Cleanup

src/measure/measurable-maps/metric.tex

@@ -7,13 +7,15 @@
     \[
     X \to Z \quad x \mapsto F(f_1(x), \cdots, f_n(x))
     \]
+
     is $(\cm, \cn)$-measurable.
 \end{proposition}
 \begin{proof}
-    By \ref{proposition:product-sigma-algebra-metric}, $\cb_{\prod_{j = 1}^n Y_j} = \bigotimes_{j = 1}^n \cb_{Y_j}$, so
+    By \autoref{proposition:product-sigma-algebra-metric}, $\cb_{\prod_{j = 1}^n Y_j} = \bigotimes_{j = 1}^n \cb_{Y_j}$, so
     \[
     X \to \prod_{j = 1}^n Y_j \quad x \mapsto (f_1(x), \cdots, f_n(x))
     \]
+
     is $(\cm, \cb_{\prod_{j = 1}^n Y_j})$-measurable. Therefore the composition is $(\cm, \cn)$-measurable.
 \end{proof}
 
@@ -27,7 +29,7 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    By \ref{proposition:metric-measurable-fibre-product}.
+    By \autoref{proposition:metric-measurable-fibre-product}.
 \end{proof}
 
 \begin{proposition}
@@ -43,7 +45,8 @@
     \[
     d_{U^c}: X \to [0, 1] \quad x \mapsto d(x, U^c) \wedge 1
     \]
-    is continuous by \ref{proposition:set-distance-continuous}. By \ref{proposition:fattening-closure}, $\bracsn{d_{U^c} > 0} = U$. Thus $\bracs{f \in U} = \bracsn{d_{U^c} \circ f > 0}$.
+
+    is continuous by \autoref{proposition:set-distance-continuous}. By \autoref{proposition:fattening-closure}, $\bracsn{d_{U^c} > 0} = U$. Thus $\bracs{f \in U} = \bracsn{d_{U^c} \circ f > 0}$.
 \end{proof}
 
 \begin{proposition}
@@ -59,9 +62,10 @@
     \[
     \bracs{\limv{n}f_n \text{ exists}} = \bigcap_{k \in \natp}\bigcup_{N \in \natp}\bigcap_{m \in \natp}\bigcap_{n \in \natp}\bracsn{d(f_m, f_n) < 1/k}
     \]
-    is measurable by \ref{proposition:metric-measurables}.
 
-    (2): For each $\phi \in C(X; [0, 1])$, $\phi \circ f = \limv{n}\phi \circ f_n$ is $(\cm, \cb_\real)$-measurable by \ref{proposition:limit-measurable}. Thus $f$ is $(\cm, \cb_\real)$-measurable by \ref{proposition:metric-measurable-compose}.
+    is measurable by \autoref{proposition:metric-measurables}.
+
+    (2): For each $\phi \in C(X; [0, 1])$, $\phi \circ f = \limv{n}\phi \circ f_n$ is $(\cm, \cb_\real)$-measurable by \autoref{proposition:limit-measurable}. Thus $f$ is $(\cm, \cb_\real)$-measurable by \autoref{proposition:metric-measurable-compose}.
 \end{proof}
 
 
@@ -89,18 +93,22 @@
     \[
     C(N, x) = \bracs{1 \le n \le N|y_n \in N(f(x))}
     \]
+
     By assumption (b), $1 \in C(N, x) \ne \emptyset$. Let
     \[
     k(N, x) = \min\bracs{n \in C(N, x) \bigg | d(f(x), y_n) = \min_{m \in C(N, x)}d(f(x), y_m)}
     \]
+
     then for any $k \in \natp$,
     \[
     \bracs{x \in X|k(n, x) \le k} = \bigcup_{j = 1}^k\bracs{x \in X \bigg |y_j \in C(n, x), d(f(x), y_j) = \min_{m \in C(N, x)}d(f(x), y_m)}
     \]
-    For each $1 \le m \le N$, $y \mapsto d(y, y_m)$ is continuous. Thus $x \mapsto d(f(x), y_m)$ and $x \mapsto \min_{m \in C(N, x)}d(f(x), y_m)$ are $(\cm, \cb_\real)$-measurable by \ref{proposition:limit-measurable} and assumption (c). By \ref{proposition:metric-measurables},
+
+    For each $1 \le m \le N$, $y \mapsto d(y, y_m)$ is continuous. Thus $x \mapsto d(f(x), y_m)$ and $x \mapsto \min_{m \in C(N, x)}d(f(x), y_m)$ are $(\cm, \cb_\real)$-measurable by \autoref{proposition:limit-measurable} and assumption (c). By \autoref{proposition:metric-measurables},
     \[
     \bracs{x \in X \bigg | d(f(x), y_j) = \min_{m \in C(N, x)}d(f(x), y_m)} \in \cm
     \]
+
     for each $1 \le j \le k$. Combining this with assumption (c) shows that $\bracs{x \in X|k(n, x) \le k} \in \cm$.
 
     Let $f_N(x) = y_{k(N, x)}$, then for each $N \in \natp$, $f_N(x) \subset \bracs{y_n|1 \le n \le N}$, so $f_N(x)$ is finitely valued. Since $x \mapsto k{(N, x)}$ is $(\cm, 2^\nat)$-measurable, $f_N$ is $(\cm, \cb_Y)$-measurable.
@@ -109,9 +117,10 @@
     \[
     f(x) \in \ol{N^o(f(x))} = \ol{\bracs{y_n| n \in \natp, y_n \in N^o(f(x))}}
     \]
-    by assumption (a) and \ref{definition:dense}. Thus for any $\eps > 0$, there exists $N \in \nat$ such that $y_N \in N^o(f(x))$ and $d(f(x), y_N) < \eps$. In which case, $d(f(x), f_n(x)) < \eps$ for all $n \ge N$. Therefore $f_n \to f$ pointwise.
 
-    (3) $\Rightarrow$ (1): By \ref{proposition:metric-measurable-limit}.
+    by assumption (a) and \autoref{definition:dense}. Thus for any $\eps > 0$, there exists $N \in \nat$ such that $y_N \in N^o(f(x))$ and $d(f(x), y_N) < \eps$. In which case, $d(f(x), f_n(x)) < \eps$ for all $n \ge N$. Therefore $f_n \to f$ pointwise.
+
+    (3) $\Rightarrow$ (1): By \autoref{proposition:metric-measurable-limit}.
 \end{proof}
 
 \begin{proposition}
@@ -127,6 +136,7 @@
     \[
     N: E \to 2^E \quad y \mapsto B_E(0, \norm{y}_E)
     \]
+
     then
     \begin{enumerate}
         \item[(a)] $y \in \ol{B_E(0, \norm{y}_E)}$.
@@ -135,6 +145,7 @@
         \[
         \bracs{y \in E|y_0 \in N(y)} = \bracs{y \in E|\norm{y_0}_E \le \norm{y}_E} \in \cb_E
         \]
+
     \end{enumerate}
-    By \ref{proposition:measurable-simple-separable}, (1) and (2) are equivalent.
+    By \autoref{proposition:measurable-simple-separable}, (1) and (2) are equivalent.
 \end{proof}