[Theorem 1] Vague convergence implies convergence of ch.f.
Let $\{\mu_n,\,1\leq n\leq\infty\}$ be probability measures on $\mathbb{R}$ with ch.f.'s $\{f_n,\,1\leq n\leq\infty\}$. We have $$\mu_n\overset{v}{\rightarrow}\mu_\infty\implies f_n\rightarrow f_\infty\mbox{ uniformly in every finite interval.}$$
[Theorem 2] Convergence of ch.f. implies vague convergence.
Let $\{\mu_n,\,1\leq n<\infty\}$ be probability measures on $\mathbb{R}$ with ch.f.'s $\{f_n,\,1\leq n<\infty\}$. Suppose that
(a1) $f_n$ converges everywhere in $\mathbb{R}$, say $f_n\rightarrow f_\infty$.
(a2) $f_\infty$ is continuous at $t=0$.
Then we have
(b1) $\mu_n\overset{v}{\rightarrow}\mu_\infty$, where $\mu_\infty$ is a probability measure.
(b2) $f_\infty$ is the ch.f. of $\mu_\infty$.