Added the Vitali convergence theorem.

src/fa/lp/ineq.tex

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+\section{$L^p$ Inequalities}
+\label{section:lp-inequalities}
+
+
+\begin{theorem}[Markov's Inequality]
+\label{theorem:markov-inequality}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $f: X \to E$ be a Borel measurable function, then
+    \begin{enumerate}
+        \item For any $\alpha > 0$, 
+        \[
+        \mu\bracs{\norm{f}_E \ge \alpha} \le \frac{1}{\alpha}\norm{f}_{L^1(X; E)}
+        \]
+        \item For $\alpha > 0$ and strictly increasing function $\phi: [0, \infty) \to [0, \infty)$,
+        \[
+        \mu\bracs{\norm{f}_E \ge \alpha} \le \frac{1}{\phi(\alpha)}\norm{\phi \circ |f|}_{L^1(X; E)}
+        \]
+        \item For any $\alpha > 0$,
+        \[
+        \mu\bracs{\norm{f}_E \ge \alpha} \le \frac{1}{\alpha^p}\norm{f}_{L^p(X; E)}^p
+        \]
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): For any $\alpha > 0$, 
+    \begin{align*}
+        \mu\bracs{\norm{f}_E \ge \alpha} &= \int_{\bracs{\norm{f}_E \ge \alpha}} d\mu \le \frac{1}{\alpha}\int |\norm{f}_Ed\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, $\bracs{\norm{f}_E \ge \alpha} = \bracs{\phi \circ \norm{f}_E \ge \phi(\alpha)}$, and the result holds by applying (1) to $\phi \circ \norm{f}_E$. 
+
+    (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{lemma}
+\label{lemma:lp-p-diff}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, $p \in [1, \infty)$, and $f, g \in L^p(X; E)$, then
+    \[
+    |\norm{f}_{L^p(X; E)}^p - \norm{g}_{L^p(X; E)}^p| \le p\norm{f - g}_{L^p(X; E)}(\norm{f}_{L^p(X; E)} \vee \norm{g}_{L^p(X; E)})^{p-1}
+    \]
+\end{lemma}
+\begin{proof}
+    By \autoref{lemma:power-difference}, 
+    \begin{align*}
+    |\norm{f}_{L^p(X; E)}^p - \norm{g}_{L^p(X; E)}^p| &\le p|\norm{f}_{L^p(X; E)} - \norm{g}_{L^p(X; E)}|\\
+    &\times (\norm{f}_{L^p(X; E)} \vee \norm{g}_{L^p(X; E)})^{p-1} \\
+    &\le p\norm{f - g}_{L^p(X; E)}(\norm{f}_{L^p(X; E)} \vee \norm{g}_{L^p(X; E)})^{p-1}
+    \end{align*}
+    
+\end{proof}