How to show that $\int x \,d\nu = 0$ using a pseudo-weak convergence of measures?

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 $$$$\label{0}\tag{0} \sup_n \int_{\mathbb R^p} |x|^2 \, d\nu_n(x) \leq C< \infty$$$$

3. Let $$\mathcal C_\#$$ be the class of continuous and bounded functions vanishing on a neighborhood of $$0$$. Then: $$$$\label{I}\tag{I} \int f \, d\nu_n \to \int f \, d\nu \quad (n \to \infty),\quad \forall f \in \mathcal C_\#$$$$

4. First, for any $$\epsilon>0$$, define the symmetric non-neg-definite matrix $$\sigma_{n,\epsilon}$$ as: $$$$\label{II}\tag{II} \langle u, \sigma_{n,\epsilon}u \rangle := \int_{|x|\leq \epsilon} \langle u ,x\rangle^2 \, d\nu_n(x), \quad u \in \mathbb R^p$$$$ Then: $$$$\label{III}\tag{III} \lim_{\epsilon \downarrow 0} \limsup_{n \to \infty} \left| \langle u, \sigma_{n,\epsilon}u \rangle - \langle u, \sigma u \rangle \right|=0$$$$ (Where $$\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

$$$$\label{IV}\tag{IV} \int_{\mathbb R^p} x \, d\nu = 0$$$$ 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: $$$$\label{V}\tag{V} \int f \, d\nu_n \to \int f \, d\nu \quad (n \to \infty), \quad \forall f \in \mathcal C$$$$ 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), 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:

$$\newcommand\de\delta$$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 $$n\to\infty$$).

• My dear, thank you for the counterexample. I apologize for the omission, but I forgot to write some assumptions related to $S_n'$. I have updated the question. I have been trying to solve this for a long time. Any help is appreciated.
– PSE
Commented Mar 9, 2023 at 20:02
• @PSE : I have modified the counterexample to satisfy the conditions you have added. Commented Mar 9, 2023 at 21:52
• @PSE : Do you have a response to the updated answer? Commented Mar 10, 2023 at 15:32
• I apologize for the delay. I had an issue and only saw it now. Answer accepted!
– PSE
Commented Mar 10, 2023 at 21:16