From 37a5ce14bffd52ced820c2e50536aa0f54044e7b Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Tue, 17 Mar 2026 15:16:13 -0400 Subject: [PATCH] Added the Bochner integral. --- src/dg/derivative/taylor.tex | 2 +- src/fa/lc/continuous.tex | 1 + src/fa/lp/definition.tex | 28 +++-- src/fa/norm/index.tex | 1 + src/fa/norm/linear.tex | 22 ++++ src/fa/norm/normed.tex | 14 +++ src/fa/norm/separable.tex | 48 +++++++++ src/fa/tvs/continuous.tex | 27 +++++ src/fa/tvs/spaces-of-linear.tex | 24 ++++- src/measure/bochner-integral/bochner.tex | 60 +++++++++++ src/measure/bochner-integral/index.tex | 1 + src/measure/bochner-integral/strongly.tex | 28 ++++- src/measure/lebesgue-integral/simple.tex | 2 +- src/measure/measurable-maps/metric.tex | 2 +- src/measure/measure/kolmogorov.tex | 2 +- src/measure/sets/algebra.tex | 90 ++-------------- src/measure/sets/borel.tex | 125 ++++++++++++++++++++++ src/measure/sets/index.tex | 1 + 18 files changed, 379 insertions(+), 99 deletions(-) create mode 100644 src/fa/norm/separable.tex create mode 100644 src/measure/bochner-integral/bochner.tex create mode 100644 src/measure/sets/borel.tex diff --git a/src/dg/derivative/taylor.tex b/src/dg/derivative/taylor.tex index 9e14199..4f4ad71 100644 --- a/src/dg/derivative/taylor.tex +++ b/src/dg/derivative/taylor.tex @@ -8,7 +8,7 @@ f(b) - f(a) - \sum_{k = 1}^n \frac{1}{k!}D^kf(a)(b - a)^k \] - is contained in the closed convex hull\footnote{It may be possible to sharpen the below claim to include the $1/(n+1)!$ factor. However, I was not able to follow the proof for this.} of + is contained in the closed convex hull of \[ \bracs{D^{n+1}f(s)(t - a)^{n+1} | s \in (a, b) \setminus N, t \in [a, b]} \] diff --git a/src/fa/lc/continuous.tex b/src/fa/lc/continuous.tex index e258547..ebedf43 100644 --- a/src/fa/lc/continuous.tex +++ b/src/fa/lc/continuous.tex @@ -35,3 +35,4 @@ \begin{proof} $(1) \Rightarrow (2)$: By continuity of $T$, there exists continuous seminorms $\seqf{[\cdot]_j}$ on $\seqf{E_j}$ such that for any $x \in \prod_{j = 1}^n E_j$, $\max_{1 \le j \le n}[x_j]_{E_j} < 1$ implies that $[Tx]_F < 1$. In which case, the inequality follows from linearity. \end{proof} + diff --git a/src/fa/lp/definition.tex b/src/fa/lp/definition.tex index 8ec5d19..4678824 100644 --- a/src/fa/lp/definition.tex +++ b/src/fa/lp/definition.tex @@ -100,19 +100,31 @@ By \hyperref[Minkowski's Inequality]{theorem:minkowski}. \end{proof} + +\begin{proposition} +\label{proposition:dct-lp} + Let $(X, \cm, \mu)$ be a measure space, $E$ be a Banach space over $K \in \RC$, $p \in [1, \infty)$, $\seq{f_n} \subset L^p(X; E)$, and $f \in L^p(X; E)$. If + \begin{enumerate} + \item[(a)] $f_n \to f$ strongly pointwise. + \item[(b)] There exists $g \in L^p(X) \cap L^+(X)$ such that $\norm{f_n}_E \le g$ for all $n \in \natp$. + \end{enumerate} + + then $f_n \to f$ in $L^p(X; E)$. +\end{proposition} +\begin{proof} + By assumptions $a$ and $b$, $\norm{f_n - f}_E \le 2g$ for all $n \in \natp$. Since $\seq{f_n} \subset L^p(X; E)$, $f \in L^p(X; E)$, and $g \in L^p(X)$, $\norm{f_n - f}_E^p, g^p \in L^1(X)$ for all $n \in \natp$. By the \hyperref[Dominated Convergence Theorem]{theorem:dct}, + \[ + \limv{n}\norm{f_n - f}_{L^p(X; E)}^p = \limv{n}\int \norm{f_n - f}_E d\mu = 0 + \] +\end{proof} + + \begin{proposition} \label{proposition:lp-simple-dense} Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $p \in [1, \infty)$, then $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$. \end{proposition} \begin{proof} - Let $f \in L^p(X; E)$. By \autoref{definition:strongly-measurable}, there exists $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that $\norm{f_n}_E \le \norm{f}_E$ and $\norm{f_n - f}_E \to 0$ pointwise as $n \to \infty$. - - For each $n \in \nat$, $\norm{f_n}_E \le \norm{f}_E$, so $\norm{f_n}_{L^p(X; E)} \le \norm{f}_{L^p(X; E)} < \infty$, and $\norm{f_n - f}_E \le 2\norm{f}_E$. By the \hyperref[Dominated Convergence Theorem]{theorem:dct}, - \[ - \limv{n}\int \norm{f_n - f}_E^p d\mu = \int \limv{n}\norm{f_n - f}_E^p d\mu = 0 - \] - - Therefore $\norm{f_n - f}_{L^p(X; E)} \to 0$ as $n \to \infty$. + Let $f \in L^p(X; E)$. By \autoref{definition:strongly-measurable}, there exists $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that $\norm{f_n}_E \le \norm{f}_E$ and $\norm{f_n - f}_E \to 0$ strongly pointwise as $n \to \infty$. By the \hyperref[Dominated Convergence Theorem]{proposition:dct-lp}, $f_n \to f$ in $L^p(X; E)$. \end{proof} \begin{theorem}[Markov's Inequality] diff --git a/src/fa/norm/index.tex b/src/fa/norm/index.tex index e4904f8..0480ae7 100644 --- a/src/fa/norm/index.tex +++ b/src/fa/norm/index.tex @@ -4,4 +4,5 @@ \input{./normed.tex} \input{./absolute.tex} \input{./linear.tex} +\input{./separable.tex} \input{./multilinear.tex} diff --git a/src/fa/norm/linear.tex b/src/fa/norm/linear.tex index ea721c7..420affc 100644 --- a/src/fa/norm/linear.tex +++ b/src/fa/norm/linear.tex @@ -12,3 +12,25 @@ \begin{proof} By \autoref{proposition:lc-spaces-linear-map}. \end{proof} + + +\begin{theorem}[Linear Extension Theorem (Normed)] +\label{theorem:linear-extension-theorem-normed} + Let $E$ be a normed vector space over $K \in \RC$, $F$ be a Banach space over $K$, $D \subset E$ be a dense subspace, and $T \in L(D; F)$, then: + \begin{enumerate} + \item There exists an extension $\ol T \in L(E; F)$ such that $\ol T|_D = T$. + \item $\normn{\ol T}_{L(E; F)} = \normn{T}_{L(D; F)}$. + \item[(U)] For any $S \in C(E; F)$ satisfying (1), $S = \ol T$. + \end{enumerate} + +\end{theorem} +\begin{proof} + (1), (U): By \autoref{theorem:linear-extension-theorem-tvs}. + + (2): Since $\ol{T}$ is continuous, the function + \[ + N: E \to \real \quad x \mapsto \normn{\ol T}_{F} - \norm{T}_{L(D; F)}\cdot \norm{x}_E + \] + + is continuous, so $\bracs{N \le 0} \supset D$ is closed. By density of $D$, $\bracs{N \le 0} = E$. Therefore $\normn{\ol T}_{L(E; F)} = \normn{T}_{L(D; F)}$. +\end{proof} diff --git a/src/fa/norm/normed.tex b/src/fa/norm/normed.tex index a598ccf..b8231cd 100644 --- a/src/fa/norm/normed.tex +++ b/src/fa/norm/normed.tex @@ -86,3 +86,17 @@ so $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le 2n/r$. \end{proof} + + +\begin{proposition} +\label{proposition:dual-norm} + Let $E$ be a normed space, then for any $x \in E$, + \[ + \norm{x}_E = \sup_{\substack{\phi \in E^* \\ \norm{\phi}_{E^*} = 1}}\dpn{x, \phi}{E} + \] +\end{proposition} +\begin{proof} + For any $\phi \in E^*$ with $\norm{\phi}_{E^*} = 1$, $\dpn{x, \phi}{E} \le \norm{x}_E \cdot \norm{\phi}_{E^*} = \norm{x}_E$. On the other hand, by the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, there exists $x \in E^*$ such that $\dpn{x, \phi}{E} = \norm{x}_E$ and $\norm{\phi}_{E^*} \le 1$. +\end{proof} + + diff --git a/src/fa/norm/separable.tex b/src/fa/norm/separable.tex new file mode 100644 index 0000000..0e44b4e --- /dev/null +++ b/src/fa/norm/separable.tex @@ -0,0 +1,48 @@ +\section{Separable Normed Spaces} +\label{section:separable-banach-space} + +\begin{proposition} +\label{proposition:separable-dual} + Let $E$ be a separable normed space, then $E^*$ is separable with respect to the weak*-topology. +\end{proposition} +\begin{proof} + Let $\seq{x_n} \subset E$ be a dense subset and $S = \bracsn{\phi \in E^*| \norm{\phi}_{E^*} \le 1}$. For each $N \in \natp$, let + \[ + T_N: S \to \real^N \quad \phi \mapsto (\dpn{x_1, \phi}{E}, \cdots, \dpn{x_N, \phi}{E}) + \] + + then since $\real^N$ is separable, there exists $\bracs{\phi_{N, k}}_{k = 1}^\infty \subset S$ such that $\bracs{T_N\phi_{N, k}}_{k = 1}^\infty$ is dense in $T_N(S)$. + + Let $\phi \in S$, then for each $N \in \natp$, there exists $k_N \in \natp$ such that for each $1 \le n \le N$, + \[ + |\dpn{x_n, \phi_{N, k_N}}{E} - \dpn{x_n, \phi}{E}| \le \frac{1}{n} + \] + + Thus for each $N \in \natp$, $\dpn{x_n, \phi_{N, k_N}}{E} \to \dpn{x_n, \phi}{E}$ as $N \to \infty$. Since $\phi_{N, k_N} \to \phi$ pointwise on a dense subset of $E$, and $\bracsn{\phi_{N, k_N}|N \in \natp} \subset S$ is uniformly equicontinuous, $\phi_{N, k_N} \to \phi$ in the weak*-topology by \autoref{proposition:strong-operator-dense}. +\end{proof} + +\begin{proposition} +\label{proposition:separable-banach-borel-sigma-algebra} + Let $E$ be a separable normed space, then the Borel $\sigma$-algebra on $E$ is generated by the following families of sets: + \begin{enumerate} + \item Open sets in $E$ with respect to the strong topology. + \item $\bracs{B(x, r)|x \in E, r > 0}$. + \item $\bracsn{\ol{B(x, r)}|x \in E, r > 0}$. + \item Open sets in $E$ with respect to the weak topology. + \end{enumerate} + +\end{proposition} +\begin{proof} + (1) $\Leftrightarrow$ (2) $\Leftrightarrow$ (3): By \autoref{proposition:separable-metric-borel-sigma-algebra}. + + (4) $\subset$ (1): Every weakly open set is strongly open. + + (2) $\subset$ (4): By \autoref{proposition:seminorm-lsc}, $\norm{\cdot}_E: E \to [0, \infty)$ is Borel measurable with respect to the weak topology. For any $x \in E$, let + \[ + \phi_x: E \to [0, \infty) \quad y \mapsto \norm{x - y}_E + \] + + then $\phi_x$ is Borel measurable with respect to the weak topology, so $B(x, r) = \bracs{\phi_x < r}$ is a Borel set with respect to the weak topology. +\end{proof} + + diff --git a/src/fa/tvs/continuous.tex b/src/fa/tvs/continuous.tex index 826b5c1..6e83183 100644 --- a/src/fa/tvs/continuous.tex +++ b/src/fa/tvs/continuous.tex @@ -51,3 +51,30 @@ \end{enumerate} The uniformity $\fU$ and its induced topology are the \textbf{product uniformity/topology}, and $E$ equipped with $\fU$ is the \textbf{product TVS} of $\seqi{E}$. \end{definition} + +\begin{theorem}[Linear Extension Theorem (TVS)] +\label{theorem:linear-extension-theorem-tvs} + Let $E$ be a TVS over $K \in \RC$, $F$ be a complete Hausdorff TVS over $K$, $D \subset E$ be a dense subspace, and $T \in L(D; F)$, then: + \begin{enumerate} + \item There exists an extension $\ol T \in L(E; F)$ such that $\ol T|_D = T$. + \item[(U)] For any $S \in C(E; F)$ satisfying (1), $S = \ol T$. + \end{enumerate} + +\end{theorem} +\begin{proof} + By (3) of \autoref{definition:continuous-linear}, $T \in UC(D; F)$. By \autoref{theorem:uniform-continuous-extension}, there exists a unique $\ol T \in C(E; F)$ such that $\ol T|_D = T$. + + It remains to show that $\ol T$ is linear. Since $\ol T$ is continuous, the maps + \[ + A: E \times E \to E \quad (x, y) \mapsto \ol Tx + \ol Ty - \ol T(x + y) + \] + + and + \[ + M: K \times E \to E \quad (\lambda x) \mapsto \lambda \ol Tx - Tx + \] + + are continuous. Thus $\bracs{A = 0} \supset D \times D$ and $\bracs{M = 0} \supset K \times D$ are both closed. By density of $D$, $\bracs{A = 0} = E \times E$ and $\bracs{M = 0} = K \times E$. Therefore $T$ is linear. +\end{proof} + + diff --git a/src/fa/tvs/spaces-of-linear.tex b/src/fa/tvs/spaces-of-linear.tex index 027547f..da2e76c 100644 --- a/src/fa/tvs/spaces-of-linear.tex +++ b/src/fa/tvs/spaces-of-linear.tex @@ -153,14 +153,34 @@ \begin{definition}[Strong Operator Topology] \label{definition:strong-operator-topology} - Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $L(E; F)$ is the \textbf{strong operator topology}. + Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $F^E$ is the \textbf{strong operator topology}. The space $L_s(E; F)$ denotes $L(E; F)$ equipped with the strong operator topology. \end{definition} + +\begin{proposition} +\label{proposition:strong-operator-dense} + Let $E, F$ be TVSs over $K \in \RC$ and $\net{T} \subset L(E; F)$ and $T \in L_s(E; F)$. If + \begin{enumerate} + \item[(a)] There exists a dense subset $S \subset E$ such that $T_\alpha x \to Tx$ strongly for all $x \in S$. + \item[(b)] $\bracs{T_\alpha|\alpha \in A}$ is uniformly equicontinuous. + \end{enumerate} + + then $T_\alpha \to T$ in $L_s(E; F)$. +\end{proposition} +\begin{proof} + Let $x \in E$, $U \in \cn_F(Tx)$, and $V \in \cn_F(Tx)$ be balanced such that $V + V + V \subset U$. By (b), there exists a balanced neighbourhood $W \in \cn_E(0)$ such that $T(W) \cup \bigcup_{\alpha \in A}T_\alpha(W) \subset V$. By (a), there exists $y \in S \cap (x + W)$ and $\alpha_0 \in A$ such that for all $\alpha \ge \alpha_0$, $T_\alpha y - Ty \in V$. In which case, for any $\alpha \ge \alpha_0$, + \[ + T_\alpha x - Tx = \underbrace{T_\alpha x - T_\alpha y}_{\in V} + \underbrace{T_\alpha y - Ty}_{\in V} + \underbrace{Ty - Tx}_{\in V} \in U + \] +\end{proof} + + + \begin{definition}[Weak Operator Topology] \label{definition:weak-operator-topology} - Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $L(E; F_w)$ is the \textbf{weak operator topology}. + Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $F_w^E$ is the \textbf{weak operator topology}. The space $L_w(E; F) = L_s(E; F_w)$ denotes $L(E; F)$ equipped with the weak operator topology. \end{definition} diff --git a/src/measure/bochner-integral/bochner.tex b/src/measure/bochner-integral/bochner.tex new file mode 100644 index 0000000..78791af --- /dev/null +++ b/src/measure/bochner-integral/bochner.tex @@ -0,0 +1,60 @@ +\section{The Bochner Integral} +\label{section:bochner-integral} + +\begin{definition}[Bochner Integral] +\label{definition:bochner-integral} + Let $(X, \cm, \mu)$ be a measure space and $E$ be a Banach space over $K \in \RC$, then there exists a unique $I \in L(L^1(X; E); E)$ such that: + \begin{enumerate} + \item For any $x \in E$ and $A \in \cm$ with $\mu(A) < \infty$, $I(x \cdot \one_A) = x \cdot \mu(A)$. + \item For all $f \in L^1(X; E)$, $\norm{If}_E \le \int \norm{f}_E d\mu$. + \end{enumerate} + + For any $f \in L^1(X; E)$, $If = \int f d\mu$ is the \textbf{Bochner integral} of $f$. +\end{definition} +\begin{proof} + (1): For any $\phi \in \Sigma(X; E) \cap L^1(X; E)$, let + \[ + I\phi = \sum_{y \in \phi(X)}^n y \cdot \mu\bracs{\phi = y} + \] + + For any $\lambda \in K$, if $\lambda \ne 0$, then the mapping $x \mapsto \lambda x$ is a bijection, so + \[ + I(\lambda \phi) = \sum_{\lambda y \in \lambda \phi(X)}^n (\lambda y) \cdot \mu\bracs{\lambda \phi = \lambda y} + = \lambda \sum_{y \in \phi(X)}y \cdot \mu\bracs{\phi = y} = \lambda I\phi + \] + + If $\lambda = 0$, then $I(\lambda \phi) = I(0) = 0$. + + Let $\phi, \psi \in \Sigma(X; E) \cap L^1(X; E)$, then + \begin{align*} + I\phi + I\psi &= \sum_{y \in \phi(X)}y \cdot \mu\bracs{\phi = y} + \sum_{z \in \psi(X)}z \cdot \mu\bracs{\psi = z} \\ + &= \sum_{y \in \phi(X)} \sum_{z \in \psi(X)} (y + z) \cdot \mu\bracs{\phi = y, \psi = z} \\ + &= \sum_{y \in (\phi + \psi)(X)}\sum_{{z \in \phi(X) \atop {z' \in \psi(X) \atop z + z' = y}}}(z + z') \cdot \mu\bracsn{\phi = g, \psi = z'} \\ + &= \sum_{y \in (\phi + \psi)(X)}y \cdot \mu(\bracs{\phi + \psi = y}) = I\phi + I\psi + \end{align*} + + so $I$ is a linear operator on $\Sigma(X; E) \cap L^1(X; E)$ that satisfies (1). + + (2): For any $\phi \in \Sigma(X; E) \cap L^1(X; E)$, + \[ + \norm{I\phi}_E \le \sum_{y \in \phi(X)}\norm{y}_E \cdot \mu\bracs{\phi = y} = \int \norm{\phi}_E d\mu = \norm{\phi}_{L^1(X; E)} + \] + + By \autoref{proposition:lp-simple-dense}, $\Sigma(X; E) \cap L^1(X; E)$ is dense in $L^1(X; E)$. Therefore by the \hyperref[Linear Extension Theorem]{theorem:linear-extension-theorem-normed}, $I$ admits a unique norm-preserving extension to $L^1(X; E)$. + +\end{proof} + +\begin{theorem}[Dominated Convergence Theorem] +\label{theorem:dct-bochner} + Let $(X, \cm, \mu)$ be a measure spacs, $E$ be a Banach space over $K \in \RC$, $\seq{f_n} \subset L^1(X; E)$, and $f \in L^1(X; E)$. If + \begin{enumerate} + \item[(a)] $f_n \to f$ strongly pointwise. + \item[(b)] There exists $g \in L^1(X) \cap L^+(X)$ such that $\norm{f_n}_E \le g$ for all $n \in \natp$. + \end{enumerate} + + then $\int f d\mu = \limv{n}\int f_n d\mu$. +\end{theorem} +\begin{proof} + By the classical \hyperref[Dominated Convergence Theorem]{proposition:dct-lp}, $f_n \to f$ in $L^1(X; E)$. Since $h \mapsto \int h d\mu$ is a bounded linear operator, $\int f d\mu = \limv{n}\int f_n d\mu$. +\end{proof} + diff --git a/src/measure/bochner-integral/index.tex b/src/measure/bochner-integral/index.tex index 51ef936..8d90c0c 100644 --- a/src/measure/bochner-integral/index.tex +++ b/src/measure/bochner-integral/index.tex @@ -2,3 +2,4 @@ \label{chap:bochner-integral} \input{./strongly.tex} +\input{./bochner.tex} diff --git a/src/measure/bochner-integral/strongly.tex b/src/measure/bochner-integral/strongly.tex index 2b7ea33..d3a569d 100644 --- a/src/measure/bochner-integral/strongly.tex +++ b/src/measure/bochner-integral/strongly.tex @@ -6,16 +6,22 @@ Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, and $f: X \to E$, then the following are equivalent: \begin{enumerate} \item For each $\phi \in E^*$, $\phi \circ f$ is $(\cm, \cb_K)$-measurable and $f(X) \subset E$ is separable. - \item $f$ is $(\cm, \cb_E)$ measurable and $f(X) \subset E$ is separable. + \item $f$ is $(\cm, \cb_E)$-measurable and $f(X) \subset E$ is separable. \item There exists a sequence $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that \begin{enumerate} \item[(a)] For each $n \in \natp$, $\norm{f_n}_E \le \norm{f}_E$. \item[(b)] $\norm{f_n(x) - f(x)}_E \to 0$ pointwise as $n \to \infty$. \end{enumerate} \end{enumerate} + + If the above holds, then $f$ is a \textbf{strongly measurable} function. \end{definition} \begin{proof} - (1) $\Rightarrow$ (2): TODO + (1) $\Rightarrow$ (2): First suppose that $E$ is separable. By \autoref{proposition:separable-banach-borel-sigma-algebra}, the Borel $\sigma$-algebra on $E$ coincides with the $\sigma$-algebra on $E$ generated by the weak topology. Thus if $\phi \circ f$ is $(\cm, \cb_K)$-measurable for all $\phi \in E^*$, then $f$ is $(\cm, \cb_E)$-measurable. + + Now suppose that $E$ is arbitrary. Let $F \subset E$ be the closure of the linear span of $f(X)$, then $F$ is a separable closed subspace of $E$. For any $\phi \in F^*$, by the \hyperref[Hahn-Banch Theorem]{theorem:hahn-banach}, there exists an extension $\Phi \in E^*$ of $\phi$. In which case, since $f(X) \subset F$, for any Borel set $B \in \cb_{K}$, $\bracs{\phi \circ f \in B} = \bracs{\Phi \circ f \in B} \in \cm$. Thus $\phi \circ f$ is $(\cm, \cb_K)$-measurable for all $\phi \in E^*$. + + By the separable case, $f$ is $(\cm, \cb_{F})$-measurable. Let $B \in \cb_E$, then $B \cap F \in \cb_F$ by \autoref{lemma:borel-induced}. Therefore $\bracs{f \in B} = \bracs{f \in B \cap F} \in \cm$, and $f$ is $(\cm, \cb_E)$-measurable. (2) $\Rightarrow$ (3): By \autoref{proposition:measurable-simple-separable-norm}. @@ -26,3 +32,21 @@ and each $f_n$ is finitely-valued, $f(X)$ is separable. \end{proof} + +\begin{proposition} +\label{proposition:strongly-measurable-properties} + Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, then: + \begin{enumerate} + \item For any strongly measurable functions $f, g: X \to E$ and $\lambda \in K$, $\lambda f + g$ is strongly measurable. + \item For any strongly measurable functions $\bracs{f_n: X \to E|n \in \natp}$ and $f: X \to E$, if $f_n \to f$ strongly pointwise, then $f$ is strongly measurable. + \end{enumerate} + +\end{proposition} +\begin{proof} + (1): Since $x \mapsto \lambda$ is continuous, $\lambda f$ is strongly measurable by (2) of \autoref{definition:strongly-measurable}. + + By (3) of \autoref{definition:strongly-measurable}, there exists $\seq{f_n}, \seq{g_n} \subset \Sigma(X, \cm; E)$ such that $f_n \to f$ and $g_n \to g$ strongly pointwise. In which case, $\seq{f_n + g_n} \subset \Sigma(X, \cm; E)$ and $f_n + g_n \to f + g$ strongly pointwise. Therefore $f + g$ is also strongly measurable. + + (2): By \autoref{proposition:metric-measurable-limit}, $f$ is $(\cm, \cb_E)$-measurable. Since $f(X) \subset \overline{\bigcup_{n \in \natp}}f_n(X)$, $f(X)$ is also separable, so $f$ is strongly measurable by (1) of \autoref{definition:strongly-measurable}. +\end{proof} + diff --git a/src/measure/lebesgue-integral/simple.tex b/src/measure/lebesgue-integral/simple.tex index 66d09be..53c8769 100644 --- a/src/measure/lebesgue-integral/simple.tex +++ b/src/measure/lebesgue-integral/simple.tex @@ -8,7 +8,7 @@ \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 $0 \cdot \infty = 0$.} + 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 (with the convention that $0 \cdot \infty = 0$) \[ \int f d\mu = \int f(x) \mu(dx) = \sum_{y \in f(X)}y \cdot \mu(\bracs{f = y}) \] diff --git a/src/measure/measurable-maps/metric.tex b/src/measure/measurable-maps/metric.tex index b3c735c..776c4c1 100644 --- a/src/measure/measurable-maps/metric.tex +++ b/src/measure/measurable-maps/metric.tex @@ -71,7 +71,7 @@ \begin{proposition} \label{proposition:measurable-simple-separable} - Let $(X, \cm)$ be a measurable space, $Y$ be a separable metric space, and $N: Y \to 2^Y$\footnote{This mapping is typically obtained as slices of the level sets of a continuous function $Y \times Y \to \real$.} such that + Let $(X, \cm)$ be a measurable space, $Y$ be a separable metric space, and $N: Y \to 2^Y$ such that \begin{enumerate} \item[(a)] For each $y \in Y$, $y \in \ol{N(y)^o}$. \item[(b)] $\bigcap_{y \in Y}N(y) \ne \emptyset$. diff --git a/src/measure/measure/kolmogorov.tex b/src/measure/measure/kolmogorov.tex index 0b8fab5..87c5f1a 100644 --- a/src/measure/measure/kolmogorov.tex +++ b/src/measure/measure/kolmogorov.tex @@ -13,7 +13,7 @@ \label{lemma:kolmogorov-compact-sequence} Let $\seq{X_n}$ be topological spaces where for each $n \in \natp$, \begin{enumerate} - \item[(a)] Every finite measure on $\prod_{j = 1}^n X_j$ is regular\footnote{A potential sufficient condition for this is that each $X_n$ is LCH where every open set is $\sigma$-compact. However, I have yet to verify if this condition persists over products.}. + \item[(a)] Every finite measure on $\prod_{j = 1}^n X_j$ is regular. \item[(b)] $X_n$ is Hausdorff. \item[(c)] $X_n$ is separable. \end{enumerate} diff --git a/src/measure/sets/algebra.tex b/src/measure/sets/algebra.tex index 1ac0a65..592a552 100644 --- a/src/measure/sets/algebra.tex +++ b/src/measure/sets/algebra.tex @@ -72,90 +72,14 @@ Let $X$ be a set and $\ce \subset 2^X$, then the $\sigma$-algebra $\sigma(\ce)$ \textbf{generated by} $\ce$ is the smallest $\sigma$-algebra on $X$ containing $\ce$. \end{definition} +\begin{definition}[Induced $\sigma$-Algebra] +\label{definition:} + Let $X$ be a set, $\cm \subset 2^X$ be a $\sigma$-algebra over $X$, and $E \subset X$, then the collection + \[ + \cm_E = \bracs{A \cap E|A \in \cm} + \] -\begin{definition}[Borel $\sigma$-Algebra] -\label{definition:borel-sigma-algebra} - Let $(X, \topo)$ be a topological space, then the \textbf{Borel $\sigma$-algebra} $\cb_X$ on $X$ is the $\sigma$-algebra generated by $\topo$. + is the \textbf{$\sigma$-algebra on $E$ induced by $\cm$}. \end{definition} -\begin{definition}[Borel $\sigma$-Algebra on $\ol{\real}$] -\label{definition:borel-sigma-algebra-extended} - The family - \[ - \cb_{\ol{\real}} = \bracsn{E \subset \ol \real| E \cap \real \in \cb_\real} - \] - - is the \textbf{Borel $\sigma$-algebra} on $\ol{\real}$. -\end{definition} - - -\begin{proposition} -\label{proposition:borel-sigma-real-generators} - The following families of sets generate the Borel $\sigma$-algebra on $\real$: - \begin{enumerate} - \item $\bracs{(-\infty, a]| a \in \real}$. - \item $\bracs{(a, \infty)|a \in \real}$. - \item $\bracs{[a, \infty)| a \in \real}$. - \item $\bracs{(-\infty, a)| a \in \real}$. - \item $\bracs{[a, b)| -\infty < a < b < \infty}$. - \item $\bracs{[a, b]| -\infty < a < b < \infty}$. - \item $\bracs{(a, b]| -\infty < a < b < \infty}$. - \item $\bracs{(a, b)| -\infty < a < b < \infty}$. - \end{enumerate} -\end{proposition} -\begin{proof} - It is sufficient to show that the $\sigma$-algebra generated by any of the above two families coincide, and that the resulting $\sigma$-algebra is the Borel $\sigma$-algebra on $\real$. - - (1) $\to$ (2): For any $a \in \real$, $(a, \infty) = (-\infty, a)^c$. - - (2) $\to$ (3): For any $a \in \real$, $[a, \infty) = \bigcap_{n \in \natp}(a - 1/n, \infty)$. - - (3) $\to$ (4): For any $a \in \real$, $(-\infty, a) = [a, \infty)^c$. - - (4) $\to$ (5): For any $a, b \in \real$, $[a, b) = (-\infty, b) \cap (-\infty, a)^c$. - - (5) $\to$ (6): For any $a, b \in \real$, $[a, b]= \bigcap_{n \in \natp}(a - 1/n, b]$. - - (6) $\to$ (7): For any $a, b \in \real$, $(a, b] = \bigcup_{n \in \natp}[a + 1/n, b]$. - - (7) $\to$ (8): For any $a, b \in \real$, $(a, b) = \bigcup_{n \in \natp}(a, b - 1/n]$. - - (8) $\to$ (1): For any $a \in \real$, $(-\infty, a] = \bigcup_{n \in \natp}\bigcap_{k \in \natp}(-n, a + 1/k]$. - - For any $U \subset X$ open and $q \in U \cap \rational$, there exists $r_q > 0$ such that $(q - r_q, q + r_q) \subset U$. In which case, - \[ - U = \bigcup_{q \in U \cap \rational}(q - r_q, q + r_q) - \] - - 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} - -\begin{proposition} -\label{proposition:borel-sigma-extended-generators} - The following families of sets generate the Borel $\sigma$-algebra on $\ol \real$: - \begin{enumerate} - \item $\bracs{[-\infty, a]| a \in \real}$. - \item $\bracs{(a, \infty]|a \in \real}$. - \item $\bracs{[a, \infty]| a \in \real}$. - \item $\bracs{[-\infty, a)| a \in \real}$. - \end{enumerate} -\end{proposition} -\begin{proof} - (1) $\to$ (2): For any $a \in \real$, $(a, \infty] = [-\infty, a]^c$. - - (2) $\to$ (3): For any $a \in \real$, $[a, \infty] = \bigcap_{n \in \natp}(a - 1/n, \infty]$. - - (3) $\to$ (4): For any $a \in \real$, $[-\infty, a) = [a, \infty]^c$. - - (4) $\to$ (1): For any $a \in \real$, $[-\infty, a] = \bigcap_{n \in \natp}[-\infty, a + 1/n)$. - - By definition, all elements of (1), (2), (3), and (4) belong to $\cb_{\ol{\real}}$. Let $\cm$ be the $\sigma$-algebra generated by (1), (2), (3), and (4), then - \[ - \bracs{\infty} = \bigcap_{n \in \nat}[n, \infty] \quad \bracs{-\infty} = \bigcap_{n \in \nat}[-\infty, n] - \] - - are elements of $\cm$. For any $a \in \real$, $(a, \infty) = (a, \infty] \setminus \bracs{\infty}$, so $\cm \supset \cb_\real$ by \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} diff --git a/src/measure/sets/borel.tex b/src/measure/sets/borel.tex new file mode 100644 index 0000000..bdf5402 --- /dev/null +++ b/src/measure/sets/borel.tex @@ -0,0 +1,125 @@ +\section{The Borel $\sigma$-Algebra} +\label{section:borel-sigma-algebra} + + +\begin{definition}[Borel $\sigma$-Algebra] +\label{definition:borel-sigma-algebra} + Let $(X, \topo)$ be a topological space, then the \textbf{Borel $\sigma$-algebra} $\cb_X$ on $X$ is the $\sigma$-algebra generated by $\topo$. +\end{definition} + + +\begin{definition}[Borel $\sigma$-Algebra on $\ol{\real}$] +\label{definition:borel-sigma-algebra-extended} + The family + \[ + \cb_{\ol{\real}} = \bracsn{E \subset \ol \real| E \cap \real \in \cb_\real} + \] + + is the \textbf{Borel $\sigma$-algebra} on $\ol{\real}$. +\end{definition} + + +\begin{proposition} +\label{proposition:borel-sigma-real-generators} + The following families of sets generate the Borel $\sigma$-algebra on $\real$: + \begin{enumerate} + \item $\bracs{(-\infty, a]| a \in \real}$. + \item $\bracs{(a, \infty)|a \in \real}$. + \item $\bracs{[a, \infty)| a \in \real}$. + \item $\bracs{(-\infty, a)| a \in \real}$. + \item $\bracs{[a, b)| -\infty < a < b < \infty}$. + \item $\bracs{[a, b]| -\infty < a < b < \infty}$. + \item $\bracs{(a, b]| -\infty < a < b < \infty}$. + \item $\bracs{(a, b)| -\infty < a < b < \infty}$. + \end{enumerate} +\end{proposition} +\begin{proof} + It is sufficient to show that the $\sigma$-algebra generated by any of the above two families coincide, and that the resulting $\sigma$-algebra is the Borel $\sigma$-algebra on $\real$. + + (1) $\to$ (2): For any $a \in \real$, $(a, \infty) = (-\infty, a)^c$. + + (2) $\to$ (3): For any $a \in \real$, $[a, \infty) = \bigcap_{n \in \natp}(a - 1/n, \infty)$. + + (3) $\to$ (4): For any $a \in \real$, $(-\infty, a) = [a, \infty)^c$. + + (4) $\to$ (5): For any $a, b \in \real$, $[a, b) = (-\infty, b) \cap (-\infty, a)^c$. + + (5) $\to$ (6): For any $a, b \in \real$, $[a, b]= \bigcap_{n \in \natp}(a - 1/n, b]$. + + (6) $\to$ (7): For any $a, b \in \real$, $(a, b] = \bigcup_{n \in \natp}[a + 1/n, b]$. + + (7) $\to$ (8): For any $a, b \in \real$, $(a, b) = \bigcup_{n \in \natp}(a, b - 1/n]$. + + (8) $\to$ (1): For any $a \in \real$, $(-\infty, a] = \bigcup_{n \in \natp}\bigcap_{k \in \natp}(-n, a + 1/k]$. + + For any $U \subset X$ open and $q \in U \cap \rational$, there exists $r_q > 0$ such that $(q - r_q, q + r_q) \subset U$. In which case, + \[ + U = \bigcup_{q \in U \cap \rational}(q - r_q, q + r_q) + \] + + 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} + +\begin{proposition} +\label{proposition:borel-sigma-extended-generators} + The following families of sets generate the Borel $\sigma$-algebra on $\ol \real$: + \begin{enumerate} + \item $\bracs{[-\infty, a]| a \in \real}$. + \item $\bracs{(a, \infty]|a \in \real}$. + \item $\bracs{[a, \infty]| a \in \real}$. + \item $\bracs{[-\infty, a)| a \in \real}$. + \end{enumerate} +\end{proposition} +\begin{proof} + (1) $\to$ (2): For any $a \in \real$, $(a, \infty] = [-\infty, a]^c$. + + (2) $\to$ (3): For any $a \in \real$, $[a, \infty] = \bigcap_{n \in \natp}(a - 1/n, \infty]$. + + (3) $\to$ (4): For any $a \in \real$, $[-\infty, a) = [a, \infty]^c$. + + (4) $\to$ (1): For any $a \in \real$, $[-\infty, a] = \bigcap_{n \in \natp}[-\infty, a + 1/n)$. + + By definition, all elements of (1), (2), (3), and (4) belong to $\cb_{\ol{\real}}$. Let $\cm$ be the $\sigma$-algebra generated by (1), (2), (3), and (4), then + \[ + \bracs{\infty} = \bigcap_{n \in \nat}[n, \infty] \quad \bracs{-\infty} = \bigcap_{n \in \nat}[-\infty, n] + \] + + are elements of $\cm$. For any $a \in \real$, $(a, \infty) = (a, \infty] \setminus \bracs{\infty}$, so $\cm \supset \cb_\real$ by \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} + + + +\begin{proposition} +\label{proposition:separable-metric-borel-sigma-algebra} + Let $X$ be a separable metric space, then the Borel $\sigma$-algebra on $X$ is generated by the following families of sets: + \begin{enumerate} + \item Open sets of $X$. + \item $\bracs{B(x, r)|x \in X, r > 0}$. + \item $\bracsn{\ol{B(x, r)}|x \in X, r > 0}$. + \end{enumerate} + +\end{proposition} +\begin{proof} + (1) $\subset$ (2): Let $U \subset X$ be open. By \autoref{definition:dense}, there exists a countable dense subset $S \subset U$. For each $x \in S$, let $r_x > 0$ such that $B(x, r) \subset U$, then $U = \bigcup_{x \in S}B(x, r_x)$ is a countable union of open balls. + + (2) $\subset$ (3): For any $x \in X$ and $r > 0$, $B(x, r) = \bigcup_{n \in \natp}\overline{B(x, r - 1/n)}$ is a countable union of closed balls. + + (3) $\subset$ (1): For each $x \in X$ and $r > 0$, $\overline{B(x, r)}$ is closed. +\end{proof} + + +\begin{lemma} +\label{lemma:borel-induced} + Let $X$ be a topological space and $Y \subset X$ be a subspace, then the Borel $\sigma$-algebra on $Y$ coincides with the $\sigma$-algebra on $Y$ induced by $\cb_X$. +\end{lemma} +\begin{proof} + Since $\bracsn{A \in \cb_X|A \cap Y \in \cb_Y}$ is a $\sigma$-algebra that contains all open sets in $X$, $\cb_Y$ contains the induced $\sigma$-algebra. + + On the other hand, the induced $\sigma$-algebra contains all open sets in $Y$, so it contains $\cb_Y$. + + Therefore the two $\sigma$-algebras coincide. +\end{proof} + + diff --git a/src/measure/sets/index.tex b/src/measure/sets/index.tex index 7d2deb9..6cf6021 100644 --- a/src/measure/sets/index.tex +++ b/src/measure/sets/index.tex @@ -2,6 +2,7 @@ \label{chap:set-system} \input{./algebra.tex} +\input{./borel.tex} \input{./lambda.tex} \input{./elementary.tex} \input{./limits.tex}