Suppose $g(x):\mathbb R^p\to \mathbb R$ is a bounded continuous function such that:
\begin{equation}\label{8.3}
\quad g(x)=1+o(|x|),\quad \hbox{ as } |x| \rightarrow 0 \quad \hbox{ and } \quad g(x)=O(1 /|x|),\quad \hbox{ as } |x| \rightarrow \infty.
\end{equation}
i.e.:
\begin{equation}\label{8.4}
\lim_{x \to 0} \frac{g(x)-1}{|x|} =0 \quad \hbox{ and } \quad |g(x)| \leq M \frac{1}{|x|},\, \forall\, x \geq x_0.
\end{equation}
Let $(\mu_n)_{n \in \mathbb{N}}$ and $\mu$ be p-dimensional Levy measures such that:
\begin{equation}\label{L}\tag{L}
\mu_n(E) \longrightarrow \mu(E),\quad \hbox{ as }n \to \infty\quad\left( \forall E \in \mathcal C_\mu, \hbox{class of continuity sets, and }\,0 \notin \bar E.\right)
\end{equation}
Remember that $\mu$ is a Levy measure if $\int_{|x|\leq \epsilon} |x|^2 d \mu < 0$ and $\int_{|x|> \epsilon} d\mu< \infty$, for all $\epsilon >0$ ( There are other equivalent definitions).
Question
Suppose
\begin{equation}\label{H}\tag{H}
\int_{\mathbb R^p} |x|^2d\mu_n \leq C< \infty, \quad \forall\, n
\end{equation}
(Given (\ref{L}), it is strightforward to show that (\ref{H}) implies $\int_{\mathbb R^p} |x|^2d\mu \leq C$).
So, assuming (\ref{L}) and (\ref{H}), how to show that the following convergence:
\begin{equation}\label{abc}\tag{I}
\int_{\mathbb R^p} x[1-g(x)] d\mu_n \longrightarrow \int_{\mathbb R^p} x[1-g(x)] d\mu < \infty, \quad (n \to \infty)
\end{equation}
Remark
If (\ref{H}) is not true, we have a counterexample. Suppose $p=1$, $g(x)=\frac{1}{1+|x|^2}$. Take $d\mu_n(x)=\frac{1}{x^2}\mathbf{1}_{(0,n)}dx$ and $d\mu(x)=\frac{1}{x^2}\mathbf{1}_{(0,\infty)}dx$. Note that for any $\epsilon>0$:
$$\int_{|x|>\epsilon}d\mu_n = \int_{\epsilon}^n\frac{1}{x^2}dx=\frac{1}{\epsilon}- \frac{1}{n} \longrightarrow \frac{1}{\epsilon} = \int_{|x|>\epsilon} d\mu(x), \quad(n \to \infty)$$
I think that this is sufficient to prove (\ref{L}).
Now, note that $x[1-g(x)]= \frac{x^3}{1+x^2}$. Thus, as $n \to \infty$:
$$\int_{\mathbb R} x[1-g(x)] d\mu_n = \int_{0}^n \frac{x}{1+x^2} dx= \frac{1}{2}\ln(1+n^2)\longrightarrow \infty = \int_{\mathbb R} x[1-g(x)] d\mu$$
Help!
(The criteria for choosing to give the bounty will be for whoever answers the correct answer the fastest.)
Best Answer
In order to obtain
\begin{equation} \label{WC}\tag{WC} \int_{\mathbb R^p} x[1-g(x)] d\mu_n \longrightarrow \int_{\mathbb R^p} x[1-g(x)] d\mu , \quad (n \to \infty) \end{equation}
from the narrow convergence of $\mu_n$ to $\mu$, you need that $|x|$ to be uniformly integrable w.r.t. $(\mu_n)_{n\geq 1}$. This can be obtained from
\begin{equation} \int_{\mathbb R^p} |x|^2d\mu_n \leq C< \infty, \quad \forall\, n. \end{equation}
Indeed, let $f_M$ be a bounded continuous function statisfying $0\leq f\leq 1$ and $f(x)=0$ for all $x$ s.t. $|x|\leq M$ and $f(x)=1$ for all $x$ s.t. $|x|\geq 2M$. Hence,
$$ \int_{\mathbb R^p} |x|^2d\mu_n\geq \int_{\mathbb R^p} f_M(x) |x|^2d\mu_n \geq M \int_{\mathbb R^p} f_M(x) |x|d\mu_n. $$ Thus $$ \sup_{n\geq 1}\int_{\mathbb R^p} f_M(x) |x|d\mu_n \leq \frac{1}{M}\int_{\mathbb R^p} |x|^2d\mu_n\leq \frac{C}{M}. $$
Take $\varepsilon >0$ and $M$ large enough so that the
\begin{equation} \label{abs}\tag{I} C/M<\varepsilon\qquad\text{ and }\qquad\int_{\mathbb R^p} f_M(x) |x|d\mu <\varepsilon. \end{equation}
We then have $$ \int_{\mathbb R^p} x[1-g(x)] d\mu_n=\int_{\mathbb R^p}f_M(x) x[1-g(x)] d\mu_n+\int_{\mathbb R^p} (1-f_M(x))x[1-g(x)] d\mu_n. $$ It follows from the definition of $f_M$ that $x\to (1-f_M(x)) x[1-g(x)]$ is bounded continuous so \begin{equation}\label{C}\tag{C} \int_{\mathbb R^p} (1-f_M(x))x[1-g(x)] d\mu_n\xrightarrow[n\to\infty]{}\int_{\mathbb R^p} (1-f_M(x))x[1-g(x)] d\mu \end{equation}
Finally,by \eqref{abs}, $$ \begin{multline*} \left|\int_{\mathbb R^p} x[1-g(x)] d(\mu_n-\mu)\right|\leq\left|\int_{\mathbb R^p} f_M(x)x[1-g(x)] d(\mu_n-\mu)\right|+\left|\int_{\mathbb R^p} (1-f_M(x))x[1-g(x)] d(\mu_n-\mu)\right|\\ \leq 2\varepsilon\|1-g\|_{\infty}+\left|\int_{\mathbb R^p} (1-f_M(x))x[1-g(x)] d(\mu_n-\mu)\right| \end{multline*} $$ but $$ \left|\int_{\mathbb R^p} (1-f_M(x))x[1-g(x)] d(\mu_n-\mu)\right| $$ goes to $0$ as $n$ goes to infinity. This gives the result.
Addendum: If the narrow converge is not assumed but rather that $$ \mu_n(E)\xrightarrow[n\to\infty]{}\mu(E) $$ for all measurable $E$ such that $\mu(\partial E)=0$ and $0\notin E$, I think one can proceed as follows:
First, note that this condition implies the narrow convergence of the sequence $\mu_n\left(B(0,\delta)^c\cap \cdot \right)$ to $\mu\left(B(0,\delta)^c\cap \cdot \right)$ for all $\delta\in \mathbb{R}_{+}\setminus D$ where $D$ is some denumberable subset of $\mathbb{R}_{+}$. So the previous argument gives \eqref{WC} for integrals over $B(0,\delta)^c$.
Now, Hypotheses on $g$ implies that, for any $\varepsilon>0$, there exists $\delta\in \mathbb{R}_{+}\setminus D$ s.t., for all $x$ with $|x|<\delta$ $$ |1-g(x)|\leq \varepsilon |x| $$ Hence, $$ \begin{align*} \left|\int_{\mathbb R^p} x[1-g(x)] d(\mu_n-\mu)\right|&\leq\left|\int_{B(0,\delta)} x[1-g(x)] d(\mu_n-\mu)\right|+\left|\int_{B(0,\delta)^c} x[1-g(x)] d(\mu_n-\mu)\right|\\ &\leq \varepsilon\int_{B(0,\delta)} |x|^2 d(\mu_n+\mu)+\left|\int_{B(0,\delta)^c} x[1-g(x)] d(\mu_n-\mu)\right|\\ &\leq 2\varepsilon C+\left|\int_{B(0,\delta)^c} x[1-g(x)] d(\mu_n-\mu)\right| \end{align*} $$ which allows to conclude. Just a last remark: