Added the Vitali convergence theorem.

src/cat/tricks/product.tex

@@ -26,3 +26,34 @@
     \]
 
 \end{proof}
+
+\begin{lemma}
+\label{lemma:power-difference}
+    Let $p \in [1, \infty)$, then for each $a, b \ge 0$, 
+    \[
+    |a^p - b^p| \le p|a - b|(a \vee b)^{p - 1}
+    \]
+\end{lemma}
+\begin{proof}
+    Assume without loss of generality that $0 < a < b$, then
+    \[
+        b^p - a^p = p\int_{a}^b t^{p - 1}dt \le p|a - b|(a \vee b)^{p - 1}
+    \]
+\end{proof}
+
+\begin{lemma}
+\label{lemma:power-sum}
+    Let $p \in [1, \infty)$, then for each $a, b \ge 0$, 
+    \[
+    (a + b)^p \le 2^{p-1}(a^p + b^p)
+    \]
+\end{lemma}
+\begin{proof}
+    Since $t \mapsto t^p$ is convex, 
+    \begin{align*}
+        \braks{\frac{a + b}{2}}^p &\le \frac{a^p}{2} + \frac{b^p}{2} \\
+        \frac{(a+b)^p}{2^p} &\le \frac{a^p}{2} + \frac{b^p}{2} \\
+        (a + b)^p &\le 2^{-1}(a^p + b^p)
+    \end{align*}
+    
+\end{proof}

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}
 
+
+

src/fa/lp/index.tex

@@ -2,5 +2,7 @@
 \label{chap:lp}
 
 \input{./definition.tex}
+\input{./ineq.tex}
 \input{./duality.tex}
+\input{./ui.tex}
 \input{./seq.tex}

src/fa/lp/ineq.tex

@@ -0,0 +1,50 @@
+\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}

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}
+

src/measure/measurable-maps/in-measure.tex

@@ -3,18 +3,15 @@
 
 \begin{definition}[Convergence in Measure]
 \label{definition:convergence-in-measure}
-    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$, then $f_n \to f$ \textbf{in measure} if for every $\eps > 0$,
+    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and $\fF$ be a filter of $(\cm, \cb_Y)$-measurable functions, and $f$ be a $(\cm, \cb_Y)$-measurable function, then $\fF \to f$ \textbf{in measure} if for each $\eps > 0$, 
     \[
-    \lim_{n \to \infty}\mu(\bracs{d(f_n, f) > \eps}) = 0
+    \lim_{g, \fF}\mu(\bracs{d(f, g) > \eps}) = 0
     \]
 \end{definition}
 
 \begin{definition}[Cauchy in Measure]
 \label{definition:cauchy-in-measure}
-    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ be Borel measurable functions from $X$ to $Y$, then $\seq{f_n}$ is \textbf{Cauchy in measure} if for every $\eps > 0$,
+    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and $\fF$ be a filter of $(\cm, \cb_Y)$-measurable functions, then $\fF$ is \textbf{Cauchy in measure} if for every $\eps, \delta > 0$, there exists $A \in \fF$ such that $\mu(\bracs{d(f, g) > \delta}) < \eps$ for all $f, g \in A$. 
-    \[
-    \mu(\bracs{d(f_m, f_n) > \eps}) \to 0 \quad \text{as}\ m, n \to \infty
-    \]
 \end{definition}
 
 \begin{lemma}

src/measure/vector/ac.tex

@@ -38,6 +38,8 @@
 \begin{definition}[Uniformly Absolutely Continuous]
 \label{definition:u-ac}
     Let $(X, \cm)$ be a measurable space, $\mu$ be a positive measure on $(X, \cm)$, $E$ be a normed space over $K \in \RC$, and $\mathcal{U} \subset M(X, \cm; E)$ be a family of $E$-valued vector measures on $(X, \cm)$, then $\mathcal{U}$ is \textbf{uniformly absolutely continuous} with respect to $\mu$ if for every $\eps > 0$, there exists $\delta > 0$ such that for all $A \in \cm$ with $\mu(A) < \delta$, $\norm{\nu(A)}_E < \eps$ for all $\nu \in \mathcal{U}$. 
+
+    For any family $\cf$ of measurable functions, $\cf$ is \textbf{uniformly absolutely continuous} with respect to $\mu$ if the measures $\mathcal{U} = \bracs{fd\mu|f \in \cf}$ are uniformly absolutely continuous with respect to $\mu$. 
 \end{definition}
 
 \begin{theorem}[Vitali-Hahn-Saks]