Let $\mu_n$ be a probability measure on $(\mathbb R,\mathcal B(\mathbb R))$ with characteristic function $\varphi_n$ for $n\in\mathbb N$ and $\varphi:\mathbb R\to\mathbb C$ be continuous at $0$ with $$\varphi_n\xrightarrow{n\to\infty}\varphi\tag1.$$
Using $(1)$, we are able to show that $(\mu_n)_{n\in\mathbb N}$ is tight, i.e. for all $\varepsilon>0$, there is a compact $K\subseteq\mathbb R$ with $\sup_{n\in\mathbb N}\mu_n(K^c)<\varepsilon$.
Question 1: Why is this enough to conclude that there is a probability measure $\mu$ on $(\mathbb R,\mathcal B(\mathbb R))$ such that $\mu_n$ weakly converges to $\mu$ as $n\to\infty$?
I know that it is a consequence of Prohorov's theorem that a sequence of probability measures $\nu_n$ weakly converges to some probability measure $\nu$ if and only if $(\nu_n)_{n\in\mathbb N}$ is tight and there is a separating family$^1$ $\mathcal C$ of bounded continuous functions $f:\mathbb R\to\mathbb R$ with $$\int f\:{\rm d}\nu_n\xrightarrow{n\to\infty}\int f\:{\rm d}\nu\tag2\;\;\;\text{for all }f\in\mathcal C.$$
So, aren't we missing the existence of the separating family?
Question 2: Does the same conclusion hold if, instead of $(1)$, we have pointwise convergence of the "Laplace transforms" $$\psi_n(t):=\int e^{-tx}\:\mu_n({\rm d}x)\;\;\;\text{for }t\ge0$$ for $n\in\mathbb N$?
$^1$ i.e. $$\forall m,n\in\mathbb N:\left(\forall f\in\mathcal C:\int f\:{\rm d}\nu_m=\int f\:{\rm d}\nu_n\right)\Rightarrow\nu_m=\nu_n\tag3.$$