[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 $\mu_\infty$.
$\bullet$ Counterexample 1. When $\mu_\infty$ is not a probability measure in Theorem 1, what would happen on the convergence of ch.f.?
Let $\mu_n$ have mass $\frac{1}{2}$ at $0$ and mass $\frac{1}{2}$ at $n$, i.e. $$\begin{array}{c}\mathscr{P}\{X_n=0\}=\frac{1}{2};\\ \mathscr{P}\{X_n=n\}=\frac{1}{2}. \end{array}$$ Then $\mu_n\rightarrow\mu_\infty$, where $\mu_\infty$ has mass $\frac{1}{2}$ at $0$ and is not a probability measure.
For each $\mu_n$, we have its ch.f. $$f_n(t)=\frac{1}{2}+\frac{1}{2}e^{int}.$$As $n\rightarrow\infty$, $f_n(t)$ does not converge, except when $t$ is a multiple of $2\pi$.
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$\bullet$ Counterexample 2. When conditions of Theorem 2 are not satisfied, what would happen on the convergence of $\mu_n$?
Let $\mu_n$ be the uniform distribution on $[-n,n]$. Then $\mu_n\rightarrow\mu_\infty$, where $\mu_\infty$ is identically zero. For each $\mu_n$, we have $$f_n(t)=\begin{cases}\frac{\sin{nt}}{nt},&\mbox{if }t\neq0;\\1,&\mbox{if }t=0,\end{cases}$$ and it converges to $$f_n(t)\rightarrow f(t)=\begin{cases}0,&\mbox{if }t\neq0;\\1,&\mbox{if }t=0.\end{cases}$$Although the condition (a1) is satisfied, the failure of condition (a2) is the key point that $f_\infty$ is not the ch.f. of $\mu_\infty$.
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