Let $\{\mu_n\}_n$ be a sequence of infinitely divisible probability measures on $\mathbb{R}^d$ with $\mu_n \xrightarrow{\text{weak}} \mu$ for some probability measure $\mu$. I want to prove that $\mu$ is also infinitely divisible.
My attempt:
If $\mu_n \xrightarrow{\text{weak}} \mu$, then (by definition) $\lim_{n \rightarrow \infty} \int f(x) \mu_n(d x) = \int f(x) \mu(d x)$ for any continuous, bounded $f$ on $\mathbb{R}^d$. In particular, this holds for $f(x) = e^{i(u, x)}$ for any $u \in \mathbb{R}^d$. This implies that \begin{equation} \lim_{n \rightarrow \infty}\phi_{\mu_n}(u) = \phi_\mu(u). \end{equation} Then, for any $m \geq 1$, \begin{equation} \lim_{n \rightarrow \infty} \big( \phi_{\mu_n} \big)^{1/m}(u) = \big( \phi_\mu (u)\big)^{1/m} \end{equation} Since each $\phi_{\mu_n}$ is infinitely divisible, $(\phi_{\mu_n})^{1/m}$ is a characteristic function for any $m \in \mathbb{N}, m \geq 1$.
It is clear to see that $\phi^{1/m}_\mu$ is continuous at zero ($\phi_\mu$ is continuous at zero). Hence by Levy's continuity theorem, $\phi_\mu^{1/m}$ is a characteristic function for any $m \in \mathbb{N}$. So $\mu$ is infintely divisible.
Is my proof correct? This is an exercise (1.2.12) in Applebaum's book on Levy processes.