Suppose that $X$ is a locally compact topological space. Let $M(X)$ denote the Banach space of regular Borel measures on $X$. It is known that the bidual of $C\_0(X)$ is a commutative $C^\*-$algebra. Denote the spectrum of $C\_0(X)^{\*\*}$ by $\tilde{X}$. 


Since $C\_0(X)^{\*\*}$ is a unital $C^\*-$algebra it can be shown that $C(X^{\infty})$ is a subspace of $C\_0(X)^{\*\*}$, where $X^{\infty}$ is the one-point compactification for $X$. Now for each $f\in C\_0(X)$ we have that $f(\infty)=0$. 


Let $x\in\tilde{X}$ and consider the point mass $\delta\_x$. Then $\delta\_x|\_{C\_0(X)}$ is a multiplicative linear functional on $C\_0(X)$ and therefore it belongs to $X^{\infty}$. Let $\pi(x)$ be the corresponding element in $X^{\infty}$. Therefore we can define a mapping $\pi:\tilde{X}\to X^{\infty}$. 


If $x\in X$ is an isolated point then it can be shown that $\pi^{-1}(x)=\{x\}$. What about the converse, if $\pi^{-1}(x)=\{x\}$ can we conclude that $x$ is an isolated point of $X$?



On the relation between the set of extreme points of the unit ball of $M(X)$ and $M(X)^{**}$
Yes, you can conclude this. Let ${\rm Bor}(X)$ be the set of bounded Borel measurable scalar functions on $X$, equipped with sup norm. Then ${\rm Bor}(X)$ embeds isometrically as a subspace of $C\_0(X)^{\*\*}$. Now let $x$ be any non-isolated point of $X$ and find a net $(x\_\lambda)$ in $X$ which converges to $x$, such that $x\_\lambda \neq x$ for all $\lambda$. Now each $\delta\_{x\_\lambda}$ belongs to $M(X)$, so by the natural embedding of $M(X)$ in $M(X)^{\*\*}$ it gives rise to an element $\tilde{x}\_\lambda$ of $\tilde{X}$. Finally, since $\tilde{X}$ is compact we can find a cluster point $y$ of the net $(\tilde{x}\_\lambda)$ in $\tilde{X}$. Then $\pi(y) = \pi(\tilde{x}) = x$, but $y \neq \tilde{x}$; we know this because the function $\chi\_{\{x\}} \in {\rm Bor}(X)$ which is $1$ at $x$ and $0$ elsewhere, when realized as an element of $C\_0(X)^{\*\*} \cong C(\tilde{X})$, evaluates to $1$ at $\tilde{x}$ and to $0$ at $y$.


