Added integration of non-negative functions.

src/measure/lebesgue-integral/index.tex

@@ -2,3 +2,4 @@
 \label{chap:lebesgue-integral}
 
 \input{./src/measure/lebesgue-integral/simple.tex}
+\input{./src/measure/lebesgue-integral/non-negative.tex}

src/measure/lebesgue-integral/non-negative.tex

@@ -0,0 +1,114 @@
+\section{Integration of Non-Negative Functions}
+\label{section:lebesgue-non-negative}
+
+\begin{definition}
+\label{definition:measurable-non-negative}
+    Let $(X, \cm)$ be a measure space, then
+    \[
+    L^+(X, \cm) = \bracs{f: X \to \real| f \ge 0, f \text{ is } (\cm, \cb_\real) \text{-measurable}}
+    \]
+    is the space of non-negative $\real$-valued measurable functions on $(X, \cm)$.
+\end{definition}
+
+\begin{definition}[Integral of Non-Negative Function]
+\label{definition:lebesgue-non-negative}
+    Let $(X, \cm, \mu)$ be a measure space and $f \in L^+(X, \cm)$, then
+    \[
+    \int f d\mu = \int f(x)\mu(dx) = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm), \phi \le f}
+    \]
+    is the \textbf{Lebesgue integral} of $f$.
+\end{definition}
+
+\begin{lemma}
+\label{lemma:lebesgue-non-negative-strict}
+    Let $(X, \cm, \mu)$ be a measure space and $f \in L^+(X, \cm)$, then
+    \[
+    \int f d\mu = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm), \phi|_{\bracs{f > 0}} < f|_{\bracs{f > 0}}, \phi \le f}
+    \]
+\end{lemma}
+\begin{proof}
+    Let $\phi \in \Sigma^+(X, \cm)$ with $\phi \le f$, then for any $\alpha \in (0, 1)$, $\alpha \phi|_{\bracs{f > 0}} < f|_{\bracs{f > 0}}$. Since
+    \[
+    \int \phi d\mu = \sup_{\alpha \in (0, 1)}\alpha \int \phi d\mu = \sup_{\alpha \in (0, 1)}\int \alpha \phi d\mu
+    \]
+    the two sides are equal.
+\end{proof}
+
+\begin{theorem}[Monotone Convergence Theorem, {{\cite[Theorem 2.14]{Folland}}}]
+\label{theorem:mct}
+    Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n} \subset L^+(X, \cm)$, and $f \in L^+(X, \cm)$ such that $f_n \upto f$ pointwise, then
+    \[
+    \limv{n}\int f_n d\mu = \int f d\mu
+    \]
+\end{theorem}
+\begin{proof}
+    By \ref{definition:lebesgue-non-negative}, $\int f_n d\mu \le \int f d\mu$ for each $n \in \natp$.
+
+    Let $\phi \in \Sigma^+(X, \cm)$ with $\phi|_{\bracs{f > 0}} < f|_{\bracs{f > 0}}$ and $\phi \le f$. Since $f_n \upto f$, $\bracs{f_n \ge \phi} \upto X$. In which case, since $A \mapsto \int \one_A \phi d\mu$ is a measure (\ref{proposition:lebesgue-simple-properties}),
+    \[
+    \limv{n}\int f_n d\mu \ge \limv{n}\int \one_{\bracs{f_n \ge \phi}} \cdot \phi d\mu = \int \phi d\mu
+    \]
+    by continuity from below (\ref{proposition:measure-properties}). Therefore $\limv{n}\int f_n d\mu \ge \int f$ by \ref{lemma:lebesgue-non-negative-strict}.
+\end{proof}
+
+\begin{lemma}[Fatou, {{\cite[Lemma 2.18]{Folland}}}]
+\label{lemma:fatou}
+    Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n} \subset L^+(X, \cm)$, then
+    \[
+    \int \liminf_{n \to \infty}f_n d\mu \le \liminf_{n \to \infty}\int f_nd\mu
+    \]
+\end{lemma}
+\begin{proof}
+    For each $n \in \natp$, $\inf_{k \ge n}f_k \le f_n$. By the monotone convergence theorem,
+    \[
+    \int \liminf_{n \to \infty}f_n d\mu = \limv{n}\int \inf_{k \ge n}f_k d\mu
+    \le \liminf_{n \to \infty}\int f_n d\mu
+    \]
+\end{proof}
+
+\begin{lemma}
+\label{lemma:lebesgue-simple-monotone}
+    Let $(X, \cm)$ be a measurable space and $f \in L^+(X, \cm)$, then there exists $\seq{f_n} \subset \Sigma^+(X, \cm)$ such that $f_n \upto f$ pointwise.
+\end{lemma}
+\begin{proof}
+    By \ref{proposition:measurable-simple-separable-norm}, there exists $\seq{g_n} \subset \Sigma^+(X, \cm)$ such that $0 \le g_n \le f$ for each $n \in \natp$, and $g_n \to f$ pointwise. For each $n \in \natp$, let $f_n = \max_{1 \le k \le n}g_k$, then $f_n \in \Sigma^+(X, \cm)$, $0 \le f_n \le f$, and $f_n \upto f$ pointwise.
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:lebesgue-non-negative-properties}
+    Let $(X, \cm, \mu)$ be a measure space, then
+    \begin{enumerate}
+        \item For any $f, g \in L^+(X, \cm)$ and $\alpha \ge 0$, $\int \alpha f + g d\mu = \alpha \int f d\mu + \int g d\mu$.
+        \item For any $\seq{f_n} \subset L^+(X, \cm)$, $\sum_{n \in \natp}\int f_nd\mu = \int \sum_{n \in \natp}f_n d\mu$.
+        \item For any $f, g \in L^+(X, \cm)$ with $f \le g$, $\int f d\mu \le \int g d\mu$.
+        \item For any $f \in L^+(X, \cm)$ with $\int f d\mu < \infty$, $\mu(\bracs{f = \infty}) = 0$.
+        \item For any $f \in L^+(X, \cm)$ with $\int f d\mu < \infty$, $\bracs{f > 0}$ is $\sigma$-finite.
+        \item For any $f \in L^+(X, \cm)$ with $\int f d\mu = 0$, $f = 0$ $\mu$-almost everywhere.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): By \ref{lemma:lebesgue-simple-monotone}, there exists $\seq{f_n}, \seq{g_n} \subset \Sigma^+(X, \cm)$ with $0 \le f_n \le f$ and $0 \le g_n \le g$ for each $n \in \natp$, $f_n \upto f$, and $g_n \upto g$. By \ref{proposition:lebesgue-simple-properties} and the Monotone Convergence Theorem (\ref{theorem:mct}),
+    \begin{align*}
+    \int \alpha f + g d\mu = \limv{n}\int \alpha f_n + g_n d\mu = \alpha\limv{n}\int f_n d\mu + \limv{n}\int g_n d\mu \\
+    &= \alpha \int f d\mu + \int g d\mu
+    \end{align*}
+
+
+    (2): By (1) and the Monotone Convergence Theorem (\ref{theorem:mct}),
+    \[
+    \int \sum_{n \in \natp}f_n d\mu = \int \limv{N}\sum_{n = 1}^N f_n d\mu = \limv{N}\sum_{n = 1}^N \int f_nd\mu = \sum_{n \in \natp}\int f_n d\mu
+    \]
+
+    (3): By \ref{definition:lebesgue-non-negative}.
+
+    (4): Since $\int f d\mu \ge \int \infty \cdot \one_{\bracs{f = \infty}}d\mu = \infty \cdot \mu(\bracs{f = \infty})$, $\int f d\mu < \infty$ implies that $\mu(\bracs{f = \infty}) = 0$.
+
+    (5): For each $\eps > 0$,
+    \[
+    \int f d\mu \ge \int \eps \cdot \one_{\bracs{f \ge \eps}} = \eps \cdot \mu(\bracs{f \ge \eps})
+    \]
+    so $\int f d\mu < \infty$ implies that $\mu(\bracs{f \ge \eps}) < \infty$. Since $\bracs{f > 0} = \bigcup_{n \in \natp}\bracs{f_n \ge 1/n}$, $\bracs{f > 0}$ is $\sigma$-finite.
+
+    (6): As in the proof of (5), $\mu(\bracs{f \ge \eps}) = 0$ for all $\eps > 0$. Thus $\bracs{f > 0} = \bigcup_{n \in \natp}\bracs{f_n \ge 1/n}$ is a $\mu$-null set, so $f = 0$ $\mu$-almost everywhere.
+\end{proof}

src/measure/lebesgue-integral/simple.tex

@@ -12,7 +12,7 @@
     \[
     \int f d\mu = \int f(x) \mu(dx) = \sum_{y \in f(X)}y \cdot \mu(\bracs{f = y})
     \]
-    is the \textbf{(Lebesgue) integral} of $f$.
+    is the \textbf{Lebesgue integral} of $f$.
 \end{definition}
 
 \begin{proposition}[{{\cite[Proposition 2.13]{Folland}}}]