Added the Singer representation theorem.

refs.bib

@@ -103,3 +103,12 @@
   year={1983},
   publisher={Hermann}
 }
+@article{HensgenSinger,
+  title={A simple proof of Singer’s representation theorem},
+  author={Hensgen, Wolfgang},
+  journal={Proceedings of the American Mathematical Society},
+  volume={124},
+  number={10},
+  pages={3211--3212},
+  year={1996}
+}

src/fa/norm/normed.tex

@@ -100,3 +100,13 @@
 \end{proof}
 
 
+% TODO: Replace this with a more general version involving polars in the future. 
+\begin{theorem}[Alaoglu's Theorem]
+\label{theorem:alaoglu}
+    Let $E$ be a normed vector space over $K \in \RC$, then $B^* = \bracsn{\phi \in E^*| \norm{x}_{E^*} \le 1}$ is compact in the weak*-topology. 
+\end{theorem}
+\begin{proof}
+    For each $x \in E$, $I_x = \bracsn{\dpn{x, \phi}{E}|\phi \in B^*}$ is compact. By \autoref{proposition:operator-space-completeness}, the closure of $B^*$ in $\prod_{x \in E}I_x$ is a subset of $\hom(E; K)$. Since $B^*$ is bounded, $I_x \subset \overline{B_K(0, 1)}$ for all $x \in E$, so the closure of $B^*$ is contained in $B^*$. By \autoref{theorem:tychonoff}, $\prod_{x \in E}I_x$ is compact. Therefore $B^*$ is compact with respect to the weak*-topology.
+\end{proof}
+
+

src/measure/measure/measure.tex

@@ -149,3 +149,26 @@
     \]
 \end{proof}
 
+\begin{definition}[Pushforward Measure]
+\label{definition:pushforward-measure}
+    Let $(X, \cm, \mu)$ be a measure space, $(Y, \cn)$ be a measurable space, and $f: X \to Y$ be a $(\cm, \cn)$-measurable map, then:
+    \begin{enumerate}
+        \item The mapping
+        \[
+        f_*\mu: \cn \to [0, \infty] \quad A \mapsto \mu\bracs{f \in A}
+        \]
+
+        is a measure on $(Y, \cn)$. 
+        \item For any Banach space $E$ and $g \in L^1(f_*\mu; E)$, 
+        \[
+        \int g df_*\mu = \int g \circ f d\mu
+        \]
+    \end{enumerate}
+    
+\end{definition}
+\begin{proof}
+    (1): Preimage commutes with unions, intersections, and complements. 
+
+    (2): By definition and linearity, (2) holds for $L^1(f_*\mu; E) \cap \Sigma(f^*\mu; E)$, which is dense in $L^1(f_*\mu; E)$ by \autoref{proposition:lp-simple-dense}. By continuity of the integral, (2) holds on $L^1(\mu_*\mu; E)$. 
+\end{proof}
+

src/measure/radon/c0.tex

@@ -55,7 +55,7 @@
 
     so $\norm{I_\mu}_{C_0(X; \complex)} \le \norm{\mu}_{\text{var}}$. 
     
-    On the other hand, let $\seqf{A_j} \subset \cb_X$ such that $X = \bigsqcup_{j = 1}^nA_j$. Let $\eps > 0$, then by \autoref{proposition:radon-measurable-description} of $|\mu|$, there exists compact sets $\seqf{K_j}$ such that for each $1 \le j \le n$, $K_j \subset A_j$ and $|\mu(K_j) - \mu(A_j)| < \eps$. By outer regularity and \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\seqf{\phi_j} \subset C_c(X; [0, 1])$ with disjoint support such that for each $1 \le j \le n$, $\norm{\phi_j - \one_{K_j}}_{L^1(|\mu|)} < \eps/|\mu(K_j)|$. Let $\phi = \sum_{j = 1}^n \ol{\sgn(\mu(K_j))}\phi_j$, then $\norm{\phi}_u \le 1$ and
+    On the other hand, let $\seqf{A_j} \subset \cb_X$ such that $X = \bigsqcup_{j = 1}^nA_j$. Let $\eps > 0$, then by \autoref{proposition:radon-regular-sigma-finite} applied to $|\mu|$, there exists compact sets $\seqf{K_j}$ such that for each $1 \le j \le n$, $K_j \subset A_j$ and $|\mu(K_j) - \mu(A_j)| < \eps$. By outer regularity and \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\seqf{\phi_j} \subset C_c(X; [0, 1])$ with disjoint support such that for each $1 \le j \le n$, $\norm{\phi_j - \one_{K_j}}_{L^1(|\mu|)} < \eps$. Let $\phi = \sum_{j = 1}^n \ol{\sgn(\mu(K_j))}\phi_j$, then $\norm{\phi}_u \le 1$ and
     \[
         \abs{\sum_{j = 1}^n |\mu(A_j)| - \int \phi d\mu} \le \sum_{j = 1}^n |\mu(A_j) - \mu(K_j)| + \sum_{j = 1}^n \norm{\phi_j - \one_{K_j}}_{L^1(|\mu|)} < 2n\eps
     \]
@@ -78,7 +78,7 @@
     &+ i\dpn{f, I_i^+}{C_0(X; \real)} - i\dpn{f, I_i^-}{C_0(X; \real)}
     \end{align*}
 
-    Thus by the Riesz representation theorem, there exists finite Radon measures $\mu_r^+, \mu_r^-, \mu_i^+, \mu_i^- \in M_R(X; \complex)$ such that for any $f \in C_0(X; \real)$, 
+    Thus by the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon}, there exists finite Radon measures $\mu_r^+, \mu_r^-, \mu_i^+, \mu_i^- \in M_R(X; \complex)$ such that for any $f \in C_0(X; \real)$, 
     \[
     \dpn{f, I}{C_0(X; \real)} = \int f d\mu_r^+ - \int f d\mu_r^- + i\int f d\mu_i^+ - i\int f d\mu_i^-
     \]
@@ -89,5 +89,85 @@
 \end{proof}
 
 
+\begin{theorem}[Singer's Representation Theorem, {{\cite{HensgenSinger}}}]
+\label{theorem:singer-representation}
+    Let $X$ be an LCH space and $E$ be a normed space over $K \in \RC$. For each $\mu \in M_R(X; E^*)$, let
+    \[
+    I_\mu: C_0(X; E) \to K \quad \dpn{f, I_\mu}{C_0(X; E)} = \int \dpn{f, d\mu}{E}
+    \]
+    
+    then the map
+    \[
+    M_R(X; E^*) \to C_0(X; E)^* \quad \mu \mapsto I_\mu
+    \]
+
+    is an isometric isomorphism. 
+\end{theorem}
+\begin{proof}
+    (Isometric): Let $\mu \in M_R(X; E^*)$, then for any $f \in C_0(X; E)$, 
+    \[
+    |\dpn{f, I_\mu}{C_0(X; E)}| \le \int \norm{f}_E d|\mu| \le \norm{f}_u \cdot \norm{\mu}_{\text{var}}
+    \]
+
+    so $\norm{I_\mu}_{C_0(X; E)^*} \le \norm{\mu}_{\text{var}}$. 
+    
+    On the other hand, let $\seqf{A_j} \subset \cb_X$ such that $\bigsqcup_{j = 1}^n A_j = X$ and $\eps > 0$.
+
+    By \autoref{proposition:radon-regular-sigma-finite} applied to $|\mu|$, there exists $\seqf{K_j}$ compact such that for each $1 \le j \le n$, $K_j \subset A_j$ and $\norm{\mu(K_j) - \mu(A_j)}_{E^*} < \eps$. By outer regularity and \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\seqf{\phi_j} \subset C_c(X; [0, 1])$ with disjoint support such that for each $1 \le j \le n$, $\norm{\phi_j - \one_{K_j}}_{L^1(|\mu|)} < \eps$. 
+    
+    Let $\seqf{x_j} \subset \overline{B_E(0, 1)}$ such that for each $1 \le j \le n$, $\dpn{x_j, \mu(K_j)}{E} > \norm{\mu(K_j)}_{E^*} - \eps$. Define $\phi = \sum_{j = 1}^n x_j \phi_j$, then $\norm{\phi}_u \le 1$ and
+    \begin{align*}
+    \abs{\sum_{j = 1}^n \norm{\mu(A_j)}_{E^*} - \int \dpn{\phi, d\mu}{E}} &\le \sum_{j = 1}^n \norm{\mu(A_j) - \mu(K_j)}_{E^*} \\
+    &+ \sum_{j = 1}^n \norm{\phi_j - \one_{K_j}}_{L^1(|\mu|)} + n\eps < 3n\eps
+    \end{align*}
+    
+    
+    As such a $\phi \in C_0(X; E)$ exists for all $\eps > 0$ and $\seqf{A_j}$, $\norm{I_\mu}_{C_0(X; E)^*} \ge \norm{\mu}_{\text{var}}$. Therefore the map $\mu \mapsto I_\mu$ is isometric. 
+
+    (Surjective): Let $B = \bracsn{\phi \in E^*|\norm{\phi}_{E^*} \le 1}$ and equip it with the weak*-topology and
+    \[
+    T: C_0(X; E) \to C_0(X \times B; K) \quad (Tf)(x, \phi) = \dpn{f(x), \phi}{E}
+    \]
+
+    then $T$ is maps $C_0(X; E)$ continuously into a subspace of $C_0(X \times B; K)$. 
+
+    Let $I \in C_0(X; E)^*$, then by the \hyperref[Hahn-Banach theorem]{theorem:hahn-banach}, there exists $\ol{I} \in C_0(X \times B; K)^*$ such that $\ol I \circ T = I$. By \hyperref[Alaoglu's Theorem]{theorem:alaoglu}, $B$ is a compact Hausdorff space. Therefore $X \times B$ is a LCH space by \autoref{proposition:lch-product}. By the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0}, there exists $\mu \in M_R(X \times B; K)$ such that for any $f \in C_0(X \times B; K)$, 
+    \[
+    \dpn{f, \ol I}{C_0(X \times B; K)} = \int_{X \times B} f d\mu
+    \]
+    
+    Now, let
+    \[
+    \nu: \cb_X \to E^* \quad \dpn{y,\nu(A)}{E} = \int_{X \times B} \one_A(x) \cdot \dpn{y, \phi}{E} \mu(dx, d\phi)
+    \]
+
+
+    then for each $A \in \cb_X$ and $y \in E$, 
+    \[
+    |\dpn{y,\nu(A)}{E}| \le \int_{X \times B} \one_A(x) \cdot \norm{y}_E |\mu|(dx, d\phi) \le \norm{y}_E \cdot |\mu|(A \times B)
+    \]
+
+    As the above holds for all $y \in E$, $\norm{\nu(A)}_{E^*} \le |\mu|(A \times B)$. Moreover, for any pairwise disjoint sequence $\seq{A_n} \subset \cb_X$ and $A \in \cb_X$ such that $A = \bigsqcup_{n \in \nat}A_n$, 
+    \[
+    \limv{n}\norm{\nu(A) - \sum_{j = 1}^n \nu(A_n)}_{E^*} \le \limv{n}|\mu|\paren{\bigcup_{k > n}A_k \times B} = 0
+    \]
+
+    so $\nu$ is a vector measure on $\cb_X$. 
+    
+    Since $\norm{\nu(A)}_{E^*} \le |\mu|(A \times B)$ for all $A \in \cb_X$, $|\nu|(A) \le |\mu|(A \times B)$ for all $A \in \cb_X$, and $\nu$ is a Radon measure by \autoref{lemma:radon-compact-project}.
+
+    Finally, let $f \in C_0(X; K)$ and $y \in E$, then
+    \begin{align*}
+        \int_X \dpn{y \cdot f, d\nu}{E} &= \int_{X \times B} f(x)\dpn{y, \phi}{E} \mu(dx, d\phi) \\
+        &= \int_{X \times B}T(y \cdot f)(x, \phi)\mu(dx, d\phi) = \dpn{y \cdot f, I}{C_0(X; E)}
+    \end{align*}
+
+    Therefore for any $f \in C_0(X; K) \otimes E$, $\int_X \dpn{f, d\nu}{E} = \dpn{f, I}{C_0(X; E)}$. By \autoref{proposition:c0-tensor}, $C_0(X; K) \otimes E$ is a dense subspace of $C_0(X; E)$, so
+    \[
+    \int_X \dpn{f, d\nu}{E} = \dpn{f, I}{C_0(X; E)} \quad \forall f \in C_0(X; E)
+    \]
+    
+\end{proof}
+
 
 

src/measure/radon/radon.tex

@@ -148,6 +148,27 @@
     Since $X$ is $\sigma$-compact, $\mu$ is $\sigma$-finite, so $\mu$ is regular by \autoref{proposition:radon-regular-sigma-finite}.
 \end{proof}
 
+\begin{lemma}
+\label{lemma:radon-compact-project}
+    Let $X$ be a LCH space, $Y$ be a compact Hausdorff space, $\mu$ be a finite Radon measure on $X \times Y$, and $\nu$ be a measure on $X$. If for each $A \in \cb_X$, $\nu(A) \le \mu(A \times Y)$, then $\nu$ is also a Radon measure. 
+\end{lemma}
+\begin{proof}
+    Let $A \in \cb_X$ and $\eps > 0$. By outer regularity of $\mu$, there exists $U \in \cn_{X \times Y}(A \times Y)$ such that $\mu(U \setminus (A \times Y)) < \eps$. By the \hyperref[Tube Lemma]{lemma:tube-lemma}, there exists $V \in \cn_X(A)$ such that $V \times Y \subset U$. In which case, 
+    \[
+    \nu(V \setminus A) \le \mu((V \setminus A) \times Y) \le \mu(U \setminus (A \times Y)) < \eps
+    \]
+
+    so $\nu$ is outer regular on $A$.
+
+    On the other hand, by \autoref{proposition:radon-regular-sigma-finite}, there exists $K \subset A \times Y$ compact such that $\mu((A \times Y) \setminus K) < \eps$. By \autoref{proposition:compact-extensions}, $\pi_1(K) \subset A$ is also compact. Thus
+    \[
+    \nu(A \setminus \pi_1(K)) \le \mu((A \setminus \pi_1(K)) \times Y) \le \mu((A \times Y) \setminus K)
+    \]
+
+    so $\nu$ is inner regular on $A$. 
+\end{proof}
+
+
 \begin{proposition}
 \label{proposition:radon-cc-dense}
     Let $X$ be a LCH space, $\mu: \cb_X \to [0, \infty]$ be a Radon measure, $E$ be a normed vector space, and $p \in [1, \infty)$, then $C_c(X; E)$ is dense in $L^p(X; E)$.

src/topology/main/compact.tex

@@ -82,3 +82,28 @@
 
     By compactness of $A$, there exists $A_0 \subset A$ finite such that $\bigcup_{x \in A_0}U_x \in \cn(A)$. Let $U = \bigcup_{x \in A_0}U_x \in \cn(A)$ and $V = \bigcap_{x \in A_0}V_x \in \cn(B)$, then $A \cap B = \emptyset$.
 \end{proof}
+
+\begin{lemma}[Tube Lemma]
+\label{lemma:tube-lemma}
+    Let $X$ be a topological space and $Y$ be a compact space, then
+    \begin{enumerate}
+        \item For any $x \in X$ and $U \in \cn_{X \times Y}^o(\bracs{x} \times Y)$, there exists $V \in \cn_X(x)$ such that $V \times Y \subset U$. 
+        \item For any $A \subset X$ and $U \in \cn_{X \times Y}^o(A \times Y)$, there exists $V \in \cn_X(A)$ such that $V \times Y \subset U$. 
+    \end{enumerate}
+    
+\end{lemma}
+\begin{proof}
+    (1): For each $y \in Y$, let $U_y \in \cn_X(x)$ and $V_y \in \cn_Y(y)$ such that $(x, y) \in U_y \times V_y \subset U$, then $\bracs{V_y|y \in Y}$ is an open cover of $Y$. By compactness, there exists $\seqf{y_j} \subset Y$ such that $\bigcup_{j = 1}^nV_{y_j} = Y$. Let $V = \bigcap_{j = 1}^n U_{y_j}$, then $V \in \cn_X(x)$ and $V \times Y \subset U$. 
+
+    (2): For each $x \in A$, there exists $U_x \in \cn_X(x)$ such that $U_x \times Y \subset U$. Let $V = \bigcup_{x \in A}U_x$, then $V \times Y \subset U$. 
+\end{proof}
+
+
+\begin{theorem}[Tychonoff]
+\label{theorem:tychonoff}
+    Let $\seqi{X}$ be compact topological spaces, then $X = \prod_{i \in I}X_i$ is compact. 
+\end{theorem}
+\begin{proof}
+    Let $\fU \subset 2^X$ be an ultrafilter. By \autoref{proposition:imagefilterbase}, for each $i \in I$, $\pi_i(\fU)$ is an ultrafilter base. Since $X_i$ is compact, there exists $x_i \in X_i$ such that $\pi_i(\fU) \to x_i$. By the axiom of choice, there exists $x \in X$ such that $\pi_i(\fU) \to \pi_i(x)$ for all $i \in I$. Therefore $\fU \to x$ by \autoref{proposition:productfilterconvergence}.
+\end{proof}
+

src/topology/main/filters.tex

@@ -48,12 +48,19 @@
 
 \begin{proposition}
 \label{proposition:imagefilterbase}
-    Let $X, Y$ be sets, $f: X \to Y$ be a function, and $\fB \subset 2^X$ be a filter base, then $f(\fB) = \bracs{f(E)| E \in \fB}$ is also a filter base.
+    Let $X, Y$ be sets, $f: X \to Y$ be a function, and $\fB \subset 2^X$ be a filter base, then
+    \begin{enumerate}
+        \item $f(\fB) = \bracs{f(E)| E \in \fB}$ is also a filter base.
+        \item If $\fB$ is an ultrafilter base, then $f(\fB)$ is also an ultrafilter base. 
+    \end{enumerate}
+    
 \end{proposition}
 \begin{proof}
-    (FB1): Let $E, F \in \fB$, then there exists $G \in \fB$ such that $G \subset E \cap F$. In which case, $f(G) \subset f(E \cap F) \subset f(E) \cap f(F)$.
+    (1): Let $E, F \in \fB$, then there exists $G \in \fB$ such that $G \subset E \cap F$. In which case, $f(G) \subset f(E \cap F) \subset f(E) \cap f(F)$.
     
-    (FB2): For any $f(E) \in f(\fB)$, $E \ne \emptyset$ implies that $f(E) \ne \emptyset$.
+    For any $f(E) \in f(\fB)$, $E \ne \emptyset$ implies that $f(E) \ne \emptyset$.
+
+    (2): Let $E \subset X$, then either $f^{-1}(E)$ or $f^{-1}(E^c)$ contains an element of $\fB$. Therefore either $E$ or $E^c$ contains an element of $f(\fB)$. 
 \end{proof}
 
 \begin{proposition}

src/topology/main/lch.tex

@@ -237,3 +237,13 @@
 
     Let $x \in X$, then there exists $n \in \natp$ such that $x \in U_n \setminus U_{n-1}$. In which case, if $n > 1$, then $x \in U_{n} \setminus \ol{U_{n - 2}} = V_{n-1}$. If $n = 1$, then $x \in U_{2} = V_1$. Thus $\seq{V_n}$ is an open cover of $X$. In addition, for any $m, n \in \natp$ with $m \le n$, $V_m \cap V_n \ne \emptyset$ implies that $n - m < 2$, so $\seq{V_n}$ is locally finite. By (2) of \autoref{proposition:lch-paracompact}, $X$ is paracompact.
 \end{proof}
+
+\begin{proposition}
+\label{proposition:lch-product}
+    Let $\seqi{X}$ be a family of LCH spaces. If $X_i$ is compact for all but finitely many $i \in I$, then $X = \prod_{i \in I}X_i$ is a LCH space. 
+\end{proposition}
+\begin{proof}
+    By \autoref{proposition:product-hausdorff}, $\prod_{i \in I}X_i$ is Hausdorff. Let $x \in \prod_{i \in I}X_i$ and $i \in I$. If $X_i$ is not compact, let $U_i \in \cn_{X_i}(\pi_i(x))$ be compact. Otherwise, let $U_i = X_i$. Let $U = \prod_{i \in I}U_i$, then since $U_i \ne X_i$ for only finitely many $i \in I$, $U \in \cn_X(x)$. By \hyperref[Tychonoff's Theorem]{theorem:tychonoff}, $U$ is compact. Therefore $X$ is locally compact. 
+\end{proof}
+
+

src/topology/main/product.tex

@@ -112,8 +112,8 @@ Let $\bracsn{(X_i, \topo_i)}_{i \in I}$ be a family of topological spaces and $\
     (1): By (U) of \autoref{definition:product-topology}/\autoref{definition:product-uniform}.
 
     (2): Consider the following diagram
-    \[
     % https://darknmt.github.io/res/xypic-editor/#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
+    \[
     \xymatrix{
     \prod_{i \in I}Y_i \ar@{->}[rd]^{\pi_i} &  \\
     \iota_P(\prod_{i \in I}X_i) \ar@{->}[u] & Y_i \\