Let $Y$ be an infinitely divisible random variable. 

Suppose that $Y \sim (0, \nu,0)_0$ according to the notation on Page 39, Equation (8.7), from <cite authors="Sato, Ken-Iti">_Sato, Ken-Iti_, Lévy processes and infinitely divisible distributions, [ZBL0973.60001](https://zbmath.org/?q=an:0973.60001).</cite> That is, the Lévy-Khintchine representation of the characteristic function is given by:
\begin{equation}\label{I}\tag{I}
\varphi_Y(z) = \exp\left\{ \int_\mathbb R [e^{izx} - 1] d\nu(x)   \right\}
\end{equation}

Note that if $\nu(\mathbb R)< \infty$, we can set $\lambda:= \nu(\mathbb R)$ and writte:
$$ \varphi_Y(z) = \exp\left\{\lambda \int_\mathbb R [e^{izx} - 1] d\eta(x)   \right\}, \quad d\eta(x):= d\nu(x)/\lambda$$
In this case, we have that $Y$ is a compound Poisson random variable  $Y \sim CP(\lambda, \eta)$( See page 18, Equation (4.1) from the Sato's book (reference above)).

However, in general, we don't have $\nu(\mathbb  R)< \infty$. For example, see [ this question][1] and this [other question][2], where we have infinite mass around zero. 

By the other hand, we know thar every random variable $Y$ is infinitely divisible if and only if there is a sequence $(X_n)_{n \in \mathbb N}$ of compound Poisson random variables such that, in  weak limits: 
\begin{equation}X_n \Longrightarrow Y, \quad (n\to \infty)
\end{equation}
C.f. Theorem 16.5, page 333, from <cite authors="Klenke, Achim">_Klenke, Achim_, [**Probability theory. A comprehensive course**](http://dx.doi.org/10.1007/978-1-4471-5361-0). [ZBL1295.60001](https://zbmath.org/?q=an:1295.60001).</cite>    

*So my question is*:  suppose I have a sequence $(X_n)_{n \in \mathbb N}$ of compound Poisson random variables with $X_n \sim CP(1, \eta_n)$ such that
\begin{equation}\label{II}\tag{II}
X_n \Longrightarrow Y, \quad (n\to \infty)
\end{equation}
where $Y$ has characterization given by (\ref{I}). 

**in what situations ( assumptions about $\eta_n$ ) the convergence given in (\ref{II}) implies that $Y$, with characterization given by (\ref{I}),  is in fact a compound Poisson random variable?** Or in other sufficient way, when we have $\nu(\mathbb  R)< \infty$?.

One trivial case is when $(X_n)$ has the same distribution. I.e. $\eta_n = \eta$ for all $n$. So we exclude this case.

My intuition tells me that it is not possible, given (\ref{II}), to have $\nu(\mathbb  R)< \infty$.

Help

**Update**

Using the theorem 8.7, page 41, from the Sato's book (reference above)), we have that if $f \in C_\#$ (bounded continuous function vanishing on a neighborhood of $0$), then 

$$\lim_{n \to \infty} \int_{\mathbb R} f(x) \eta_n (dx) = \int_{\mathbb R} f(x) \nu (dx)$$

So, for  any $\epsilon>0$, taking the indicator function $f_\epsilon(x) = \chi_{|x|>\epsilon}(x)= 1 $ if $|x|>\epsilon$ and $0$ other wise, we have

$$\eta_n( E_\epsilon ) \to \nu( E_\epsilon ), \quad E_\epsilon = \{x: |x|>\epsilon\}, \quad (n \to \infty) $$

Could this be a useful way?

  [1]: https://math.stackexchange.com/questions/4561024/example-of-a-levy-measure-with-infinite-mass-around-zero
  [2]: https://math.stackexchange.com/questions/4561609/example-of-a-levy-measure-with-infinite-mass-around-zero-but-with-finite-second