Metrization of the weak topology on the set of radon measures Let $\mathcal{M}$ denote the set of Radon measures on $\mathbb{R}$. We endow $\mathcal{M}$ with the the weakest topology such that $\mu \to \int f \, \mathrm{d} \mu$ is continuous for all $f \in C_c(\mathbb{R})$, where $C_c(\mathbb{R})$ denotes the space of all real-valued continuous compactly supported functions. Is this topology metrizable? If we require that it holds for all $f$ bounded continuous real-valued functions, then I know that it is possible. But they don't generate the same topology...
 A: The weak* topology of a dual Banach space is metrisable if and only if the underlying space is finite-dimensional. It seems to me, however, that you are interested in metrisability of the dual unit ball rather than the whole space. In this case, the answer is yes in the case of compactly supported continuous functions, and no in the case of all bounded continuous functions.
More specifically, the Banach space $C_0(\mathbb{R})$ of all continuous functions vanishing at infinity endowed with the supremum norm is separable, so the dual unit ball is weak*-metrisable. Note that by the Riesz–Markov–Kakutani representation theorem the weak*-topology of $C_0(\mathbb{R})^*$ is precisely the weak topology you have introduced above as $C_0(\mathbb{R})$ is the completion of $C_c(\mathbb{R})$ under the supremum norm.
When you consider all bounded continuous functions, you end up with the dual unit ball of a massive Banach space, namely $C(\beta \mathbb{R})$, which is very much non-separable, so the unit ball of $C(\beta\mathbb{R})^*$ is not metrisable.
