Show that a Levy measure $\nu$ (which arises from a convergence of Infinitely Divisible random vectors) is such that $\int x d\nu(x)=0$ Let $(X_{jn})_{1\leq j \leq n}$ be a triangular array of $p-$dimensional random vectors (row independent). Suppose $X_{jn} \sim \mu_{jn}$ and
 1. $\,\, E X_{jn}= \int_{\mathbb R^p} x d \mu_{jn}=0$
 
2. $\,\,\lim_{n \to \infty} \max_{1\leq j \leq n} P(|X_{jn}|> \epsilon)=0$, for all $\epsilon > 0$
 3. $\,\,var(S_n):=\sum_{j=1}^n \int_{\mathbb R^p} |x|^2 d\mu_{jn} \leq C < \infty$, for all $n \in \mathbb N$.
Assume that $S_n := \sum_{j=1}^n X_{jn} \Longrightarrow X $, form some $X$.
Now, consider $Y_{jn} \sim CP(1,\mu_{jn})$ [compound Poisson distribution, where the paramenter of the Poisson r.v. is $\lambda =1$ for all $(j,n)$ and the coumpounded vectors  are copies of $X_{jn}$]. Define
$$S_n' := \sum_{j=1}^n Y_{jn}$$
It is easy to show that $E[S_n']=E[S_n]=0$ and $var[S_n']=var[S_n]$. Moreover, we can show that the characteristic function of $S'_n$ is given by:
$$\varphi_{S_n'}(u)=\exp\left\{ \int_{\mathbb R^p} \left[e^{iu'x} - 1 \right] d\nu_n \right\} = \exp\left\{ \int_{\mathbb R^p} \left[e^{iu'x} - 1 - iu'x \right] d\nu_n \right\}, \quad \nu_n(E):= \sum_{j=1}^n \int_E d\mu_{jn}, \quad E\, \,\hbox{ borelian set.}$$
By an argument of Accompanying Law (section 3.7 from the Varadhan'lecture notes), we have that $S_n = \sum_{j=1}^n X_{jn} \Longrightarrow X $ if and only if
$$S_n'= \sum_{j=1}^n Y_{jn} \Longrightarrow X $$  
Using the theorem 8.7, page 41, from the  Sato's book, we have $X$ is  Infinitely Divisible (I.D.) and its characteristic function is:
$$\varphi_{X}(u) = \exp\left\{ \frac{- u'\sigma u}{2} + \int_{\mathbb R^p} \left[e^{iu'x} - 1 - iu'x \right] d\nu \right\}.$$
Moreover,
$$\int f d\nu_n \to \int f d\nu \quad (n \to \infty),\quad \forall f \in \mathcal C_\#$$
($\mathcal C_\#$ is the class of continuous and bounded functions vanishing on a neighborhood of $0$ ). The mentioned theorem has another implication involving $\sigma$, but I don't think it will be useful to mention it. According to this question, the last integral convergence is equivalent to
\begin{equation}\label{asd}\tag{I}
\nu_n(E) \to \nu(E), \quad \forall E \in \mathcal{C}_\nu, \,\, 0 \notin \bar E
\end{equation}
Where $\bar E$ is clousure of the borelian $E$.
Question:
Since $\int_{\mathbb R^p} x d\nu_n = \sum_{j=1}^n \int_{\mathbb R^p} x d\mu_{jn} = 0$ for all $n$, I suspect that $\int_{\mathbb R^p} x d\nu = 0$.  How to show this?
Although each $\nu_n$ is not a probability measure (since $\nu_n$ is a sum of $n$ probability measures), convergence in (\ref{asd}) looks a lot like a weak convergence of measures. Furthermore, given that $\sup_n \int x^2d\nu_n(x) < C $, I could apply some similar uniform integrability result to conclude that $\int x d\nu_n(x) \to \int x d\nu (x)$. Given that $\int x d\nu_n(x) =0$, I would have the desired result. But I don't know how to do this rigorously.
 A: For simplicity, we only consider the case $p=1$.
At first, from (I)($\nu_n\to\nu$) and
\begin{equation*}
 \sup_n\sum_{j=1}^{n} \int_{\mathbb{R}} |x|^2\mathrm{d}\mu_{jn}\le C <\infty,
\end{equation*}
we could get
\begin{equation*}
 \int_{\mathbb{R}} |x|^2\mathrm{d}\nu<\infty,\qquad \int_{\mathbb{R}} |x|\mathrm{d}\nu<\infty.
\end{equation*}
Now we could prove following equality,
\begin{equation*}
 \lim_{n\to\infty} \int_{\mathbb{R}} x\,\mathrm{d}\nu_n=\int_{\mathbb{R}} x\,\mathrm{d}\nu. \quad \tag{1}
\end{equation*}
Let
\begin{equation*}
 f_k(x)=(k+1-|x|)^+\wedge 1, 
\end{equation*}
then
\begin{gather*}
 1-f_k(x)=(|x|-k)^+\wedge 1 \leq 1_{\{|x|\geq k\}} . \\
 \int_{\mathbb{R}} x\,\mathrm{d}\nu_n= \int_{\mathbb{R}} xf_k(x)\,\mathrm{d}\nu_n +
 \int_{\mathbb{R}} x(1-f_k(x))\,\mathrm{d}\nu_n
\end{gather*}
Now from (I),
\begin{equation*}
 \lim_{n\to\infty}\int_{\mathbb{R}} xf_k(x)\,\mathrm{d}\nu_n
 = \int_{\mathbb{R}} xf_k(x)\,\mathrm{d}\nu, \quad \forall k\ge 1. \tag{2}
\end{equation*}
Meanwhile,
\begin{align*}
 &\Big|\int_{\mathbb{R}} x(1-f_k(x))\,\mathrm{d}\nu_n\Big|
        \le \int_{\mathbb{R}} |x|\,|1-f_k(x)|\,\mathrm{d}\nu_n \\
 &\quad \le \int_{\mathbb{R}} |x|1_{\{|x|\geq k\}}\,\mathrm{d}\nu_n
  \le \frac1k \int_{\mathbb{R}} |x|^2\,\mathrm{d}\nu_n \le \frac Ck, 
\end{align*}
and
\begin{equation*}
\lim_{k\to\infty}\sup_n\Big|\int_{\mathbb{R}} x(1-f_k(x))\,\mathrm{d}\nu_n\Big|=0.    \tag{3} 
\end{equation*}
Form (3), for each $\varepsilon>0$, there exists a $k_0$ such that $\forall k\ge k_0 $,
\begin{gather*}
 -\varepsilon< \int_{\mathbb{R}} x\,\mathrm{d}\nu_n-\int_{\mathbb{R}} xf_k(x)\,\mathrm{d}\nu_n  < \varepsilon,\\
-\varepsilon + \int_{\mathbb{R}} xf_k(x)\,\mathrm{d}\nu_n 
< \int_{\mathbb{R}} x\,\mathrm{d}\nu_n  
< \int_{\mathbb{R}} xf_k(x)\,\mathrm{d}\nu_n +\varepsilon \tag{4}
\end{gather*}
Let $n\to\infty$ in (4), get
\begin{align*}
  -\varepsilon + \int_{\mathbb{R}} xf_k(x)\,\mathrm{d}\nu
  &< \liminf_{n\to\infty}\int_{\mathbb{R}} x\,\mathrm{d}\nu_n \\ 
  &< \limsup_{n\to\infty}\int_{\mathbb{R}} x\,\mathrm{d}\nu_n
  < \int_{\mathbb{R}} xf_k(x)\,\mathrm{d}\nu_n +\varepsilon  
\end{align*}
At last, letting $k\to\infty$, get (1).
