Added Lusin's theorem.

document.tex

@@ -1,4 +1,4 @@
-\documentclass{report}
+%\documentclass{report}
 \usepackage{amssymb, amsmath, hyperref}
 \usepackage{preamble}
 

src/fa/lp/definition.tex

@@ -114,3 +114,49 @@
 
     Therefore $\norm{f_n - f}_{L^p(X; E)} \to 0$ as $n \to \infty$.
 \end{proof}
+
+\begin{theorem}[Markov's Inequality]
+\label{theorem:markov-inequality}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $f: X \to E$ be a Borel measurable function, then
+    \begin{enumerate}
+        \item For any $\alpha > 0$, 
+        \[
+        \mu\bracs{|f| \ge \alpha} \le \frac{1}{\alpha}\norm{f}_{L^1(X; E)}
+        \]
+        \item For $\alpha > 0$ and strictly increasing function $\phi: [0, \infty) \to [0, \infty)$,
+        \[
+        \mu\bracs{|f| \ge \alpha} \le \frac{1}{\phi(\alpha)}\norm{\phi \circ |f|}_{L^1(X; E)}
+        \]
+        \item For any $\alpha > 0$,
+        \[
+        \mu\bracs{|f| \ge \alpha} \le \frac{1}{\alpha^p}\norm{f}_{L^p(X; E)}^p
+        \]
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): For any $\alpha > 0$, 
+    \begin{align*}
+        \mu\bracs{|f| \ge \alpha} &= \int_{\bracs{f \ge \alpha}} d\mu \le \frac{1}{\alpha}\int |f|d\mu = \frac{\norm{f}_{L^1(X; E)}}{\alpha}
+    \end{align*}
+    
+    (2): By \autoref{lemma:monotone-measurable}, $\phi: [0, \infty) \to [0, \infty)$ is Borel measurable. Since $\phi$ is strictly increasing, $\bracs{|f| \ge \alpha} = \bracs{\phi \circ |f| \ge \phi(\alpha)}$, and the result holds by applying (1) to $\phi \circ |f|$. 
+
+    (3): Since $x \mapsto x^p$ is strictly increasing on $[0, \infty)$, applying (2) to $\phi(x) = x^p$ yields the desired result.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:lp-in-measure}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, $p \in [1, \infty]$, $\seq{f_n} \subset L^p(X; E)$, and $f \in L^p(X; E)$ such that $f_n \to f$ in $L^p$, then $f_n \to f$ in measure.
+\end{proposition}
+\begin{proof}
+    Let $\eps > 0$. If $p < \infty$, then by \hyperref[Markov's Inequality]{theorem:markov-inequality},
+    \[
+    \limv{n}\mu\bracs{|f_n - f| \ge \eps} \le \limv{n}\frac{1}{\eps^p}\norm{f_n - f}_{L^p(X; E)}^p = 0
+    \]
+    
+    If $p = \infty$, then there exists $N \in \natp$ such that $\norm{f_n - f}_{L^\infty(X; E)} < \eps$ for all $n \ge N$. In which case, $\mu\bracs{|f_n - f| \ge \eps} = 0$ for all $n \ge N$. 
+\end{proof}
+
+
+
+

src/measure/measurable-maps/real-valued.tex

@@ -50,3 +50,12 @@
 
     Finally, if the above functions are $\real$-valued, then they are $(\cm, \cb_\real)$-measurable by \autoref{lemma:extended-real-measurable}.
 \end{proof}
+
+\begin{lemma}
+\label{lemma:monotone-measurable}
+    Let $f: \real \to \real$ be a non-decreasing or non-increasing function, then $f$ is Borel measurable.
+\end{lemma}
+\begin{proof}
+    By taking $-f$, assume without loss of generality that $f$ is non-decreasing. In which case, for any $a \in \real$, $x \in f^{-1}((-\infty, a])$, and $y \le x$, $f(y) \le f(x) \le a$, so $y \in f^{-1}((-\infty, a])$. Thus $f^{-1}((-\infty, a))$ is an interval and hence measurable. Since the open rays generate $\cb_\real$ (\autoref{proposition:borel-sigma-real-generators}), $f$ is Borel measurable.
+\end{proof}
+

src/measure/measure/radon.tex

@@ -159,5 +159,47 @@
     \[
     \norm{x \cdot \one_A - x \cdot f}_{L^p(X; E)} \le \norm{x}_E \mu(\bracs{f \ne \one_A})^{1/p} \le \norm{x}_E\mu(U \setminus K)^{1/p} < \eps^{1/p}\norm{x}_E
     \]
-
 \end{proof}
+
+\begin{theorem}[Lusin, {{\cite[Theorem 7.10]{Folland}}}]
+\label{theorem:lusin}
+    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 $\eps > 0$,
+    \begin{enumerate}
+        \item There exists $A \subset \bracs{f \ne 0}$ such that $f|_A$ is continuous and $\mu(\bracs{f \ne 0} \setminus A) < \eps$
+        \item If $E = \complex$, then there exists $\phi \in C_c(X; E)$ such that $\mu\bracs{f \ne \phi} < \eps$.
+        \item If $E = \complex$ and $f$ is bounded, then $\phi$ can be taken such that $\norm{\phi}_u \le \norm{f}_u$.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    First assume that $f$ is bounded.
+
+    (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. 
+
+    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 $\mu(A \setminus K) < \eps/3$. 
+
+    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 $\mu(\bracs{f \ne 0} \setminus K) < \eps$.
+
+    (2, bounded): By outer regularity, there exists $U \in \cn^o(\bracs{f \ne 0})$ such that $\mu(U \setminus \bracs{f \ne 0}) < \eps/3$. 
+
+    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, 
+    \[
+    \mu\bracs{\phi \ne f} \le \underbrace{\mu(\bracs{f \ne 0} \setminus K)}_{< 2\eps/3} + \underbrace{\mu(U \setminus \bracs{f \ne 0})}_{< \eps/3} < \eps
+    \]
+
+    (3): Let
+    \[
+    \psi: \complex \to \complex \quad z \mapsto \begin{cases}
+        z &|z| \le \norm{f}_u \\
+        z\norm{f}_u/|z| &|z| > \norm{f}_u
+    \end{cases}
+    \]
+    then $\psi$ is continuous with $\bracs{\phi = f} = \bracs{\psi \circ \phi = f}$ and $\norm{\psi \circ \phi}_u \le \norm{f}_u$. 
+
+    Now assume that $f$ is arbitrary.
+
+    (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. 
+
+    (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 $\mu(\bracs{\phi \ne f}) < \eps$.
+\end{proof}
+
+