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Bokuan Li
2026-03-06 14:06:15 -05:00
parent 173727665b
commit 5034bc4220
109 changed files with 1184 additions and 410 deletions
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19
src/measure/lebesgue-integral/complex.tex
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@@ -7,10 +7,12 @@
\[
\int \abs{f} d\mu < \infty
\]
The set
\[
\mathcal{L}^1(X, \cm, \mu; \complex) = \mathcal{L}^1(X, \cm, \mu) = \mathcal{L}^1(X; \complex) = \mathcal{L}^1(X) = \mathcal{L}^1(\mu; \complex) = \mathcal{L}^1(\mu)
\]
is the vector space of \textbf{$\mu$-integrable functions} on $X$.
\end{definition}
\begin{proof}
@@ -18,7 +20,8 @@
\[
\int \abs{\lambda f + g}d\mu \le \int \abs{\lambda} \abs{f} + \abs{g}d\mu = \lambda \int \abs{f}d\mu + \int \abs{g}d\mu
\]
by \ref{proposition:lebesgue-non-negative-properties}.
by \autoref{proposition:lebesgue-non-negative-properties}.
\end{proof}
\begin{definition}[Positive and Negative Parts]
@@ -27,6 +30,7 @@
\[
f^+ = \max(f, 0) \quad f^- = -\min(f, 0)
\]
are the \textbf{positive} and \textbf{negative} parts of $f$, and $f = f^+ - f^-$.
\end{definition}
@@ -37,10 +41,12 @@
\[
\int f d\mu = \int f^+ d\mu - \int f^- d\mu
\]
is the \textbf{integral} of $f$. If $f$ is $\complex$-valued, then
\[
\int f d\mu = \int \text{Re}(f)d\mu + i\int \text{Im}(f)d\mu
\]
is the \textbf{integral} of $f$.
\end{definition}
@@ -58,7 +64,8 @@
-\lambda\int f^- d\mu + \lambda\int f^+ d\mu &\lambda < 0
\end{cases}
\]
by \ref{proposition:lebesgue-non-negative-properties}, so $\int \lambda f d\mu = \lambda \int f d\mu$.
by \autoref{proposition:lebesgue-non-negative-properties}, so $\int \lambda f d\mu = \lambda \int f d\mu$.
Let $h = f + g$, then $h = h^+ - h^- = f^+ + g^+ - f^- - g^-$, so
\begin{align*}
@@ -67,7 +74,7 @@
\int h^+ d\mu - \int h^- d\mu &= \int f^+ d\mu - \int f^- d\mu + \int g^+ d\mu - \int g^- d\mu \\
&= \int f d\mu + \int g d\mu
\end{align*}
by \ref{proposition:lebesgue-non-negative-properties}, so $\int f + g d\mu = \int f d\mu + \int g d\mu$.
by \autoref{proposition:lebesgue-non-negative-properties}, so $\int f + g d\mu = \int f d\mu + \int g d\mu$.
Now suppose that $f, g$ are $\complex$-valued and $\lambda = \alpha + \beta i \in \complex$ with $\alpha, \beta \in \real$, then
\begin{align*}
@@ -88,6 +95,7 @@
\[
\int f = \int f^+ d\mu - \int f^-d\mu \le \int f^+ + \int f^-d\mu = \int \abs{f}d\mu
\]
and if $f$ is $\complex$-valued and $\alpha = \ol{\sgn(\int f d\mu)}$, then
\begin{align*}
\abs{\int f d\mu} &= \alpha \int f d\mu = \int \alpha f \\
@@ -113,7 +121,7 @@
then $\int fd\mu = \limv{n}\int f_n d\mu$.
\end{theorem}
\begin{proof}
By (1) and (2), $f$ is measurable with $\int \abs{f}d\mu \le \int \abs{g}d\mu < \infty$, so $f \in \mathcal{L}^1(X)$. Now, since $g + f, g - f \ge 0$, by Fatou's lemma (\ref{lemma:fatou}) and \ref{proposition:lebesgue-integral-properties},
By (1) and (2), $f$ is measurable with $\int \abs{f}d\mu \le \int \abs{g}d\mu < \infty$, so $f \in \mathcal{L}^1(X)$. Now, since $g + f, g - f \ge 0$, by \hyperref[Fatou's lemma]{lemma:fatou} and \autoref{proposition:lebesgue-integral-properties},
\begin{align*}
\int g d\mu + \int f d\mu &\le \liminf_{n \to \infty}\int g + f_n d\mu = \int g d\mu + \liminf_{n \to \infty}f_n d\mu \\
\int g d\mu - \int f d\mu &\le \liminf_{n \to \infty}\int g - \int f_n d\mu = \int g d\mu - \limsup_{n \to \infty}\int f_n d\mu
@@ -122,12 +130,13 @@
\[
\limsup_{n \to \infty}\int f_n d\mu \le \int f d\mu \le \liminf_{n \to \infty}\int f_n d\mu
\]
and $\int f d\mu = \limv{n}\int f_n d\mu$.
\end{proof}
\begin{remark}[There is no dominated convergence theorem for nets]
\label{remark:dct-no-net}
In analysis, one frequently encounters places where only sequential continuity is provided or required. It is my opinion that a good portion of this comes from the lack of an extension of the dominated convergence theorem (\ref{theorem:dct}) to nets. This limitation arises from the monotone convergence theorem (\ref{theorem:mct}), where continuity from below is used.
In analysis, one frequently encounters places where only sequential continuity is provided or required. It is my opinion that a good portion of this comes from the lack of an extension of the \hyperref[dominated convergence theorem]{theorem:dct} to nets. This limitation arises from the \hyperref[monotone convergence theorem]{theorem:mct}, where continuity from below is used.
For an example, consider the Lebesgue measure on $[0, 1]$. Let $A$ be the net of all finite subsets of $[0, 1]$, directed by inclusion, then $\lim_{\alpha \in A}\one_\alpha = 1$ pointwise. However, $\int \one_\alpha = 0$ for all $\alpha \in A$.
\end{remark}

6
src/measure/lebesgue-integral/index.tex
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@@ -1,6 +1,6 @@
\chapter{The Lebesgue Integral}
\label{chap:lebesgue-integral}
\input{./src/measure/lebesgue-integral/simple.tex}
\input{./src/measure/lebesgue-integral/non-negative.tex}
\input{./src/measure/lebesgue-integral/complex.tex}
\input{./simple.tex}
\input{./non-negative.tex}
\input{./complex.tex}

24
src/measure/lebesgue-integral/non-negative.tex
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@@ -7,6 +7,7 @@
\[
\mathcal{L}^+(X, \cm) = \bracs{f: X \to \real| f \ge 0, f \text{ is } (\cm, \cb_\real) \text{-measurable}}
\]
is the space of non-negative $\real$-valued measurable functions on $(X, \cm)$.
\end{definition}
@@ -16,6 +17,7 @@
\[
\int f d\mu = \int f(x)\mu(dx) = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm), \phi \le f}
\]
is the \textbf{Lebesgue integral} of $f$.
\end{definition}
@@ -25,12 +27,14 @@
\[
\int f d\mu = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm), \phi|_{\bracs{f > 0}} < f|_{\bracs{f > 0}}, \phi \le f}
\]
\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}
@@ -40,15 +44,17 @@
\[
\limv{n}\int f_n d\mu = \int f d\mu
\]
\end{theorem}
\begin{proof}
By \ref{definition:lebesgue-non-negative}, $\int f_n d\mu \le \int f d\mu$ for each $n \in \natp$.
By \autoref{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}),
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 (\autoref{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}.
by continuity from below (\autoref{proposition:measure-properties}). Therefore $\limv{n}\int f_n d\mu \ge \int f$ by \autoref{lemma:lebesgue-non-negative-strict}.
\end{proof}
\begin{lemma}[Fatou, {{\cite[Lemma 2.18]{Folland}}}]
@@ -57,6 +63,7 @@
\[
\int \liminf_{n \to \infty}f_n d\mu \le \liminf_{n \to \infty}\int f_nd\mu
\]
\end{lemma}
\begin{proof}
For each $n \in \natp$, $\inf_{k \ge n}f_k \le f_n$. By the monotone convergence theorem,
@@ -64,6 +71,7 @@
\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}
@@ -71,7 +79,7 @@
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.
By \autoref{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}
@@ -88,19 +96,20 @@
\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}),
(1): By \autoref{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 \autoref{proposition:lebesgue-simple-properties} and the \hyperref[Monotone Convergence Theorem]{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}),
(2): By (1) and the \hyperref[Monotone Convergence Theorem]{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}.
(3): By \autoref{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 $\mu(\bracs{f = \infty}) = 0$.
@@ -108,6 +117,7 @@
\[
\int f d\mu \ge \int \eps \cdot \one_{\bracs{f \ge \eps}} = \eps \cdot \mu(\bracs{f \ge \eps})
\]
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.

6
src/measure/lebesgue-integral/simple.tex
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@@ -12,6 +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 $f$.
\end{definition}
@@ -30,11 +31,13 @@
\[
\alpha f = \sum_{y \in f(X)} (\alpha y) \cdot \one_{\bracs{f = y}}
\]
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}) \\
@@ -53,12 +56,15 @@
\[
\one_A \cdot f = \sum_{y \in f(X)}y \cdot \one_{A \cap \bracs{f = y}}
\]
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}