Let $\mathcal{C}:=\mathcal{C}(\mathbb{R})$ be the space of continuous functions on $\mathbb{R}$ and $\mathcal{C}_b$ its subspace consisting of bounded elements. Define for $\phi(x):=1+|x|$,

$$ \mathcal{C}_{\phi}=\big\{f\in\mathcal{C}: \frac{f}{\phi}\in\mathcal{C}_b\big\} $$

Let $ M=M(\mathbb{R})$ be the space of bounded measures on $\mathbb{R}$ and denote by $M_1$

$$M_1=\{\mu\in M: \int_{\mathbb{R}} |x|d|\mu|<\infty\}$$

Define $\mathbf{P}\subset M$ and $\mathbf{P}_1\subset M_1$ as spaces of probability measures.

Let $M$ be endowed with the topology generated by the sets

$$ U_{f_1, f_2,..., f_n, r}(\mu)=\big\{ \nu\in M: |\int_{\mathbb{R}}f_id\mu-\int_{\mathbb{R}}f_id\nu|<r|\big\} $$

wherer $f_i\in\mathcal{C}_b$ and $r>0$.

A sequence $\{\mu_n\}\subset \mathbf{P}$ is called weakly convergent to a measure $\mu\in \mathbf{P}$ if for every $f\in\mathcal{C}_b$, one has

$$ \lim_{n\to\infty}\int_{\mathbb{R}}fd\mu_n=\int_{\mathbb{R}}fd\mu $$

The weak convergence is denoted by $\Rightarrow$. Then by classical theory about probability(such as Large deviations(Deuschel-Stroock)), we know that the topology induced by $\Rightarrow$ is consistent with the restriction of the previous topology to $\mathbf{P}$. Now I would like to know if we have the similar result for $\mathbf{P}_1$.

Let $M_1$ be endowed with the topology generated by the sets

$$ U_{f_1, f_2,..., f_n, r}(\mu)=\big\{ \nu\in M_1: |\int_{\mathbb{R}}f_id\mu-\int_{\mathbb{R}}f_id\nu|<r|\big\} $$

wherer $f_i\in\mathcal{C}_{\phi}$ and $r>0$.

A sequence $\{\mu_n\}\subset \mathbf{P}_1$ is called weakly convergent to a measure $\mu\in \mathbf{P}_1$ if for every $f\in\mathcal{C}_{\phi}$, one has

$$ \lim_{n\to\infty}\int_{\mathbb{R}}fd\mu_n=\int_{\mathbb{R}}fd\mu $$

The weak convergence is denoted again by $\Rightarrow$ and I would like to know if the topology induced by $\Rightarrow$ is consistent with the restriction of the previous topology to $\mathbf{P}_1$? Many thanks!

Let $\mathcal{C}:=\mathcal{C}(\mathbb{R})$ be the space of continuous functions on $\mathbb{R}$ and $\mathcal{C}_b$ its subspace consisting of bounded elements. Define for $\phi(x):=1+|x|$,

$$ \mathcal{C}_{\phi}=\left\{f\in\mathcal{C}: \frac{f}{\phi}\in\mathcal{C}_b\right\} $$

Let $ M=M(\mathbb{R})$ be the space of bounded measures on $\mathbb{R}$ and denote by $M_1$

$$M_1=\left\{\mu\in M: \int_{\mathbb{R}} |x|d|\mu|<\infty\right\}$$

Define $\mathbf{P}\subset M$ and $\mathbf{P}_1\subset M_1$ as spaces of probability measures.

Let $M$ be endowed with the topology generated by the sets

$$ U_{f_1, f_2,\dots, f_n, r}(\mu)=\left\{ \nu\in M: |\int_{\mathbb{R}}f_id\mu-\int_{\mathbb{R}}f_id\nu|<r|\right\} $$

where $f_i\in\mathcal{C}_b$ and $r>0$.

A sequence $\{\mu_n\}\subset \mathbf{P}$ is called weakly convergent to a measure $\mu\in \mathbf{P}$ if for every $f\in\mathcal{C}_b$, one has

$$ \lim_{n\to\infty}\int_{\mathbb{R}}fd\mu_n=\int_{\mathbb{R}}fd\mu $$

The weak convergence is denoted by $\Rightarrow$. Then by classical theory about probability(such as Large deviations(Deuschel-Stroock)), we know that the topology induced by $\Rightarrow$ is consistent with the restriction of the previous topology to $\mathbf{P}$. Now I would like to know if we have the similar result for $\mathbf{P}_1$.

Let $M_1$ be endowed with the topology generated by the sets

$$ U_{f_1, f_2,\dots, f_n, r}(\mu)=\left\{ \nu\in M_1: \left|\int_{\mathbb{R}}f_id\mu-\int_{\mathbb{R}}f_id\nu\right|<r|\right\} $$

wherer $f_i\in\mathcal{C}_{\phi}$ and $r>0$.

A sequence $\{\mu_n\}\subset \mathbf{P}_1$ is called weakly convergent to a measure $\mu\in \mathbf{P}_1$ if for every $f\in\mathcal{C}_{\phi}$, one has

$$ \lim_{n\to\infty}\int_{\mathbb{R}}fd\mu_n=\int_{\mathbb{R}}fd\mu $$

The weak convergence is denoted again by $\Rightarrow$ and I would like to know if the topology induced by $\Rightarrow$ is consistent with the restriction of the previous topology to $\mathbf{P}_1$? Many thanks!