From 2e81d6f87a0cfc430ed1eacec6b49eec5831675d Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Thu, 22 Jan 2026 11:36:14 -0500 Subject: [PATCH] Added integration of simple functions. --- src/measure/lebesgue-integral/index.tex | 2 +- .../lebesgue-integral/non-negative.tex | 0 src/measure/lebesgue-integral/simple.tex | 64 +++++++++++++++++++ src/measure/measurable-maps/simple.tex | 2 +- 4 files changed, 66 insertions(+), 2 deletions(-) create mode 100644 src/measure/lebesgue-integral/non-negative.tex create mode 100644 src/measure/lebesgue-integral/simple.tex diff --git a/src/measure/lebesgue-integral/index.tex b/src/measure/lebesgue-integral/index.tex index abacd27..e8e2c57 100644 --- a/src/measure/lebesgue-integral/index.tex +++ b/src/measure/lebesgue-integral/index.tex @@ -1,4 +1,4 @@ \chapter{The Lebesgue Integral} \label{chap:lebesgue-integral} -%\input{./src/measure/lebesgue-integral/simple.tex} +\input{./src/measure/lebesgue-integral/simple.tex} diff --git a/src/measure/lebesgue-integral/non-negative.tex b/src/measure/lebesgue-integral/non-negative.tex new file mode 100644 index 0000000..e69de29 diff --git a/src/measure/lebesgue-integral/simple.tex b/src/measure/lebesgue-integral/simple.tex new file mode 100644 index 0000000..5e1b508 --- /dev/null +++ b/src/measure/lebesgue-integral/simple.tex @@ -0,0 +1,64 @@ +\section{Integration of Simple Functions} +\label{section:lebesgue-simple} + +\begin{definition}[Space of Simple Functions] +\label{definition:simple-function-scalar} + Let $(X, \cm)$ be a measurable space, then $\Sigma(X, \cm) = \Sigma(X, \cm; \complex)$ is the space of $\complex$-valued simple functions on $(X, \cm)$, and $\Sigma^+(X, \cm)$ is the space of non-negative simple functions. +\end{definition} + +\begin{definition}[Integral of Non-Negative Simple Functions] +\label{definition:lebesgue-simple} + Let $(X, \cm, \mu)$ be a measure space and $f = \sum_{y \in f(X)}y \cdot \one_{\bracs{f = y}} \in \Sigma^+(X, \cm)$ be a non-negative simple function in standard form, then\footnote{With the convention that $0 \cdot \infty = 0$.} + \[ + \int f d\mu = \int f(x) \mu(dx) = \sum_{y \in f(X)}y \cdot \mu(\bracs{f = y}) + \] + is the \textbf{(Lebesgue) integral} of $f$. +\end{definition} + +\begin{proposition}[{{\cite[Proposition 2.13]{Folland}}}] +\label{proposition:lebesgue-simple-properties} + Let $(X, \cm, \mu)$ be a measure space and $f, g \in \Sigma^*(X, \cm)$, then: + \begin{enumerate} + \item For any $\alpha \ge 0$, $\int \alpha f d\mu = \alpha \int f d\mu$. + \item $\int f + g d\mu = \int f d\mu + \int g d\mu$. + \item If $f \le g$, then $\int f \le \int g$. + \item The mapping $A \mapsto \int \one_A \cdot f d\mu$ is a measure on $(X, \cm)$. + \end{enumerate} +\end{proposition} +\begin{proof} + (1): If $\alpha = 0$, then $\int \alpha f d\mu = \int 0 d\mu = 0 = 0 \cdot \int f d\mu$. Otherwise, the mapping $y \mapsto \alpha y$ is a bijection. Hence + \[ + \alpha f = \sum_{y \in f(X)} (\alpha y) \cdot \one_{\bracs{f = y}} + \] + is the standard form of $\alpha f$. Therefore + \[ + \int \alpha f d\mu = \sum_{y \in f(X)}(\alpha y) \cdot \mu({\bracs{f = y}}) = \alpha \int f d\mu + \] + + (2): Since $\mu$ is finitely additive, + \begin{align*} + \int f d\mu + \int g d\mu &= \sum_{y \in f(X)}y \cdot \mu(\bracs{f = y}) + \sum_{y \in g(X)}y \cdot \mu(\bracs{g = y}) \\ + &= \sum_{y \in f(X)}\sum_{z \in g(X)}(y + z) \cdot \mu(\bracs{f = y, g = z}) \\ + &= \sum_{y \in (f + g)(X)}\sum_{{z \in f(X) \atop {z' \in g(X) \atop z + z' = y}}}(z + z') \cdot \mu(\bracs{f = z, g = z'}) \\ + &= \sum_{y \in (f + g)(X)}y \cdot \mu(\bracs{f + g = y}) = \int f + g d\mu + \end{align*} + + (3): Since $\mu$ is finitely additive, + \begin{align*} + \int f d\mu &= \sum_{y \in f(X)}y \cdot \mu(\bracs{f = y}) = \sum_{y \in f(X)}\sum_{z \in g(X)}y \cdot \mu(\bracs{f = y, g = z}) \\ + &\le \sum_{y \in f(X)}\sum_{z \in g(X)}z \cdot \mu(\bracs{f = y, g = z}) = \sum_{z \in g(X)}z \cdot \mu\paren{\bracs{g = z}} = \int g d\mu + \end{align*} + + (4): Firstly, for any $A \in \cm$, + \[ + \one_A \cdot f = \sum_{y \in f(X)}y \cdot \one_{A \cap \bracs{f = y}} + \] + so by (1), + \[ + \int \one_A \cdot f d\mu = \sum_{y \in f(X)}y \cdot \mu(A \cap \bracs{f = y}) + \] + Since $\mu$ is countably additive, for any $\seq{E_n} \subset \cm$ pairwise disjoint with $E = \bigcup_{n \in \natp}E_n$, + \[ + \int \one_E \cdot f d\mu = \sum_{y \in f(X)}y \cdot \one_{E \cap \bracs{f = y}} = \sum_{y \in f(X)}\sum_{n \in \natp}y \cdot \one_{E_n \cap \bracs{f = y}} = \sum_{n \in \natp}\int \one_{E_n} \cdot f d\mu + \] +\end{proof} diff --git a/src/measure/measurable-maps/simple.tex b/src/measure/measurable-maps/simple.tex index fd81329..6da479b 100644 --- a/src/measure/measurable-maps/simple.tex +++ b/src/measure/measurable-maps/simple.tex @@ -30,5 +30,5 @@ \] is the \textbf{standard form} of $f$. - The set $\Sigma(X, \cm, V)$ is the vector space of all simple functions from $X$ to $Y$. + The set $\Sigma(X, \cm; V)$ is the space of $V$-valued simple functions on $(X, \cm)$, which forms a vector space. \end{definition}