Added the Vitali convergence theorem.

src/fa/lp/ui.tex

@@ -0,0 +1,140 @@
+\chapter{Uniform Integrability}
+\label{chap:uniform-integrable}
+
+\begin{definition}[Uniform Integrability]
+\label{definition:uniform-integrable}
+    Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, and $\cf \subset L^p(X; E)$, then $\cf$ is \textbf{uniformly $p$-integrable} if
+    \[
+    \lim_{M \to \infty}\sup_{f \in \cf} \int_{\bracs{\norm{f}_E \ge M}} \norm{f}^p d\mu = 0
+    \]
+\end{definition}
+
+\begin{proposition}
+\label{proposition:ui-gymnastics}
+    Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, $\cf \subset L^p(X; K)$, and let $|\cf|^p = \bracsn{\norm{f}^p| f \in \cf}$, then:
+    \begin{enumerate}
+        \item If $\cf$ is uniformly $p$-integrable, then $|\cf|^p$ is uniformly absolutely continuous with respect to $\mu$. 
+        \item If $\sup_{f \in \cf}\norm{f}_{L^p(X; E)} < \infty$ and $|\cf|^p$ is uniformly absolutely continuous with respect to $\mu$, then $\cf$ is uniformly $p$-integrable.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $\eps > 0$, then there exists $M \ge 0$ such that
+    \[
+    \sup_{f \in \cf}\int_{\bracs{\norm{f}_E \ge M}} \norm{f}^p d\mu < \eps/2
+    \]
+
+    Thus for any $A \in \cm$, 
+    \begin{align*}
+        \sup_{f \in \cf}\int_{A}\norm{f}^p d\mu &\le \sup_{f \in \cf}\int_{\bracs{\norm{f}_E \le M} \cap A} \norm{f}^p d\mu+ \sup_{f \in \cf}\int_{\bracs{\norm{f}_E \ge M} \cap A} \norm{f}^p d\mu \\
+        &\le M \mu(A) + \eps/2
+    \end{align*}
+
+    so if $\mu(A) \le M\eps/2$, then $\sup_{f \in \cf}\int_{A}\norm{f}^p d\mu \le \eps$. 
+    
+    (2): Let $\eps > 0$, then there exists $\delta > 0$ such that for any $A \in \cm$ with $\mu(A) < \delta$, $\sup_{f \in \cf}\int_A \norm{f}^p d\mu < \eps$. By \hyperref[Markov's inequality]{theorem:markov-inequality}, 
+    \[
+    \sup_{f \in \cf}\mu(\bracs{\norm{f}_E \ge M}) \le \frac{\sup_{f \in \cf}\norm{f}_{L^p(X; E)}^p}{M^p}
+    \]
+
+    Thus for sufficiently large $M$, $\sup_{f \in \cf}\mu(\bracs{\norm{f}_E \ge M}) < \delta$, and 
+    \[
+    \sup_{f \in \cf}\int_{\bracs{\norm{f}_E \ge M}} \norm{f}^p d\mu < \eps
+    \]
+\end{proof}
+
+
+
+\begin{theorem}[Vitali Convergence Theorem]
+\label{theorem:vitali-convergence}
+    Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, and $\fF \subset 2^{L^p(X; E)}$ be a filter, then $\fF$ is Cauchy in $L^p(X; E)$ if and only if:
+    \begin{enumerate}
+        \item[(M)] $\fF$ is Cauchy in measure. 
+        \item[(UI)] For each $\eps > 0$, there exists $M \ge 0$ and $F \in \fF$ such that
+        \[
+        \sup_{f \in F}\int_{\bracs{\norm{f}_E \ge M}}\norm{f}_E^p d\mu < \eps
+        \]
+        \item[(T)] For each $\eps > 0$, there exists $A \in \cm$ and $F \in \fF$ with $\mu(A) < \infty$ and $\sup_{f \in F}\int_{A^c}\norm{f}_E^p < \eps$.
+    \end{enumerate}
+
+    % M stands for measure, UI stands for Uniform Integrability, T stands for tightness. 
+\end{theorem}
+\begin{proof}
+    ($L^p$) $\Rightarrow$ (M): By \hyperref[Markov's inequality]{theorem:markov-inequality}. 
+
+    ($L^p$) $\Rightarrow$ (UI): Let $F \in \fF$ such that for each $f, g \in F$, $\norm{f - g}_{L^p(X; E)} < \eps$. Fix $f \in F$, then by the \hyperref[Dominated Convergence Theorem]{theorem:dct}, there exists $M > 0$ such that $\int_{\bracs{\norm{f}_E \ge M}}\norm{f}_E^p d\mu < \eps$. For any $g \in F$, 
+    \[
+    \bracs{\norm{g}_E \ge 2M} \subset \bracs{\norm{f}_E \ge M} \cup \bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M}
+    \]
+
+    so by \autoref{lemma:lp-p-diff},
+    \begin{align*}
+        \int_{\bracs{\norm{g}_E \ge 2M}}\norm{g}_E^pd\mu &\le \int_{\bracs{\norm{f}_E \ge M}}\norm{g}_E^pd\mu + \int_{\bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M}}\norm{g}_E^pd\mu \\
+        &\le \int_{\bracs{\norm{f}_E \ge M}}\norm{f}_E^pd\mu + \int_{\bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M}}\norm{f}_E^pd\mu \\
+        &+ 2p\norm{f - g}_{L^p(X; E)}(\norm{f}_{L^p(X; E)} \vee \norm{g}_{L^p(X; E)})^{p-1}
+    \end{align*}
+
+    Since $\bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M} \subset \bracs{\norm{f - g}_E \ge M}$, by \hyperref[Markov's inequality]{theorem:markov-inequality}, 
+    \[
+    \int_{\bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M}}\norm{f}_E^pd\mu \le M^p\mu\bracs{\norm{f - g}_E \ge M} \le \norm{f - g}_{L^p(X; E)}^p
+    \]
+
+    Therefore
+    \[
+    \sup_{g \in F}\int_{\bracs{\norm{g}_E \ge 2M}}\norm{g}_E^p d\mu \le 2p \eps (\norm{f}_{L^p(X; E)} + \eps)^{p - 1} + \eps + \eps^p
+    \]
+
+    ($L^p$) $\Rightarrow$ (T): Let $F \in \fF$ such that for each $f, g \in F$, $\norm{f - g}_{L^p(X; E)} < \eps$. Fix $f \in F$, then by the \hyperref[Dominated Convergence Theorem]{theorem:dct}, there exists $A \in \cm$ such that $\mu(A) < \infty$ and $\norm{\one_{A^c}f}_{L^p(X; E)} < \eps$. In which case, for any $g \in F$, 
+    \[
+        \norm{\one_{A^c}g}_{L^p(X; E)} \le \norm{\one_{A^c}f}_{L^p(X; E)} + \norm{f - g}_{L^p(X; E)}
+        \le \norm{\one_{A^c}f}_{L^p(X; E)} + \eps
+    \]
+    
+
+
+    (M) + (UI) + (T) $\Rightarrow$ ($L^p$): Let $\eps > 0$. By (T), there exists $A \in \cm$ and $F_1 \in \fF$ with $\mu(A) < \infty$ and $\sup_{f \in F_1}\int_{A^c}\norm{f}_E^p < \eps^p$. Thus for every $f, g \in F_1$, 
+    \begin{align*}
+        \norm{f - g}_{L^p(X; E)} &\le \norm{\one_A(f - g)}_{L^p(X; E)}  + \norm{\one_{A^c}f}_{L^p(X; E)}  + \norm{\one_{A^c}g}_{L^p(X; E)} \\
+        &\le \norm{\one_A(f - g)}_{L^p(X; E)} + 2\eps
+    \end{align*}
+
+    By (UI), there exists $M > 0$ and $F_2 \in \fF$ with $F_2 \subset F_1$ such that
+    \[
+    \sup_{h \in F_2} \int_{\bracs{\norm{h}_E \ge M}}\norm{h}_E^p < \eps^p
+    \]
+    
+    Assume without loss of generality that $\mu(A) > 0$ and let $\delta = \eps\mu(A)^{-1/p}$. By (M), there exists $F_3 \in \fF$ with $F_3 \subset F_2$, such that for any $f, g \in F_3$, 
+    \[
+    \mu\bracsn{\norm{f - g}_E \ge \delta} \le \paren{\frac{\eps}{2M}}^p
+    \]
+    
+    In which case,
+    \begin{align*}
+        \norm{\one_{A}(f - g)}_{L^p(X; E)} &\le \normn{\one_{A \cap \bracs{\norm{f - g}_E \le \delta}}(f - g)}_{L^p(X; E)} \\
+        &+ \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}(f - g)}_{L^p(X; E)} \\
+        &\le \delta\mu(A)^{1/p} + \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}(f - g)}_{L^p(X; E)}  \\
+        &\le \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}(f - g)}_{L^p(X; E)} + \eps
+    \end{align*}
+
+    then for any $f, g \in F_3$, 
+    \begin{align*}
+        \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}(f - g)}_{L^p(X; E)} &\le \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}f}_{L^p(X; E)} \\
+        &+ \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}g}_{L^p(X; E)}
+    \end{align*}
+    
+    Now, 
+    \begin{align*}
+        \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}f}_{L^p(X; E)} &\le \normn{\one_{\bracsn{\norm{f}_E \ge M}}f}_{L^p(X; E)} \\
+        &+ \normn{\one_{\bracsn{\norm{f - g}_E \ge \delta, \norm{f}_E \le M}}f}_{L^p(X; E)} \\
+        &\le \eps + \eps/2 = 3\eps/2
+    \end{align*}
+
+    Similarly, $\normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}g}_{L^p(X; E)} \le 3\eps/2$. Thus
+    \[
+    \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}(f - g)}_{L^p(X; E)} \le 3\eps
+    \]
+    
+    
+    Therefore for any $f, g \in F_3$, $\norm{f - g}_{L^p(X; E)} \le 6 \eps$, so $\fF$ is Cauchy in $L^p(X; E)$.  
+    
+\end{proof}
+