Show that a Levy measure (which arises from a convergence of Infinitely Divisible random vectors) has infinite total mass

Question

Given an array of probability measures $(\mu_{jn})_{1\leq j \leq n}$, all defined on borelians of $\mathbb R^p$. Let $S_n'$ a sequence of $p-$dimensional random vectors with $S_n' \Longrightarrow X$ where: $$\varphi_{S_n'}(u) = \exp\left\{ \int_{\mathbb R^p} \left[e^{iu'x} - 1 - iu'x \right] d\nu_n \right\}, \quad \nu_n(E)= \sum_{j=1}^n \int_E d\mu_{jn}$$ Using the theorem 8.7, page 41, from the from the Sato's book, we have $X$ is Infinitely Divisible (I.D.) and: $$\varphi_{X}(u) = \exp\left\{ \frac{- u'\sigma u}{2} + \int_{\mathbb R^p} \left[e^{iu'x} - 1 - iu'x \right] d\nu \right\}$$ Moreover, $$\int f d\nu_n \to \int f d\nu \quad (n \to \infty),\quad \forall f \in \mathcal C_\#$$ ($C_\#$ is the class of continuous and bounded functions vanishing on a neighborhood of $0$ ). 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} So I want to show that $\nu(\mathbb R^p)= \infty$ or $\nu(\mathbb R^p)< \infty$?

I think that the former is true, because $$\nu_n(\mathbb R^p)=\sum_{j=1}^n \int_{\mathbb R^p} d\mu_{jn}=n \to \infty$$ Note that I can'n use (\ref{asd}), because $0 \in \overline{\mathbb{R}^p}= \mathbb{R}^p$.

Answer 1

$\nu$ can be a finite measure. In fact, it can be $0$. Take $p=1$. Let $\mu_{jn}=\delta_{1/n}$ for all $j$. Then $\nu_n=n\delta_{1/n}$ and $v_n \to 0$. Note that $S_n'\to 0$ in distribution since the characteristic function of $S_n'$ is $exp(n[e^{iu/n}-1-iu/n])$ which tends to $1$ for all $u$. Hence, the hypothesis is satisfied by $(\mu_{jn})$.