Is the limit of compound Poisson random variables a compound Poisson r.v.? Let $Y$ be an infinitely divisible (I.D.) random variable.
Let $\nu$ be any measure not necessarily finite: $\nu(\mathbb R)\leq \infty$. 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}
$\underline{Remark\,\,1:}$

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 $\eta$ is a probability measure ($\eta(\mathbb R)=1$) and $Y$ is a compound Poisson random variable  $Y \sim CP(\lambda, \eta)$( See page 18, Equation (4.1) from the Sato's book (reference above) or Updated remark 2 below.)

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.
So my question is: Given a sequence $(X_n)_{n \in \mathbb N}$ of I.D. random variables with
$$ \varphi_{X_n}(z) = \int_{\mathbb R} [e^{izx}-1 ]d\nu_n(x), \quad \nu_n(\mathbb R)< \infty$$
By $\underline{Remark\,\,1}$ above, we have that $X_n \sim CP(\lambda_n, \eta_n)$ where $\lambda_n = \nu_n(\mathbb R)$ and $d\eta_n(x):= d\nu_n(x)/\lambda_n$. Now, supppose that
\begin{equation}\label{II}\tag{II}
X_n \Longrightarrow Y, \quad (n\to \infty)
\end{equation}
where $Y \sim (0, \nu,0)_0$ has characterization given by (\ref{I}).
So, in what situations ( assumptions about $\eta_n$ and $\lambda_n$ ) 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. $\eta_n = \eta$  and $\lambda_n = \lambda$ 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
$$\lim_{n \to \infty} \int_{\mathbb R} f(x) \underbrace{\lambda_n \eta_n (dx)}_{= \nu_n (dx)} = \int_{\mathbb R} f(x) \nu (dx)$$
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?
After the comment from Christophe Leuridan, we have to take a convenient continuous function and not the indicator function, because it is not continuous.
$2.-$ After the comment from Christophe Leuridan, I think it is necessary to specify that, in general, given a probability measure $\eta$,  $Y \sim CP(\lambda, \eta)$ means that:
\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:
$$\varphi_{Y}(z) = \exp\left\{\lambda \int_\mathbb R [e^{izx} - 1] d\eta(x)   \right\} = \exp\left\{ \lambda \int_\mathbb R e^{izx} d\eta(x) - 1  \right\}= \exp\left\{ \lambda [\varphi_{\eta}(z) - 1]  \right\} $$
$3.-$ I had posted this result above, but I don't know if it will help much. However, I will register it here. 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.
 A: I hope I did not make a mistake, but I think it works.
The convergence
$$\exp\Big(\int_\mathbb{R} (e^{izx}-1)d\eta_n(x) \Big) \to \exp\Big(\int_\mathbb{R} (e^{izx}-1)d\nu(x)\Big)$$ yields the convergence $$\int_\mathbb{R} (e^{izx}-1)d\eta_n(x) \to  \int_\mathbb{R} (e^{izx}-1)d\nu(x).$$
I take the real parts and change the signs to have non-negative functions.
$$\int_\mathbb{R} (1-\cos(zx))d\eta_n(x) \to \exp \int_\mathbb{R} (1-\cos(zx))d\nu(x).$$
By Fubini's theorem and Fatou's lemma, for every $T>0$,
\begin{eqnarray*}
\int_\mathbb{R} (1-(Tx)^{-1}\sin(Tx))d\nu(x) 
&=& \frac{1}{T}\int_0^T\Big(\int_\mathbb{R} (1-\cos(zx))d\nu(x)\Big)dz \\
&=& \frac{1}{T} \int_0^T \lim_n\Big(\int_\mathbb{R} (1-\cos(zx))d\eta_n(x)\Big)dz \\
&\le& \liminf_n \frac{1}{T}\int_0^T \Big(\int_\mathbb{R} (1-\cos(zx))d\eta_n(x)\Big)dz \\
&=& \liminf_n \int_\mathbb{R} (1-(Tx)^{-1}\sin(Tx))d\eta_n(x) \\
&\le& 1+1/\pi,
\end{eqnarray*}
since the function sinc is bounded below by $-1/\pi$. Applying Fatou's lemma again,
\begin{eqnarray*}
\int_\mathbb{R} 1d\nu(x) 
&\le& \liminf_{T \to +\infty} \int_\mathbb{R} (1-(Tx)^{-1}\sin(Tx))d\nu(x) \\
&\le& \liminf_{T \to +\infty} 1+1/\pi.
\end{eqnarray*}
Thus $\nu$ is finite. I guess that a refinement of this argument shows that $\nu(\mathbb{R}) \le 1$.
