Every closed subspace $A$ of $C_0(K)$ can be regarded as a subspace of continuous functions on $A^*$? We consider a locally compact Hausdorff space $X$ and the Banach space $C_0(X)$ of continuous functions on $X$ taking values at $\mathbb K = \mathbb R$ or $\mathbb C$, equipped with the supremum norm.
I'm studying this paper: http://www.siue.edu/MATH/kj_papers/Into.pdf. At page 375, in the proof of Lemma 2, it says the following:


"If $A$ is a closed subspace of $C_0(X)$, then any functional $F\in A^\ast$ can be extended, with the same norm, to a regular Borel measure $\mu_F$ on $X$."


Until this point, there is no big deal. We use Hahn-Banach to extend $F$ to $\tilde F\in(C_0(X))^*$ with the same norm then use Riesz Representation Theorem to get a regular Borel measure $\mu_F$ at $X$ such that
$$
\tilde F(f)=\int f d\mu_F,\ \forall f \in C_0(X).
$$
Continuing:


"On the other hand, $A$ can be regarded as a subspace of continuous functions on $A^*$ with the weak* topology, and any such measure $\mu_F$ can be regarded as a regular Borel measure defined on a closed subset of $A^*$."


I'm having difficult to understand it. 
In the first statement, I came to the conclusion that he reffers to the image of $A$ by the James map $J: A \to A^{**}$, and since the weak* topology preserves the continuity of the functionals in $J(A)$, every element of $J(A)$ is a continuous function on the topological space $(A^*,\omega^*)$. The map $J$ is linear (so $J(A)$ is a subspace of $(A^*,\omega^*)$).
I guess there is a second possibility to interpret the first statement, that is to consider the map $\delta: X \to A^*$ such that $\delta(x)$ is the evaluation map on $x$ restricted to $A$. But then I don't know where to go from here.
In the second statement, I have no idea what meant. Please help me.
 A: I think I got it now. My interpretation of the first statement is right. Let us recall it:

"A  can be regarded as a subspace of continuous functions on A∗ with the weak* topology"

Consider the map $J: A \to A^{**}$ defined by $ a\in A \mapsto \hat a\in A^{**}$. Since the weak* topology preserves the continuity of the functionals $\hat a$, it follows that every $\hat a$ is a continuous real function on $(A^*,\omega^*)$. Moreover, the map $J$ is injective so we can identify A with $J(A)$, and $J$ is linear, so $J(A)$ is a subspace of the spaces of continuous real function on $(A^*,\omega^*)$.
Now, let us interpret the secon statement:

"any such measure $\mu_F$ can be regarded as a regular Borel measure defined on a closed subset of $A^∗$."

Consider the map $\delta: X \to (A^*,\omega^*)$ such that
$$
   \delta(x)(f)=f(x),\ \ \forall f\in A.
$$
It is simple to see that $\delta$ is a continuous map, so for every Borel set $E\subset (A^*,\omega^*)$, the set $\delta^{-1}(E)$ is a Borel set of $X$, and then we can define a measure $\tilde \mu_F$ on $(A^*,\omega^*)$ by
$$
    \tilde \mu_F(E) = \mu_F(\delta^{-1}(E)).
$$
So we may consider that $\mu_F$ is defined on the Borel sets of $(A^*,\omega^*)$. Since $\|\delta(x)\|=1$ for any $x\in X$, we have that such measure is concentrated at the unit sphere of $A^*$, which is a closed subset of $(A^*,\omega^*)$.
