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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 $m\in\mathbb{R}$)?

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(1) Note that $\lim_{K \to \infty} \mu_n([-K,K]^c) = 0$. Thus choosing $K$ sufficiently large ensures that $\mu_n([-K,K]^c) \leq \epsilon$.

(2) If you have $\mu_n([-a,a] ^c) \leq \epsilon$ for all $n$, then this means that $\epsilon$ is an upper bound for $\{\mu_n([-a,a]^c): n \geq 1\}$. By definition of sup as the LEAST upper bound, we get $$\sup \{\mu_n([-a,a]^c): n \geq 1\} \leq \epsilon$$

(3) Yes, you can do that. Recall that $\liminf_n a _n \le \limsup_n a_n$.

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  • $\begingroup$ Thank you a lot!! First two points are perfectly clear to me now. As to the third one, just to be sure that the way in which I get to the conclusion is correct...I have 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 $\sup\limits_{n}\mu_n\left(\left[-a,a\right]^c\right)\le\varepsilon$" [CONTINUE] @ε-δ $\endgroup$ – Strictly_increasing Jun 1 at 13:27
  • $\begingroup$ [CONTINUE] From this, I deduce that, by definition of limits, there exists a limit for $\{\sup\limits_{n}\mu_n\left(\left[-a,a\right]^c\right): n\ge1\}$, i.e. $\limsup\limits_{m\to\infty}\sup\limits_{n}\mu_n\left(\left[-m,m\right]^c\right)=\liminf\limits_{m\to\infty}\sup\limits_{n}\mu_n\left(\left[-m,m\right]^c\right)=0$ @ε-δ $\endgroup$ – Strictly_increasing Jun 1 at 13:27
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    $\begingroup$ Yes, (4) just says that $\lim_m \sup_n \mu_n([-m,m]^c) = 0$. $\endgroup$ – ε-δ Jun 1 at 13:28
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    $\begingroup$ $\lim_n a_n = a \iff \limsup_n a_n = \liminf_n a_n = a$. Here, we have $0 \leq \liminf_n a_n \leq \limsup_n a_n = 0$ so $\liminf_n a_n = \limsup_n a_n = 0$ and we conclude $\lim_n a_n = 0.$ $\endgroup$ – ε-δ Jun 1 at 13:31
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    $\begingroup$ Thanks! You too :) $\endgroup$ – Strictly_increasing Jun 1 at 13:34

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