# Some doubts about Levy's Continuity Theorem proof - Convergence results

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 $$n there exists an $$\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 $$n there exists an $$\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 $$m\in\mathbb{R}$$)?

(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 $$\liminf_n a _n \le \limsup_n a_n$$.
• 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 Jun 1, 2020 at 13:27
• [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 Jun 1, 2020 at 13:27
• Yes, (4) just says that $\lim_m \sup_n \mu_n([-m,m]^c) = 0$. Jun 1, 2020 at 13:28
• $\lim_n a_n = a \iff \limsup_n a_n = \liminf_n a_n = a$. 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 $\lim_n a_n = 0.$ Jun 1, 2020 at 13:31