I quote a part of Levy's Continuity Theorem and its proof.
Theorem
Let $\left(\mu_n\right)_{n\geq1}$ be a sequence of probability measures on $\mathbb{R}^d$, and let
$\left(\hat{\mu}_n\right)_{n\geq1}$ denote their Fourier transforms (aka characteristic functions). If
$\mu_n$ converges weakly (that is, in distribution) to a probability measure $\mu$, then
$\hat{\mu}_n(u)\rightarrow\hat{\mu}(u)$ for all $u\in\mathbb{R}^d.$Proof
Suppose $\mu_n$ converges weakly to $\mu$. Since $e^{iux}$ is continuous and bounded in modulus,
$$\hat{\mu}_n(u)={\displaystyle \int e^{iux}\mu_n(dx)}$$
converges to
$$\hat{\mu}(u)={\displaystyle \int e^{iux}\mu(dx)}$$
My question is:
which is the result implicitly used so as to state that:
"Since $f=e^{iux}$ is continous and bounded in modulus
$$\mu_n\xrightarrow{\mathcal{D}}\mu\Rightarrow\hat{\mu}_n(u)={\displaystyle \int e^{iux}\mu_n(dx)}\to\hat{\mu}(u)={\displaystyle \int e^{iux}\mu(dx)}\hspace{0.5cm}\text{"}\,?$$
Best Answer
The Portmanteau theorem says, among other things, that $$\mu_n \to \mu \mathrm{\ weakly}$$ iff $$\forall f \in C_b: \int fd \mu_n \to \int fd \mu$$
What you are asking for uses this with your particular choice for $f$.