diff --git a/src/measure/measurable-maps/approx.tex b/src/measure/measurable-maps/approx.tex index e334f86..5ad7186 100644 --- a/src/measure/measurable-maps/approx.tex +++ b/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 $k_N$, + \[ + d(x, I_N(x)) = d(x, x_{k_N(x)}) = \min\bracs{d(x, x_n)| 1 \le n \le N, x_n \in \mathcal{A}(x)} + \] + + 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)-(3). + 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}