convergence in distribution for Poisson $\newcommand{\Poi}{\operatorname{Poi}}$I try to show that if $X_n$ converges to $Z$ in distribution, that
$\Poi(X_n)$ converges in distribution to $\Poi(Z)$,
with Poi, i mean a Poisson distribution.
So I need to show that $\lim_{n \to \infty} E[f(\Poi(X_n))]=E[f(\Poi(Z))]$.
I thought about using tower property but I am not sure if
$E[E[f(\Poi(X_n))\mid X_n]]=E[f(X_n)]$ holds. How can I show that?
 A: Outline/Sketch:
My first instinct would be to use characteristic functions: you want to show, that, for any $t\in\mathbb{R}$, $$\mathbb{E}[e^{it Y_n}] \xrightarrow[n\to\infty]{} \mathbb{E}[e^{it Y}]$$ where $Y_n\sim\operatorname{Poi}(X_n)$ and  $Y\sim\operatorname{Poi}(Z)$ (this is because of Lévy's convergence theorem). You can use conditioning to get $$\mathbb{E}[e^{it Y_n}] = \mathbb{E}[\mathbb{E}[e^{it Y_n}\mid X_n]] = \mathbb{E}[ e^{X_n(e^{it}-1)} ]$$ using the expression the characteristic function of a Poisson r.v.; and similarly $$\mathbb{E}[e^{it Y}] = \mathbb{E}[\mathbb{E}[e^{it Y}\mid Z]] = \mathbb{E}[ e^{Z(e^{it}-1)} ]$$so it only remains to use the convergence assumption on $(X_n)_n$ to show that $$\mathbb{E}[ e^{X_n(e^{it}-1)} ] \xrightarrow[n\to\infty]{} \mathbb{E}[ e^{Z(e^{it}-1)} ]$$ for every fixed $t\in\mathbb{R}$. 
(This last part would be trivial if $i(e^{it}-1)$ were a real number, as it would then be a direct consequence of $X_n \xrightarrow[n\to\infty]{\mathcal{D}} Z$ by Lévy's theorem; but even without this, it should not be too much of a problem.)
