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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...

user154277
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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.

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Tomasz Kania
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  • $C_c(\mathbb{R})$ is not a Banach space under any natural norm. The completion with respect to the $\sup$-norm is $C_0(\mathbb{R})$, the space of continuous functions vanishing at infinity. – Daniel Fischer Jul 03 '14 at 14:04
  • Sure, I meant the dual under the sup-norm, which is the same as of $C_0(\mathbb{R})$. I'll clarify this. – Tomasz Kania Jul 03 '14 at 14:06
  • Thanks. But I'm actually interested in the space of all Radon measures. Or at least all Radon measures with mass 1 on the unit ball. – user154277 Jul 03 '14 at 14:44
  • Then what does Radon measure mean for you? I think that it is commonly agreed that a Radon measure on a locally compact space $X$ is a measure which pops out in the Riesz–Markov–Kakutani representation of the dual space $C_0(X)^$. Purely measure-theoretic definitions are rather tedious. As I said, if you consider metrisability of $C_0(X)^$ in the weak*-topology, then there is no chance for that as $C_0(X)$ is infinite-dimensional. (This is a general fact in basic Banach space theory.) – Tomasz Kania Jul 03 '14 at 14:49
  • I'm fine with this definition of Radon measure. But I only consider positive measures, and so this only "corresponds" to positive linear functionals on $C_c(X)$. The dual of $C_0(X)$ does also contain signed measure, so I had hope that the space of positive Radon measures is metrizable in the above mentioned topology. Perhaps I'm just confused, but I don't see why we can conclude by this result for the space of positive Radon measures. – user154277 Jul 03 '14 at 15:15
  • The space of all positive Radon measures won't be metrisable either. The reason is that it is not first-countable. You see this modifying this proof: http://math.stackexchange.com/questions/623642/the-weak-topology-on-x-is-not-first-countable-if-x-has-uncountable-dim – Tomasz Kania Jul 03 '14 at 15:22
  • @Tomek Kania: The space of positive Radon measures is metrisable (and even Polish) in contrast to the space of all Radon measures. – yada Feb 14 '15 at 08:30