Added Lusin's theorem.

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}
+
+
+
+