Moved Radon measures to their own section.
This commit is contained in:
Bokuan Li
6
src/measure/radon/index.tex
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src/measure/radon/index.tex
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\chapter{Radon Measures}
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\label{chap:radon}
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\input{./radon.tex}
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\input{./riesz.tex}
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205
src/measure/radon/radon.tex
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205
src/measure/radon/radon.tex
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\section{Radon Measures}
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\label{section:radon-measure}
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\begin{definition}[Radon Measure]
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\label{definition:radon-measure}
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Let $X$ be a LCH space and $\mu: \cb_X \to [0, \infty]$ be a Borel measure, then $\mu$ is a \textbf{Radon measure} if:
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\begin{enumerate}
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\item[(R1)] For any $K \subset X$ compact, $\mu(K) < \infty$.
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\item[(R2)] $\mu$ is outer regular on all Borel sets.
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\item[(R3')] $\mu$ is inner regular on all open sets.
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\end{enumerate}
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\end{definition}
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\begin{proposition}
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\label{proposition:radon-measure-cc}
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Let $X$ be a LCH space and $\mu: \cb_X \to [0, \infty]$ be a Radon measure, then
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\begin{enumerate}
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\item For any $U \subset X$ open,
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\[
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\mu(U) = \sup\bracs{\int f d\mu \bigg | f \in C_c(X; [0, 1]), \supp{f} \subset U}
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\]
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\item For any $K \subset X$ compact,
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\[
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\mu(K) = \inf\bracs{\int f d\mu \bigg | f \in C_c(X; [0, 1]), f \ge \one_K}
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\]
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1): Let $U \subset X$ be open. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, for any $K \subset U$ compact, there exists $f_K \in C_c(X; [0, 1])$ such that $f_K|_K = 1$ and $\supp{f_K} \subset U$. In which case,
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\[
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\mu(K) \le \int f_K d\mu \le \mu(U)
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\]
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By (R3'),
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\[
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\mu(U) = \sup_{K \subset U \text{ compact}}\mu(K) \le \sup_{K \subset U \text{ compact}}\int f_K d\mu \le \mu(U)
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\]
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(2): Let $K \subset X$ be compact and $U \in \cn^o(K)$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f_U \in C_c(X; [0, 1])$ such that $f_U|_K = 1$ and $\supp{f_U} \subset U$. In which case,
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\[
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\mu(K) \le \int f_U d\mu \le \mu(U)
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\]
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By (R2),
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\[
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\mu(K) = \inf_{U \in \cn^o(K)}\mu(U) \ge \inf_{U \in \cn^o(K)}\int f_U d\mu \ge \mu(K)
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\]
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\end{proof}
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\begin{proposition}[{{\cite[Proposition 7.5]{Folland}}}]
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\label{proposition:radon-regular-sigma-finite}
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Let $X$ be a LCH space and $\mu: \cb_X \to [0, \infty]$ be a Radon measure, then
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\begin{enumerate}
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\item $\mu$ is inner regular on all its $\sigma$-finite sets.
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\item If $X$ is $\sigma$-compact or $\mu$ is $\sigma$-finite, then $\mu$ is regular.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1): Let $E \in \cb_X$ with $\mu(E) < \infty$ and $\eps > 0$, then there exists
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\begin{itemize}
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\item $U \in \cn^o(E)$ with $\mu(U \setminus E) < \eps/2$.
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\item $K \subset U$ compact with $\mu(K) > \mu(U) - \eps/2$.
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\item $V \in \cn^o(U \setminus E)$ with $\mu(V) < \eps/2$.
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\end{itemize}
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In which case, $K \setminus V \subset E$ is compact with
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\[
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\mu(K \setminus V) \ge \mu(K) - \mu(V) \ge \mu(U) - \eps/2 - \eps/2 \ge \mu(E) - \eps
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\]
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so $\mu$ is inner regular on $E$.
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Now suppose that $E$ is $\sigma$-finite, then there exists $\seq{E_n} \subset \cb_X$ with $\mu(E_n) < \infty$ for each $n \in \natp$ and $E = \bigsqcup_{n \in \natp}E_n = E$. Let $N \in \natp$ and $\seqf{K_n} \subset 2^X$ compact with $K_n \subset E_n$ for each $n \in \natp$, then $\bigcup_{n = 1}^N K_n \subset E$ is compact. Hence
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\[
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\sup_{\substack{K \subset E \\ \text{compact}}}\mu(K) \ge \sum_{n = 1}^N\sup_{\substack{K_n \subset E_n \\ \text{compact}}}\mu(K_n) = \sum_{n = 1}^N \mu(E_n)
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\]
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As the above holds for all $N \in \natp$,
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\[
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\sup_{\substack{K \subset E \\ \text{compact}}}\mu(K) \ge \sum_{n \in \natp}\mu(E_n) = \mu(E)
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\]
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\end{proof}
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\begin{proposition}[{{\cite[Proposition 7.7]{Folland}}}]
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\label{proposition:radon-measurable-description}
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Let $X$ be a LCH space, $\mu: \cb_X \to [0, \infty]$ be a $\sigma$-finite Radon measure, and $E \in \cb_X$, then
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\begin{enumerate}
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\item For every $\eps > 0$, there exists $U \in \cn^o(E)$ and $F \subset E$ closed such that $\mu(U \setminus F) < \eps$.
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\item There exists a $F_\sigma$ set $A$ and a $G_\delta$ set $B$ such that $A \subset E \subset B$ and $\mu(B \setminus A) = 0$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1): Let $\seq{E_n} \subset \cb_X$ such that $\mu(E_n) < \infty$. By outer regularity, there exists $\seq{U_n}$ open such that $U_n \in \cn^o(E_n)$ and $\mu(U_n) < \mu(E_n) + \eps/2^n$ for all $n \in \natp$. In which case,
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\[
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\mu\paren{\bigcup_{n \in \natp}U_n \setminus E} \le \sum_{n \in \natp}\mu(U_n \setminus E_n) < \eps
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\]
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so $U = \bigcup_{n \in \natp}U_n$ is the desired open set.
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Applying the above result to $E^c$, there exists $V \in \cn^o(E^c)$ such that $\mu(V \setminus E^c) < \eps$. Let $F = V^c$, then $F \subset E$ is closed and
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\[
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\mu(E \setminus F) = \mu(E \cap F^c) = \mu(E \cap V) = \mu(V \setminus E^c) < \eps
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\]
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\end{proof}
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\begin{proposition}
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\label{proposition:finite-compact-regular}
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Let $X$ be a LCH space and $\mu: \cb_X \to [0, \infty]$ be a Borel measure such that:
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\begin{enumerate}
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\item[(a)] For any $U \subset X$ open, $U$ is $\sigma$-compact.
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\item[(b)] For any $K \subset X$ compact, $\mu(K) < \infty$.
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\end{enumerate}
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then $\mu$ is a regular measure on $X$.
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\end{proposition}
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\begin{proof}
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By assumption (b), $f \mapsto \int f d\mu$ is a positive linear functional on $C_c(X; \real)$. By the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon}, there exists a Radon measure $\nu: \cb_X \to [0, \infty]$ such that for any $f \in C_c(X; \real)$, $\int f d\mu = \int f d\mu$.
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Let $U \subset X$ be open, then by \autoref{proposition:radon-measure-cc},
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\[
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\nu(U) = \sup_{\substack{f \in C_c(X; [0, 1]) \\ f \le \one_U}} \int f d\nu = \sup_{\substack{f \in C_c(X; [0, 1]) \\ f \le \one_U}} \int f d\mu \le \mu(U)
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\]
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By assumption (a), there exists $\seq{K_n} \subset 2^X$ compact such that $K_n \upto U$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\seq{f_n} \subset C_c(X; [0, 1])$ such that $\one_{K_n} \le f_n \le \one_U$ for all $n \in \natp$, and $f_n \upto f$ pointwise. Using the \hyperref[Monotone Convergence Theorem]{theorem:mct},
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\[
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\mu(U) = \limv{n} \int f_n d\mu \le \sup_{\substack{f \in C_c(X; [0, 1]) \\ f \le \one_U}}\int f d\mu = \nu(U)
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\]
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Therefore $\mu(U) = \nu(U)$.
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Now let $E \in \cb_X$ be arbitrary and $\eps > 0$. By \autoref{proposition:radon-measurable-description}, there exists $U \in \cn^o(E)$ and $V \subset E$ closed such that $\nu(U \setminus E), \nu(E \setminus V) < \eps$. In which case, since $U \setminus V$ is open,
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\[
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\mu(U \setminus E) \le \mu(U \setminus V) = \nu(U \setminus V) < \eps
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\]
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so $\mu(U) = \mu(U \setminus E) + \mu(E) \le \mu(E) + \eps$. Therefore
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\[
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\mu(E) = \inf_{U \in \cn^o(E)}\mu(U) = \inf_{U \in \cn^o(E)}\nu(U) = \nu(E)
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\]
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and $\mu = \nu$.
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Since $X$ is $\sigma$-compact, $\mu$ is $\sigma$-finite, so $\mu$ is regular by \autoref{proposition:radon-regular-sigma-finite}.
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\end{proof}
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\begin{proposition}
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\label{proposition:radon-cc-dense}
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Let $X$ be a LCH space, $\mu: \cb_X \to [0, \infty]$ be a Radon measure, $E$ be a normed space, and $p \in [1, \infty)$, then $C_c(X; E)$ is dense in $L^p(X; E)$.
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\end{proposition}
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\begin{proof}
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By \autoref{proposition:lp-simple-dense}, $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$. Using linearity, it is sufficient to approximate indicator functions of Borel sets with finite measure.
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Let $A \in \cb_X$ and $\eps > 0$. By \autoref{proposition:radon-regular-sigma-finite}, there exists $U \in \cn^o(A)$ and $K \subset A$ compact such that $\mu(U \setminus A), \mu(A \setminus K) < \eps/2$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f \in C_c(X; [0, 1])$ such that $f|_K = 1$ and $\supp{f} \subset U$. In which case, for any $y \in E$,
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\[
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\norm{x \cdot \one_A - x \cdot f}_{L^p(X; E)} \le \norm{x}_E \mu(\bracs{f \ne \one_A})^{1/p} \le \norm{x}_E\mu(U \setminus K)^{1/p} < \eps^{1/p}\norm{x}_E
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\]
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\end{proof}
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\begin{theorem}[Lusin, {{\cite[Theorem 7.10]{Folland}}}]
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\label{theorem:lusin}
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Let $X$ be a LCH space, $\mu$ be a Radon measure on $X$, $E$ be a normed vector space, and $f: X \to E$ be a measurable function with $\mu\bracs{f \ne 0} < \infty$, then for any $\eps > 0$,
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\begin{enumerate}
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\item There exists $A \subset \bracs{f \ne 0}$ such that $f|_A$ is continuous and $\mu(\bracs{f \ne 0} \setminus A) < \eps$
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\item If $E = \complex$, then there exists $\phi \in C_c(X; E)$ such that $\mu\bracs{f \ne \phi} < \eps$.
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\item If $E = \complex$ and $f$ is bounded, then $\phi$ can be taken such that $\norm{\phi}_u \le \norm{f}_u$.
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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First assume that $f$ is bounded.
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(1, bounded): If $f$ is bounded, then $f \in L^1(X; E)$. By \autoref{proposition:radon-cc-dense}, there exists $\seq{\phi_n} \subset C_c(X)$ such that $\phi_n \to f$ in $L^1(\mu)$. Since $\phi_n \to f$ in $L^1(\mu)$, $\phi_n \to f$ in measure by \autoref{proposition:lp-in-measure}. By taking a subsequence using \ref{proposition:cauchy-in-measure-limit}, assume without loss of generality that $\phi_n \to f$ almost everywhere.
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By \hyperref[Egoroff's Theorem]{theorem:egoroff}, there exists $A \subset \bracs{f \ne 0}$ such that $\phi_n \to f$ uniformly and $\mu(\bracs{f \ne 0} \setminus A) < \eps/3$. By \autoref{proposition:radon-regular-sigma-finite}, there exists $K \subset A$ compact such that $\mu(A \setminus K) < \eps/3$.
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Let $\phi_0 = \lim_{n \to \infty}\phi_n$ on $K$, then $\phi_0 \in C(K; E)$ by \autoref{proposition:uniform-limit-continuous}, $\phi_0 = f$ almost everywhere on $K$, and $\mu(\bracs{f \ne 0} \setminus K) < \eps$.
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(2, bounded): By outer regularity, there exists $U \in \cn^o(\bracs{f \ne 0})$ such that $\mu(U \setminus \bracs{f \ne 0}) < \eps/3$.
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By the \hyperref[Tietze Extension Theorem]{theorem:lch-tietze}, there exists $\phi \in C_c(X; \complex)$ such that $\phi|_K = \phi_0$ and $\supp{\phi} \subset U$. In which case,
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\[
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\mu\bracs{\phi \ne f} \le \underbrace{\mu(\bracs{f \ne 0} \setminus K)}_{< 2\eps/3} + \underbrace{\mu(U \setminus \bracs{f \ne 0})}_{< \eps/3} < \eps
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\]
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(3): Let
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\[
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\psi: \complex \to \complex \quad z \mapsto \begin{cases}
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z &|z| \le \norm{f}_u \\
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z\norm{f}_u/|z| &|z| > \norm{f}_u
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\end{cases}
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\]
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then $\psi$ is continuous with $\bracs{\phi = f} = \bracs{\psi \circ \phi = f}$ and $\norm{\psi \circ \phi}_u \le \norm{f}_u$.
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Now assume that $f$ is arbitrary.
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(1, unbounded): Since $\mu\bracs{f \ne 0} < \infty$, there exists $\alpha > 0$ such that $\mu\bracs{|f| > \alpha} < \eps/2$. Let $g \in L^\infty(X; E)$ such that $g = f$ on $\bracs{|f| \le \alpha}$. By (1) applied to $g$, there exists $A \subset \bracs{|f| \le \alpha}$ such that $f|_A$ is continuous and $\mu(\bracs{|f| \le \alpha} \setminus A) < \eps/2$. In which case, $\mu(\bracs{f \ne 0} \setminus A) < \eps$, as desired.
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(2, unbounded): By (1) applied to $g$, there exists $\phi \in C_c(X; \complex)$ such that $\mu\bracs{\phi \ne g} < \eps/2$, and $\mu(\bracs{\phi \ne f}) < \eps$.
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\end{proof}
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189
src/measure/radon/riesz.tex
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src/measure/radon/riesz.tex
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\section{The Riesz Representation Theorem}
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\label{section:riesz-radon}
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\begin{definition}[Positive Linear Functional on $C_c$]
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\label{definition:positive-linear-functional-cc}
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Let $X$ be a topological space and $I \in \hom(C_c(X; \real); \real)$, then $\phi$ is \textbf{positive} if for any $f \in C_c(X; [0, \infty))$, $\dpb{f, I}{C_c(X; \real)} \ge 0$.
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\end{definition}
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\begin{proposition}
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\label{proposition:positive-linear-functional-cc-property}
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Let $X$ be a LCH space and $I \in \hom(C_c(X; \real); \real)$ be a positive linear functional, then
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\begin{enumerate}
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\item For any $f, g \in C_c(X; \real)$ with $f \le g$, $\dpb{f, I}{C_c(X; \real)} \le \dpb{g, I}{C_c(X; \real)}$.
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\item For any $K \subset X$ compact, there exists $C_K \ge 0$ such that for all $f \in C_c(X; \real)$ with $\supp{f} \subset K$, $|{\dpb{f, I}{C_c(X; \real)}}| \le \norm{f}_u$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1): $\dpb{g - f, I}{C_c(X; \real)} \ge 0$.
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(2): By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $g \in C_c(X; [0, 1])$ such that $g|_K = 1$. In which case,
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\[
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-\norm{f}_u\dpn{g, I}{C_c(X; \real)} \le \dpn{f, I}{C_c(X; \real)} \le \norm{f}_u\dpn{g, I}{C_c(X; \real)}
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\]
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so $C_K = \dpn{g, I}{C_c(X; \real)}$ is a desired constant.
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\end{proof}
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\begin{theorem}[Riesz Representation Theorem, {{\cite[Theorem 7.2]{Folland}}}]
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\label{theorem:riesz-radon}
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Let $(X, \topo)$ be a LCH space and $I \in \hom(C_c(X; \real); \real)$ be a positive linear functional, then there exists a Borel measure $\mu: \cb_X \to [0, \infty]$ such that:
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\begin{enumerate}
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\item For any $U \subset X$ open, $\mu(U) = \sup_{f \prec U}\dpb{f, I}{C_c(X; \real)}$.
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\item For any $K \subset X$ compact, $\mu(K) = \inf_{f \in C_c(X; [0, 1]), f \ge \one_K}\dpb{f, I}{C_c(X; \real)}$.
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\item For any $f \in C_c(X; \real)$, $\int f d\mu = \dpb{f, I}{C_c(X; \real)}$.
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\item $\mu$ is a Radon measure.
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\item[(U)] If $\nu: \cb_X \to [0, \infty]$ is a Borel measure that satisfies (3) and (4), then $\mu = \nu$.
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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(1): For any $U \in \topo$ and $f \in C_c(X; \real)$, denote $f \prec U$ if $f \in C_c(X; [0, 1])$ and $\supp{f} \subset U$. Let
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\[
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\mu_0: \topo \to [0, \infty] \quad U \mapsto \sup_{f \prec U}\dpb{f, I}{C_c(X; \real)}
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\]
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and
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\[
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\mu^*: 2^X \to [0, \infty] \quad E \mapsto \inf\bracsn{\mu_0(U)|U \in \cn^o(E)}
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\]
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then
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\begin{enumerate}
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\item[(OM1)] Since $\emptyset \in \topo$ and $\mu_0(\emptyset) = 0$, $\mu^*(\emptyset) = 0$.
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\item[(OM2)] Let $E, F \subset X$ with $E \subset F$, then $\cn^o(E) \subset \cn^o(F)$, so $\mu^*(E) \le \mu^*(F)$.
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\item[(OM3)] Let $\seq{E_n} \subset 2^X$, $E = \bigcup_{n \in \natp}E_n$, $U \in \cn^o(E)$, and $\seq{U_n} \subset \topo$ such that $U_n \in \cn^o(E_n)$ for each $n \in \natp$.
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Let $f \prec U$, then by compactness of $\supp{f}$, there exists $N \in \natp$ such that $\supp{f} \subset \bigcup_{n = 1}^N U_n$. By \autoref{proposition:lch-partition-of-unity}, there exists a partition of unity $\seqf{\phi_j} \subset C_c(X; [0, 1])$ on $K$ subordinate to $\seqf[N]{U_n}$. In which case,
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\[
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\dpb{f, I}{C_c(X; \real)} = \sum_{n = 1}^N \dpb{\phi_n f, I}{C_c(X; \real)} \le \sum_{n = 1}^N \mu_0(U_n) \le \sum_{n \in \natp}\mu_0(U_n)
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\]
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Since this holds for all $f \prec U$,
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\[
|
||||
\mu^*(E) \le \mu_0(U) \le \sum_{n \in \natp}\mu_0(U_n)
|
||||
\]
|
||||
|
||||
|
||||
Let $\eps > 0$, then there exists $\seq{U_n} \subset \topo$ such that $U_n \supset E_n$ and $\mu_0(U_n) \le \mu^*(E_n) + \eps/2^n$ for each $n \in \natp$. In which case,
|
||||
\[
|
||||
\mu^*(E) \le \sum_{n \in \natp}\mu_0(U_n) \le \eps + \sum_{n \in \natp}\mu^*(E_n)
|
||||
\]
|
||||
|
||||
As this holds for all $\eps > 0$, $\mu^*(E) \le \sum_{n \in \natp}\mu^*(E_n)$.
|
||||
\end{enumerate}
|
||||
Therefore $\mu^*: 2^E \to [0, \infty]$ is an outer measure.
|
||||
|
||||
To see that all Borel sets are $\mu^*$-measurable, let $E \subset X$ and $U \in \topo$. First suppose that $E$ is open. Let $f \prec E \cap U$, then for any $g \prec E \setminus \supp{f}$, $\supp{f} \cap \supp{g} = \emptyset$ and $f + g \prec E$, so
|
||||
\[
|
||||
\dpb{f, I}{C_c(X; \real)} + \dpb{g, I}{C_c(X; \real)} \le \mu_0(E)
|
||||
\]
|
||||
|
||||
Since this holds for all $g \prec E \setminus \supp{f}$,
|
||||
\[
|
||||
\dpb{f, I}{C_c(X; \real)} + \mu_0(E \setminus \supp{f}) \le \mu_0(E)
|
||||
\]
|
||||
|
||||
As $E \setminus \supp{f} \supset E \setminus U$,
|
||||
\[
|
||||
\dpb{f, I}{C_c(X; \real)} + \mu_0(E \setminus U) \le \mu_0(E)
|
||||
\]
|
||||
|
||||
Finally, the above holds for all $f \prec E \cap U$,
|
||||
\[
|
||||
\mu_0(E \cap U) + \mu_0(E \setminus U) \le \mu_0(E)
|
||||
\]
|
||||
|
||||
Now suppose that $E$ is arbitrary. Let $V \in \cn^o(E)$, then
|
||||
\[
|
||||
\mu^*(E \cap U) + \mu^*(E \setminus U) \le \mu_0(V \cap U) + \mu_0(V \setminus U) \le \mu_0(V)
|
||||
\]
|
||||
|
||||
As this holds for all such $V$,
|
||||
\[
|
||||
\mu^*(E) = \mu^*(E \cap U) + \mu^*(E \setminus U)
|
||||
\]
|
||||
|
||||
|
||||
By \hyperref[Carathéodory's Extension Theorem]{theorem:caratheodory-extension}, there exists a Borel measure $\mu: \cb_X \to [0, \infty]$ such that for all $U \in \topo$, $\mu(U) = \mu_0(U)$.
|
||||
|
||||
(2): Let $K \subset X$ be compact and $U \in \cn^o(K)$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f \prec U$ with $f \ge \one_K$. In which case,
|
||||
\[
|
||||
\inf_{\substack{f \in C_c(X; \real) \\ f \ge \one_K}}\dpb{f, I}{C(X; \real)} \le \inf_{U \in \cn^o(K)}\mu(U) = \mu(K)
|
||||
\]
|
||||
|
||||
On the other hand, let $f \in C_c(X; [0, 1])$ with $f \ge \one_K$. For any $r \in (0, 1)$, let $g \prec \bracs{f > r}$, then $r^{-1}f \ge g$ and
|
||||
\[
|
||||
\dpb{g, I}{C_c(X; \real)} \le r^{-1}\dpb{f, I}{C_c(X; \real)}
|
||||
\]
|
||||
|
||||
As this holds for all $g \prec \bracs{f > r}$,
|
||||
\[
|
||||
\mu(K) \le \mu(\bracs{f > r}) \le r^{-1}\dpb{f, I}{C_c(X; \real)}
|
||||
\]
|
||||
|
||||
Since the above holds for all $r \in (0, 1)$,
|
||||
\[
|
||||
\mu(K) \le \inf_{\substack{f \in C_c(X; \real) \\ f \ge \one_K}}\dpb{f, I}{C_c(X; \real)}
|
||||
\]
|
||||
|
||||
|
||||
(3): Let $f \in C_c(X; \real)$. Using linearity, assume without loss of generality that $f \in C_c(X; [0, 1])$. Let $N \in \natp$. For each $1 \le n \le N$, let
|
||||
\[
|
||||
f_n = \min\paren{\max\paren{f - \frac{n - 1}{N}, 0}, \frac{1}{N}}
|
||||
\]
|
||||
|
||||
For any $x \in X$, there exists $1 \le n \le N$ such that $f(x) \in [(n-1)/N, n/N]$. In which case,
|
||||
\begin{itemize}
|
||||
\item For each $1 \le j < n$, $f_j(x) = 1/N$.
|
||||
\item $f_n(x) = f(x) - \frac{n - 1}{N}$.
|
||||
\item For each $n < j \le N$, $f_j(x) = 0$.
|
||||
\end{itemize}
|
||||
so $\sum_{j = 1}^Nf_j(x) = \frac{n-1}{N} + f(x) - \frac{n - 1}{N} = f(x)$.
|
||||
|
||||
Let $K_0 = \supp{f}$, and for each $1 \le n \le N$, let $K_n = \bracs{f \ge n/N}$.
|
||||
\[
|
||||
\frac{1}{N}\mu\paren{K_n} \le \int f_n d\mu \le \frac{1}{N}\mu\paren{K_{n-1}}
|
||||
\]
|
||||
|
||||
Since $\supp{f_n} \subset \bracs{f \ge (n-1)/N}$, for any $U \supset \bracs{f_n \ge (n - 1)/N}$, $f_n \prec U$, so
|
||||
\[
|
||||
\dpb{f_n, I}{C_c(X; \real)} \le \frac{1}{N}\mu\paren{K_{n-1}}
|
||||
\]
|
||||
|
||||
On the other hand, $f_n \ge \frac{1}{N}\one_{K_n}$,
|
||||
\[
|
||||
\mu(K_n) \le \dpb{f_n, I}{C_c(X; \real)}
|
||||
\]
|
||||
|
||||
so
|
||||
\[
|
||||
\frac{1}{N}\sum_{n = 1}^N \mu(K_n) \le \dpb{f, I}{C_c(X; \real)}, \int f d\mu \le \frac{1}{N}\sum_{n = 1}^N \mu(K_{n-1})
|
||||
\]
|
||||
|
||||
|
||||
Therefore
|
||||
\[
|
||||
\abs{\int f d\mu - \dpb{f, I}{C_c(X; \real)}} \le \frac{\mu(K_{0}) - \mu(K_N)}{N} \le \frac{\mu(\supp{f})}{N}
|
||||
\]
|
||||
|
||||
As this holds for all $N \in \natp$, $\int f d\mu = \dpb{f, I}{C_c(X; \real)}$.
|
||||
|
||||
(4):
|
||||
\begin{enumerate}
|
||||
\item[(R1)] For any $K \subset X$ compact, by \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f \in C_c(X; [0, 1])$ with $f \ge \one_K$. In which case, $\mu(K) \le \int f d\mu = \dpb{f, I}{C_c(X; \real)}$.
|
||||
\item[(R2)] By definition of $\mu^*$, $\mu$ is outer regular.
|
||||
\item[(R3')] For any $U \in \topo$,
|
||||
\[
|
||||
\mu(U) = \sup_{f \prec U}\dpb{f, I}{C_c(X; \real)} \le \sup_{f \prec U}\mu(\supp{f}) \le \sup_{\substack{K \subset U \\ \text{compact}}}\mu(K) \le \mu(U)
|
||||
\]
|
||||
|
||||
\end{enumerate}
|
||||
|
||||
(U): By \autoref{proposition:radon-measure-cc}, $\nu$ also satisfies (2). Thus for any $E \in \cb_X$, by (R2) and (R3'),
|
||||
\[
|
||||
\nu(E) = \inf_{U \in \cn^o(E)}\sup_{\substack{K \subset U \\ \text{compact}}}\inf_{\substack{f \in C_c(X; [0, 1]) \\ f \ge \one_K}}\dpb{f, I}{C_c(X; \real)}
|
||||
\]
|
||||
|
||||
so $\nu$ is uniquely determined by $I$.
|
||||
\end{proof}
|
||||
Reference in New Issue
Block a user