
 \begin{lemma}
 \label{lemma:uniform-first-countable}
-    Let $X$ be a uniform space with a countable fundamental system of entourages, then there exists a countable fundamental system of symmetric entourages $\fB = \bracs{U_{1/2^n}|n \in \nat^+}$ such that $U_{1/2^{n+1}} \circ U_{1/2^{n+1}} \subset U_{1/2^n}$ for all $n \in \nat$.
+    Let $X$ be a uniform space with a countable fundamental system of entourages, then there exists a countable fundamental system of symmetric entourages $\fB = \bracs{U_{1/2^n}|n \in \nat^+}$ such that $U_{1/2^{n+1}} \circ U_{1/2^{n+1}} \subset U_{1/2^n}$ for all $n \in \nat^+$.
 \end{lemma}
 \begin{proof}
     Let $\seq{V}$ be a countable fundamental system of entourages. By \ref{lemma:symmetricfundamentalentourage}, there exists a symmetric entourage $U_{1/2} \subset V_1$. Suppose inductively that $U_{1/2^n}$ has been constructed, then by (U2) and \ref{lemma:symmetricfundamentalentourage}, there exists a symmetric entourage $U_{1/2^{n+1}} \subset U_{1/2^n} \cap V$. Thus $\fB = \bracs{U_{1/2^n}|n \in \nat^+}$ is the desired family.