Fleshed out the dual of measure spaces.

refs.bib

@@ -215,3 +215,12 @@
   year    = {1971},
   doi     = {10.1007/BF02771592}
 }
+
+@MISC {StackRadonDual,
+    TITLE = {How to understand C(X)'' = bounded Borel measurable functions?},
+    AUTHOR = {GEdgar (https://math.stackexchange.com/users/442/gedgar)},
+    HOWPUBLISHED = {Mathematics Stack Exchange},
+    NOTE = {URL:https://math.stackexchange.com/q/392719 (version: 2013-05-15)},
+    EPRINT = {https://math.stackexchange.com/q/392719},
+    URL = {https://math.stackexchange.com/q/392719}
+}

src/fa/lp/seq.tex

@@ -63,5 +63,38 @@
     \]
 \end{proof}
 
+\begin{theorem}
+\label{theorem:lp-sum-dual}
+    Let $\seqi{X}$ be normed vector spaces over $K \in \RC$ and $p \in [1, \infty)$ and $q \in (1, \infty]$ be Hölder conjugates. For each $y \in [l^q(I); X_i^*]$, let
+    \[
+    \phi_y: [l^p(I); X_i] \to K \quad x \mapsto \sum_{i \in I}\dpn{x_i, y_i}{X_i}
+    \]
 
+    then the mapping
+    \[
+    [l^q(I); X_i^*] \to [l^p(I); X_i]^* \quad y \mapsto \phi_y
+    \]
+
+    is an isometric isomorphism. 
+\end{theorem}
+\begin{proof}
+    Let $\phi \in [l^p(I); X_i]^*$, then there exists $y \in [l^\infty(I); X_i^*]$ such that for each $i \in I$ and $x \in X_i$, $\dpn{x_i \cdot \one_{\bracs{i}}, \phi}{[l^p(I); X_i]} = \dpn{x_i, y_i}{X_i}$. 
+
+    Since the $q = \infty$ case has been ruled out, assume that $q \in (1, \infty)$. For each $\alpha \in (0, 1)$, there exists $x \in [l^\infty(I); X_i]$ with $\norm{x_i}_{X_i} \le 1$ and $\dpn{x_i, y_i}{X_i} \ge \alpha \norm{y_i}_{X_i^*}$. For each $J \subset I$ finite and $i \in I$, let $F_J(i) = \one_{J}(i) \cdot \norm{y_i}_{X_i^*}^{q - 1}$, then by \autoref{lemma:holder-conjugate-gymnastics},
+    \[
+    \norm{F_J x}_{[l^p(I); X_i]}^p \le \sum_{j \in J}\norm{y_j}_{X_j^*}^{p(q - 1)}= \sum_{j \in J}\norm{y_j}_{X_j^*}^{q}
+    \]
+
+    so 
+    \begin{align*}
+    \alpha \sum_{j \in J} \norm{y_j}_{X_j^*}^q & \le 
+    \sum_{i \in I}F_J(i)\dpn{x_i, y_i}{X_i} = 
+    \dpn{F_J x, \phi}{[l^p(I); X_i]} \\
+    &\le \norm{\phi}_{[l^p(I); X_i]^*} \cdot \braks{\sum_{j \in J}\norm{y_j}_{X_j^*}^{q}}^{1/p} \\
+    \alpha \braks{\sum_{j \in J}\norm{y_j}_{X_j^*}^q}^{1/q} &\le \norm{\phi}_{[l^p(I); X_i]^*}
+    \end{align*}
+
+    As the above holds for all $\alpha \in (0, 1)$ and $J \subset I$ finite, $y \in [l^q(I); X_i^*]$ with $\norm{y}_{[l^q(I); X_i^*]} = \norm{\phi}_{[l^p(I); X_i]^*}$. 
+    
+\end{proof}
 

src/measure/vector/fin.tex

@@ -0,0 +1,101 @@
+\section{Spaces of Finite Measures}
+\label{section:space-finite-measure}
+
+
+\begin{definition}[Space of Finite Measures]
+\label{definition:vector-measure-finite-space}
+    Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, and $M(X, \cm; E)$ be the set of all finite $E$-valued vector measures on $(X, \cm)$. For each $\mu \in M(X, \cm; E)$, let $\norm{\mu}_{\text{var}} = |\mu|(X)$, then:
+    \begin{enumerate}
+        \item $M(X, \cm; E)$ equipped with $\norm{\cdot}_{\text{var}}$ is a normed vector space over $K$.
+        \item If $E$ is a Banach space, then so is $M(X, \cm; E)$.
+    \end{enumerate}
+
+\end{definition}
+\begin{proof}
+    (1): For any $\mu, \nu \in M(X, \cm; E)$ and $\seqf{A_j} \subset \cm$ such that $X = \bigsqcup_{j = 1}^n A_j$, 
+    \[
+    |\mu|(X) + |\nu|(X) \ge \sum_{j = 1}^n \norm{\mu(A_j)}_E + \norm{\nu(A_j)}_E \ge \sum_{j = 1}^n \norm{(\mu + \nu)(A_j)}_E
+    \]
+
+    As this holds for all choices of $\seqf{A_j}$, $\norm{\mu + \nu}_{\text{var}} \le \norm{\mu}_{\text{var}} + \norm{\nu}_{\text{var}}$. 
+
+    (2): Let $\seq{\mu_n} \subset M(X, \cm; E)$ such that $\sum_{n \in \natp}\norm{\mu_n}_{\text{var}} < \infty$. For each $A \in \cm$, since $E$ is complete, let
+    \[
+    \mu(A) = \sum_{n = 1}^\infty \mu_n(A)
+    \]
+
+    then for any $\seq{A_k} \subset \cm$ and $A \in \cm$ with $A = \bigsqcup_{k \in \natp}A_k$, 
+    \[
+        \mu(A) = \sum_{n = 1}^\infty \mu_n(A) = \sum_{n = 1}^\infty \sum_{k = 1}^\infty \mu_n(A_k) = \sum_{k = 1}^\infty \mu(A_k)
+    \]
+
+    by \hyperref[Fubini's theorem]{theorem:fubini-tonelli}.
+\end{proof}
+
+
+While the space of bounded Borel functions on $(X, \cm)$ forms a subspace of the dual of $M(X, \cm; \complex)$, it may not be immediately clear that they are insufficient. Before moving on to an explicit description of this dual, it is beneficial to consider a simple "example". 
+
+Let $X = [0, 1]$, equipped with its Borel $\sigma$-algebra, and $\mu$ be the Lebesgue measure on $X$, then for any $x \in [0, 1]$, $\mu$ is mutually singular with the delta mass at $x$. Therefore the closure of $\text{span}\bracs{\delta_x|x \in X}$ in $M_R(X, \cm; \complex)$ is a proper closed subspace of $M(X, \cm; \complex)$. By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, there exists $\phi \in M(X, \cm; \complex)^*$ such that $\dpn{\delta_x, \phi}{M(X, \cm; \complex)} = 0$ for all $x \in X$, but $\dpn{\mu, \phi}{M(X, \cm; \complex)} = 1$. If there exists a bounded Borel function $f: X \to \complex$ such that $\dpn{\nu, \phi}{M(X, \cm; \complex)} = \int f d\nu$ for all $\nu \in M(X, \cm; \complex)$, then $f(x) = \dpn{\delta_x, \phi}{M(X, \cm; \complex)} = 0$ for all $x \in [0, 1]$, which is impossible. Therefore $\phi$ cannot be represented as a bounded Borel function. 
+
+Aside from abusing the Hahn-Banach theorem, a more explicit construction of such a functional is via the atomic decomposition. However, it is not included here due to time constraints. 
+
+In any case, the above example shows that a linear functional on $M(X, \cm; \complex)$ may act as \textit{different} Borel functions on different families of measures that are mutually singular to each other. The following theorem will make this relation precise. 
+
+\begin{theorem}
+\label{theorem:hilbert-measures-dual}
+    Let $(X, \cm)$ be a measurable space, $H$ be a Hilbert space over $K \in \RC$, and $\mathscr{M} \subset M(X, \cm; H)$ be a closed subspace such that:
+    \begin{enumerate}
+        \item[(A)] For each $\mu \in \mathscr{M}$ and $\nu \in M(X, \cm; H)$ with $\nu \ll \mu$, $\nu \in \mathscr{M}$.
+    \end{enumerate}
+
+    Let $|\mathscr{M}| = \bracsn{|\mu|: \mu \in \mathscr{M}}$, then for any maximal family $\seqi{\mu} \subset |\mathscr{M}|$ of pairwise mutually singular measures, 
+    \begin{enumerate}
+        \item For each $f \in [l^1(I); L^1(\mu_i; H)]$, let
+        \[
+        \mu_f: \cm \to H \quad A \mapsto \sum_{i \in I} \int_A f_i d\mu_i
+        \]
+
+        then the mapping 
+        \[
+        [l^1(I); L^1(\mu_i; H)] \to \mathscr{M} \quad f \mapsto \mu_f
+        \]
+
+        is an isometric isomorphism. 
+        \item $\mathscr{M}^*$ is isometrically isomorphic to $[l^\infty(I); L^\infty(\mu_i; H)]$. 
+    \end{enumerate}
+    
+    In other words,
+    \begin{enumerate}[start=2]
+        \item There exists a semifinite measure space $(\Omega, \cn, \mu)$ such that $\mathscr{M} \iso L^1(\Omega; H)$ and $\mathscr{M}^* \iso L^\infty(\Omega; H)$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite{StackRadonDual}}}. ]
+    (1): By (A), $\mu_f \in \mathscr{M}$ for all $f \in [l^1(I); L^1(\mu_i; H)]$. 
+    
+    On the other hand, let $\nu \in \mathscr{M}$. For each $i \in I$, let $\nu = \nu_a^{(i)} + \nu_s^{(i)}$ be the Lebesgue decomposition of $\nu$ with respect to $\mu_i$, then by the \hyperref[Radon-Nikodym theorem]{theorem:lebesgue-radon-nikodym}, there exists $f_i \in L^1(\mu_i; H)$ such that $\nu_a^{(i)}(dx) = f_i \mu_i(dx)$. 
+    
+    For each countable $J \subset I$, define
+    \[
+    \nu_J: \cm \to H \quad A \mapsto \sum_{j \in J}\int_A f_j d\mu_j
+    \]
+
+    Let $M = \sup\bracs{\norm{\nu_J}_{\text{var}}| J \subset I \text{ countable}}$, then there exists $J \subset I$ countable such that $\norm{\nu_J}_{\text{var}} = M$. For each $i \in I \setminus J$, 
+    \[
+    \norm{\nu_J}_{\text{var}} + \normn{\nu_a^{(i)}}_{\text{var}}
+    = \norm{\nu_{J} + \nu_a^{(i)}}_{\text{var}} = \normn{\nu_{J \cup \bracs{i}}}_{\text{var}} \le M
+    \]
+
+    By maximality of $M$, $\normn{\nu_a^{(i)}}_{\text{var}} = 0$, and by maximality of $\seqi{\mu}$, $\nu - \nu_J = 0$. Let $g \in [l^1(I); L^1(\mu_i; H)]$ be defined by $g_i = f_i$ for each $i \in I$, then $\nu = \nu_J = \mu_g$, and the mapping is surjective. 
+
+    (2): By \autoref{theorem:lp-sum-dual} and \autoref{theorem:lp-duality}, 
+    \[
+    [l^1(I); L^1(\mu_i; H)]^* \iso [l^\infty(I); L^1(\mu_i; H)^*] \iso [l^\infty(I); L^\infty(\mu_i; H)] 
+    \]
+\end{proof}
+
+\begin{proposition}
+\label{proposition:measure-l-infinity-dominated-convergence}
+    Let $
+\end{proposition}
+
+

src/measure/vector/index.tex

@@ -7,4 +7,5 @@
 \input{./ac.tex}
 \input{./ms.tex}
 \input{./rn.tex}
+\input{./fin.tex}
 

src/measure/vector/variation.tex

@@ -90,34 +90,3 @@
     Let $(X, \cm)$ be a measurable space and $\mu$ be a signed/vector measure, then $\mu$ is \textbf{finite} if $|\mu|$ is finite. 
 \end{definition}
 
-\begin{definition}[Space of Finite Measures]
-\label{definition:vector-measure-finite-space}
-    Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, and $M(X, \cm; E)$ be the set of all finite $E$-valued vector measures on $(X, \cm)$. For each $\mu \in M(X, \cm; E)$, let $\norm{\mu}_{\text{var}} = |\mu|(X)$, then:
-    \begin{enumerate}
-        \item $M(X, \cm; E)$ equipped with $\norm{\cdot}_{\text{var}}$ is a normed vector space over $K$.
-        \item If $E$ is a Banach space, then so is $M(X, \cm; E)$.
-    \end{enumerate}
-
-    
-\end{definition}
-\begin{proof}
-    (1): For any $\mu, \nu \in M(X, \cm; E)$ and $\seqf{A_j} \subset \cm$ such that $X = \bigsqcup_{j = 1}^n A_j$, 
-    \[
-    |\mu|(X) + |\nu|(X) \ge \sum_{j = 1}^n \norm{\mu(A_j)}_E + \norm{\nu(A_j)}_E \ge \sum_{j = 1}^n \norm{(\mu + \nu)(A_j)}_E
-    \]
-
-    As this holds for all choices of $\seqf{A_j}$, $\norm{\mu + \nu}_{\text{var}} \le \norm{\mu}_{\text{var}} + \norm{\nu}_{\text{var}}$. 
-
-    (2): Let $\seq{\mu_n} \subset M(X, \cm; E)$ such that $\sum_{n \in \natp}\norm{\mu_n}_{\text{var}} < \infty$. For each $A \in \cm$, since $E$ is complete, let
-    \[
-    \mu(A) = \sum_{n = 1}^\infty \mu_n(A)
-    \]
-
-    then for any $\seq{A_k} \subset \cm$ and $A \in \cm$ with $A = \bigsqcup_{k \in \natp}A_k$, 
-    \[
-        \mu(A) = \sum_{n = 1}^\infty \mu_n(A) = \sum_{n = 1}^\infty \sum_{k = 1}^\infty \mu_n(A_k) = \sum_{k = 1}^\infty \mu(A_k)
-    \]
-
-    by \hyperref[Fubini's theorem]{theorem:fubini-tonelli}.
-\end{proof}
-