Can weak convergence of probability measures be characterized by countably many functions without having a limit a priori? Let $(\mu_n)_{n \geq 1}$ be a sequence of Borel probability measures on $\mathbb{R}^d$. I'd like to know the following: Does there exist a countable family $(f_k)_{k \geq 1}$ of continuous, bounded real-valued functions with the following property:
If $\text{lim}_{n}\int f_k d\mu_n$ exists in $\mathbb{R}$ for each $k \geq 1$, then there exists a unique Borel probability measure $\mu$ such that $\mu_n \underset{n \to \infty}{\longrightarrow} \mu$ weakly?
Clearly, it'd be sufficient to take a countable, dense subset of $C_b(\mathbb{R}^d)$ - only problem, such set doesn't exist ;-). On the other hand, the Riesz-Markov-representation theorem shows that a dense countable subset of $C_0(\mathbb{R}^d)$ (the continuous functions vanishing at infinity) [which exists - $C_0$ is separable] is "too small" in the sense that it allows mass to spread out at infinity, which yields that the limit measure $\mu$ is in general only a sub-probability measure. Next, I was thinking about the uniformly continuous bounded functions - but again: not separable. Next thought: Consider the vector space spanned by $C_0$ and $1$. But for this vector lattice, the positive, linear, normalized functional $J: f \mapsto \text{lim}_n\int f d\mu_n$ is not continuous (also called $\sigma$-continuous), meaning it does not hold 
$f_l \to 0$ pointwise decreasing from above $\implies$ $J(f) \to 0$
(which is, however, true for the vector lattice $C_0$, which is essential for the proof of Riesz-Markov representation). Hence the classical Daniell-Stone theory does not apply, so we cannot obtain the desired limit measure (at least not by this method).
Any comment or help on this is much appreciated!
 A: No such family exists.
Let $f_k$ be a countable subset of $C_b(\mathbb{R}^d)$ and consider the Banach space $X \subset C_b(\mathbb{R}^d)$ which is the closed linear span of the $f_k$.  Note that $X$ is separable.  Choose your favorite sequence $x_n \in \mathbb{R}^d$ with $|x_n| \to \infty$.  The point mass measures $\mu_n = \delta_{x_n}$ can be viewed as bounded linear functionals on $X$ of norm $1$.  Since $X$ is separable, the unit ball of $X^*$ is weak-* compact and metrizable.  Therefore, passing to a subsequence, we can suppose that the sequence $\mu_n$ is weak-* convergent in $X^*$, and in particular, $\lim_n \int f_k\,d\mu_n = 
\lim_n f_k(x_n)$ exists for every $k$.  But the sequence of measures $\mu_n = \delta_{x_n}$ clearly does not converge weakly to any probability measure (and indeed the sequence converges vaguely to 0).
To say the same thing in a different way, we could suppose without loss of generality that $0 \le f_k \le 1$ for every $k$, and then identify each $\delta_{x_n}$ with the sequence $(f_1(x_n), f_2(x_n), \dots)$ in the Hilbert cube $[0,1]^{\mathbb{N}}$.  Since the latter is compact metrizable, we can pass to a subsequence so that $f_k(x_n)$ converges for every $k$.
