From 61a8f3d37bd28aa9a214e0dd5a0d1e35d397bf86 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Thu, 22 Jan 2026 18:12:10 -0500 Subject: [PATCH] Added integration of non-negative functions. --- src/measure/lebesgue-integral/index.tex | 1 + .../lebesgue-integral/non-negative.tex | 114 ++++++++++++++++++ src/measure/lebesgue-integral/simple.tex | 2 +- 3 files changed, 116 insertions(+), 1 deletion(-) diff --git a/src/measure/lebesgue-integral/index.tex b/src/measure/lebesgue-integral/index.tex index e8e2c57..4d8cdf3 100644 --- a/src/measure/lebesgue-integral/index.tex +++ b/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} diff --git a/src/measure/lebesgue-integral/non-negative.tex b/src/measure/lebesgue-integral/non-negative.tex index e69de29..5e624a9 100644 --- a/src/measure/lebesgue-integral/non-negative.tex +++ b/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} diff --git a/src/measure/lebesgue-integral/simple.tex b/src/measure/lebesgue-integral/simple.tex index 5e1b508..214400a 100644 --- a/src/measure/lebesgue-integral/simple.tex +++ b/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}}}]