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Given $T>0$, let us denote by $\mathcal G$ the set of the Borel measures $\nu$ on $\mathbb R^d$ bounded by a constant $G$ (i.e. $\int_{\mathbb R^d}d\nu\leq G$) and endowed with the weak* convergence topology (see The wide or weak* topology what I mean for weak * convergence topology) and, if necessary, with finite $p^{\text{th}}$ moment.

I want to know if the space $C^0([0, T], \mathcal G)$ is metrizable.

I know that it depends on the metrizability of the space $\mathcal G$. Indeed, once $\mathcal G$ is metrizable, $C^0([0, T], \mathcal G)$ has a natural topology and a natural metric, which is the sup norm.

In the web page above, it is written that the weak* convergence topology is metrizable but I do not know if it can be applied to the particular space $\mathcal G$ too. Moreover it is not specified how could be the metric (Wasserstein?).

So my question is: is $\mathcal G$ metrizable? If not, what are the conditions for which $\mathcal G$ can be metrizable? And what is the metric?

Thank You

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The Levy-Prohorov metric $d$ metrizes weak* convergence of Borel probability measures on $\mathbb R^{d}$ [ https://en.wikipedia.org/wiki/L%C3%A9vy%E2%80%93Prokhorov_metric ]. We can metrize $\mathcal G$ by $D(\mu, \nu)=|\mu_1 (\mathbb R^{d})-\nu_1 (\mathbb R^{d})+|d(\mu, \nu)|$ where $\mu_1=\frac {\mu} {\mu (\mathbb R^{d})}$ and $\nu_1=\frac {\nu} {\nu (\mathbb R^{d})}$.

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  • $\begingroup$ Ok, thank you. The Wikipedia page says that it metrizes the weak convergence topology, not the weak* convergence one. Is it the same? Moreover, is it true also for bounded Borel measures (in the sense above)? $\endgroup$
    – Redeldio
    Sep 16, 2020 at 9:46
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    $\begingroup$ Probabilists say weak convergence for weak* convergence! Also, I have edited my answer. @Redeldio $\endgroup$ Sep 16, 2020 at 9:49
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    $\begingroup$ @Redeldio The metric I have defined does give weak* convergence. You only need the fact that $\mu_n \to \mu$ iff $\mu_n (\mathbb R^{d}) \to \mu (\mathbb R^{d})$ and $\frac {\mu_n} {\mu_n (\mathbb R^{d})} \to \frac {\mu } { \mu (\mathbb R^{d})}$ $\endgroup$ Sep 16, 2020 at 9:54
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    $\begingroup$ @Redeldio As mentioned in the following link convergence in Wassersetin metric is not equivalent to weak* convergence: en.wikipedia.org/wiki/Wasserstein_metric $\endgroup$ Sep 16, 2020 at 10:17
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    $\begingroup$ @Redeldio If $\mu, \mu_n$ are finite measures on $X=\mathbb R^{d}$ then $\mu_n \to mu$ in weak* topology iff $\mu_n(X) \to \mu(X)$ and the probability measures $\frac {\mu_n} {\mu_n(X)}$ converge to the probability measure $\frac {\mu} {\mu(x)}$ in weak* topology. This is immediate from definition of weak* convergence. $\endgroup$ Sep 16, 2020 at 23:22

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