# Convergence of integrals of all smooth functions implies weak convergence of measures

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?

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

$$\lim_{n \to \infty} |\mu_n(g)-\mu(g)|=0.$$