Refined the approximation argument.

src/measure/measurable-maps/approx.tex

@@ -23,9 +23,13 @@
 
 \begin{definition}[Approximation of the Identity]
 \label{definition:approximation-id-measure}
-    Let $X$ be a topological space, $\mathcal{A}: X \to 2^X$ be an \hyperref[admissible approximant function]{definition:admissible-approximant-function} and $\net{I} \subset X^X$ be a net, then $\net{I}$ is an \textbf{$\mathcal{A}$-admissible approximation of the identity} if:
+    Let $X$ be a topological space and $\net{I} \subset X^X$ be a net, then $\net{I}$ is an \textbf{approximation of the identity} if:
     \begin{enumerate}[label=(AI\arabic*)]
         \item For each $x \in X$, $I_\alpha(x) \to x$. 
+    \end{enumerate}
+
+    For any \hyperref[admissible approximant function]{definition:admissible-approximant-function} $\mathcal{A}: X \to 2^X$, $\net{I}$ is \textbf{$\mathcal{A}$-admissible} if:
+    \begin{enumerate}[label=(AI\arabic*), start=1]
         \item For each $x \in X$ and $\alpha \in A$, $I_\alpha(x) \in \mathcal{A}(x)$. 
     \end{enumerate}
 
@@ -36,14 +40,12 @@
 \label{lemma:separable-metric-space-approx-identity}
     Let $X$ be a separable metric space, $\mathcal{A}: X \to 2^X$ be a Borel measurable \hyperref[admissible approximant function]{definition:admissible-approximant-function}, and $\seq{x_n} \subset X$ be a dense subset with $x_1 \in \bigcap_{x \in X}\mathcal{A}(x)$, then there exists $\seq{I_n} \subset X^X$ such that:
     \begin{enumerate}
-        \item $\seq{I_n}$ is an $\mathcal{A}$-admissible \hyperref[approximation of the identity]{definition:approximation-id-measure}. 
-        \item For each $N \in \natp$, $I_N$ is Borel measurable with $I_N(X) \subset \bracsn{x_n|1 \le n \le N}$. 
+        \item $\seq{I_n}$ is a Borel measurable, $\mathcal{A}$-admissible \hyperref[approximation of the identity]{definition:approximation-id-measure}. 
+        \item For each $N \in \natp$, $I_N(X) \subset \bracsn{x_n|1 \le n \le N}$. 
         \item For each $N \in \natp$ and $x \in X$, 
         \[
         d(x, I_N(x)) = \min\bracs{d(x, x_n)| 1 \le n \le N, x_n \in \mathcal{A}(x)}
         \]
-
-        In particular, $I_N(x) \to x$ as $N \to \infty$. 
     \end{enumerate}
 \end{lemma}
 \begin{proof}
@@ -59,7 +61,21 @@
     k_N(x) = \min\bracs{n \in C_N(x) \bigg | d(x, x_n) = \min_{m \in C_N(x)}d(x, x_m)}
     \]
 
-    be the minimum $n \in C_N(x)$ on which the minimal distance from $x$ to $\bracs{x_m|m \in C_N(x)}$ is achieved. Then, for each $n \in \natp$, 
+    be the minimum $n \in C_N(x)$ on which the minimal distance from $x$ to $\bracs{x_m|m \in C_N(x)}$ is achieved. Define
+    \[
+    I_N: X \to X \quad x \mapsto x_{k_N(x)}
+    \]
+
+    (2): For each $x \in X$, $k_N(x) \in [N]$, so $I_N(x) \in \bracsn{x_n|1 \le n \le N}$. 
+
+    (3): Let $x \in X$, then by definition of $k_N$ and $C_N$, 
+    \begin{align*}
+    d(x, I_N(x))& = d(x, x_{k_N(x)}) =  \min_{n \in C_N(x)}d(x, x_n) \\
+    &= \min\bracsn{d(x, x_n)|1 \le n \le N, x_n \in \mathcal{A}(x)}
+    \end{align*}
+    
+    
+    (1, Borel Measurable): Fix $N \in \natp$, then for each $n \in \natp$, 
     \begin{align*}
         \bracs{k_N \le n} &= \bigcup_{j = 1}^n \bracs{x \in X \bigg | j \in C_N(x), d(x, x_j) = \min_{m \in C_N(x)}d(x, x_m)} \\
         &= \bigcup_{j = 1}^n \bracs{j \in C_N} \cap \bracs{x \in X \bigg | d(x, x_j) = \min_{m \in C_N(x)}d(x, x_m)} \\
@@ -76,23 +92,17 @@
     I_N: X \to \bracsn{x_n|1 \le n \le N} \quad x \mapsto x_{k_N(x)}
     \]
 
-    then for each $1 \le n \le N$, $\bracs{I_N = x_n} = \bracs{k_N = n}$ is a Borel set. Thus $I_N$ is Borel measurable. 
+    By assumption that $\seq{x_n}$ are distinct, $\bracs{I_N = x_n} = \bracs{k_N = n}$ is a Borel set for each $1 \le n \le N$. Therefore $I_N$ is Borel measurable. 
 
-    In addition, for each $x \in X$, $k_N(x) \in C_N(x)$, so
+    (1, $\mathcal{A}$-Admissible): Let $N \in \natp$ and $x \in X$, then
     \[
     I_N(x) = x_{k_N(x)} \in \bracs{x_n|n \in C_N(x)} \subset \mathcal{A}(x)
     \]
 
-    and $\seq{I_N}$ satisfies (AI2).
-
-    Finally, let $x \in X$, then by definition of $k_N$, 
+    (1, Approximation): Let $x \in X$ and $\eps > 0$. Since $x \in \ol{\mathcal{A}(x)^o}$ and $\seq{x_n}$ is dense in $X$, there exists $N_0 \in \natp$ such that $x_{N_0} \in \mathcal{A}(x)$ and $d(x, x_{N_0}) < \eps$. By (3), 
     \[
-        d(x, I_N(x)) = d(x, x_{k_N(x)}) = \min\bracs{d(x, x_n)| 1 \le n \le N, x_n \in \mathcal{A}(x)}
+    \limv{N}d(x, I_N(x)) = \limv{N}\min\bracs{d(x, x_n)| 1 \le n \le N, x_n \in \mathcal{A}(x)} = 0
     \]
-
-    For any $\eps > 0$, since $x \in \ol{\mathcal{A}(x)^o}$, there exists $N_0 \in \natp$ such that $x_{N_0} \in \mathcal{A}(x)$ and $d(x, x_{N_0}) < \eps$. In which case, for any $N \ge N_0$, $N_0 \in C_N(x)$ and $d(x, I_N(x)) \le d(x, x_{N_0}) < \eps$. Thus $I_N(x) \to x$ as $N \to \infty$, and $\seq{I_N}$ satisfies (AI1). 
-    
-    Therefore $\seq{I_N}$ is an approximation of the identity satisfying (1) and (2). 
 \end{proof}
 
 \begin{remark}