One more remark.

src/measure/measurable-maps/approx.tex

@@ -38,7 +38,12 @@
     \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 For each $n \in \natp$ and $x \in X$, $d(x, I_{n+1}(x)) \le d(x, I_n(x))$. 
+        \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, $I_N(x) \to I$ as $n \to \infty$. 
     \end{enumerate}
 \end{lemma}
 \begin{proof}
@@ -80,12 +85,20 @@
 
     and $\seq{I_N}$ satisfies (AI2).
 
-    Finally, for each $x \in X$ and $M, N \in \natp$ with $M \le N$, $C_M(x) \subset C_N(x)$, so $d(x, I_M(x)) \ge d(x, I_N(x))$, and $\seq{I_n}$ satisfies (3). In particular, 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). 
+    Finally, let $x \in X$, then by definition of $k_N$, 
+    \[
+        d(x, I_N(x)) = d(x, x_{k_N(x)}) = \min\bracs{d(x, x_n)| 1 \le n \le N, x_n \in \mathcal{A}(x)}
+    \]
 
-    Therefore $\seq{I_N}$ is an approximation of the identity satisfying (1)-(3). 
+    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}
+\label{remark:separable-metric-space-approx-identity}
+    In \autoref{lemma:separable-metric-space-approx-identity}, if $X$ is compact and $\mathcal{A} \equiv X$, then $I_N \to I$ \textit{uniformly}. 
+\end{remark}
 
 \begin{corollary}
 \label{corollary:measurable-simple-separable}