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I am reading Kallenberg's Random Measures, Theory and Applications, 1st ed. 2017.

Background:

Let $S$ be a Polish space equipped with its Borel sigma algebra. Denote by $\mathcal{M}_S$ the space of locally finite measures on $S$. We equip $\mathcal{M}_S$ with the vague topology, defined as the topology generated by maps $\pi_f : \mu \rightarrow \mu f = \int f d\mu$ continuous for $f \in \hat{\mathcal{C}}(S)$ the continuous bounded functions $f : S \rightarrow \mathbb{R}_+$ with bounded support. Similarly, letting $\hat{\mathcal{M}_S}$ denote the set of bounded measures on $S$, we define the weak topology as the topology on $\hat{\mathcal{M}_S}$ as the topology generated by maps $\pi_f$ for all $f \in \mathcal{C}(S)$ the set of continuous and bounded functions $f : S \rightarrow \mathbb{R}_+$.

In the proof of lemma 4.6, we want to show that the vague topology is metrizable. To do this, we fix some point $s_0 \in S$, and define $B_n = \{s \in S ~:~ d(s,s_0) < n\}$. Then for every $k \in \mathbb{N}$ we choose an $f_k \in \hat{\mathcal{C}}(S)$ with $1_{B_k} \leq f_k \leq 1_{B_{k+1}}$, such that $\mu_n \overset{v}{\rightarrow} \mu$ if and only if $f_k \cdot \mu_n \overset{w}{\rightarrow} f_k \cdot \mu$ for all $k \in \mathbb{N}$. Using that the Prokhorov metric $\hat{\rho}$ metrizes the weak topology, it is then stated that the metric $$ \rho(\mu, \nu) = \sum_k 2^{-k} \{\hat{\rho}(f_k \cdot \mu, f_k \cdot \nu) \wedge 1 \} $$ metrizes the vague topology.

Question:

I am confused as to why the above implies that $\rho$ metrizes the vague topology, and I believe that there must be some fact that I am missing. I understand that $\rho(\mu_n, \mu) \rightarrow 0$ if and only if $f_k \cdot \mu_n \overset{w}{\rightarrow} f_k \cdot \mu$ for all $k \in \mathbb{N}$ if and only if $\mu_n \overset{v}{\rightarrow} \mu$. However, at this stage it is not clear to me why the vague topology is necessarily sequential, see for example Is the topology that has the same sequential convergence with a metrizable topology equivalent as that topology?.

Any help is much appreciated.

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1 Answers1

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I believe I have found an answer:

With same notation as in the question, consider $(\hat{\mathcal{M}}_S)^{\mathbb{N}}$ equipped with the product topology. Then we can define a map $T : \mathcal{M}_S \rightarrow (\hat{\mathcal{M}}_S)^{\mathbb{N}}$ by $T(\mu) = (f_k \cdot \mu)_{k \in \mathbb{N}}$. One can show that $T$ is continuous and injective, and so when restricted to its image $T$ is bijective and continuous. Further, since $\hat{\mathcal{M}}_S$ is Hausdorff, we have that $(\hat{\mathcal{M}}_S)^{\mathbb{N}}$ is Hausdorff. This gives us that $T$ is proper. Similarly, we have that $(\hat{\mathcal{M}}_S)^{\mathbb{N}}$ is first-countable, which is enough to show that $T$ is a homeomorphism into it's image. We then obtain that $\mathcal{M}_S$ is first-countable, in particular it is sequential.

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