Image of collection of probability measures in $C_b(S)'$ Let $(S,d)$ be a Polisch space (i.e. a complete and separable metric space) and $\mathcal{P}$ the collection of probability  measures on the borel sigma algebra of $(S,d)$ which we denote by $\mathcal{B}$ then we can look at the space of continuous bounded functions from $S$ to $\mathbb{R}$ which we denote by $C_b(S)$ which we equip with the supremum norm turning this space into a Banach space. We define the mapping:
$$ \iota: \mathcal{P} \rightarrow C_b(S)': P \mapsto[\iota_P:C_b(S) \rightarrow \mathbb{R}: f \mapsto \int f dP] $$
where $C_b(S)'$ is the dual of $C_b(S)$ equipped with the weak$^*$ topology.
Then the statement is: $\iota$ is a bijection between $\mathcal{P}$ and the dual unit sphere of $C_b(S)'$. That $\iota$ is injective and that $\iota(\mathcal{P})$ is contained in the dual unit sphere is easy. I don't really know how to prove the other inclusion (nor do I know if this inclusion is correct).
An idea is to define $P$ by $P(A) := \lim_{n\rightarrow \infty} \psi(f_n)$ with $f_n := (1-nd(x,A))\vee 0$ for $A \subseteq S$ closed and then prove that $\psi = \iota_P$ but I don't think this is the way to go..
 A: This is false for three reasons, two trivial and one interesting.


*

*The zero linear functional is the unit ball of $C_b(S)'$, but not in the image of $\iota$.  You probably meant to say "unit sphere".

*If $P$ is any probability measure on $S$ then $f \mapsto -\int f\,dP$ is in the unit sphere of $C_b(S)'$ but is not the image of any probability measure, since it maps the constant function 1 to the number -1.  Either you should replace "probability measure" with "signed measure of total variation 1", or you should replace "unit sphere" by the "positive unit sphere", consisting of all continuous linear functionals $\ell$ such that $\|\ell\|=1$ and $\ell(f) \ge 0$ for all nonnegative $f \in C_b(S)$.  Let's denote the set of such functionals by $D^+$.
It's clear that $\iota(\mathcal{P}) \subset D^+$.  We could ask if equality holds.  The answer is:

$\iota(\mathcal{P}) = D^+$ if and only if $S$ is compact.

If $S$ is compact, then $\iota(\mathcal{P}) = D^+$ is the Riesz-Markov-Kakutani representation theorem.
For the converse, suppose $S$ is not compact; then there is a countable set $E = \{x_1, x_2, \dots\} \subset S$ that has no limit point.  Let $\delta_{x_n}$ be the probability measure that puts a point mass at $x_n$.  By Alaoglu's theorem, the unit ball of $C_b(S)'$ is weak-* compact, so the sequence $\{\iota(\delta_{x_n})\}$ has at least one weak-* cluster point; let $\ell$ be one of them.  By definition of the weak-* topology, this means that for every $f \in C_b(S)$, $\ell(f)$ is a cluster point of the sequence $\{f(x_n)\}$. In particular, if $\{f(x_n)\}$ converges then $\ell(f) = \lim_{n to \infty} f(x_n)$ since a convergent sequence has only one cluster point.
It's not hard to check that $\ell \in D^+$.  Indeed, if $f \in C_b(S)$ with $f \ge 0$ then $\{f(x_n)\}$ is a sequence of nonnegative numbers, hence any cluster point of that sequence must also be nonnegative, since $[0,\infty)$ is closed.
But $\ell$ is not in the image of $\iota$.  Suppose that it were, so that $\ell = \iota(\mu)$ for some probability measure $\mu$.  For each $k \ge 1$, let $E_k = \{x_k, x_{k+1}, \dots\} \subset E$.  Fix a compatible metric $d$ on $S$ and set
$$f_k(x) = \max(0, 1 - k d(E_k, x)).$$
Note that $f_k$ is continuous, $0 \le f_k \le 1$, and $f_k(x) = 0$ if $d(x, E) \ge 1/k$.  Moreover, since $f_k(x_n) = 1$ for $n \ge k$, we have $\ell(f_k) = \lim_{n \to \infty} f_k(x_n) = 1$.  So we have
$$\lim_{k \to \infty} \int f_k\,d\mu = \lim_{k \to \infty} \ell(f_k) = \lim_{k \to \infty} 1 = 1. \tag{*}$$
Now I claim that $f_k \to 0$ pointwise.  Fix $x \in S$.  Since $\bigcap_k E_k = \emptyset$, we can choose $M$ so large that for all $k \ge M$ we have $x \notin E_k$.   Since $E$ has no limit points, neither does $E_M$.  Thus for all $k \ge M$ we have $d(x, E_k) \ge d(x, E_M) > 0$.  Suppose $k > \max(M, 1/d(x,E_M))$.  Then $1/k < d(x,E_M) \le d(x, E_k)$, so we have $f_k(x) = 0$.  In particular, $\lim_{k \to \infty} f_k(x) = 0$, and $x$ was arbitrary.
By the dominated convergence theorem, we have $\lim_{k \to \infty} \int f_k\,d\mu = \int 0\,d\mu = 0$ which contradicts (*).
