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Bokuan Li
2026-03-06 14:06:15 -05:00
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2
src/measure/bochner-integral/index.tex
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@@ -1,4 +1,4 @@
\chapter{The Bochner Integral}
\label{chap:bochner-integral}
\input{./src/measure/bochner-integral/strongly.tex}
\input{./strongly.tex}

5
src/measure/bochner-integral/strongly.tex
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@@ -17,11 +17,12 @@
\begin{proof}
(1) $\Rightarrow$ (2): TODO
(2) $\Rightarrow$ (3): By \ref{proposition:measurable-simple-separable-norm}.
(2) $\Rightarrow$ (3): By \autoref{proposition:measurable-simple-separable-norm}.
(3) $\Rightarrow$ (1): For each $\phi \in E^*$, $\phi \circ f = \limv{n}\phi \circ f_n$ is measurable by \ref{proposition:limit-measurable}. Since
(3) $\Rightarrow$ (1): For each $\phi \in E^*$, $\phi \circ f = \limv{n}\phi \circ f_n$ is measurable by \autoref{proposition:limit-measurable}. Since
\[
f(X) \subset \ol{\bigcup_{n \in \natp}f_n(X)}
\]
and each $f_n$ is finitely-valued, $f(X)$ is separable.
\end{proof}

10
src/measure/index.tex
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@@ -1,8 +1,8 @@
\part{Measure Theory and Integration}
\label{part:measure}
\input{./src/measure/sets/index.tex}
\input{./src/measure/measure/index.tex}
\input{./src/measure/measurable-maps/index.tex}
\input{./src/measure/lebesgue-integral/index.tex}
\input{./src/measure/bochner-integral/index.tex}
\input{./sets/index.tex}
\input{./measure/index.tex}
\input{./measurable-maps/index.tex}
\input{./lebesgue-integral/index.tex}
\input{./bochner-integral/index.tex}

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}

10
src/measure/measurable-maps/index.tex
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@@ -1,8 +1,8 @@
\chapter{Measurable Functions}
\label{chap:measurable-maps}
\input{./src/measure/measurable-maps/measurable-maps.tex}
\input{./src/measure/measurable-maps/product.tex}
\input{./src/measure/measurable-maps/real-valued.tex}
\input{./src/measure/measurable-maps/simple.tex}
\input{./src/measure/measurable-maps/metric.tex}
\input{./measurable-maps.tex}
\input{./product.tex}
\input{./real-valued.tex}
\input{./simple.tex}
\input{./metric.tex}

1
src/measure/measurable-maps/measurable-maps.tex
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@@ -31,5 +31,6 @@
\[
\sigma(\bracs{f_i| i \in I}) = \sigma\paren{f_i^{-1}(\cn_i)|i \in I}
\]
is the smallest $\sigma$-algebra on $X$ such that each $\seqi{f}$ is measurable.
\end{definition}

29
src/measure/measurable-maps/metric.tex
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@@ -7,13 +7,15 @@
\[
X \to Z \quad x \mapsto F(f_1(x), \cdots, f_n(x))
\]
is $(\cm, \cn)$-measurable.
\end{proposition}
\begin{proof}
By \ref{proposition:product-sigma-algebra-metric}, $\cb_{\prod_{j = 1}^n Y_j} = \bigotimes_{j = 1}^n \cb_{Y_j}$, so
By \autoref{proposition:product-sigma-algebra-metric}, $\cb_{\prod_{j = 1}^n Y_j} = \bigotimes_{j = 1}^n \cb_{Y_j}$, so
\[
X \to \prod_{j = 1}^n Y_j \quad x \mapsto (f_1(x), \cdots, f_n(x))
\]
is $(\cm, \cb_{\prod_{j = 1}^n Y_j})$-measurable. Therefore the composition is $(\cm, \cn)$-measurable.
\end{proof}
@@ -27,7 +29,7 @@
\end{enumerate}
\end{proposition}
\begin{proof}
By \ref{proposition:metric-measurable-fibre-product}.
By \autoref{proposition:metric-measurable-fibre-product}.
\end{proof}
\begin{proposition}
@@ -43,7 +45,8 @@
\[
d_{U^c}: X \to [0, 1] \quad x \mapsto d(x, U^c) \wedge 1
\]
is continuous by \ref{proposition:set-distance-continuous}. By \ref{proposition:fattening-closure}, $\bracsn{d_{U^c} > 0} = U$. Thus $\bracs{f \in U} = \bracsn{d_{U^c} \circ f > 0}$.
is continuous by \autoref{proposition:set-distance-continuous}. By \autoref{proposition:fattening-closure}, $\bracsn{d_{U^c} > 0} = U$. Thus $\bracs{f \in U} = \bracsn{d_{U^c} \circ f > 0}$.
\end{proof}
\begin{proposition}
@@ -59,9 +62,10 @@
\[
\bracs{\limv{n}f_n \text{ exists}} = \bigcap_{k \in \natp}\bigcup_{N \in \natp}\bigcap_{m \in \natp}\bigcap_{n \in \natp}\bracsn{d(f_m, f_n) < 1/k}
\]
is measurable by \ref{proposition:metric-measurables}.
(2): For each $\phi \in C(X; [0, 1])$, $\phi \circ f = \limv{n}\phi \circ f_n$ is $(\cm, \cb_\real)$-measurable by \ref{proposition:limit-measurable}. Thus $f$ is $(\cm, \cb_\real)$-measurable by \ref{proposition:metric-measurable-compose}.
is measurable by \autoref{proposition:metric-measurables}.
(2): For each $\phi \in C(X; [0, 1])$, $\phi \circ f = \limv{n}\phi \circ f_n$ is $(\cm, \cb_\real)$-measurable by \autoref{proposition:limit-measurable}. Thus $f$ is $(\cm, \cb_\real)$-measurable by \autoref{proposition:metric-measurable-compose}.
\end{proof}
@@ -89,18 +93,22 @@
\[
C(N, x) = \bracs{1 \le n \le N|y_n \in N(f(x))}
\]
By assumption (b), $1 \in C(N, x) \ne \emptyset$. Let
\[
k(N, x) = \min\bracs{n \in C(N, x) \bigg | d(f(x), y_n) = \min_{m \in C(N, x)}d(f(x), y_m)}
\]
then for any $k \in \natp$,
\[
\bracs{x \in X|k(n, x) \le k} = \bigcup_{j = 1}^k\bracs{x \in X \bigg |y_j \in C(n, x), d(f(x), y_j) = \min_{m \in C(N, x)}d(f(x), y_m)}
\]
For each $1 \le m \le N$, $y \mapsto d(y, y_m)$ is continuous. Thus $x \mapsto d(f(x), y_m)$ and $x \mapsto \min_{m \in C(N, x)}d(f(x), y_m)$ are $(\cm, \cb_\real)$-measurable by \ref{proposition:limit-measurable} and assumption (c). By \ref{proposition:metric-measurables},
For each $1 \le m \le N$, $y \mapsto d(y, y_m)$ is continuous. Thus $x \mapsto d(f(x), y_m)$ and $x \mapsto \min_{m \in C(N, x)}d(f(x), y_m)$ are $(\cm, \cb_\real)$-measurable by \autoref{proposition:limit-measurable} and assumption (c). By \autoref{proposition:metric-measurables},
\[
\bracs{x \in X \bigg | d(f(x), y_j) = \min_{m \in C(N, x)}d(f(x), y_m)} \in \cm
\]
for each $1 \le j \le k$. Combining this with assumption (c) shows that $\bracs{x \in X|k(n, x) \le k} \in \cm$.
Let $f_N(x) = y_{k(N, x)}$, then for each $N \in \natp$, $f_N(x) \subset \bracs{y_n|1 \le n \le N}$, so $f_N(x)$ is finitely valued. Since $x \mapsto k{(N, x)}$ is $(\cm, 2^\nat)$-measurable, $f_N$ is $(\cm, \cb_Y)$-measurable.
@@ -109,9 +117,10 @@
\[
f(x) \in \ol{N^o(f(x))} = \ol{\bracs{y_n| n \in \natp, y_n \in N^o(f(x))}}
\]
by assumption (a) and \ref{definition:dense}. Thus for any $\eps > 0$, there exists $N \in \nat$ such that $y_N \in N^o(f(x))$ and $d(f(x), y_N) < \eps$. In which case, $d(f(x), f_n(x)) < \eps$ for all $n \ge N$. Therefore $f_n \to f$ pointwise.
(3) $\Rightarrow$ (1): By \ref{proposition:metric-measurable-limit}.
by assumption (a) and \autoref{definition:dense}. Thus for any $\eps > 0$, there exists $N \in \nat$ such that $y_N \in N^o(f(x))$ and $d(f(x), y_N) < \eps$. In which case, $d(f(x), f_n(x)) < \eps$ for all $n \ge N$. Therefore $f_n \to f$ pointwise.
(3) $\Rightarrow$ (1): By \autoref{proposition:metric-measurable-limit}.
\end{proof}
\begin{proposition}
@@ -127,6 +136,7 @@
\[
N: E \to 2^E \quad y \mapsto B_E(0, \norm{y}_E)
\]
then
\begin{enumerate}
\item[(a)] $y \in \ol{B_E(0, \norm{y}_E)}$.
@@ -135,6 +145,7 @@
\[
\bracs{y \in E|y_0 \in N(y)} = \bracs{y \in E|\norm{y_0}_E \le \norm{y}_E} \in \cb_E
\]
\end{enumerate}
By \ref{proposition:measurable-simple-separable}, (1) and (2) are equivalent.
By \autoref{proposition:measurable-simple-separable}, (1) and (2) are equivalent.
\end{proof}

7
src/measure/measurable-maps/product.tex
View File

@@ -12,6 +12,7 @@
\[
\ce = \bracs{\bigcap_{j \in J}\pi_j^{-1}(E_j) \bigg | J \subset I \text{ finite}, E_j \in \cm_j}
\]
then $\ce$ is an elementary family, and $\sigma(\ce) = \bigotimes_{i \in I}\cm_i$.
\end{lemma}
\begin{proof}
@@ -21,15 +22,18 @@
\[
\bigcap_{j \in J}\pi_j^{-1}(E_j), \bigcap_{j \in J'}\pi_j^{-1}(F_j) \in \ce
\]
Assume without loss of generality that $J = J'$, then
\[
\bigcap_{j \in J}\pi_j^{-1}(E_j) \cap \bigcap_{j \in J'}\pi_j^{-1}(F_j) = \bigcap_{j \in J}\pi_j^{-1}(E_j \cap F_j) \in \ce
\]
\item[(E)] For any $j \in I$, $\prod_{i \in I}X_i = \pi_j^{-1}(X_j) \in \ce$. Thus it is sufficient to show that $\ce$ is closed under complements. Let $\bigcap_{j \in J}\pi_j^{-1}(E_j) \in \ce$, then
\[
\braks{\bigcap_{j \in J}\pi_j^{-1}(E_j)}^c = \bigcup_{j \in J}\pi_j^{-1}(E_j^c) = \bigsqcup_{\emptyset \subsetneq K \subset J}
\underbrace{\paren{\bigcap_{k \in K}\pi_j^{-1}(E_k^c)} \cap \paren{\bigcap_{j \in J \setminus K}\pi_j^{-1}(E_k)}}_{\in \ce}
\]
\end{enumerate}
\end{proof}
@@ -42,11 +46,12 @@
\end{enumerate}
\end{proposition}
\begin{proof}
(2): By \ref{proposition:separable-product}, $X$ is also separable. Let $U \subset X$, then there exists $\seq{x_j} \subset X$ such that $\overline{\bracs{x_j|j \in \natp}} \supset U$. For each $j \in \nat$, let $\bracs{U_{j, k}}$ such that $U_{j, k} \in \cn_{X_k}(\pi_k(x_j))$ for each $1 \le k \le n$ and $x_j \in \bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k}) \subset U$.
(2): By \autoref{proposition:separable-product}, $X$ is also separable. Let $U \subset X$, then there exists $\seq{x_j} \subset X$ such that $\overline{\bracs{x_j|j \in \natp}} \supset U$. For each $j \in \nat$, let $\bracs{U_{j, k}}$ such that $U_{j, k} \in \cn_{X_k}(\pi_k(x_j))$ for each $1 \le k \le n$ and $x_j \in \bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k}) \subset U$.
Since $\bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k}) \in \bigotimes_{j = 1}^n \cb_{X_j}$ and
\[
U = \bigcup_{j \in \natp}\bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k})
\]
The open set $U$ is in $\bigotimes_{j = 1}^n \cb_{X_j}$.
\end{proof}

6
src/measure/measurable-maps/real-valued.tex
View File

@@ -33,12 +33,14 @@
\bracs{F > \alpha} = \bigcup_{n \in \natp}\bracs{f_n > \alpha}
\]
(2): Let $\alpha \in \real$, then
\[
\bracs{f < \alpha} = \bigcup_{n \in \natp}\bracs{f_n < \alpha}
\]
By \ref{proposition:borel-sigma-extended-generators} and \ref{lemma:measurable-function-generating-set}, $F$ and $f$ are both $(\cm, \cb_{\ol \real})$-measurable.
By \autoref{proposition:borel-sigma-extended-generators} and \autoref{lemma:measurable-function-generating-set}, $F$ and $f$ are both $(\cm, \cb_{\ol \real})$-measurable.
(3): $\limsup_{n \to \infty}f_n = \inf_{n \in \natp}\sup_{k \ge n}f_k$.
@@ -46,5 +48,5 @@
(5): If $\limv{n}f_n$ exists, then it is equal to (3) and (4). In which case, it is measurable.
Finally, if the above functions are $\real$-valued, then they are $(\cm, \cb_\real)$-measurable by \ref{lemma:extended-real-measurable}.
Finally, if the above functions are $\real$-valued, then they are $(\cm, \cb_\real)$-measurable by \autoref{lemma:extended-real-measurable}.
\end{proof}

2
src/measure/measurable-maps/simple.tex
View File

@@ -10,6 +10,7 @@
0 &x \not\in E
\end{cases}
\]
is the \textbf{characteristic/indicator function} of $E$.
\end{definition}
@@ -28,6 +29,7 @@
\[
f = \sum_{y \in f(X)}y \cdot \one_{\bracs{f = y}}
\]
is the \textbf{standard form} of $f$.
The set $\Sigma(X, \cm; V)$ is the space of $V$-valued simple functions on $(X, \cm)$, which forms a vector space.

8
src/measure/measure/complete.tex
View File

@@ -12,6 +12,7 @@
\[
\ol{\cm} = \bracs{E \cup F| E \in \cm, F \subset N, N \in \cn}
\]
then
\begin{enumerate}
\item For any $A \in \ol{\cm}$, there exists $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $E \cap N = \emptyset$ and $A = E \sqcup F$.
@@ -35,22 +36,27 @@
\bigcup_{n \in \natp}A_n = \underbrace{\braks{\bigcup_{n \in \natp}E_n}}_{\in \cm} \cup \underbrace{\braks{\bigcup_{n \in \natp}F_n}}_{\subset \bigcup_{n \in \natp}N_n} \in \ol{\cm}
\]
(3): Let $A \in \ol{\cm}$, $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $A = E \cup F$. Since $A \setminus N = E \setminus N \in \cm$ and $A \cup N = E \cup N \in \cm$,
\[
\mu(A \setminus N) \le \mu(E) \le \mu(A \cup N)
\]
Let $E' \in \cm$, $N' \in \cn$, and $F' \subset N'$ such that $A = E' \cup F'$, then
\[
\mu(A \setminus (N \cup N')) \le \mu(A \setminus N') \le \mu(E') \le \mu(A \cup N') \le \mu(A \cup (N \cup N'))
\]
Now, since $N$ and $N'$ are null,
\[
\mu(A \cup (N \cup N')) = \mu(A \setminus (N \cup N')) + \mu(N \cup N') = \mu(A \setminus (N \cup N'))
\]
so
\[
\mu(A \setminus (N \cup N')) = \mu(E) = \mu(E') = \mu(A \cup (N \cup N'))
\]
Thus for any decomposition $A = E \cup F$ where $E \in \cm$ and $F$ is contained in a null set, $\ol{\mu}(A) = \mu(E)$ is well-defined.
To see that $\ol{\mu}$ is a measure, let $\seq{A_n} \subset \ol{\cm}$ be pairwise disjoint. For each $n \in \nat$, let $E_n \in \cm$, $N_n \in \cn$, and $F_n \subset N_n$ such that $A_n = E_n \cup F_n$, then
@@ -58,6 +64,7 @@
\ol{\mu}\paren{\bigsqcup_{n \in \natp}A_n} = \mu\paren{\bigsqcup_{n \in \natp}E_n} = \sum_{n \in \natp}\mu(E_n) = \sum_{n \in \natp}\mu(A_n)
\]
(4): Let $A \in \ol{\cm}$ with $\ol{\mu}(A) = 0$, $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $A = E \cup F$, then $\mu(E) = \mu(E \cup N) = 0$. By definition of $\ol{\cm}$, any subset of $E \cup N$ is also in $\ol{\cm}$. Thus $\ol{\mu}$ is complete.
(U): Let $N \in \cn \subset \cm \subset \cf$, then for any $F \subset N$, $\nu(F) \le \nu(N) = \mu(N)$, and $F \in \cf$ by completeness of $(X, \cf, \nu)$. Thus $\cf \supset \ol{\cm}$.
@@ -66,4 +73,5 @@
\[
\nu(A) = \nu(E \cup F) = \nu(E) = \mu(E) = \ol{\mu}(A)
\]
\end{proof}

22
src/measure/measure/index.tex
View File

@@ -1,14 +1,14 @@
\chapter{Positive Measures}
\label{chap:measures}
\input{./src/measure/measure/measure.tex}
\input{./src/measure/measure/complete.tex}
\input{./src/measure/measure/semifinite.tex}
\input{./src/measure/measure/sigma-finite.tex}
\input{./src/measure/measure/regular.tex}
\input{./src/measure/measure/radon.tex}
\input{./src/measure/measure/outer.tex}
\input{./src/measure/measure/lebesgue-stieltjes.tex}
\input{./src/measure/measure/radon.tex}
\input{./src/measure/measure/riesz.tex}
\input{./src/measure/measure/kolmogorov.tex}
\input{./measure.tex}
\input{./complete.tex}
\input{./semifinite.tex}
\input{./sigma-finite.tex}
\input{./regular.tex}
\input{./radon.tex}
\input{./outer.tex}
\input{./lebesgue-stieltjes.tex}
\input{./radon.tex}
\input{./riesz.tex}
\input{./kolmogorov.tex}

6
src/measure/measure/kolmogorov.tex
View File

@@ -36,10 +36,12 @@
\[
K_n = \bigcap_{j = 1}^n \pi_{[n], [j]}^{-1}(C_n)
\]
Since each $X_j$ is Hausdorff, $K_n \subset \prod_{j = 1}^n X_j$ is compact with $K_n \subset B_n$ and $K_{n+1} \subset K_n \times X_{n+1}$. Moreover,
\[
\mu_{[n]}(B_n \setminus K_n) \le \sum_{j = 1}^n\mu_{[n]}\braks{\pi_{[n], [j]}^{-1}(B_j \setminus C_j)} \le \sum_{j = 1}^n\mu_{[j]}(B_j \setminus C_j) \le \eps/2
\]
Thus $\mu_{[n]}(K_n) \ge \eps/2$.
\end{proof}
@@ -58,11 +60,12 @@
\[
\alg = \bracs{\pi_J^{-1}(B)| B \in \bigotimes_{j \in J}\cb_{X_j}, J \subset I \text{ finite}}
\]
then $\alg$ is an algebra. For any $B \in \bigotimes_{j \in J}\cb_{X_j}$, define $\mu_0(\pi_J^{-1}(B)) = \mu_J(B)$, then $\mu_0: \alg \to [0, 1]$ is well-defined and finitely additive by the consistency of $\bracs{\mu_J}$.
To show that $\mu_0$ is a premeasure, it is sufficient to show that for any $\{\pi_{J_n}^{-1}(B_n)\}_1^\infty$ with $\pi_{J_n}^{-1}(B_n) \downto \emptyset$, $\mu_0(\pi_{J_n}^{-1}(B_n)) \downto 0$. To this end, suppose for contradiction that $\limv{n}\mu_0(\pi_{J_n}^{-1}(B_n)) = \eps > 0$.
By inserting additional elements into the sequence and relabeling the indices, assume without loss of generality that $J_n = [n]$ for all $n \in \natp$. By \ref{lemma:kolmogorov-compact-sequence}, there exists $\seq{K_n}$ such that:
By inserting additional elements into the sequence and relabeling the indices, assume without loss of generality that $J_n = [n]$ for all $n \in \natp$. By \autoref{lemma:kolmogorov-compact-sequence}, there exists $\seq{K_n}$ such that:
\begin{enumerate}
\item $K_n \subset \prod_{j = 1}^n X_j$ is compact.
\item $K_n \subset B_n$.
@@ -73,4 +76,5 @@
\[
x \in \bigcap_{n \in \natp}\pi_{[n]}^{-1}(K_n) \subset \bigcap_{n \in \natp}\pi_{[n]}^{-1}(B_n) \ne \emptyset
\]
\end{proof}

15
src/measure/measure/lebesgue-stieltjes.tex
View File

@@ -20,6 +20,7 @@
\[
\alg = \bracs{\bigsqcup_{j = 1}^n I_j \bigg | \seqf{I_j} \subset \ci, n \in \natp}
\]
is a ring.
\end{enumerate}
\end{lemma}
@@ -30,7 +31,7 @@
\item[(E)] $(a, b] \setminus (c, d] = (a, \min(b, c)] \sqcup (\max(a, d), b]$.
\end{enumerate}
(2): By \ref{proposition:elementary-family-algebra}, $\alg$ is a ring.
(2): By \autoref{proposition:elementary-family-algebra}, $\alg$ is a ring.
\end{proof}
\begin{definition}[Lebesgue-Stieltjes Measure, {{\cite[Theorem 1.16]{Folland}}}]
@@ -41,6 +42,7 @@
\[
\mu((a, b]) = F(b) - F(a)
\]
\item[(U)] For any function $G: \real \to \real$ satisfying (1), $F - G$ is constant.
\end{enumerate}
Conversely, if $F: \real \to \real$ is a Stieltjes function, then there exists a unique Borel measure $\mu_F: \cb_\real \to [0, \infty]$ satisfying (1), and $\mu_F$ is the \textbf{Lebesgue-Stieltjes measure} associated with $F$.
@@ -53,7 +55,8 @@
-\mu((x, 0]) &x \le 0
\end{cases}
\]
then $F$ is a Stieltjes function by monotonicity and continuity from above (\ref{proposition:measure-properties}). For any $-\infty < a < b < \infty$,
then $F$ is a Stieltjes function by monotonicity and continuity from above (\autoref{proposition:measure-properties}). For any $-\infty < a < b < \infty$,
\[
\mu((a, b]) = \begin{cases}
\mu((0, b]) - \mu((0, a]) &a, b \ge 0 \\
@@ -61,6 +64,7 @@
\mu((a, 0]) - \mu((b, 0]) &a, b \le 0
\end{cases}
\]
In all three cases, $\mu((a, b]) = F(b) - F(a)$.
(U): If $G: \real \to \real$ satisfies (1), then $F(x) = G(x) - G(0)$. Hence $F - G$ is constant.
@@ -70,10 +74,12 @@
\[
\alg = \bracs{\bigsqcup_{j = 1}^n (a_j, b_j] \bigg | \seqf{(a_j, b_j]} \subset \ci, n \in \natp}
\]
Define
\[
\mu_0: \alg \to [0, \infty) \quad \bigsqcup_{j = 1}^n (a_j, b_j] \mapsto \sum_{j = 1}^n [F(b_j) - F(a_j)]
\]
Let $(a, b] \in \ci$ with $(a, b] = \bigsqcup_{j = 1}^n (a_j, b_j]$. Assume without loss of generality that $\seqf{a_j}$ is non-decreasing, then $b_j = a_{j+1}$ for each $1 \le j \le n - 1$, $b = b_n$, and $a = a_1$. Thus
\begin{align*}
F(b) - F(a) &= F(b_n) + \sum_{j = 1}^{n-1} [F(b_{j}) - F(b_j)] - F(a_1) \\
@@ -85,10 +91,12 @@
\[
\sum_{k = 1}^n \mu_0((a_k, b_k]) = \mu_0 \paren{\bigsqcup_{k = 1}^n (a_k, b_k]} \le \mu_0((a, b])
\]
Let $\eps > 0$. By right-continuity of $F$, there exists $\delta > 0$ such that $F(a + \delta) - F(a) < \eps/2$, and $\seq{\delta_n} \subset \real_{> 0}$ such that $F(b_n + \delta_n) - F(b_n) < \eps/2^{n+1}$ for all $n \in \natp$. The intervals $\seq{(a_n, b_n + \delta_n)}$ forms an open cover for $[a + \delta, b]$. By compactness, there exists $N \in \natp$ such that
\[
[a + \delta, b] \subset \bigcup_{n = 1}^N (a_n, b_n + \delta_n) \subset \bigcup_{n = 1}^N (a_n, b_n + \delta_n]
\]
Thus
\begin{align*}
\mu_0((a, b]) &\le F(a + \delta) - F(a) + \mu_0([a + \delta, b]) \\
@@ -100,5 +108,6 @@
\[
\mu_0((a, b]) = \sum_{n \in \natp}\mu_0((a_n, b_n])
\]
Therefore $\mu_0$ is a premeasure on $\alg$. By Carathéodory's Extension Theorem (\ref{theorem:caratheodory-extension}), $\mu_0$ extends uniquely to a Borel measure on $\real$, which satisfies (1).
Therefore $\mu_0$ is a premeasure on $\alg$. By \hyperref[Carathéodory's Extension Theorem]{theorem:caratheodory-extension}, $\mu_0$ extends uniquely to a Borel measure on $\real$, which satisfies (1).
\end{proof}

16
src/measure/measure/measure.tex
View File

@@ -46,10 +46,12 @@
\[
\mu\paren{\liminf_{n \to \infty}E_n} \le \liminf_{n \to \infty}\mu(E_n)
\]
\item For any $\seq{E_n} \subset \cm$ with $\mu\paren{\bigcup_{n \in \nat}E_n} < \infty$,
\[
\mu\paren{\limsup_{n \to \infty}E_n} \ge \limsup_{n \to \infty}\mu(E_n)
\]
\end{enumerate}
\end{proposition}
\begin{proof}
@@ -60,15 +62,18 @@
\mu\paren{\bigcup_{n \in \nat}E_n} = \mu\paren{\bigsqcup_{n \in \nat}F_n} = \sum_{n \in \natp}\mu(F_n) \le \sum_{n \in \natp}\mu(E_n)
\]
(3): Denote $E_0 = \emptyset$. For each $n \in \natp$, let $F_n = E_n \setminus E_{n - 1}$, then $\seq{F_n}$ is pairwise disjoint with $\bigsqcup_{n \in \natp}F_n = \bigcup_{n \in \natp}E_n$. In which case,
\[
\mu\paren{\bigcup_{n \in \natp}E_n} = \mu\paren{\bigsqcup_{n \in \natp}F_n} = \limv{n}\sum_{k = 1}^n \mu(F_n) = \limv{n}\mu(E_n)
\]
(4): Since $\mu(E_1) < \infty$,
\[
\limv{n}\mu(E_n) = \mu(E_1) - \limv{n}\mu(E_1 \setminus E_n) = \mu(E_1) - \mu\paren{E_1 \setminus \bigcap_{n \in \natp}E_n} = \mu\paren{\bigcap_{n \in \natp}E_n}
\]
by (3).
(5): Using (1) and (3),
@@ -99,6 +104,7 @@
\[
\alg(F) = \bracs{E \in \cm|\mu(E \cap F) = \nu(E \cap F)}
\]
then $\alg(F) \supset \mathcal{P}$ by (b), and
\begin{enumerate}
\item[(L2)] For any $E, E' \in \alg(F)$ with $E \subset E'$ and $F \in \mathcal{P}$,
@@ -110,12 +116,14 @@
\[
\mu\paren{\bigcup_{n \in \nat}E_n \cap F} = \limv{n}\mu(E_n \cap F) = \limv{n}\nu(E_n \cap F) = \nu\paren{\bigcup_{n \in \nat}E_n \cap F}
\]
by continuity from below (\ref{proposition:measure-properties}).
\end{enumerate}
so $\alg(F)$ is a $\lambda$-system. By (a) and Dynkin's $\pi$-$\lambda$ theorem (\ref{theorem:pi-lambda}), $\alg(F) = \cm$.
Let $\seq{E_n}$ as in assumption (c), then $\mu(E_n \cap F) = \mu(E_n \cap F)$ for all $n \in \natp$ and $F \in \cm$. Thus by continuity from below (\ref{proposition:measure-properties}),
by continuity from below (\autoref{proposition:measure-properties}).
\end{enumerate}
so $\alg(F)$ is a $\lambda$-system. By (a) and Dynkin's $\pi$-$\lambda$ theorem (\autoref{theorem:pi-lambda}), $\alg(F) = \cm$.
Let $\seq{E_n}$ as in assumption (c), then $\mu(E_n \cap F) = \mu(E_n \cap F)$ for all $n \in \natp$ and $F \in \cm$. Thus by continuity from below (\autoref{proposition:measure-properties}),
\[
\mu(F) = \limv{n}\mu(E_n \cap F) = \limv{n}\nu(E_n \cap F) = \nu(F)
\]
\end{proof}

21
src/measure/measure/outer.tex
View File

@@ -11,6 +11,7 @@
\[
\mu^*\paren{\bigcup_{n \in \natp}E_n} \le \sum_{n \in \natp}\mu^*(E_n)
\]
\end{enumerate}
\end{definition}
@@ -20,6 +21,7 @@
\[
\mu^*: 2^X \to [0, \infty] \quad E \mapsto \inf\bracs{\sum_{n \in \natp}\rho(E_n) \bigg | \seq{E_n} \subset \ce, \bigcup_{n \in \natp}E_n \supset E}
\]
then $\mu^*$ is an outer measure.
\end{proposition}
\begin{proof}
@@ -39,6 +41,7 @@
\[
\mu^*(F) = \mu^*(E \cap F) + \mu^*(E \setminus F)
\]
\end{definition}
\begin{theorem}[Carathéodory, {{\cite[Theorem 1.11]{Folland}}}]
@@ -49,6 +52,7 @@
\[
\mu^*\paren{F \cap \bigsqcup_{n \in \natp}E_n} = \sum_{n \in \natp}\mu^*(F \cap E_n)
\]
\item $\cm$ is a $\sigma$-algebra.
\item $(X, \cm, \mu^*|_\cm)$ is a complete measure space.
\end{enumerate}
@@ -58,11 +62,13 @@
\[
\mu^*\paren{F \cap \bigsqcup_{n \in \natp}E_n} \ge \mu^*\paren{F \cap \bigsqcup_{n =1}^N E_n} = \sum_{n = 1}^N \mu^*(F \cap E_n)
\]
As this holds for all $N \in \nat$,
\[
\mu^*\paren{F \cap \bigsqcup_{n \in \natp}E_n} \ge \sum_{n \in \natp}\mu^*(F \cap E_n)
\]
(2): Firstly, $\emptyset, X \in \cm$ since $\mu^*(\emptyset) = 0$. For any $A \in \cm$, $A^c \in \cm$ as well because the definition of $\mu^*$-measurability is symmetric.
Let $A, B \in \cm$ and $F \subset X$, then
@@ -72,7 +78,7 @@
&= \mu^*(F \cap B) + \mu^*(F \cap A \setminus B) + \mu^*(F \setminus (A \cup B)) \\
&= \mu^*(F \cap (A \cup B)) + \mu^*(F \setminus (A \cup B))
\end{align*}
so $\cm$ is an algebra. Since (1) implies that $\cm$ is closed under countable disjoint unions, $\cm$ is a $\sigma$-algebra by \ref{lemma:sigma-algebra-condition}.
so $\cm$ is an algebra. Since (1) implies that $\cm$ is closed under countable disjoint unions, $\cm$ is a $\sigma$-algebra by \autoref{lemma:sigma-algebra-condition}.
(3): By (1), $\mu|_\cm$ is a measure. Let $N \in \cm$ with $\mu^*(N) = 0$, then for any $E \subset N$ and $F \subset X$,
\begin{align*}
@@ -93,6 +99,7 @@
\[
\mu_0\paren{\bigsqcup_{n \in \natp}A_n} = \sum_{n \in\natp}\mu_0(A_n)
\]
\end{enumerate}
\end{definition}
@@ -116,28 +123,33 @@
\[
\mu^*: 2^X \to [0, \infty] \quad E \mapsto \inf\bracs{\sum_{n \in \natp}\mu_0(E_n) \bigg | \seq{E_n} \subset \alg, \bigcup_{n \in \natp}E_n \supset E}
\]
then $\mu^*$ is an outer measure by \ref{proposition:outer-measure-inf}. By Carathéodory's theorem (\ref{theorem:caratheodory}), there exists a measure space $(X, \cm, \mu)$ such that $\mu = \mu^*|_{\cm}$.
then $\mu^*$ is an outer measure by \autoref{proposition:outer-measure-inf}. By \hyperref[Carathéodory's theorem]{theorem:caratheodory}, there exists a measure space $(X, \cm, \mu)$ such that $\mu = \mu^*|_{\cm}$.
(1): Let $E \in \alg$ and $F \subset X$. If there exists $\seq{F_n} \subset \alg$ with $\bigcup_{n \in \natp}F_n \supset F$, then
\[
\sum_{n \in \natp}\mu_0(F_n) = \sum_{n \in \natp}\mu_0(F_n \cap E) + \sum_{n \in \natp}\mu_0(F_n \setminus E) \ge \mu^*(F \cap E) + \mu^*(F \setminus E)
\]
for all such $\seq{F_n}$, so $\mu^*(F) = \mu^*(F \cap E) + \mu^*(F \setminus E)$. Otherwise, $\mu^*(F) = \infty$ and the result holds directly.
(2): Let $E \in \alg$ and $\seq{E_n} \subset \alg$ such that $\bigcup_{n \in \natp}E_n \supset E$, then
\[
\sum_{n \in \natp}\mu_0(E_n) = \sum_{n \in \natp}\mu_0(E_n \cap E) + \sum_{n \in \natp}\mu_0(E_n \setminus E) \ge \sum_{n \in \natp}\mu_0(E_n \cap E) = \mu_0(E)
\]
As this holds for all such $\seq{E_n}$, $\mu^*(E) = \mu_0(E)$.
(U): Let $\seq{E_n} \subset \alg$, then by continuity from below (\ref{proposition:measure-properties}),
(U): Let $\seq{E_n} \subset \alg$, then by continuity from below (\autoref{proposition:measure-properties}),
\[
\nu\paren{\bigcup_{n \in \natp}E_n} = \limv{n}\nu\paren{\bigcup_{k = 1}^nE_k} = \limv{n}\mu\paren{\bigcup_{k = 1}^n E_k} = \mu\paren{\bigcup_{n \in \natp}E_n}
\]
Thus if $E \in \cm \cap \cn$ with $\bigcup_{n \in \natp}E_n \supset E$, then
\[
\nu(E) \le \nu\paren{\bigcup_{n \in \natp}E_n} = \mu\paren{\bigcup_{n \in \natp}E_n} \le \sum_{n \in \natp}\mu_0(E_n)
\]
As this holds for all such $\seq{E_n}$, $\nu(E) \le \mu^*(E) = \mu(E)$.
Suppose that $\mu(E) < \infty$, then there exists $\seq{E_n} \subset \alg$ such that $\bigcup_{n \in \natp}E_n \supset E$ and $\sum_{n \in \natp}\mu_0(E_n) < \infty$. In which case,
@@ -148,9 +160,10 @@
\end{align*}
so $\nu(E) = \mu(E)$ whenever $\mu(E) < \infty$.
Finally, if $\mu$ is $\sigma$-finite, then there exists $\seq{E_n} \subset \cm$ such that $E_n \upto X$ and $\mu(E_n) < \infty$ for all $n \in \natp$. For each $n \in \natp$, there exists $\bracsn{E_{n, k}}_{k = 1}^\infty \subset \alg$ such that $E_n \subset \bigcup_{k \in \natp}E_{n, k}$ and $\mu\paren{\bigcup_{k \in \natp}E_{n, k}} < \infty$. Let $F_n = \bigcup_{j = 1}^n \bigcup_{k \in \natp}E_{j, k}$, then $\seq{F_n} \subset \sigma(\alg) \subset \cm \cap \cn$ with $F_n \upto X$ and $\mu(F_n) < \infty$ for all $n \in \natp$. For any $E \in \cm \cap \cn$, by continuity from below (\ref{proposition:measure-properties}),
Finally, if $\mu$ is $\sigma$-finite, then there exists $\seq{E_n} \subset \cm$ such that $E_n \upto X$ and $\mu(E_n) < \infty$ for all $n \in \natp$. For each $n \in \natp$, there exists $\bracsn{E_{n, k}}_{k = 1}^\infty \subset \alg$ such that $E_n \subset \bigcup_{k \in \natp}E_{n, k}$ and $\mu\paren{\bigcup_{k \in \natp}E_{n, k}} < \infty$. Let $F_n = \bigcup_{j = 1}^n \bigcup_{k \in \natp}E_{j, k}$, then $\seq{F_n} \subset \sigma(\alg) \subset \cm \cap \cn$ with $F_n \upto X$ and $\mu(F_n) < \infty$ for all $n \in \natp$. For any $E \in \cm \cap \cn$, by continuity from below (\autoref{proposition:measure-properties}),
\[
\nu(E) = \limv{n}\nu(E \cap F_n) = \limv{n}\mu(E \cap F_n) = \mu(E)
\]
so $\nu|_{\cm \cap \cn} = \mu|_{\cm \cap \cn}$.
\end{proof}

34
src/measure/measure/radon.tex
View File

@@ -19,30 +19,36 @@
\[
\mu(U) = \sup\bracs{\int f d\mu \bigg | f \in C_c(X; [0, 1]), \supp{f} \subset U}
\]
\item For any $K \subset X$ compact,
\[
\mu(K) = \inf\bracs{\int f d\mu \bigg | f \in C_c(X; [0, 1]), f \ge \one_K}
\]
\end{enumerate}
\end{proposition}
\begin{proof}
(1): Let $U \subset X$ be open. By Urysohn's lemma (\ref{lemma:lch-urysohn}), for any $K \subset U$ compact, there exists $f_K \in C_c(X; [0, 1])$ such that $f_K|_K = 1$ and $\supp{f_K} \subset U$. In which case,
(1): Let $U \subset X$ be open. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, for any $K \subset U$ compact, there exists $f_K \in C_c(X; [0, 1])$ such that $f_K|_K = 1$ and $\supp{f_K} \subset U$. In which case,
\[
\mu(K) \le \int f_K d\mu \le \mu(U)
\]
By (R3'),
\[
\mu(U) = \sup_{K \subset U \text{ compact}}\mu(K) \le \sup_{K \subset U \text{ compact}}\int f_K d\mu \le \mu(U)
\]
(2): Let $K \subset X$ be compact and $U \in \cn^o(K)$. By Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $f_U \in C_c(X; [0, 1])$ such that $f_U|_K = 1$ and $\supp{f_U} \subset U$. In which case,
(2): Let $K \subset X$ be compact and $U \in \cn^o(K)$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f_U \in C_c(X; [0, 1])$ such that $f_U|_K = 1$ and $\supp{f_U} \subset U$. In which case,
\[
\mu(K) \le \int f_U d\mu \le \mu(U)
\]
By (R2),
\[
\mu(K) = \inf_{U \in \cn^o(K)}\mu(U) \ge \inf_{U \in \cn^o(K)}\int f_U d\mu \ge \mu(K)
\]
\end{proof}
\begin{proposition}[{{\cite[Proposition 7.5]{Folland}}}]
@@ -64,16 +70,19 @@
\[
\mu(K \setminus V) \ge \mu(K) - \mu(V) \ge \mu(U) - \eps/2 - \eps/2 \ge \mu(E) - \eps
\]
so $\mu$ is inner regular on $E$.
Now suppose that $E$ is $\sigma$-finite, then there exists $\seq{E_n} \subset \cb_X$ with $\mu(E_n) < \infty$ for each $n \in \natp$ and $E = \bigsqcup_{n \in \natp}E_n = E$. Let $N \in \natp$ and $\seqf{K_n} \subset 2^X$ compact with $K_n \subset E_n$ for each $n \in \natp$, then $\bigcup_{n = 1}^N K_n \subset E$ is compact. Hence
\[
\sup_{\substack{K \subset E \\ \text{compact}}}\mu(K) \ge \sum_{n = 1}^N\sup_{\substack{K_n \subset E_n \\ \text{compact}}}\mu(K_n) = \sum_{n = 1}^N \mu(E_n)
\]
As the above holds for all $N \in \natp$,
\[
\sup_{\substack{K \subset E \\ \text{compact}}}\mu(K) \ge \sum_{n \in \natp}\mu(E_n) = \mu(E)
\]
\end{proof}
\begin{proposition}[{{\cite[Proposition 7.7]{Folland}}}]
@@ -89,12 +98,14 @@
\[
\mu\paren{\bigcup_{n \in \natp}U_n \setminus E} \le \sum_{n \in \natp}\mu(U_n \setminus E_n) < \eps
\]
so $U = \bigcup_{n \in \natp}U_n$ is the desired open set.
Applying the above result to $E^c$, there exists $V \in \cn^o(E^c)$ such that $\mu(V \setminus E^c) < \eps$. Let $F = V^c$, then $F \subset E$ is closed and
\[
\mu(E \setminus F) = \mu(E \cap F^c) = \mu(E \cap V) = \mu(V \setminus E^c) < \eps
\]
\end{proof}
@@ -108,29 +119,33 @@
then $\mu$ is a regular measure on $X$.
\end{proposition}
\begin{proof}
By assumption (b), $f \mapsto \int f d\mu$ is a positive linear functional on $C_c(X; \real)$. By the Riesz Representation Theorem (\ref{theorem:riesz-radon}), there exists a Radon measure $\nu: \cb_X \to [0, \infty]$ such that for any $f \in C_c(X; \real)$, $\int f d\mu = \int f d\mu$.
By assumption (b), $f \mapsto \int f d\mu$ is a positive linear functional on $C_c(X; \real)$. By the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon}, there exists a Radon measure $\nu: \cb_X \to [0, \infty]$ such that for any $f \in C_c(X; \real)$, $\int f d\mu = \int f d\mu$.
Let $U \subset X$ be open, then by \ref{proposition:radon-measure-cc},
Let $U \subset X$ be open, then by \autoref{proposition:radon-measure-cc},
\[
\nu(U) = \sup_{\substack{f \in C_c(X; [0, 1]) \\ f \le \one_U}} \int f d\nu = \sup_{\substack{f \in C_c(X; [0, 1]) \\ f \le \one_U}} \int f d\mu \le \mu(U)
\]
By assumption (a), there exists $\seq{K_n} \subset 2^X$ compact such that $K_n \upto U$. By Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $\seq{f_n} \subset C_c(X; [0, 1])$ such that $\one_{K_n} \le f_n \le \one_U$ for all $n \in \natp$, and $f_n \upto f$ pointwise. Using the Monotone Convergence Theorem (\ref{theorem:mct}),
By assumption (a), there exists $\seq{K_n} \subset 2^X$ compact such that $K_n \upto U$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\seq{f_n} \subset C_c(X; [0, 1])$ such that $\one_{K_n} \le f_n \le \one_U$ for all $n \in \natp$, and $f_n \upto f$ pointwise. Using the \hyperref[Monotone Convergence Theorem]{theorem:mct},
\[
\mu(U) = \limv{n} \int f_n d\mu \le \sup_{\substack{f \in C_c(X; [0, 1]) \\ f \le \one_U}}\int f d\mu = \nu(U)
\]
Therefore $\mu(U) = \nu(U)$.
Now let $E \in \cb_X$ be arbitrary and $\eps > 0$. By \ref{proposition:radon-measurable-description}, there exists $U \in \cn^o(E)$ and $V \subset E$ closed such that $\nu(U \setminus E), \nu(E \setminus V) < \eps$. In which case, since $U \setminus V$ is open,
Now let $E \in \cb_X$ be arbitrary and $\eps > 0$. By \autoref{proposition:radon-measurable-description}, there exists $U \in \cn^o(E)$ and $V \subset E$ closed such that $\nu(U \setminus E), \nu(E \setminus V) < \eps$. In which case, since $U \setminus V$ is open,
\[
\mu(U \setminus E) \le \mu(U \setminus V) = \nu(U \setminus V) < \eps
\]
so $\mu(U) = \mu(U \setminus E) + \mu(E) \le \mu(E) + \eps$. Therefore
\[
\mu(E) = \inf_{U \in \cn^o(E)}\mu(U) = \inf_{U \in \cn^o(E)}\nu(U) = \nu(E)
\]
and $\mu = \nu$.
Since $X$ is $\sigma$-compact, $\mu$ is $\sigma$-finite, so $\mu$ is regular by \ref{proposition:radon-regular-sigma-finite}.
Since $X$ is $\sigma$-compact, $\mu$ is $\sigma$-finite, so $\mu$ is regular by \autoref{proposition:radon-regular-sigma-finite}.
\end{proof}
\begin{proposition}
@@ -138,10 +153,11 @@
Let $X$ be a LCH space, $\mu: \cb_X \to [0, \infty]$ be a Radon measure, $E$ be a normed space, and $p \in [1, \infty)$, then $C_c(X; E)$ is dense in $L^p(X; E)$.
\end{proposition}
\begin{proof}
By \ref{proposition:lp-simple-dense}, $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$. Using linearity, it is sufficient to approximate indicator functions of Borel sets with finite measure.
By \autoref{proposition:lp-simple-dense}, $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$. Using linearity, it is sufficient to approximate indicator functions of Borel sets with finite measure.
Let $A \in \cb_X$ and $\eps > 0$. By \ref{proposition:radon-regular-sigma-finite}, there exists $U \in \cn^o(A)$ and $K \subset A$ compact such that $\mu(U \setminus A), \mu(A \setminus K) < \eps/2$. By Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $f \in C_c(X; [0, 1])$ such that $f|_K = 1$ and $\supp{f} \subset U$. In which case, for any $y \in E$,
Let $A \in \cb_X$ and $\eps > 0$. By \autoref{proposition:radon-regular-sigma-finite}, there exists $U \in \cn^o(A)$ and $K \subset A$ compact such that $\mu(U \setminus A), \mu(A \setminus K) < \eps/2$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f \in C_c(X; [0, 1])$ such that $f|_K = 1$ and $\supp{f} \subset U$. In which case, for any $y \in E$,
\[
\norm{x \cdot \one_A - x \cdot f}_{L^p(X; E)} \le \norm{x}_E \mu(\bracs{f \ne \one_A})^{1/p} \le \norm{x}_E\mu(U \setminus K)^{1/p} < \eps^{1/p}\norm{x}_E
\]
\end{proof}

2
src/measure/measure/regular.tex
View File

@@ -7,6 +7,7 @@
\[
\mu(E) = \sup\bracs{\mu(K)| K \subset E, K \text{ compact}}
\]
\end{definition}
\begin{definition}[Outer Regular]
@@ -15,6 +16,7 @@
\[
\mu(E) = \sup\bracs{\mu(U)| U \in \cn^o(A)}
\]
\end{definition}
\begin{definition}[Regular]

36
src/measure/measure/riesz.tex
View File

@@ -17,10 +17,11 @@
\begin{proof}
(1): $\dpb{g - f, I}{C_c(X; \real)} \ge 0$.
(2): By Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $g \in C_c(X; [0, 1])$ such that $g|_K = 1$. In which case,
(2): By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $g \in C_c(X; [0, 1])$ such that $g|_K = 1$. In which case,
\[
-\norm{f}_u\dpn{g, I}{C_c(X; \real)} \le \dpn{f, I}{C_c(X; \real)} \le \norm{f}_u\dpn{g, I}{C_c(X; \real)}
\]
so $C_K = \dpn{g, I}{C_c(X; \real)}$ is a desired constant.
\end{proof}
@@ -41,29 +42,34 @@
\[
\mu_0: \topo \to [0, \infty] \quad U \mapsto \sup_{f \prec U}\dpb{f, I}{C_c(X; \real)}
\]
and
\[
\mu^*: 2^X \to [0, \infty] \quad E \mapsto \inf\bracsn{\mu_0(U)|U \in \cn^o(E)}
\]
then
\begin{enumerate}
\item[(OM1)] Since $\emptyset \in \topo$ and $\mu_0(\emptyset) = 0$, $\mu^*(\emptyset) = 0$.
\item[(OM2)] Let $E, F \subset X$ with $E \subset F$, then $\cn^o(E) \subset \cn^o(F)$, so $\mu^*(E) \le \mu^*(F)$.
\item[(OM3)] Let $\seq{E_n} \subset 2^X$, $E = \bigcup_{n \in \natp}E_n$, $U \in \cn^o(E)$, and $\seq{U_n} \subset \topo$ such that $U_n \in \cn^o(E_n)$ for each $n \in \natp$.
Let $f \prec U$, then by compactness of $\supp{f}$, there exists $N \in \natp$ such that $\supp{f} \subset \bigcup_{n = 1}^N U_n$. By \ref{proposition:lch-partition-of-unity}, there exists a partition of unity $\seqf{\phi_j} \subset C_c(X; [0, 1])$ on $K$ subordinate to $\seqf[N]{U_n}$. In which case,
Let $f \prec U$, then by compactness of $\supp{f}$, there exists $N \in \natp$ such that $\supp{f} \subset \bigcup_{n = 1}^N U_n$. By \autoref{proposition:lch-partition-of-unity}, there exists a partition of unity $\seqf{\phi_j} \subset C_c(X; [0, 1])$ on $K$ subordinate to $\seqf[N]{U_n}$. In which case,
\[
\dpb{f, I}{C_c(X; \real)} = \sum_{n = 1}^N \dpb{\phi_n f, I}{C_c(X; \real)} \le \sum_{n = 1}^N \mu_0(U_n) \le \sum_{n \in \natp}\mu_0(U_n)
\]
Since this holds for all $f \prec U$,
\[
\mu^*(E) \le \mu_0(U) \le \sum_{n \in \natp}\mu_0(U_n)
\]
Let $\eps > 0$, then there exists $\seq{U_n} \subset \topo$ such that $U_n \supset E_n$ and $\mu_0(U_n) \le \mu^*(E_n) + \eps/2^n$ for each $n \in \natp$. In which case,
\[
\mu^*(E) \le \sum_{n \in \natp}\mu_0(U_n) \le \eps + \sum_{n \in \natp}\mu^*(E_n)
\]
As this holds for all $\eps > 0$, $\mu^*(E) \le \sum_{n \in \natp}\mu^*(E_n)$.
\end{enumerate}
Therefore $\mu^*: 2^E \to [0, \infty]$ is an outer measure.
@@ -72,50 +78,61 @@
\[
\dpb{f, I}{C_c(X; \real)} + \dpb{g, I}{C_c(X; \real)} \le \mu_0(E)
\]
Since this holds for all $g \prec E \setminus \supp{f}$,
\[
\dpb{f, I}{C_c(X; \real)} + \mu_0(E \setminus \supp{f}) \le \mu_0(E)
\]
As $E \setminus \supp{f} \supset E \setminus U$,
\[
\dpb{f, I}{C_c(X; \real)} + \mu_0(E \setminus U) \le \mu_0(E)
\]
Finally, the above holds for all $f \prec E \cap U$,
\[
\mu_0(E \cap U) + \mu_0(E \setminus U) \le \mu_0(E)
\]
Now suppose that $E$ is arbitrary. Let $V \in \cn^o(E)$, then
\[
\mu^*(E \cap U) + \mu^*(E \setminus U) \le \mu_0(V \cap U) + \mu_0(V \setminus U) \le \mu_0(V)
\]
As this holds for all such $V$,
\[
\mu^*(E) = \mu^*(E \cap U) + \mu^*(E \setminus U)
\]
By Carathéodory's Extension Theorem (\ref{theorem:caratheodory-extension}), there exists a Borel measure $\mu: \cb_X \to [0, \infty]$ such that for all $U \in \topo$, $\mu(U) = \mu_0(U)$.
(2): Let $K \subset X$ be compact and $U \in \cn^o(K)$. By Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $f \prec U$ with $f \ge \one_K$. In which case,
By \hyperref[Carathéodory's Extension Theorem]{theorem:caratheodory-extension}, there exists a Borel measure $\mu: \cb_X \to [0, \infty]$ such that for all $U \in \topo$, $\mu(U) = \mu_0(U)$.
(2): Let $K \subset X$ be compact and $U \in \cn^o(K)$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f \prec U$ with $f \ge \one_K$. In which case,
\[
\inf_{\substack{f \in C_c(X; \real) \\ f \ge \one_K}}\dpb{f, I}{C(X; \real)} \le \inf_{U \in \cn^o(K)}\mu(U) = \mu(K)
\]
On the other hand, let $f \in C_c(X; [0, 1])$ with $f \ge \one_K$. For any $r \in (0, 1)$, let $g \prec \bracs{f > r}$, then $r^{-1}f \ge g$ and
\[
\dpb{g, I}{C_c(X; \real)} \le r^{-1}\dpb{f, I}{C_c(X; \real)}
\]
As this holds for all $g \prec \bracs{f > r}$,
\[
\mu(K) \le \mu(\bracs{f > r}) \le r^{-1}\dpb{f, I}{C_c(X; \real)}
\]
Since the above holds for all $r \in (0, 1)$,
\[
\mu(K) \le \inf_{\substack{f \in C_c(X; \real) \\ f \ge \one_K}}\dpb{f, I}{C_c(X; \real)}
\]
(3): Let $f \in C_c(X; \real)$. Using linearity, assume without loss of generality that $f \in C_c(X; [0, 1])$. Let $N \in \natp$. For each $1 \le n \le N$, let
\[
f_n = \min\paren{\max\paren{f - \frac{n - 1}{N}, 0}, \frac{1}{N}}
\]
For any $x \in X$, there exists $1 \le n \le N$ such that $f(x) \in [(n-1)/N, n/N]$. In which case,
\begin{itemize}
\item For each $1 \le j < n$, $f_j(x) = 1/N$.
@@ -128,38 +145,45 @@
\[
\frac{1}{N}\mu\paren{K_n} \le \int f_n d\mu \le \frac{1}{N}\mu\paren{K_{n-1}}
\]
Since $\supp{f_n} \subset \bracs{f \ge (n-1)/N}$, for any $U \supset \bracs{f_n \ge (n - 1)/N}$, $f_n \prec U$, so
\[
\dpb{f_n, I}{C_c(X; \real)} \le \frac{1}{N}\mu\paren{K_{n-1}}
\]
On the other hand, $f_n \ge \frac{1}{N}\one_{K_n}$,
\[
\mu(K_n) \le \dpb{f_n, I}{C_c(X; \real)}
\]
so
\[
\frac{1}{N}\sum_{n = 1}^N \mu(K_n) \le \dpb{f, I}{C_c(X; \real)}, \int f d\mu \le \frac{1}{N}\sum_{n = 1}^N \mu(K_{n-1})
\]
Therefore
\[
\abs{\int f d\mu - \dpb{f, I}{C_c(X; \real)}} \le \frac{\mu(K_{0}) - \mu(K_N)}{N} \le \frac{\mu(\supp{f})}{N}
\]
As this holds for all $N \in \natp$, $\int f d\mu = \dpb{f, I}{C_c(X; \real)}$.
(4):
\begin{enumerate}
\item[(R1)] For any $K \subset X$ compact, by Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $f \in C_c(X; [0, 1])$ with $f \ge \one_K$. In which case, $\mu(K) \le \int f d\mu = \dpb{f, I}{C_c(X; \real)}$.
\item[(R1)] For any $K \subset X$ compact, by \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f \in C_c(X; [0, 1])$ with $f \ge \one_K$. In which case, $\mu(K) \le \int f d\mu = \dpb{f, I}{C_c(X; \real)}$.
\item[(R2)] By definition of $\mu^*$, $\mu$ is outer regular.
\item[(R3')] For any $U \in \topo$,
\[
\mu(U) = \sup_{f \prec U}\dpb{f, I}{C_c(X; \real)} \le \sup_{f \prec U}\mu(\supp{f}) \le \sup_{\substack{K \subset U \\ \text{compact}}}\mu(K) \le \mu(U)
\]
\end{enumerate}
(U): By \ref{proposition:radon-measure-cc}, $\nu$ also satisfies (2). Thus for any $E \in \cb_X$, by (R2) and (R3'),
(U): By \autoref{proposition:radon-measure-cc}, $\nu$ also satisfies (2). Thus for any $E \in \cb_X$, by (R2) and (R3'),
\[
\nu(E) = \inf_{U \in \cn^o(E)}\sup_{\substack{K \subset U \\ \text{compact}}}\inf_{\substack{f \in C_c(X; [0, 1]) \\ f \ge \one_K}}\dpb{f, I}{C_c(X; \real)}
\]
so $\nu$ is uniquely determined by $I$.
\end{proof}

7
src/measure/measure/semifinite.tex
View File

@@ -10,6 +10,7 @@
\[
\mu(E) = \sup\bracs{\mu(F)| F \in \cm, F \subset E, \mu(F) < \infty}
\]
\end{enumerate}
If the above holds, then $\mu$ is a \textbf{semifinite measure}.
\end{definition}
@@ -18,6 +19,7 @@
\[
M = \sup\bracs{\mu(F)| F \in \cm, F \subset E, \mu(F) < \infty} < \infty
\]
Let $\seq{F_n} \subset \cm$ such that $F \subset E$, $\mu(F) < \infty$, and $\mu(F_n) \upto M$. Let $F = \bigcup_{n \in \nat}F_n$, then $\mu(F) = M$, and $\mu(E \setminus F) = \infty$. By (1), there exists $G \in \cm$ with $G \subset E \setminus F$ and $0 < \mu(G) < \infty$. Thus $F \cup G \subset E$ with $M < \mu(F) + \mu(G) = \mu(F \cup G) < \infty$. This contradicts the fact that $M$ is the supremum.
\end{proof}
@@ -27,6 +29,7 @@
\[
\mu_0: \cm \to [0, \infty] \quad E \mapsto \sup\bracs{\mu(F)| F \in \cm, F \subset E, \mu(F) < \infty}
\]
then $\mu_0$ is a semifinite measure, and the \textbf{semifinite part} of $\mu$.
\end{definition}
\begin{proof}
@@ -36,18 +39,22 @@
\[
\mu(F) = \sum_{n \in \natp}\mu(F \cap E_n) \le \sum_{n \in \natp}\mu_0(E_n)
\]
Thus $\mu_0\paren{\bigcup_{n \in \natp}E_n} \le \sum_{n \in \natp}\mu_0(E_n)$.
On the other hand, let $n \in \natp$ and $\seqf{F_k} \subset \cm$ such that $F_k \subset E_k$ for each $1 \le k \le n$ and $\mu(F_k) < \infty$, then $F = \bigsqcup_{k = 1}^n F_k \subset \bigsqcup_{k \in \natp}E_k$. Thus
\[
\sum_{k = 1}^n \mu(F_k) = \mu(F) \le \mu_0\paren{\bigsqcup_{k \in \natp}E_k}
\]
As this holds for all such $\seqf{F_k}$,
\[
\sum_{k = 1}^n \mu_0(E_k) \le \mu_0\paren{\bigsqcup_{k \in \natp}E_k}
\]
Since this holds for all $n \in \natp$,
\[
\sum_{n \in \natp}\mu_0(E_n) \le \mu_0\paren{\bigsqcup_{k \in \natp}E_k}
\]
\end{proof}

5
src/measure/sets/algebra.tex
View File

@@ -85,6 +85,7 @@
\[
\cb_{\ol{\real}} = \bracsn{E \subset \ol \real| E \cap \real \in \cb_\real}
\]
is the \textbf{Borel $\sigma$-algebra} on $\ol{\real}$.
\end{definition}
@@ -126,6 +127,7 @@
\[
U = \bigcup_{q \in U \cap \rational}(q - r_q, q + r_q)
\]
so (8) generates all open sets in $\real$. Conversely, every element of (8) is open, so the $\sigma$-algebra generated by (8) is the Borel $\sigma$-algebra on $\real$.
\end{proof}
@@ -152,7 +154,8 @@
\[
\bracs{\infty} = \bigcap_{n \in \nat}[n, \infty] \quad \bracs{-\infty} = \bigcap_{n \in \nat}[-\infty, n]
\]
are elements of $\cm$. For any $a \in \real$, $(a, \infty) = (a, \infty] \setminus \bracs{\infty}$, so $\cm \supset \cb_\real$ by \ref{proposition:borel-sigma-real-generators}.
are elements of $\cm$. For any $a \in \real$, $(a, \infty) = (a, \infty] \setminus \bracs{\infty}$, so $\cm \supset \cb_\real$ by \autoref{proposition:borel-sigma-real-generators}.
In addition, for any $E \in \cb_{\ol{\real}}$, $E = (E \cap \real) \cup (E \setminus \real)$, where $E \cap \real \in \cb_\real$. Since $\bracs{\infty}, \bracs{-\infty} \in \cm$ and $\cb_\real \subset \cm$, $E \in \cm$ and $\cm = \cb_{\ol \real}$.
\end{proof}

5
src/measure/sets/elementary.tex
View File

@@ -21,6 +21,7 @@
\[
\alg = \bracs{\bigsqcup_{i = 1}^n E_j \bigg | \seqf{E_j} \subset \text{ pairwise disjoint}}
\]
then is a ring. If $X \in \ce$, then $\ce$ is an algebra.
\end{proposition}
\begin{proof}
@@ -28,6 +29,7 @@
\[
A \sqcup B = \bigsqcup_{j = 1}^n A_j \sqcup \bigsqcup_{j = 1}^mB_j \in \alg
\]
so $\alg$ is closed under disjoint unions.
(A1): By (P1) $\emptyset \in \ce$. By (E1), there exists $\seqf{E_j} \subset \ce$ such that $X = \emptyset^c = \bigsqcup_{j = 1}^nE_j \in \alg$.
@@ -36,14 +38,17 @@
\[
B \setminus A = \bigcap_{i = 1}^n B \setminus A_i = \bigcap_{i = 1}^n \bigsqcup_{j = 1}^m E_{i, j} = \bigsqcup_{\alpha \in [1, m]^n} \underbrace{\bigcap_{i = 1}^n E_{i, \alpha_i}}_{\in \ce} \in \ce
\]
Thus if $B \in \alg$ with $B = \bigsqcup_{j = 1}^n B_j$, then
\[
B \setminus A = \bigsqcup_{j = 1}^n B_j \setminus A \in \alg
\]
(A3): Let $A = \bigsqcup_{j = 1}^n A_j, B = \bigsqcup_{j = 1}^m B_j \in \alg$, then
\[
A \cap B = \braks{\bigsqcup_{j = 1}^n A_j} \cap \braks{\bigsqcup_{j = 1}^m B_j} = \bigsqcup_{i = 1}^n \bigsqcup_{j = 1}^m \underbrace{A_i \cap B_j}_{\in \ce} \in \alg
\]
so $\alg$ is closed under intersections. Thus using (A2), $A \cup B = A \setminus B \sqcup B \setminus A \sqcup A \cap B$.
\end{proof}

6
src/measure/sets/index.tex
View File

@@ -1,6 +1,6 @@
\chapter{Set Systems}
\label{chap:set-system}
\input{./src/measure/sets/algebra.tex}
\input{./src/measure/sets/lambda.tex}
\input{./src/measure/sets/elementary.tex}
\input{./algebra.tex}
\input{./lambda.tex}
\input{./elementary.tex}

7
src/measure/sets/lambda.tex
View File

@@ -34,7 +34,7 @@
\end{enumerate}
\end{lemma}
\begin{proof}
$(2) \Rightarrow (1)$: Let $A, B \in \alg$, then $A \cup B = (A^c \cap B^c)^c \in \alg$. Thus $\alg$ is an algebra. By \ref{lemma:sigma-algebra-condition}, $\alg$ is a $\sigma$-algebra.
$(2) \Rightarrow (1)$: Let $A, B \in \alg$, then $A \cup B = (A^c \cap B^c)^c \in \alg$. Thus $\alg$ is an algebra. By \autoref{lemma:sigma-algebra-condition}, $\alg$ is a $\sigma$-algebra.
\end{proof}
@@ -47,18 +47,21 @@
\[
\cm(\ce) = \bracs{E \in \lambda(\mathcal{P})| E \cap F \in \lambda(\mathcal{P}) \forall F \in \ce}
\]
then
\begin{enumerate}
\item[(L2)] Let $E, F \in \cm$ with $E \subset F$, then for any $G \in \ce$,
\[
(F \setminus E) \cap G = (F \cap G) \setminus (E \cap G) \in \lambda(\mathcal{P})
\]
\item[(L3)] Let $\seq{E_n} \in \cm$ with $E_n \subset E_{n+1}$ for all $n \in \natp$, and $F \in \ce$, then
\[
\braks{\bigcup_{n \in \natp}E_n} \cap F = \bigcup_{n \in \natp}E_n \cap F \in \lambda(\mathcal{P})
\]
\end{enumerate}
so $\cm(\ce)$ is a $\lambda$-system.
Since $\mathcal{P}$ is a $\pi$-system, $\cm(\mathcal{P}) \supset \mathcal{P}$, so $\cm(\mathcal{P}) = \lambda(\mathcal{P})$. Thus for any $E \in \lambda(\mathcal{P})$ and $F \in \lambda(\mathcal{P})$, $E \cap F \in \lambda(\mathcal{P})$. Therefore $\cm(\lambda(\mathcal{P})) \supset \mathcal{P}$, $\cm(\lambda(\mathcal{P})) = \lambda(\mathcal{P})$, and $\lambda(\mathcal{P})$ satisfies (P2). By \ref{lemma:pi-lambda}, $\lambda(\mathcal{P})$ is a $\sigma$-algebra.
Since $\mathcal{P}$ is a $\pi$-system, $\cm(\mathcal{P}) \supset \mathcal{P}$, so $\cm(\mathcal{P}) = \lambda(\mathcal{P})$. Thus for any $E \in \lambda(\mathcal{P})$ and $F \in \lambda(\mathcal{P})$, $E \cap F \in \lambda(\mathcal{P})$. Therefore $\cm(\lambda(\mathcal{P})) \supset \mathcal{P}$, $\cm(\lambda(\mathcal{P})) = \lambda(\mathcal{P})$, and $\lambda(\mathcal{P})$ satisfies (P2). By \autoref{lemma:pi-lambda}, $\lambda(\mathcal{P})$ is a $\sigma$-algebra.
\end{proof}