Let $(\mu_n)_{n \in \mathbb{N}}$ be a sequence of finite measures on $(\mathbb{R}^d, \mathcal{B}(\mathbb{R}^d))$ and assume that $\mu$ is also a measure on $(\mathbb{R}^d, \mathcal{B}(\mathbb{R}^d))$ which is finite outside any neighbourhood of $0$ and that
$$
\lim_{ n \rightarrow \infty } \int_{ \mathbb{R}^d} f ( x ) \mu_n (dx) = \int_{ \mathbb{R}^d} f ( x ) \mu (dx)
$$
for all bounded continuous functions $f : \mathbb{R}^d \rightarrow \mathbb{R}$ which are $0$ in some neighborhood of $0$.
Now let $g : \mathbb{R}^d \rightarrow \mathbb{R}$ be continuous and bounded and $\varepsilon > 0$. How can one show that $$ \lim_{ n \rightarrow \infty } \int_{ \{ | x | > \varepsilon\} } g ( x ) \mu_n (dx) = \int_{ \{ | x | > \varepsilon\} } g ( x ) \mu (dx) \quad ?$$
If necessary, one can assume that $\mu( x \in \mathbb{R}^d : |x| = \varepsilon)=0$. Since $\{ x \in \mathbb{R}^d : |x| = \varepsilon \}$ is the set of the possible discontinuity points of the function $g(x) 1_{ \{ | x | > \varepsilon\}}$, the latter assumption somehow reminds of the Portmanteau theorem, which does not seem to apply here.
Best Answer
Fix $\epsilon>0$ such that $\mu(\{x; |x|=\epsilon\})=0$. Then, by the dominated convergence theorem, $$\lim_{k \to \infty} \mu(\{x; \epsilon-2/k< |x|< \epsilon+2/k\})=0.$$ For given $\delta>0$ choose $k \in \mathbb{N}$ such that $$\mu(\{x; \epsilon-2/k< |x|< \epsilon+2/k\}) \leq \delta.\tag{1}$$ Take a continuous function $\varphi$ such that $0 \leq \varphi \leq 1$, $\varphi=1$ for $x$ with $\epsilon-1/k\leq|x|\leq\epsilon+1/k$ and $\varphi(x)=0$ for all $x$ with $|x|\leq \epsilon-2/k$ and $|x|\geq \epsilon+2/k$. By assumption,
$$\lim_{n \to \infty} \underbrace{\int \varphi(x) \, \mu_n(dx)}_{\geq \mu_n(\{x; \epsilon-1/k \leq |x| \leq \epsilon+1/k\})} = \underbrace{\int \varphi(x) \, \mu(dx)}_{\leq \delta} $$
and so
$$\sup_{n \geq N}\mu_n(\{x; \epsilon-1/k \leq |x| \leq \epsilon+1/k\})\leq 2 \delta \tag{2}$$ for $N =N(\delta)$ sufficiently large.
Now pick continuous functions $\chi_k$, $k \in \mathbb{N}$, with the following properties: $0 \leq \chi_k \leq 1$, $\chi_k=0$ on $B(0,\epsilon-\frac{1}{k})$ and $\chi_k=1$ on $\mathbb{R}^d \setminus B(0,\epsilon)$. If $g$ is bounded and continuous, then by $(1)$ and $(2)$
\begin{align*} \left| \int_{|x| \geq \epsilon} g(x) \, \mu_n(dx) - \int_{|x| \geq \epsilon} g(x) \, \mu(dx) \right| &\leq \left|\int \chi_k(x) g(x) \, \mu_n(dx)- \int \chi_k(x) g(x) \, \mu(dx) \right| \\ &\quad + \left| \int g(x) (1_{\{|x| \geq \epsilon\}}-\chi_k(x)) \, \mu_n(dx) \right| \\ &\quad + \left| \int g(x) (1_{\{|x| \geq \epsilon\}}-\chi_k(x)) \, \mu(dx) \right| \\ &\leq 3 \delta \|g\|_{\infty} + \left|\int \chi_k(x) g(x) \, \mu_n(dx)- \int \chi_k(x) g(x) \, \mu(dx) \right| \end{align*}
for all $n \geq N$. Since the product $g \chi_k$ is continuous, bounded and zero in a neighbourhood of zero, it follows that the second term on the right-hand side converges to $0$ as $n \to \infty$. Hence,
$$\limsup_{n \to \infty} \left| \int_{|x| \geq \epsilon} g(x) \, \mu_n(dx) - \int_{|x| \geq \epsilon} g(x) \, \mu(dx) \right| \leq 3 \delta \|g\|_{\infty}.$$
As $\delta>0$ is arbitrary, this proves
$$\lim_{n \to \infty} \int_{|x| \geq \epsilon} g(x) \, \mu_n(dx) = \int_{|x| \geq \epsilon} g(x) \, \mu(dx).$$