From 9ce4986002decd6b4a5ad5ef7ed2126a94cb1a86 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Thu, 19 Mar 2026 12:05:58 -0400 Subject: [PATCH] Added Fubini's theorem. --- src/measure/bochner-integral/bochner.tex | 6 +- src/measure/measure/index.tex | 1 + src/measure/measure/product.tex | 129 +++++++++++++++++++++++ src/measure/sets/elementary.tex | 38 ++++++- src/measure/vector/variation.tex | 2 +- src/topology/main/c0.tex | 26 ++++- src/topology/uniform/uc.tex | 12 +++ 7 files changed, 207 insertions(+), 7 deletions(-) create mode 100644 src/measure/measure/product.tex diff --git a/src/measure/bochner-integral/bochner.tex b/src/measure/bochner-integral/bochner.tex index 74a66bf..e5e5d46 100644 --- a/src/measure/bochner-integral/bochner.tex +++ b/src/measure/bochner-integral/bochner.tex @@ -78,7 +78,7 @@ \item For any $f \in L^1(X, |\mu|; E)$, $\normn{I_\lambda f}_{G} \le \norm{\lambda}_{L^2(E, F; G)} \cdot \norm{f}_{L^1(X, |\mu|; E)}$. \end{enumerate} - For any $f \in L^1(X; E)$, $I_\lambda f = \int f d\lambda\mu$ is the \textbf{Bochner integral} of $f$ with respect to $\mu$ and $\lambda$. + For any $f \in L^1(X; E)$, $I_\lambda f = \int \lambda(f, d\mu)$ is the \textbf{Bochner integral} of $f$ with respect to $\mu$ and $\lambda$. \end{definition} \begin{proof} Same as \autoref{definition:bochner-integral}. @@ -98,10 +98,10 @@ \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 $\int f d\lambda\mu = \limv{n}\int f_n d\lambda\mu$. + then $\int \lambda(f , d\mu) = \limv{n}\int \lambda(f_n, d\mu)$. \end{theorem} \begin{proof} - By the classical \hyperref[Dominated Convergence Theorem]{proposition:dct-lp}, $f_n \to f$ in $L^1(X, |\mu|; E)$. Since $h \mapsto \int h d\lambda\mu$ is a bounded linear operator, $\int f d\lambda\mu = \limv{n}\int f_n d\lambda\mu$. + By the classical \hyperref[Dominated Convergence Theorem]{proposition:dct-lp}, $f_n \to f$ in $L^1(X, |\mu|; E)$. Since $h \mapsto \int \lambda(f , d\mu)$ is a bounded linear operator, $\int \lambda(f , d\mu) = \limv{n}\int \lambda(f_n, d\mu)$. \end{proof} diff --git a/src/measure/measure/index.tex b/src/measure/measure/index.tex index 63211f3..744845e 100644 --- a/src/measure/measure/index.tex +++ b/src/measure/measure/index.tex @@ -8,4 +8,5 @@ \input{./regular.tex} \input{./outer.tex} \input{./lebesgue-stieltjes.tex} +\input{./product.tex} \input{./kolmogorov.tex} diff --git a/src/measure/measure/product.tex b/src/measure/measure/product.tex new file mode 100644 index 0000000..97702f0 --- /dev/null +++ b/src/measure/measure/product.tex @@ -0,0 +1,129 @@ +\section{Product Measures} +\label{section:product-measures} + + +\begin{definition}[Product Measure] +\label{definition:product-measure} + Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be measure spaces, then there exists a measure $\mu \otimes \nu: \cm \times \cn \to [0, \infty]$ such that: + \begin{enumerate} + \item For each $E \in \cm$ and $F \in \cn$, $\mu \otimes \nu(E \times F) = \mu(E)\nu(F)$. + \item[(U)] For any measure $\lambda: \cm \otimes \cn \to [0, \infty]$, $\lambda \le \mu$. For any $A \in \cm \otimes \cn$ with $\mu(A) < \infty$, $\lambda(A) = \mu(A)$. In particular, if $\mu$ is $\sigma$-finite, then $\lambda = \mu$. + \end{enumerate} + + The measure $\mu \otimes \nu$ is the \textbf{product} of $\mu$ and $\nu$. +\end{definition} +\begin{proof} + Let + \[ + \mathcal{R} = \bracs{E \times F|E \in \cm, F \in \cn} + \] + + then $\mathcal{R}$ is an elementary family by \autoref{proposition:rectangle-elementary-family}. Let + \[ + \alg = \bracs{\bigsqcup_{i = 1}^n E_j \times F_j \bigg | \seqf{E_j \times F_j} \subset \mathcal{R} \text{ pairwise disjoint}} + \] + + then $\alg$ is a ring over $X \times Y$. For each pairwise disjoint collection $\seqf{E_j \times F_j} \subset \mathcal{R}$, let + \[ + \mu \otimes \nu\paren{\bigsqcup_{j = 1}^n E_j \times F_j} = \sum_{j = 1}^n \mu(E_j) \mu(F_j) + \] + + then $\mu \otimes \nu$ is well-defined and finitely additive on $\alg$. + + Let $A \times B \in \mathcal{A}$ and $\seq{A_n \times B_n} \subset \mathcal{R}$ such that $A \times B = \bigsqcup_{n \in \natp}A_n \times B_n$, then for any $x \in X$ and $y \in Y$, + \[ + \one_{A \times B}(x, y) = \sum_{n \in \natp}\one_{A_n \times B_n}(x, y) = \sum_{n \in \natp}\one_{A_n}(x)\one_{B_n}(y) + \] + + By the \hyperref[Monotone Convergence Theorem]{theorem:mct}, for any $y \in Y$, + \begin{align*} + \mu(A)\one_{B}(y) &= \int_X \one_{A}(x)\one_B(y)\mu(dx) = \sum_{n \in \natp}\int_X \one_{A_n}(x)\one_{B_n}(y) \mu(dx) \\ + &= \sum_{n = 1}^\infty \mu(A_n)\one_{B_n}(y) + \end{align*} + + By the Monotone Convergence Theorem again, + \begin{align*} + \mu(A)\nu(B) &= \int_Y \mu(A)\one_B(y) \nu(dy) = \sum_{n = 1}^\infty \int_Y \mu(A_n)\one_{B_n}\nu(dy) \\ + &= \sum_{n = 1}^\infty \mu(A_n)\nu(B_n) + \end{align*} + + Therefore $\mu \otimes \nu$ is a premeasure on $\alg$. By \hyperref[Carathéodory's Extension Theorem]{theorem:caratheodory-extension}, there exists a measure $\mu \otimes \nu$ satisfying (1) and (U). + +\end{proof} + +\begin{lemma}[{{\cite[Proposition 2.34]{Folland}}}] +\label{lemma:section-measurable} + Let $(X, \cm)$ and $(Y, \cn)$ be measurable spaces, then: + \begin{enumerate} + \item For any $E \in \cm \otimes \cn$, $x \in X$, and $y \in Y$, $\bracs{z \in Y|(x, z) \in E} \in \cm$ and $\bracs{z \in X|(z, y) \in E} \in \cn$. + \item For any measure space $(Z, \cf)$, $(\cm \otimes \cn, \cf)$-measurable function $f: X \times Y \to Z$, $x \in X$, and $y \in Y$, $f(x, \cdot)$ is $(\cn, \cf)$-measurable and $f(\cdot, y)$ is $(\cm, \cf)$-measurable. + \end{enumerate} + +\end{lemma} +\begin{proof} + (1): Let $\alg \subset \cm \otimes \cn$ be the collection of all sets that satisfy (1), then + \[ + \alg \supset \bracs{E \times F| E \in \cm, F \in \cn} + \] + + For any $E \in \alg$, $\bracs{z \in Y|(y, z) \in E}^c = \bracs{z \in Y|(y, z) \in E^c}$, so $E^c \in \alg$ as well. For any $\seq{E_n} \subset \alg$, + \[ + \bracs{z \in Y \bigg| (y, z) \in \bigcup_{n \in \natp}E_n} = \bigcup_{n \in \natp}\bracs{z \in Y | (y, z) \in E_n} + \] + + so $\bigcup_{n \in \natp}E_n \in \alg$. Therefore $\alg$ is a $\sigma$-algebra, and $\alg = \cm \otimes \cn$. + + (2): For any $F \in \cf$, + \[ + \bracs{y \in Y|f(x, \cdot) \in F} = \bracs{y \in Y| (x, y) \in f^{-1}(F)} \in \cn + \] + + by (1). +\end{proof} + +\begin{theorem}[Fubini-Tonelli Theorem] +\label{theorem:fubini-tonelli} + Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be $\sigma$-finite measure spaces, then + \begin{enumerate} + \item For any $f \in L^+(X \times Y)$, $[x \mapsto \int f(x, y)\nu(dy)] \in L^+(X)$, $[y \mapsto \int f(x, y)\mu(dx)] \in L^+(Y)$, and + \begin{align*} + \int_{X \times Y}f(z)\mu \otimes \nu(dz) &= \int_X\int_Y f(x, y)\nu(dy)\mu(dx) \\ + &= \int_{Y}\int_X f(x, y)\mu(dx)\nu(dy) + \end{align*} + + + \item For any normed space $E$ and $f \in L^1(X \times Y; E)$, + \begin{enumerate} + \item[(a)] For almost every $x \in X$, $f(x, \cdot) \in L^1(Y; E)$. For almost every $y \in Y$, $f(\cdot, y) \in L^1(X; E)$. + \item[(b)] $[x \mapsto \int f(x, y)\nu(dy)] \in L^1(X; E)$ and $[y \mapsto \int f(x, y)\mu(dx)] \in L^1(Y; E)$. + \item[(c)] \begin{align*} + \int_{X \times Y}f(z)\mu \otimes \nu(dz) &= \int_X\int_Y f(x, y)\nu(dy)\mu(dx) \\ + &= \int_{Y}\int_X f(x, y)\mu(dx)\nu(dy) + \end{align*} + \end{enumerate} + \end{enumerate} + +\end{theorem} +\begin{proof} + (1): First suppose that $\mu$ and $\nu$ are both finite. Let $\alg \subset \cm \otimes \cn$ be the collection of sets whose indicator functions satisfy (1), then $\alg$ contains all rectangles with finite measure. By linearity of the integral, for any $E, F\in \alg$, $E \setminus F \in \alg$. For any $\seq{E_n} \subset \alg$ and $E \in \alg$ such that $E_n \upto E$, by the \hyperref[Monotone Convergence Theorem]{theorem:mct}, + \begin{align*} + \int_{X \times Y}\one_{E}\mu \otimes \nu(dz) &= \limv{n}\int_{X \times Y}\one_{E_n}\mu \otimes \nu(dz) \\ + &= \limv{n}\int_X\int_Y \one_{E_n}(x, y)\nu(dy)\mu(dx) \\ + &= \int_X\int_Y \one_{E_n}(x, y)\nu(dy)\mu(dx) + \end{align*} + + so $\alg$ is a $\lambda$-system. By \hyperref[Dynkin's $\pi$-$\lambda$ Theorem]{theorem:pi-lambda}, $\alg = \cm \otimes \cn$. + + Now suppose that $\mu$ and $\nu$ are $\sigma$-finite, then $\alg$ contains all sets in $\cm \otimes \cn$ with finite measure. Since $\mu$ and $\nu$ are $\sigma$-finite, there exists rectangles $\seq{A_n \times B_n} \subset \alg$ such that $\mu \otimes \nu(A_n \times B_n)< \infty$ for all $n \in \natp$ and $A_n \times B_n \upto X \times Y$. Let $A \in \cm \otimes \cn$, then by the Monotone Convergence Theorem, + \begin{align*} + \mu \otimes \nu(A) &= \limv{n}\mu \otimes \nu(A \cap (A_n \times B_n)) \\ + &= \limv{n}\int_X\int_Y \one_{A \cap (A_n \times B_n)}(x, y)\nu(dy)\mu(dx) \\ + &= \int_X\int_Y \one_{A}(x, y)\nu(dy)\mu(dx) + \end{align*} + + Therefore $\alg = \cm \otimes \cn$. + + Now let $F \subset L^+(X \times Y)$ be the set of functions satisfying (1), then $F \supset \Sigma^+(X, \cm)$ by linearity. Let $f \in L^+(X \times Y)$, then by \autoref{proposition:measurable-simple-separable}, there exists $\seq{\phi_n} \subset \Sigma^+(X \times Y)$ such that $\phi_n \upto f$. In which case, by the Monotone Convergence Theorem, (1) also holds for $f$. +\end{proof} + + diff --git a/src/measure/sets/elementary.tex b/src/measure/sets/elementary.tex index 438ce02..f03c006 100644 --- a/src/measure/sets/elementary.tex +++ b/src/measure/sets/elementary.tex @@ -19,10 +19,10 @@ \label{proposition:elementary-family-algebra} Let $X$ be a set and $\ce \subset 2^X$ be an elementary family and \[ - \alg = \bracs{\bigsqcup_{i = 1}^n E_j \bigg | \seqf{E_j} \subset \text{ pairwise disjoint}} + \alg = \bracs{\bigsqcup_{i = 1}^n E_j \bigg | \seqf{E_j} \subset \ce \text{ pairwise disjoint}} \] - then is a ring. If $X \in \ce$, then $\ce$ is an algebra. + then $\alg$ is a ring. If $X \in \ce$, then $\alg$ is an algebra. \end{proposition} \begin{proof} Firstly, let $A, B \in \alg$ with $\seqf{A_j}, \seqf[m]{B_j} \subset \ce$ and $A = \bigsqcup_{j = 1}^n A_j$ and $B = \bigsqcup_{j = 1}^mB_j$. If $A \cap B = \emptyset$, then @@ -52,3 +52,37 @@ 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} + +\begin{proposition} +\label{proposition:rectangle-elementary-family} + Let $X, Y$ be sets, $\ce \subset 2^X$, and $\cf \subset 2^Y$ be elementary families, then the collection of rectangles + \[ + \mathcal{R}(\ce, \cf) = \bracs{E \times F| E \in \ce, F \in \cf} + \] + + is an elementry family. +\end{proposition} +\begin{proof} + (P1): $\emptyset = \emptyset \times \emptyset$. + + (P2): For any $A \times B, C \times D \in \mathcal{R}(\ce, \cf)$, + \[ + (A \times B) \cap (C \times D) = \underbrace{(A \cap C)}_{\in \ce} \times \underbrace{(B \times D)}_{\in \cf} \in \mathcal{R}(\ce, \cf) + \] + + (E): Let $A \times B, C \times D \in \mathcal{R}(\ce, \cf)$, then + \begin{align*} + (A \times B) \setminus (C \times D) &= (A \setminus C) \times (B \setminus D) \sqcup (A \setminus C) \times (B \cap D) \\ + &\sqcup (A \cap C) \times (B \setminus D) + \end{align*} + + Let $\seqf{A_j} \subset \ce$ such that $A \setminus C = \bigsqcup_{j = 1}^n A_j$ and $\bracsn{B_j}_1^n \subset \cf$ such that $B \setminus D = \bigsqcup_{j = 1}^m B_j$, then + \begin{align*} + (A \setminus C) \times (B \setminus D) &= \bigsqcup_{i = 1}^n \bigsqcup_{j = 1}^m A_i \times B_j \\ + (A \setminus C) \times (B \cap D) &= \bigsqcup_{i = 1}^n A_i \times (B \cap D) \\ + (A \cap C) \times (B \setminus D) &= \bigsqcup_{j = 1}^m (A \cap C) \times B_j + \end{align*} + + are all finite disjoint unions of elements of $\mathcal{R}(\ce, \cf)$. Therefore $(A \times B) \setminus (C \times D) \in \mathcal{R}(\ce, \cf)$. + +\end{proof} diff --git a/src/measure/vector/variation.tex b/src/measure/vector/variation.tex index 905518d..de477d8 100644 --- a/src/measure/vector/variation.tex +++ b/src/measure/vector/variation.tex @@ -117,7 +117,7 @@ \mu(A) = \sum_{n = 1}^\infty \mu_n(A) = \sum_{n = 1}^\infty \sum_{k = 1}^\infty \mu_n(A_k) = \sum_{k = 1}^\infty \mu(A_k) \] - by Fubini's theorem. + by \hyperref[Fubini's theorem]{theorem:fubini-tonelli}. % TODO: Actually link Fubini once it's there. \end{proof} diff --git a/src/topology/main/c0.tex b/src/topology/main/c0.tex index bc0419f..7230112 100644 --- a/src/topology/main/c0.tex +++ b/src/topology/main/c0.tex @@ -5,7 +5,7 @@ \label{definition:vanish-at-infinity} Let $X$ be a topological space, $E$ be a TVS over $K \in \RC$, and $f \in C(X; E)$, then $f$ \textbf{vanishes at infinity} if for every $U \in \cn_E^o(0)$, $\bracs{f \not\in U}$ is compact. - The set $C_0(X; E)$ is the space of all functions that vanish at infinity. + The set $C_0(X; E)$ is the space of all functions that vanish at infinity, equipped with the uniform topology. \end{definition} \begin{proposition} @@ -38,3 +38,27 @@ so $f \in \ol{C_c(X; E)}$. \end{proof} + +\begin{proposition} +\label{proposition:c0-tensor} + Let $X$ be a LCH space and $E$ be a locally convex space over $K \in \RC$. Identify $C_0(X; K) \otimes E$ as a subspace of $C_0(X; E)$ under the natural map + \[ + C_0(X; K) \otimes E \to C_0(X; E) \quad \sum_{j = 1}^n \phi_j \otimes x_j \mapsto \sum_{j = 1}^n x_j \cdot \phi_j + \] + + then $C_0(X; K) \otimes E$ is a dense subspace of $C_0(X; E)$. +\end{proposition} +\begin{proof} + Let $\phi \in C_0(X; E)$. Using \autoref{proposition:c0-properties}, assume without loss of generality that $\phi \in C_c(X; E)$. + + Since $\supp{\phi}$ is compact, so is $\phi(X)$ by \autoref{proposition:compact-extensions}. Let $U \in \cn_E^o(0)$ be balanced, then there exists $\seqf{y_j} \subset E \setminus \bracs{0}$ such that $\bigcup_{j = 1}^n (y_j + U) \supset \phi(X)$. For each $1 \le j \le n$, let $V_j = \phi^{-1}(y_j + U)$, then $\seqf{V_j}$ is an open cover of $\supp{\phi}$ consisting of precompact open sets. By \autoref{proposition:lch-partition-of-unity}, there exists a partition of unity $\seqf{\phi_j} \subset C_c(X; [0, 1])$ on $\supp{\phi}$ subordinate to $\seqf{V_j}$. For any $x \in E$, + \begin{align*} + \phi(x) - \sum_{j = 1}^n y_j \phi_j(x) &= \sum_{j = 1}^n \phi(x) \phi_j(x) - \sum_{j = 1}^n y_j \phi_j(x) \\ + &= \sum_{j = 1}^n \phi_j(x)[\phi(x) - y_j] \in \sum_{j = 1}^n \phi_j(x)U \subset U + \end{align*} + + Therefore $(\phi - \sum_{j = 1}^n y_j \phi_j)(X) \subset U$. + +\end{proof} + + diff --git a/src/topology/uniform/uc.tex b/src/topology/uniform/uc.tex index 58551ad..7b8b8d0 100644 --- a/src/topology/uniform/uc.tex +++ b/src/topology/uniform/uc.tex @@ -163,3 +163,15 @@ \] \end{proof} + + +\begin{proposition} +\label{proposition:uniform-continuous-compact} + Let $X$ be a compact uniform space, $Y$ be a uniform space, and $f \in C(X; Y)$, then $f \in UC(X; Y)$. +\end{proposition} +\begin{proof} + Let $V$ be an entourage of $Y$. For each $x \in X$, let $U_x$ be an entourage of $X$ such that $(f(y), f(z)) \in V$ for all $y, z \in (U_x \circ U_x)(x)$. Since $X$ is compact, there exists $\seqf{x_j} \subset X$ such that $X = \bigcup_{j = 1}^nU_{x_j}(x_j)$. + + Let $U = \bigcap_{j = 1}^n U_{x_j}$, then for any $(x, y) \in U$, there exists $1 \le j \le n$ such that $x \in U_{x_j}(x_j)$. In which case, $x, y \in (U_{x_j} \circ U_{x_j})(x_j)$, so $(f(x), f(y)) \in V$. +\end{proof} +