Added duality result for spaces of measures.

src/measure/vector/fin.tex

@@ -0,0 +1,88 @@
+\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}
+
+\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}
+    (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}
+
+
+

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}
-