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Added the theory of admissible approximant functions.

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Bokuan Li
2026-06-28 19:29:22 -04:00
parent 121033cfb6
commit 3a4f4b46e8
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src/measure/measurable-maps/metric.tex
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@@ -21,7 +21,7 @@
\begin{proposition}
\label{proposition:metric-measurables}
Let $(X, \cm)$ be a measurable space, $Y$ be a metrisable topological space, $f, g: X \to Y$ be $(\cm, \cb_Y)$-measurable functions, then the following functions are measurable:
Let $(X, \cm)$ be a measurable space, $Y$ be a separable and metrisable topological space, $f, g: X \to Y$ be $(\cm, \cb_Y)$-measurable functions, then the following functions are measurable:
\begin{enumerate}
\item For any metric $d$ on $Y$, $x \mapsto d(f(x), f(y))$.
\item If $Y$ is a TVS over $K \in \RC$ and $\lambda \in K$, $\lambda f + g$. In particular, $\bracs{f = g} \in \cm$.
@@ -69,92 +69,3 @@
\end{proof}
\begin{lemma}
\label{lemma:measurable-simple-separable}
Let $(X, \cm)$ be a measurable space, $Y$ be a separable metric space, and $\mathcal{A}: Y \to 2^Y$ such that
\begin{enumerate}
\item[(a)] For each $y \in Y$, $y \in \ol{\mathcal{A}(y)^o}$.
\item[(b)] $\bigcap_{y \in Y}\mathcal{A}(y) \ne \emptyset$.
\item[(c)] For any $y_0 \in Y$, $\bracs{y \in Y|y_0 \in \mathcal{A}(y)} \in \cb_Y$.
\end{enumerate}
Then, for any $f: X \to Y$, the following are equivalent:
\begin{enumerate}
\item $f$ is $(\cm, \cb_Y)$-measurable.
\item For any dense subset $\seq{y_n} \subset Y$ with $y_1 \in \bigcap_{y \in Y}\mathcal{A}(y)$, there exists a sequence $\seq{f_n}$ of $(\cm, \cb_Y)$-measurable simple functions such that
\begin{enumerate}
\item[(i)] For each $x \in X$ and $N \in \natp$, $f_N(x) \in \mathcal{A}(f(x)) \cap \bracsn{y_n|1 \le n \le N}$.
\item[(ii)] $f_n \to f$ pointwise.
\end{enumerate}
\item There exists a sequence $\seq{f_n}$ of $(\cm, \cb_Y)$-measurable simple functions such that $f_n \to f$ pointwise.
\end{enumerate}
\end{lemma}
\begin{proof}
(1) $\Rightarrow$ (2): Let $\seq{y_n} \subset Y$ be a dense subset with $y_1 \in \bigcap_{y \in Y}\mathcal{A}(y)$. For each $N \in \natp$ and $x \in X$, define
\[
C(N, x) = \bracs{1 \le n \le N|y_n \in \mathcal{A}(f(x))}
\]
Since $y_1 \in \bigcap_{y \in Y}\mathcal{A}(y)$, $1 \in C(N, x) \ne \emptyset$. 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 $k \in \natp$, $\bracs{x \in X|k(N, x) \le k}$ is equal to
\[
\bigcup_{j = 1}^k\bracs{x \in X \bigg |j \in C(N, x), d(f(x), y_j) = \min_{m \in C(N, x)}d(f(x), y_m)}
\]
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 \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 $N \in \natp$, $f_N(X) \subset \bracs{y_n|1 \le n \le N}$. Since $x \mapsto k{(N, x)}$ is $(\cm, 2^\natp)$-measurable, $f_N$ is $(\cm, \cb_Y)$-measurable.
Fix $x \in X$, then
\[
f(x) \in \ol{\mathcal{A}(f(x))^o} = \ol{\bracs{y_n| n \in \natp, y_n \in \mathcal{A}(f(x))^o}}
\]
by assumption (a) and \autoref{definition:dense}. Thus for any $\eps > 0$, there exists $N \in \natp$ such that $y_N \in \mathcal{A}(f(x))^o$ 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{corollary}
\label{corollary:measurable-simple-separable-norm}
Let $(X, \cm)$ be a measurable space, $(E, \norm{\cdot}_E)$ be a separable normed vector space, and $f: X \to E$, then the following are equivalent:
\begin{enumerate}
\item $f$ is $(\cm, \cb_E)$-measurable.
\item There exists simple functions $\seq{f_n}$ such that $\abs{f_n} \le \abs{f}$ for all $n \in \natp$, and $f_n \to f$ pointwise.
\end{enumerate}
\end{corollary}
\begin{proof}
(1) $\Rightarrow$ (2): Let
\[
\mathcal{A}: E \to 2^E \quad y \mapsto \begin{cases}
B_E(0, \norm{y}_E) & y \ne 0 \\
E & y = 0
\end{cases}
\]
then
\begin{enumerate}
\item[(a)] For each $y \in E$, $y \in \ol{\mathcal{A}(y)^o}$.
\item[(b)] $0 \in \bigcap_{y \in E}\mathcal{A}(y)$.
\item[(c)] For any fixed $y_0 \in E \setminus \bracs{0}$,
\[
\bracs{y \in E|y_0 \in \mathcal{A}(y)} = \bracs{y \in E|\norm{y_0}_E < \norm{y}_E} \cup \bracs{0} \in \cb_E
\]
and $\bracs{y \in E|0 \in \mathcal{A}(y)} = E$.
\end{enumerate}
By (2) of \autoref{lemma:measurable-simple-separable}, there exists simple functions $\seq{f_n}$ such that $|f_n| \le |f|$ on $\bracs{f \ne 0}$ for all $n \in \natp$ and $f_n \to f$ pointwise. In which case, $|\one_{\bracs{f \ne 0}}f_n| \le |f|$ globally for all $n \in \natp$ and $\one_{\bracs{f \ne 0}}f_n \to f$ pointwise.
(2) $\Rightarrow$ (1): By \autoref{proposition:metric-measurable-limit}.
\end{proof}