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Defined the Lebesgue integral.

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Bokuan Li
2026-01-25 14:10:21 -05:00
parent 4e451700d2
commit f9d7a448bf
3 changed files with 139 additions and 12 deletions
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24
src/measure/lebesgue-integral/non-negative.tex
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@@ -5,14 +5,14 @@
\label{definition:measurable-non-negative}
Let $(X, \cm)$ be a measure space, then
\[
L^+(X, \cm) = \bracs{f: X \to \real| f \ge 0, f \text{ is } (\cm, \cb_\real) \text{-measurable}}
\mathcal{L}^+(X, \cm) = \bracs{f: X \to \real| f \ge 0, f \text{ is } (\cm, \cb_\real) \text{-measurable}}
\]
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
Let $(X, \cm, \mu)$ be a measure space and $f \in \mathcal{L}^+(X, \cm)$, then
\[
\int f d\mu = \int f(x)\mu(dx) = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm), \phi \le f}
\]
@@ -21,7 +21,7 @@
\begin{lemma}
\label{lemma:lebesgue-non-negative-strict}
Let $(X, \cm, \mu)$ be a measure space and $f \in L^+(X, \cm)$, then
Let $(X, \cm, \mu)$ be a measure space and $f \in \mathcal{L}^+(X, \cm)$, then
\[
\int f d\mu = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm), \phi|_{\bracs{f > 0}} < f|_{\bracs{f > 0}}, \phi \le f}
\]
@@ -36,7 +36,7 @@
\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
Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n} \subset \mathcal{L}^+(X, \cm)$, and $f \in \mathcal{L}^+(X, \cm)$ such that $f_n \upto f$ pointwise, then
\[
\limv{n}\int f_n d\mu = \int f d\mu
\]
@@ -53,7 +53,7 @@
\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
Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n} \subset \mathcal{L}^+(X, \cm)$, then
\[
\int \liminf_{n \to \infty}f_n d\mu \le \liminf_{n \to \infty}\int f_nd\mu
\]
@@ -68,7 +68,7 @@
\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.
Let $(X, \cm)$ be a measurable space and $f \in \mathcal{L}^+(X, \cm)$, 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.
@@ -79,12 +79,12 @@
\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.
\item For any $f, g \in \mathcal{L}^+(X, \cm)$ and $\alpha \ge 0$, $\int \alpha f + g d\mu = \alpha \int f d\mu + \int g d\mu$.
\item For any $\seq{f_n} \subset \mathcal{L}^+(X, \cm)$, $\sum_{n \in \natp}\int f_nd\mu = \int \sum_{n \in \natp}f_n d\mu$.
\item For any $f, g \in \mathcal{L}^+(X, \cm)$ with $f \le g$, $\int f d\mu \le \int g d\mu$.
\item For any $f \in \mathcal{L}^+(X, \cm)$ with $\int f d\mu < \infty$, $\mu(\bracs{f = \infty}) = 0$.
\item For any $f \in \mathcal{L}^+(X, \cm)$ with $\int f d\mu < \infty$, $\bracs{f > 0}$ is $\sigma$-finite.
\item For any $f \in \mathcal{L}^+(X, \cm)$ with $\int f d\mu = 0$, $f = 0$ $\mu$-almost everywhere.
\end{enumerate}
\end{proposition}
\begin{proof}