THEOREM (Levy's Continuity Theorem)
Let $(\mu_n)_{n\geq1}$ be a sequence of probability measures on $\mathbb{R}^d$, and let $(\hat{\mu}_n)_{n\geq1}$ denote their characteristic functions (or Fourier transforms).
If $\hat{\mu}_n(u)$ converges to a function $f(u)$ for all $u\in\mathbb{R}^d$, and if in addition $f$ is continuous at $0$, then there exists a probability $\mu$ on $\mathbb{R}^d$ such that $f(u)=\hat{\mu}(u)$, and $\mu_n$ converges weakly to $\mu$.A PART OF THE PROOF FOR $d=1$ (FIRST PART)
$(\ldots)$ let $\beta=\dfrac{2}{\alpha}$ ($\alpha$ and $\beta$ constants) and we have the useful estimate $$\mu_n\left(\left[-\beta,\beta\right]^c\right)\le\dfrac{\beta}{2}{\displaystyle \int_{-\frac{2}{\beta}}^{\frac{2}{\beta}}\left(1-\hat{\mu}_n(u)\right)du}\tag{1}$$ Let $\varepsilon>0$. Since by hypothesis $f$ is continous at $0$, there exists $\alpha>0$ such that $\left\vert1-f(u)\right\vert\le\dfrac{\varepsilon}{4}$ if $\left\vert u\right\vert\le\dfrac{2}{\alpha}$ (This is because $\hat{\mu}_n(0)=1$ for all $n$, whence $\lim\limits_{n\to\infty}\hat{\mu}_n(0)=f(0)=1$ as well.) Therefore $$\left\vert\dfrac{\alpha}{2}\displaystyle{\int_{-\frac{2}{\alpha}}^{\frac{2}{\alpha}}\left(1-f(u)\right)du}\right\vert\le\dfrac{\alpha}{2}\displaystyle{\int_{-\frac{2}{\alpha}}^{\frac{2}{\alpha}}\dfrac{\varepsilon}{4}du}=\dfrac{\varepsilon}{2}\tag{2}$$ $(\ldots)$ there exists an $N$ ($\in\mathbb{N}$) such that $n\geq\mathbb{N}$ ($n\in\mathbb{N}$) implies $$\left\vert\displaystyle{\int_{-\frac{2}{\alpha}}^{\frac{2}{\alpha}}\left(1-\hat{\mu}_n(u)\right)du} - {\displaystyle\int_{-\frac{2}{\alpha}}^{\frac{2}{\alpha}}\left(1-f(u)\right)du}\right\vert\le\dfrac{\varepsilon}{\alpha}\tag{3}$$ whence, by $(2)$, $\dfrac{\alpha}{2}{\displaystyle\int_{-\frac{2}{\alpha}}^{\frac{2}{\alpha}}\left(1-\hat{\mu}_n(u)\right)du}\le\varepsilon$. Next apply $(1)$ to conclude $\mu_n\left(\left[-\alpha, \alpha\right]^c\right)\le\varepsilon$, for all $n\ge N$.
So far so good to me. The following SECOND PART is not that clear instead.
A PART OF THE PROOF FOR $d=1$ (SECOND PART)
There are only a finite number of $n$ before $N$, and for each $n<N$ there exists an $\alpha_n$ such that $\mu_n\left(\left[-\alpha_n, \alpha_n\right]^c\right)\le\varepsilon$.
Let $a=\max(\alpha_1,\ldots,\alpha_n;\alpha)$. Then $$\mu_n\left(\left[-a, a\right]^c\right)\le\varepsilon,\hspace{0.3cm}\text{for all }n\tag{4}$$
The inequality $(4)$ means that for the sequence $(\mu_n)_{n\ge1}$ for any $\varepsilon>0$ there exists an $a\in\mathbb{R}$ such that $\sup\limits_{n}\mu_n\left(\left[-a,a\right]^c\right)\le\varepsilon$. Therefore, we have shown $$\limsup\limits_{m\to\infty}\sup\limits_{n}\mu_n\left(\left[-m,m\right]^c\right)=0\tag{5}$$ for any fixed $m\in\mathbb{R}$.
Given the first part, my doubts about SECOND PART of the proof are:
1. Why can I be sure that "for each $n<N$ there exists an $\alpha_n$ such that $\mu_n\left(\left[-\alpha_n, \alpha_n\right]^c\right)\le\varepsilon$"?;
2. Why can I state that "the inequality $(4)$ means that for the sequence $(\mu_n)_{n\ge1}$ for any $\varepsilon>0$ there exists an $a\in\mathbb{R}$ such that $\sup\limits_{n}\mu_n\left(\left[-a,a\right]^c\right)\le\varepsilon$"? More precisely, why can I draw a conclusion specifically on the $\sup\limits_n$ of the set $\mu_n\left(\left[-a,a\right]^c\right)$?;
3. Could I also state that the conclusion of all the reasoning is that $\limsup\limits_{m\to\infty}\sup\limits_{n}\mu_n\left(\left[-m,m\right]^c\right)=\liminf\limits_{m\to\infty}\sup\limits_{n}\mu_n\left(\left[-m,m\right]^c\right)=0$ (for any fixed $m\in\mathbb{R}$) and not just that $\limsup\limits_{m\to\infty}\sup\limits_{n}\mu_n\left(\left[-m,m\right]^c\right)=0$ (for any fixed $m\in\mathbb{R}$)?
