If $\int f\:{\rm d}\mu_n\xrightarrow{n\to\infty}\int f\:{\rm d}\mu$ for all $f\in C_0$, can we conclude $\mu_n\xrightarrow{n\to\infty}\mu$ weakly?

Question

Let

• $E$ be a locally compact Hausdorff space (if it's a considerably simplification, assume $E=\mathbb R$)
• $C_0(E)$ denote the space of continuous functions vanishing at infinity equipped with the supremum norm
• $\mu_n,\mu$ be probability measures on $(E,\mathcal B(E))$ for $n\in\mathbb N$

By definition, $\mu_n\xrightarrow{n\to\infty}\mu$ weakly if $$\int f\:{\rm d}\mu_n\xrightarrow{n\to\infty}\int f\:{\rm d}\mu\tag1$$ for all bounded continuous $f:E\to\mathbb R$.

If we only know that $(1)$ holds for all $f\in C_0(E)$, are we still able to conclude $\mu_n\xrightarrow{n\to\infty}\mu$ weakly?

This might be related to the notion of a separating class: Given a family $\mathcal F$ on Radon measures, a class $\mathcal C$ of measurable real-valued functions is called separating, if for all $\nu_1,\nu_2\in\mathcal F$ the implication $$\left(\int f\:{\rm d}\mu=\int f\:{\rm d}\nu\;\text{for all }f\in\mathcal C\cap\mathcal L^1(\mu)\cap\mathcal L^1(\nu)\right)\Rightarrow\mu=\nu\tag2$$ holds. Maybe we can show that $C_0(E)$ is separating. In that case, a reference to a textbook would be enough for me.

EDIT: I guess we obtain that $\mu,\mu_n$ are uniquely determined by their actions on $C_0(E)$-functions by the Riesz–Markov theorem. Can we conclude from this?

Answer 1

The convergence in question is sometimes called vague convergence. (In fact, there are two slightly different definitions of vague convergence in literature, but I digressed.) Vague convergence is in general different from weak convergence because it is too weak to detect mass escaping towards infinity (i.e. failure of tightness). On the other hand, if we can somehow guarantee that no mass escapes, vague convergence often improves to weak convergence.

Here, let's consider the case $E = \mathbb{R}^d$.

Proposition. Let $\mu_n$ and $\mu$ be probability measures on $\mathbb{R}^d$. Then the followings are equivalent:

(1) $\mu_n \to \mu$ weakly, i.e., $\int f \, \mathrm{d}\mu_n \to \int f \, \mathrm{d}\mu$ for all $f$ in the space $C_b(\mathbb{R}^d)$ of all bounded continuous functions.

(2) $\mu_n \to \mu$ vaguely, i.e., $\int f \, \mathrm{d}\mu_n \to \int f \, \mathrm{d}\mu$ for all $f$ in the space $C_0(\mathbb{R}^d)$.

Since the implication (1) $\Rightarrow$ (2) is trivial, let us prove the other direction. Assume that $\mu_n \to \mu$ vaguely. Fix $f \in C_b(\mathbb{R}^d)$. Then for any $\chi \in C_0(\mathbb{R}^d)$ with $0 \leq \chi \leq 1$, we have $\chi f \in C_0(\mathbb{R}^d)$. By splitting $f$ as the sum $f = (1-\chi)f + \chi f$ and denoting by $\|f\|_{\sup}$ the supremum-norm of $f$, we get

\begin{align*} \left| \int f \, \mathrm{d}\mu_n - \int f \, \mathrm{d}\mu \right| \leq \|f\|_{\sup}\int (1 - \chi) \, \mathrm{d}\mu_n + \|f\|_{\sup} \int (1 - \chi) \, \mathrm{d}\mu + \left| \int \chi f \, \mathrm{d}\mu_n - \int \chi f \, \mathrm{d}\mu \right|. \end{align*}

Letting $\limsup$ as $n\to\infty$ and noting that $\int (1 - \chi) \, \mathrm{d}\mu_n = 1 - \int \chi \, \mathrm{d}\mu_n \to 1 - \int \chi \, \mathrm{d}\mu$ as $n\to\infty$,

\begin{align*} \limsup_{n\to\infty} \left| \int f \, \mathrm{d}\mu_n - \int f \, \mathrm{d}\mu \right| \leq 2\|f\|_{\sup} \left( 1 - \int \chi \, \mathrm{d}\mu \right). \end{align*}

Now it is easy to check that $\sup \{ \int \chi \, \mathrm{d}\mu : \chi \in C_0(\mathbb{R}^d) \text{ and } 0 \leq \chi \leq 1 \} = 1$, and so, the above bound can be made arbitrarily small. So the limsup in the left-hand side is zero, thus proving that $\int f \, \mathrm{d}\mu_n \to \int f \, \mathrm{d}\mu$ for any $f \in C_b(\mathbb{R}^d)$. $\square$

As for the generalization to arbitrary locally compact Hausdorff $E$, I guess the same proof will work when $E$ is a Radon space (i.e. every Borel probability measure on $E$ is inner regular). But for full generality, I have no good idea as I have not enough expertise in this direction.