\section{Dual of $C_0$}
\label{section:dual-of-c0}

\begin{lemma}[Jordan Decomposition]
\label{lemma:positive-linear-jordan}
    Let $X$ be a topological space and $I \in C_0(X; \real)^*$, then there exists positive linear functionals $I^+, I^- \in C_0(X; \real)^*$ such that:
    \begin{enumerate}
        \item $I = I^+ - I^-$.
        \item $I^+ \perp I^-$. 
        \item $\norm{I^+}_{C_0(X; \real)^*}, \norm{I^-}_{C_0(X; \real)^*} \le \norm{I}_{C_0(X; \real)^*}$.
    \end{enumerate}

    
\end{lemma}
\begin{proof}
    (1), (2): Since $C_0(X; \real)$ is a Banach lattice, $C_0(X; \real)^*$ is also a Banach lattice by \autoref{proposition:banach-lattice-dual}. Therefore there exists positive linear functionals $I^+, I^- \in C(X; \real)^*$ such that $I = I^+ - I^-$ and $I^+ \perp I^-$. 

    (3): By \autoref{proposition:banach-lattice-properties}.
\end{proof}

\begin{definition}[Radon Measure]
\label{definition:radon-measure-extended}
    Let $X$ be an LCH space and $\mu$ be a Borel signed/vector measure on $X$, then $\mu$ is \textbf{Radon} if $|\mu|$ is Radon. 
\end{definition}



\begin{definition}[Space of Finite Radon Measures]
\label{definition:space-radon-measures}
    Let $X$ be an LCH space and $E$ be a normed vector space over $K \in \RC$, then $M_R(X; E)$ is the \textbf{space of finite Radon measures} on $X$, which forms a vector space over $K$. 
\end{definition}
\begin{proof}
    Let $\mu, \nu \in M_R(X; E)$, then for any $A \in \cb_X$, $|\mu + \nu|(A) \le |\mu|(A) + |\nu|(A)$. Let $\eps > 0$, then by outer regularity and \autoref{proposition:radon-measurable-description}, there exists $K \subset A$ compact and $U \in \cn^o(A)$ such that $(|\mu| + |\nu|)(A \setminus K), (|\mu| + |\nu|)(U \setminus A) < \eps$. Therefore $|\mu + \nu|$ is regular on all Borel sets, and hence Radon. 
\end{proof}


\begin{theorem}[Riesz Representation Theorem]
\label{theorem:riesz-radon-c0}
    Let $X$ be an LCH space. For each $\mu \in M_R(X; \complex)$, let
    \[
    I_\mu: C_0(X; \complex) \to \complex \quad \dpn{f, I_\mu}{C_0(X; \complex)} = \int f d\mu
    \]
    
    then the map
    \[
    M_R(X; \complex) \to C_0(X; \complex) \quad \mu \mapsto I_\mu
    \]

    is an isometric isomorphism. 
\end{theorem}
\begin{proof}[Proof {{\cite[Theorem 7.17]{Folland}}}. ]
    Let $f \in C_0(X; \complex)$, then
    \[
    \abs{\int f d\mu} \le \int \norm{f}_u d\mu \le \norm{f}_u \cdot |\mu|(X)
    \]

    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-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
    \[
        \abs{\sum_{j = 1}^n |\mu(A_j)| - \int \phi d\mu} \le \sum_{j = 1}^n |\mu(A_j) - \mu(K_j)| + \sum_{j = 1}^n \norm{\phi_j - \one_{K_j}}_{L^1(|\mu|)} < 2n\eps
    \]

    so
    \[
    \abs{\int \phi d\mu} \ge \sum_{j = 1}^n |\mu(A_j)| - 2n\eps
    \]

    As such a $\phi$ exists for all $\eps > 0$, $\norm{I_\mu}_{C_0(X; \complex)} \ge \sum_{j = 1}^n |\mu(A_j)|$. Since this holds for all such partitions, $\norm{I_\mu}_{C_0(X; \complex)} \ge |\mu|(X)$. Therefore the map $\mu \mapsto I_\mu$ is isometric. 

    Finally, let $I \in C_0(X; \complex)^*$, then there exists bounded linear functionals $I_r, I_i \in C_0(X; \real)^*$ such that for any $f \in C_0(X; \real)$, 
    \[
    \dpn{f, I}{C_0(X; \real)^*} = \dpn{f, I_r}{C_0(X; \real)} + i\dpn{f, I_i}{C_0(X; \real)}
    \]

    By \autoref{lemma:positive-linear-jordan}, there exists bounded positive linear functionals $I_r^+, I_r^-, I_i^+, I_i^-$ such that for any $f \in C_0(X; \real)$, 
    \begin{align*}
    \dpn{f, I}{C_0(X; \real)} &= \dpn{f, I_r^+}{C_0(X; \real)} - \dpn{f, I_r^-}{C_0(X; \real)} \\
    &+ i\dpn{f, I_i^+}{C_0(X; \real)} - i\dpn{f, I_i^-}{C_0(X; \real)}
    \end{align*}

    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 $f \in C_0(X; \real)$, 
    \[
    \dpn{f, I}{C_0(X; \real)} = \int f d\mu_r^+ - \int f d\mu_r^- + i\int f d\mu_i^+ - i\int f d\mu_i^-
    \]

    Let $\mu = \mu_r^+ - \mu_r^- + i\mu_i^+ - i\mu_i^-$, then $I = I_\mu$, and the map $\mu \mapsto I_\mu$ is surjective. 
    
    
\end{proof}


\begin{theorem}[Singer's Representation Theorem]
\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}[Proof {{\cite{HensgenSinger}}}. ]
    (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 an 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}

\begin{remark}
\label{remark:singer-representation}
    In the proof of \hyperref[Singer's Representation Theorem]{theorem:singer-representation}, the $E^*$-valued measure is constructed pointwise as
    \[
    \nu: \cb_X \to E^* \quad \dpn{y,\nu(A)}{E} = \int_{X \times B} \one_A(x) \cdot \dpn{y, \phi}{E} \mu(dx, d\phi)
    \]

    It may be tempting to use the strong formulation directly
    \[
    \nu: \cb_X \to E^* \quad \nu(A) = \int_{X \times B} \phi \cdot \one_A(x) \mu(dx, d\phi)
    \]

    However, without additional assumptions on $E^*$, $\phi \cdot \one_A(x)$ may not be strongly measurable, which prevents this direct use of the Bochner integral. Thus the weak formulation is a necessary complication in the proof. 
\end{remark}


