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).
i) If $\mu_n$ converges weakly to a probability measure $\mu$, then $\hat{\mu}_n(u)\rightarrow\hat{\mu}(u)$ for all $u\in\mathbb{R}^d$;
ii) 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 $\mu$.
Let $(X_n)_{n\geq1}$ be a sequence of random variables, $i$ the imaginary unit, $S_n=\sum\limits_{i=1}^nX_i$ and $u\in\mathbb{R}$. For a certain constant $L$, it holds that \begin{equation} \Bigg|\mathbb{E}\Big(e^{iu\frac{1}{\sqrt{n}}S_n}\Big)-\Big(1-\frac{u^2}{2n}\Big)^{n}\Bigg|\leq L\frac{|u|^3}{6\sqrt{n}} \end{equation} At this point, since the r.h.s of the above inequality tends to $0$ as $n\rightarrow\infty$ and since \begin{equation*} \begin{split} \lim\limits_{n\to\infty}\Big(1-\frac{u^2}{2n}\Big)^{n}=e^{-\frac{u^2}{2}} \end{split} \end{equation*} recalling that $\lim\limits_{x \to \infty} |f(x) + g(x)| = \lim\limits_{x \to \infty} f(x) + \lim\limits_{x \to \infty} g(x)$, I have that \begin{equation} \lim\limits_{n\to\infty}\mathbb{E}\Big(e^{iu\frac{S_n}{\sqrt{n}}}\Big)=\lim\limits_{n\to\infty}\Big(1-\frac{u^2}{2n}\Big)^{n}=e^{-\frac{u^2}{2}} \end{equation} Now, I read the following statement:
By Levy's Continuity Theorem, we have that $\frac{S_n}{\sqrt{n}}$ converges in law to $Z$, where the characteristic function of $Z$ is $e^{-\frac{u^2}{2}}$
My question is: HOW EXACTLY is the above-quoted Levy's Continuity Theorem applied to get the above conclusion?
You have $$\lim_n \mathbb{E}(\exp(iu S_n/\sqrt{n})) = \exp(-u^2/2) = \mathbb{E}(\exp(iuZ))$$for all $u$ and thus the characteristic function of $S_n/ \sqrt{n}$ converges to the characteristic function of $Z$. The Lévy continuity theorem says that this is equivalent with $S_n/ \sqrt{n} \stackrel{d}\to Z$.
Recall, that by definition for a sequence of random variables on a $p$-space $(\Omega, \mathcal{F}, \mathbb{P})$ we have
$$X_n \stackrel{d}\to X \iff \mathbb{P}_{X_n} \stackrel{w}\to \mathbb{P}_X$$
where the $d$ is convergence in distribution and the $w$ is weak convergence.