This is a setting from Ken Iti Sato's Levy Processes. Define $$g(z,x) = e^{i \langle z,x \rangle} -1 - i\langle z,x \rangle c(x)$$ where $c(x) = 1+o(|x|)$ as $|x|\to 0$ and $c(x)$ is some bounded measurable function from $\mathbb{R}^d$ to $\mathbb{R}$ such that $c(x) = O(\frac{1}{|x|})$ as $|x| \to \infty$.
Suppose $\nu_n$ is a measure on $\mathbb{R}^d$ such that $\nu_n(\{0\})=0$ and $$\int_{\mathbb{R}^d} (|x|^2 \wedge 1) \nu_n (dx) < \infty.$$
Define $\rho_n(dx) = (|x|^2 \wedge 1)\nu_n (dx)$. Suppose $\rho_n$ is tight in the sense that $\sup_n \rho_n(\mathbb{R}^d)<\infty$ and $\lim_{l \to \infty} \sup_n \int_{|x|>l} \rho_n(dx)=0$. Then by the selection theorem there exists some subsequence $\rho_{n_k}$ that converges to some finite measure $\rho$.
Now define $\nu$ by $\nu(\{0\})=0$ and $\nu(dx) = (|x|^2 \wedge 1)^{-1} \rho(dx)$ on $\{|x|>0\}$.
Let $E$ be the set of $\epsilon>0$ for which $\int_{|x|=\epsilon} \rho(dx)=0$. Let $g:\mathbb{R}^d \to \mathbb{R}$ that is bounded and continuous in $x$.
Questions.
Why is $$\lim_{k\to \infty} \int_{|x|>\epsilon} g(z,x) (|x|^2 \wedge 1)^{-1} \rho_{n_k}(dx) = \int_{|x|>\epsilon} g(z,x) (|x|^2 \wedge 1)^{-1} \rho(dx)?$$
Further, how do we get $$\lim_{E \ni \epsilon \downarrow 0}\lim_{k\to \infty} \int_{|x|>\epsilon} g(z,x) (|x|^2 \wedge 1)^{-1} \rho_{n_k}(dx) = \int_{\mathbb{R}^d} g(z,x)\nu(dx)?$$
Finally, why do we get $$\lim_{\epsilon \downarrow 0} \sup_n |\int_{|x|\le \epsilon} (g(z,x) + \frac{1}{2} \langle z,x\rangle^2)(|x|^2 \wedge 1)^{-1} \rho_n(dx) | =0?$$
I know that $\rho_{n_k} \to \rho$ implies that $\rho_{n_k}(|x|>\epsilon) \to \rho(|x|>\epsilon)$ for $\epsilon \in E$, but I don't know why we get the limits to pass when we integrate $g(x) (|x|^2 \wedge 1)^{-1}$ on this set. Indeed, $(|x|^2 \wedge 1)^{-1}$ is not even bounded so for the bottom limit, how do we even know that $\nu(\mathbb{R}^d)$ is finite?
