I have a sequence of $p$-dimensional infinitely divisible random vectors $S\_n'$, such that $S\_n' \Longrightarrow X$, as $n \to \infty$.


Suppose the following assumptions


1. The characteristic functions are given by:
$$\varphi\_{S\_n'}(u)=\exp\left\{ \int\_{\mathbb R^p} \left[e^{iu'x} - 1 - i u'x \right] \, d\nu\_n \right\}, \quad \varphi\_{X}(u) = \exp\left\{ \frac{- u'\sigma u}{2} + \int\_{\mathbb R^p} \left[e^{iu'x} - 1 - iu'x \right] d\nu \right\} $$
($\nu\_n$ and $\nu$ are Lévy measures)
2. $E[S\_n']=\int\_{\mathbb R^p}x \,d\nu\_n=0$ and
\begin{equation}\label{0}\tag{0}
\sup\_n \int\_{\mathbb R^p} |x|^2 \, d\nu\_n(x) \leq C< \infty
\end{equation}
3. Let $\mathcal C\_\#$ be the class of continuous and bounded functions vanishing on a neighborhood of $0$. Then:
\begin{equation}\label{I}\tag{I}
\int f \, d\nu\_n \to \int f \, d\nu \quad (n \to \infty),\quad \forall f \in \mathcal C\_\#
\end{equation}
4. First, for any $\epsilon>0$, define the symmetric non-neg-definite matrix $\sigma\_{n,\epsilon}$ as:
\begin{equation}\label{II}\tag{II}
\langle u, \sigma\_{n,\epsilon}u \rangle := \int\_{|x|\leq \epsilon} \langle u ,x\rangle^2 \, d\nu\_n(x), \quad u \in \mathbb R^p
\end{equation}
Then:
\begin{equation}\label{III}\tag{III}
\lim\_{\epsilon \downarrow 0} \limsup\_{n \to \infty} \left| \langle u, \sigma\_{n,\epsilon}u \rangle - \langle u, \sigma u \rangle \right|=0
\end{equation}
(Where $\sigma$ appears in the characteristic function $\varphi\_X$, see hypothesis 1)


**Question:**


Since $\int\_{\mathbb R^p} x \, d\nu\_n =0$ for all $n$, I suspect that


\begin{equation}\label{IV}\tag{IV}
\int\_{\mathbb R^p} x \, d\nu = 0
\end{equation}
How to show this?


*Remarks:*


* Note that (\ref{I}) is almost like a weak convergence of measures. If I could show that (\ref{I}) holds for every continuous and bounded function, I would actually have that $\nu\_n \Longrightarrow \nu$ ($n \to \infty$). In this case, together with (\ref{0}), I could conclude (\ref{IV}) via a uniform integrability argument. Thus, defining $\mathcal C$ be the class of continuous and bounded functions, I think my question boils down to showing that:
\begin{equation}\label{V}\tag{V}
\int f \, d\nu\_n \to \int f \, d\nu \quad (n \to \infty), \quad \forall f \in \mathcal C
\end{equation}
I spent a few hours trying to solve this using hypothesis 1-4, but I couldn't. It may also be that there is another way of showing (\ref{IV}).


I appreciate any help.


**Update**


So far, Iosif Pinelis gave a counterexample. I apologize for the omission, but I forgot to write some assumptions related to $S\_n'$. I will put the updates here.


The $S\_n'$ arise as follows: let $(X\_{jn})\_{1\leq j \leq n}$, $X\_{jn} \sim \mu\_{jn}$, be a triangular array of $p$-dimensional random vectors (row independent) such that:


* $E X\_{jn}= \int\_{\mathbb R^p} x \, d \mu\_{jn}=0$
* $\lim\_{n \to \infty} \max\_{1\leq j \leq n} P(|X\_{jn}|> \epsilon)=0$, for all $\epsilon > 0$
* Defining $S\_n := \sum\_{j=1}^n X\_{jn}$, we have $var(S\_n):=\sum\_{j=1}^n \int\_{\mathbb R^p} |x|^2 \, d\mu\_{jn} \leq C < \infty$, for all $n \in \mathbb N$.
* $S\_n \Longrightarrow X$, as $n \to \infty$. (Here, $X$ is the same limit of the $S\_n'$ given above)
* $\nu\_n(E):= \sum\_{j=1}^n \int\_E d\mu\_{jn}, \quad E\, \,\hbox{ borelian set.}$


Definig first $Y\_{jn}:= [X\_{jn}]^{[1]}$ (This the compound Poisson radom variable i.e $Y\_{jn} \sim CP(\mu\_{jn}, 1)$ ), we define:
$$S\_n' := \sum\_{j=1}^n Y\_{jn}$$
It is easy to show that $E[S\_n']=E[S\_n]=0$ and $var[S\_n']=var[S\_n]$. By an argument of Accompanying Law (section 3.7 from the [Varadhan'lecture notes](https://www.math.nyu.edu/%7Evaradhan/course/PROB.ch3.pdf)), we have that $S\_n'$ is such that
$$S\_n' \Longrightarrow X, \quad (n \to \infty)$$


The assumptions 1-4 given above are true using the theorem 8.7, page 41, from the  [Sato's book](https://zbmath.org/?q=an:0973.60001):



How to show that $\int x \,d\nu = 0$ using a pseudo-weak convergence of measures?
$\newcommand\de\delta$A counterexample is given by $p=1$, $\nu(dx):=|x|^{-5/2}\,1(0<|x|<1)\,dx$, and $\nu\_n(dx):=|x|^{-5/2}\,1(1/n<|x|<1)\,dx$.




---


The OP has added certain conditions. The only consequence of those additional conditions that matters in this context is that $\int d\nu\_n=n$ for all $n$ (so that $\mu\_{jn}:=\nu\_n/n$ be a probability measure for any $n$ and $j$). This extra condition on $\nu\_n$ is easy to satisfy by the following slight modification of the definition of the measure $\nu\_n$:
$$\nu\_n(dx):=|x|^{-5/2}\,1(\de\_n<|x|<1)\,dx,$$
where $\de\_n:=(1+\frac34\,n)^{-2/3}$ (so that $\de\_n\to0$ as $n\to\infty$).


