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 Sato, Ken-Iti, Lévy processes and infinitely divisible distributions, ZBL0973.60001. 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 and this other question, 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 Klenke, Achim, Probability theory. A comprehensive course. ZBL1295.60001.

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}). For the characteristic function of $X_n$, see the second Updated remark below.

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

Updated remarks

1.- 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?

2.- After the commment from Christophe Leuridan, I think it is necessary to specify that $X_n \sim CP(1, \eta_n)$, means that $\lambda_n = 1$ and $$\varphi_{X_n}(z) = \exp\left\{ \int_\mathbb R [e^{izx} - 1] d\eta_n(x) \right\} = \exp\left\{ \int_\mathbb R e^{izx} d\eta_n(x) - 1 \right\}= \exp\left\{ \varphi_{\eta_n}(z) - 1 \right\} $$ This the characeristic function of a compound poisson distrubution with $N \sim Poisson(1)$. In general, we have $Y \sim CP(\lambda, \eta)$ \begin{equation} Y = \sum_{j=1}^{N} X_j, \quad N\sim \hbox{Poisson}(\lambda), \, X_j\,\, i.i.d. \sim \eta \end{equation} For more details, see this. Moreover, the characteristic function is $$\varphi_{Y}(z) = \exp\left\{ \lambda \int_\mathbb R e^{izx} d\eta(x) - 1 \right\}= \exp\left\{ \lambda [\varphi_{\eta}(z) - 1] \right\} = \exp\left\{ \lambda [\varphi_{X_1}(z) - 1] \right\} $$