Let $\mu$ be a Borel probability measure on $\mathbb{R}^d$ and and let $(\mu_n : n \in \mathbb{N})$ be a sequence of such measures. Suppose that $\mu_n(f) \to \mu(f)$ for all smooth $f$ of compact support. How can I show that $\mu_n$ converges weakly to $\mu$ on $\mathbb{R}^d$?
I have tried to prove this result by contradiction and directly, however I am unable to make any serious progress on it. I know that if for any continuous bounded $g$ I could find smooth $f_k$ of compact support converging uniformly to $g$ then for any $\varepsilon > 0$ I have $$|\mu(g)-\mu_i(g)| \leq |\mu(g)-\mu(f_k)|+|\mu(f_k)-\mu_i(f_k)|+|\mu_i(f_k)-\mu_i(g)| $$ $$ \leq \varepsilon +|\mu(f_k)-\mu_i(f_k)|+\varepsilon$$ by uniform convergence and the fact that $\mu_i$ are probability measures so we can find some $k$ such that $\|g-f_k\|_\infty \leq \varepsilon$. Then we use the fact that $\mu_i(f_k) \to \mu(f_k)$ to get $|\mu(g)-\mu_i(g)| \leq 3\varepsilon$ for all large enough $i$, giving us weak convergence.
However I don't believe that there exist smooth $f_k$ of compact support converging uniformly to $g$ for this to work (or at least I can't prove it). I know the result is true if we had a compact subset of $\mathbb{R}^d$ by the Stone-Weierstrass theorem but that does not generalise to this case which leaves me stumped as I can't find any other way to attack the problem.
How should I proceed on this question?
Best Answer
Fix $\epsilon>0$. Since $\mu$ is a probability measure we can find $R>0$ such that $$\mu(\{y \in \mathbb{R}^d; |y| \leq R\}) \geq 1-\epsilon.$$ Let $\chi: \mathbb{R}^d \to [0,1]$ be a smooth function such that $\text{supp} \, \chi \subseteq B(0,2R)$ and $\chi(y)=1$ for $|y| \leq R$. As $$\lim_{n \to \infty} \mu_n(\chi) = \mu(\chi) \geq 1-\epsilon$$ we get $$\limsup_{n \to \infty} \int_{|y| \geq 2R} d\mu_n(y) \leq \limsup_{n \to \infty} \int (1-\chi(y)) \, \mu_n(dy) \leq \epsilon.$$This implies that we can find $M>0$ such that $$\sup_{n \in \mathbb{N}} \int_{|y| \geq M} \, \mu_n(dy) + \int_{|y| \geq M} \, d\mu(y) \leq 3 \epsilon. \tag{1}$$ This shows that the sequence $(\mu_n)_{n \in \mathbb{N}}$ is tight.
Now if $g$ is a bounded continuous function, choose a smooth function $f$ with compact support such that $\sup_{|y| \leq 2M} |f(y)-g(y)| \leq \epsilon$. Moreover, pick a smooth function $\varphi$ with compact support such that $0 \leq \varphi \leq 1$, $\varphi|_{B(0,M)}=1$ and $\varphi|_{B(0,2M)^c}=0$. Then
$$|\mu_n(g)-\mu(g)| \leq |\mu_n(g \cdot \varphi)-\mu(g \varphi)| + |\mu_n(g(1-\varphi))- \mu(g(1-\varphi))|.$$
For the first term you can proceed as you suggested in your question (because $g \cdot \varphi$ has compact support). By $(1)$, the second term on the right-hand side is bounded by
$$3 \|g\|_{\infty} \epsilon.$$
Combining both considerations we get
$$\lim_{n \to \infty} |\mu_n(g)-\mu(g)|=0.$$