Updated formulation of convergence in measure.

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

@@ -5,7 +5,7 @@
 \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
     \[
-    U(\delta, \eps) = \bracs{(f, g) \in \mathscr{M}(X; Y)| \mu\bracs{d(f, g) > \delta} < \eps}
+    U(\delta, \eps) = \bracs{(f, g) \in \mathcal{L}^0(X; Y)| \mu\bracs{d(f, g) > \delta} < \eps}
     \]
 
     then
@@ -13,7 +13,7 @@
     \fB = \bracs{U(\delta, \eps)|\eps, \delta > 0}
     \]
 
-    forms a fundamental system of entourages for a uniformity. The uniformity induced by $\fB$ is the \textbf{uniform structure of convergence in measure} on $\mathscr{M}(X; Y)$.
+    forms a fundamental system of entourages for a uniformity. The uniformity induced by $\fB$ is the \textbf{uniform structure of 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}:
@@ -22,7 +22,7 @@
         \[
         U(\delta \wedge \delta', \eps \wedge \eps') \subset U(\delta, \eps) \cap U(\delta', \eps')
         \]
-        \item[(UB3)] For each $\eps, \delta > 0$ and $f, g, h \in \mathscr{M}(X; Y)$, 
+        \item[(UB3)] For each $\eps, \delta > 0$ and $f, g, h \in \mathcal{L}^0(X; Y)$, 
         \[
         \bracs{d(f, h) > \delta} \subset \bracs{d(f, g) > \delta} \cup \bracs{d(g, h) > \delta}
         \]
@@ -35,19 +35,19 @@
 \label{definition:ky-fan}
     Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and
     \[
-    \alpha: \mathscr{M}(X; Y)^2 \to [0, \infty) \quad (f, g) \mapsto \inf\bracs{\eps > 0| \mu\bracs{d(f, g) > \eps} \le \eps}
+    \alpha: L^0(X; Y)^2 \to [0, \infty) \quad (f, g) \mapsto \inf\bracs{\eps > 0| \mu\bracs{d(f, g) > \eps} \le \eps} \wedge 1
     \]
 
     then:
     \begin{enumerate}
-        \item $\alpha$ is a metric on $\mathscr{M}(X; Y)$ modulo almost everywhere equality. 
+        \item $\alpha$ is a metric on $L^0(X; Y)$.
-        \item $\alpha$ induces the uniform structure of convergence in measure on $\mathscr{M}(X; Y)$. 
+        \item $\alpha$ induces the uniform structure of convergence in measure on $L^0(X; Y)$. 
     \end{enumerate}
 
-    The mapping $\alpha$ is the \textbf{Ky Fan metric} on $\mathscr{M}(X; Y)$. 
+    The mapping $\alpha$ is the \textbf{Ky Fan metric} on $L^0(X; Y)$. 
 \end{definition}
 \begin{proof}
-    (1): Let $f, g, h \in \mathscr{M}(X; Y)$, then
+    (1): Let $f, g, h \in \mathcal{L}^0(X; Y)$, then
     \begin{enumerate}
         \item[(M)] If $\alpha(f, g) = 0$, then by \hyperref[continuity from above]{proposition:measure-properties}, 
         \[
@@ -63,12 +63,12 @@
         so $\alpha(f, h) \le \alpha(f, g) + \alpha(g, h)$. 
     \end{enumerate}
 
-    so $\alpha$ is a metric on $\mathscr{M}(X; Y)$, modulo almost everywhere equality. 
+    so $\alpha$ is a metric on $\mathcal{L}^0(X; Y)$, modulo almost everywhere equality. 
 
-    (2): Let $f, g \in \mathscr{M}(X; Y)$. For any $\eps, \delta > 0$, if $\alpha(f, g) < \eps \wedge \delta$, then there exists $r \in (0, \eps \wedge \delta]$ such that $\mu\bracs{d(f, g) > r} \le r$. Thus
+    (2): Let $f, g \in \mathcal{L}^0(X; Y)$. For any $\eps, \delta > 0$, if $\alpha(f, g) < \eps \wedge \delta$, then there exists $r \in (0, \eps \wedge \delta]$ such that $\mu\bracs{d(f, g) > r} \le r$. Thus
     \[
-    \bracs{(f, g) \in \mathscr{M}(X; Y)|\alpha(f, g) < \eps \wedge \delta} \subset 
+    \bracs{(f, g) \in \mathcal{L}^0(X; Y)|\alpha(f, g) < \eps \wedge \delta} \subset 
-    \bracs{(f, g) \in \mathscr{M}(X; Y)|\mu\bracs{d(f, g) >\delta} < \eps}
+    \bracs{(f, g) \in \mathcal{L}^0(X; Y)|\mu\bracs{d(f, g) >\delta} < \eps}
     \]
 
     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. 
@@ -78,7 +78,7 @@
 \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
     \[
-    U(A, \delta, \eps) = \bracs{(f, g) \in \mathscr{M}(X; Y)| \mu(A \cap \bracs{d(f, g) > \delta}) < \eps}
+    U(A, \delta, \eps) = \bracs{(f, g) \in \mathcal{L}^0(X; Y)| \mu(A \cap \bracs{d(f, g) > \delta}) < \eps}
     \]
 
     then
@@ -86,7 +86,7 @@
     \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 $\mathscr{M}(X; Y)$.
+    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}:
@@ -95,7 +95,7 @@
         \[
         U(A \cup A', \delta \wedge \delta', \eps \wedge \eps') \subset U(A, \delta, \eps) \cap U(A', \delta', \eps')
         \]
-        \item[(UB3)] For each $\eps, \delta > 0$, $A \in \cm$ with $\mu(A) < \infty$, and $f, g, h \in \mathscr{M}(X; Y)$, 
+        \item[(UB3)] For each $\eps, \delta > 0$, $A \in \cm$ with $\mu(A) < \infty$, and $f, g, h \in \mathcal{L}^0(X; Y)$, 
         \[
         \bracs{d(f, h) > \delta} \subset \bracs{d(f, g) > \delta} \cup \bracs{d(g, h) > \delta}
         \]
@@ -149,18 +149,16 @@
     \]
 \end{proof}
 
-
+\begin{theorem}
-\begin{proposition}[{{\cite[Theorem 2.30]{Folland}}}]
+\label{theorem:cauchy-in-measure-limit}
-\label{proposition: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, $(Y, d)$ be a complete metric space, and $\seq{f_n} \subset Y^X$ be a sequence Borel measurable functions from $X \to Y$ that is Cauchy in measure, then:
     \begin{enumerate}
-        \item There exists a Borel measurable function $f: X \to Y$ such that $f_n \to f$ in measure.
+        \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.
-        \item There exists a subsequence $\seq{n_k}$ such that $f_{n_k} \to f$ almost everywhere.
+        \item $L^0(X; Y)$ equipped with the uniform structure of convergence in measure is complete. 
-        \item For any Borel measurable function $g: X \to Y$ such that $f_n \to g$ in measure, $f = g$ almost everywhere.
     \end{enumerate}
-\end{proposition}
+\end{theorem}
-\begin{proof}
+\begin{proof}[Proof, [{{\cite[Theorem 2.30]{Folland}}}]. ]
-    (2): Since $\seq{f_n}$ is Cauchy in measure, there exists a subsequence $\seq{n_k}$ such that for each $k \in \natp$, $\mu(\bracsn{d(f_{n_k}, f_{n_{k+1}}) > 2^{-k}}) \le 2^{-k}$.
+    (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 $j \ge k \ge K$, 
     \[
@@ -168,7 +166,7 @@
     \subset \bigcup_{\ell = K}^{\infty}\bracsn{d(f_{n_\ell}, f_{n_{\ell + 1}}) > 2^{-\ell}}
     \]
 
-    By monotonicity and subadditivity (\autoref{proposition:measure-properties}), 
+    By \hyperref[monotonicity and subadditivity]{proposition:measure-properties}, 
     \[
     \mu\paren{\bigcup_{j, k \ge K}\bracsn{d(f_{n_j}, f_{n_k}) > 2^{-K+1}}} 
     \le \sum_{\ell \ge K}\mu\bracsn{d(f_{n_\ell}, f_{n_{\ell + 1}}) > 2^{-\ell}}
@@ -187,12 +185,7 @@
 
     Thus, for almost every $x \in X$, there exists $K \in \natp$ such that $d(f_{n_j}(x), f_{n_k}(x)) < 2^{-K+1}$ for all $j, k \ge K$. Therefore $\seq{f_n(x)}$ is Cauchy for almost every $x$, and converges to a Borel measurable function $f: X \to Y$.
 
-    (1): Let $\eps, \delta > 0$, then there exists $N \in \natp$ such that $\mu(\bracs{d(f_m, f_n) > \delta/2}) < \eps/2$ for all $m, n \ge N$. Let $k \in \natp$ such that $n_k \ge N$ and $\mu(\bracs{d(f, f_{n_k}) > \delta/2}) < \eps/2$, then for any $m \ge N$,
+    (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}. 
-    \[
-    \mu\bracs{d(f_m, f) > \delta} \le \mu\bracs{d(f_m, f_{n_k}) > \delta/2} + \mu\bracs{d(f, f_{n_k}) > \delta/2} < \eps
-    \]
-
-    (3): By (2), there exists a subsequence $\seq{n_k}$ such that $f_{n_k} \to f$ and $f_{n_k} \to g$ almost everywhere, so $f = g$ almost everywhere. 
 \end{proof}
 
 \begin{theorem}[Monotone Convergence Theorem (in Measure)]

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

@@ -4,10 +4,16 @@
 \begin{definition}[Measurable Function]
 \label{definition:measurable-function}
     Let $(X, \cm)$ and $(Y, \cn)$ be measurable spaces and $f: X \to Y$ be a mapping, then $f$ is \textbf{$(\cm, \cn)$-measurable} if $f^{-1}(E) \in \cm$ for all $E \in \cn$.
-
-    The set $\mathscr{M}(X; Y)$ is the \textbf{space of measurable functions} from $X$ to $Y$. 
 \end{definition}
 
+\begin{definition}[Space of Measurable Functions]
+\label{definition:measurable-function-space}
+    Let $(X, \cm)$ and $(Y, \cn)$ be measurable spaces, then the set $\mathscr{L}^0(X, \cm; Y) = \mathcal{L}^0(X; Y)$ is the \textbf{space of measurable functions} from $X$ to $Y$. 
+
+    For any measure $\mu$ on $(X, \cm)$, the space $L^0(X, \cm, \mu; Y) = L^0(X; Y)$ is the space of measurable functions from $X$ to $Y$, modulo almost everywhere equality. 
+\end{definition}
+
+
 
 \begin{definition}[Borel Measurable]
 \label{definition:borel-measurable-function}

src/measure/notation.tex

@@ -12,7 +12,8 @@
     $\mathcal{B}_X$ & Borel $\sigma$-algebra on $X$. & \autoref{definition:borel-sigma-algebra} \\
     $\sigma(\{f_i \mid i \in I\})$ & $\sigma$-algebra generated by the maps $\{f_i\}$. & \autoref{definition:generated-sigma-algebra-function} \\
     $\bigotimes_{i \in I} \mathcal{M}_i$ & Product $\sigma$-algebra. & \autoref{definition:product-sigma-algebra} \\
-    $\mathscr{M}(X; Y)$ & Space of measurable functions from $X$ to $Y$. & \autoref{definition:measurable-function} \\
+    $\mathcal{L}^0(X; Y)$ & Space of measurable functions from $X$ to $Y$. & \autoref{definition:measurable-function-space} \\
+    $L^0(X; Y)$ & Space of measurable functions from $X$ to $Y$, modulo almost everywhere equality. & \autoref{definition:measurable-function-space} \\
     $\chi_E = \mathbf{1}_E$ & Indicator function of $E$. & \autoref{definition:indicator-function} \\
     $\Sigma(X, \mathcal{M}; E)$ & Space of $E$-valued simple functions on $(X, \mathcal{M})$. & \autoref{definition:simple-function-standard-form} \\
     $\Sigma^+(X, \mathcal{M})$ & Space of non-negative simple functions. & \autoref{definition:simple-function-scalar} \\

src/measure/radon/radon.tex

@@ -197,7 +197,7 @@
 \begin{proof}[Proof, {{\cite[Theorem 7.10]{Folland}}}. ]
     First assume that $f$ is bounded.
 
-    (1, bounded): If $f$ is bounded, then $f \in L^1(X; E)$. By \autoref{proposition:radon-cc-dense}, there exists $\seq{\phi_n} \subset C_c(X)$ such that $\phi_n \to f$ in $L^1(\mu)$. Since $\phi_n \to f$ in $L^1(\mu)$, $\phi_n \to f$ in measure by \autoref{proposition:lp-in-measure}. By taking a subsequence using \ref{proposition:cauchy-in-measure-limit}, assume without loss of generality that $\phi_n \to f$ almost everywhere. 
+    (1, bounded): If $f$ is bounded, then $f \in L^1(X; E)$. By \autoref{proposition:radon-cc-dense}, there exists $\seq{\phi_n} \subset C_c(X)$ such that $\phi_n \to f$ in $L^1(\mu)$. Since $\phi_n \to f$ in $L^1(\mu)$, $\phi_n \to f$ in measure by \autoref{proposition:lp-in-measure}. By taking a subsequence using \autoref{theorem:cauchy-in-measure-limit}, assume without loss of generality that $\phi_n \to f$ almost everywhere. 
 
     By \hyperref[Egoroff's Theorem]{theorem:egoroff}, there exists $A \subset \bracs{f \ne 0}$ such that $\phi_n \to f$ uniformly and $\mu(\bracs{f \ne 0} \setminus A) < \eps/3$. By \autoref{proposition:radon-regular-sigma-finite}, there exists $K \subset A$ compact such that $\mu(A \setminus K) < \eps/3$. 
 

src/topology/metric/metric.tex

@@ -55,6 +55,20 @@
 \end{proof}
 
 
+\begin{proposition}
+\label{proposition:complete-metric-space}
+    Let $(X, d)$ be a metric space, then the following are equivalent:
+    \begin{enumerate}
+        \item For any Cauchy sequence $\seq{x_n} \subset X$, there exists $x \in X$ such that $x = \limv{n}x_n$. 
+        \item For any Cauchy sequence $\seq{x_n} \subset X$, there exists a subsequence $\seq{n_k}$ and $x \in X$ such that $x = \limv{k}x_{n_k}$. 
+        \item For every Cauchy filter $\fF \subset 2^X$, there exists $x \in X$ such that $\fF \to x$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (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{theorem}[Banach's Fixed Point Theorem]
 \label{theorem:banach-fixed-point}
     Let $(X, d)$ be a complete metric space and $f: X \to X$. If there exists $C \in (0, 1)$ such that

src/topology/uniform/cauchy.tex

@@ -54,11 +54,11 @@
 \end{definition}
 
 
-\begin{proposition}[{{\cite[Proposition 2.3.3]{Bourbaki}}}]
+\begin{proposition}
 \label{proposition:imagecauchy}
     Let $(X, \fU)$ and $(Y, \mathfrak{V})$ be uniform spaces and $f \in UC(X; Y)$, then $f$ is Cauchy continuous.
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Proposition 2.3.3]{Bourbaki}}}. ]
     Let $V \in \mathfrak{V}$, then there exists $V' \in \fU$ such that $(f(x), f(y)) \in V$ whenever $(x, y) \in V'$. For any Cauchy filter base $\fB \subset 2^X$, there exists $E \in \fB$ such that $E \times E \subset V'$, so $f(E) \times f(E) \subset V$.
 \end{proof}
 
@@ -67,7 +67,7 @@
     Let $X$ be a uniform space and $\fF \subset 2^X$ be a Cauchy filter, then $\fF$ is \textbf{minimal} if it is minimal with respect to inclusion.
 \end{definition}
 
-\begin{proposition}[{{\cite[Proposition 2.3.5]{Bourbaki}}}]
+\begin{proposition}
 \label{proposition:minimalcauchyexistence}
     Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a Cauchy filter, then:
     \begin{enumerate}
@@ -75,7 +75,7 @@
         \item If $\fB \subset \fF$ is a base of $\fF$, and $\fV \subset \fU$ is a fundamental system of symmetric entourages, then $\mathfrak{M} = \bracs{V(M): V \in \fV, M \in \fB}$ is a base for $\fF_0$.
     \end{enumerate}
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Proposition 2.3.5]{Bourbaki}}}. ]
     Since $\fV$ and $\fB$ are both non-empty, $\mathfrak{M}$ is non-empty as well. Let $V_1, V_2 \in \fV$ and $M_1, M_2 \in \fB$. By (F2) of $\fV$ and $\fB$, there exists $V_0 \in \fV$ and $M_0 \in \fB$ such that $V_0 \subset V_1 \cap V_2$ and $M_0 \subset M_1 \cap M_2$. So $V_0(M_0) \subset V_1(M_1) \cap V_2(M_2)$, and $\mathfrak{M}$ satisfies (FB1). Given that $\emptyset \not\in \fV, \fB$, $\emptyset \not\in \mathfrak{M}$, so $\mathfrak{M}$ satisfies (FB2). By \autoref{proposition:filterbasecriterion}, $\mathfrak{M}$ is a filter base for
     \[
     \fF_0 = \bracs{E \subset X| \exists V \in \fV, M \in \fB: V(M) \subset E} \subset \fF
@@ -87,17 +87,17 @@
     Finally, let $\fF' \subset \fF$ be a Cauchy filter. Let $M \in \fB$ and $V \in \fV$, then there exists $N \in \fF'$ such that $N \times N \subset V$. Since $N \in \fF$, $M \cap N \ne \emptyset$, and there exists $y \in M \cap N$. For any $x \in N$, $N \times N \subset V$ implies that $x \in V(y)$. Hence $N \subset V(M)$ and $\fF' \supset \fF_0$.
 \end{proof}
 
-\begin{proposition}[{{\cite[Corollary 2.3.4]{Bourbaki}}}]
+\begin{proposition}
 \label{proposition:cauchyinterior}
     Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a minimal Cauchy filter, then $\fF$ admits a base consisting of open sets.
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Corollary 2.3.4]{Bourbaki}}}. ]
     Let $\fV \subset \fU$ be the set of all symmetric, open entourages. By \autoref{proposition:goodentourages}, $\fV$ is a fundamental system of entourages. By \autoref{proposition:minimalcauchyexistence}, $\mathfrak{M} = \bracs{U(M)| U \in \fV, M \in \fF}$ is a base for $\fF$.
 
     Let $V \in \fU$, then there exists $U \in \fV$ such that $U \subset V$. For any $M \in \fF$, $U(M)$ is open by \autoref{lemma:openentourageneighbourhoods}. Thus $\mathfrak{M}$ consists of open sets.
 \end{proof}
 
-\begin{proposition}[{{\cite[Corollary 2.3.1, 2.3.2]{Bourbaki}}}]
+\begin{proposition}
 \label{proposition:cauchyfilterlimit}
     Let $(X, \fU)$ be a uniform space, then:
     \begin{enumerate}
@@ -106,7 +106,7 @@
         \item For any $x \in X$, filter $\fF \subset 2^X$ converging to $x$, and Cauchy filter $\mathfrak{G} \subset \fF$, $\mathfrak{G}$ converges to $x$.
     \end{enumerate}
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Corollary 2.3.1, 2.3.2]{Bourbaki}}}. ]
     (1): Let $\fF = \bracs{E \subset X: x \in E}$, then $\fF$ is a filter generated by the filter base $\fB = \bracs{\bracs{x}}$. By \autoref{proposition:minimalcauchyexistence}, $\bracs{U(x): U \in \fU} = \cn(x)$ is the unique minimal filter contained in $\fF$.
 
     (2): Let $x \in X$ be an accumulation point of $\fF$, then $U(x) \cap E \ne \emptyset$ for all $U \in \fU$ and $E \in \fF$. By \autoref{definition:generatedfilter}, there exists a filter $\fF' \supset \fF \cup \cn(x)$. Since $\fF$ is a Cauchy filter, so is $\fF'$. By minimality of $\cn(x)$, $\fF \supset \cn(x)$.