Added the Vitali convergence theorem.

src/fa/lp/definition.tex

@@ -169,36 +169,6 @@
 \end{proof}
 
 
-
-\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{proposition}
 \label{proposition:lp-in-measure}
     Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector 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.
@@ -244,3 +214,5 @@
     
 \end{proof}
 
+
+