Added Fubini's theorem.

src/fa/lc/tensor.tex

@@ -47,6 +47,13 @@
     On the other hand, for any convex and circled set $W \in \cn_{E \otimes_\pi F}(0)$, there exists $U \in \cn_E(0)$ and $V \in \cn_F(0)$ such that $U \otimes V \subset W$. In which case, $W \supset \Gamma(U \otimes V)$, so $\fB$ is a fundamental system of neighbourhoods at $0$ for $E \otimes_\pi F$. 
 \end{proof}
 
+\begin{remark}
+\label{remark:projective-construction}
+    In constructing the \hyperref[projective tensor product]{definition:projective-tensor-product}, it may be more natural to obtain its topology as a projective topology using its universal property. However, doing so requires taking a least upper bound across \textit{all continuous linear maps defined on} $E \times F$, a collection too big to be a set. As such, constructing it as a projective topology is logically dubious, or at the very least beyond my abilities. 
+\end{remark}
+
+
+
 \begin{definition}[Cross Seminorm, {{\cite[III.6.3]{SchaeferWolff}}}]
 \label{definition:cross-seminorm}
     Let $E, F$ be locally convex spaces over $K \in \RC$. For any convex circled sets $U \in \cn_E(0)$ and $V \in \cn_F(0)$, let $p: E \to [0, \infty)$ and $q: F \to [0, \infty)$ be their \hyperref[gauges]{definition:gauge}. For any $z \in E \otimes_\pi F$, let

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}
 
 

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}

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}
+
+

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}

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}

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}
+
+

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}
+