Assume that $X_1, X_2, ...$ are independent and identically distributed random variables with mean $\mu$ and variance $1$, then let $\bar{X}_n=n^{-1}\sum_{i=1}^n X_i$ be the sample mean. We all know that by the strong law of large numbers, for all $\epsilon>0$ $$ P(|\bar{X}_n-\mu|>\epsilon)\to 0, \quad n\to \infty, $$ while, by the central limit theorem $$ \sqrt{n}(\bar{X}_n-\mu) \overset{d}{\to} \mathcal{N}(0,1). $$ In particular, the above statement is true because of the fact that Lindeberg's condition is herein satisfied, i.e. $$ 0=\lim_{n\to \infty}\int_{\{|X_1-\mu|>\epsilon \sqrt{n}\}} |X_1-\mu|^2\text{d}P= \lim_{n\to \infty}\frac{1}{n}\sum_{i=1}^n\int_{\{|X_i-\mu|>\epsilon \sqrt{n}\}}|X_i-\mu|^2\text{d}P. $$ QUESTION: is it also true that $\lim_{n\to\infty} E1(|\bar{X}_n- \mu|>\epsilon)(\sqrt{n}(\bar{X}_n -\mu))^2=0$? This would be true if, e.g., the sequence $(\sqrt{n}(\bar{X}_n -\mu))^2$ was uniformly integrable: is it the case?
MY ATTEMPT: using Minkowski inequality, I've only managed to obtain $$ \left[ E1(|\bar{X}_n- \mu|>\epsilon)(\sqrt{n}(\bar{X}_n -\mu))^2 \right]^{1/2} \leq \sum_{i=1}^n \left[E1(|\bar{X}_n- \mu|>\epsilon)(n^{-1/2}(X_i -\mu))^2 \right]^{1/2}\\ =n \left[E1(|\bar{X}_n- \mu|>\epsilon)(n^{-1/2}(X_1 -\mu))^2 \right]^{1/2}\\ = \sqrt{n} \left[E1(|\bar{X}_n- \mu|>\epsilon)(X_1 -\mu)^2 \right]^{1/2}, $$ where $1(X\in B)$ is the indicator function of the event $\{X \in B\}$, for a random variable $X$ and a measurable set $B$.
The sequence $ (n(\overline{X}_n-\mu)^2)$ is uniformly integrable, the proof is given in the following. Denote \begin{equation*} T_n=\sqrt{n}(\overline{X}_n-\mu), \qquad F_{n}(x)=\mathsf{P}(T_n\le x), \end{equation*} then \begin{equation*} \mathsf{E}[T_n]=0,\qquad \mathsf{E}[T_n^2]=1. \tag{1} \end{equation*} Using CLT, \begin{gather*} F_n\stackrel{d}{\longrightarrow}\Phi,\qquad \Phi(x)=\int_{-\infty}^{x}\phi(t)\,\mathrm{d}t,\qquad \phi(t)=\frac{e^{-t^2/2}}{\sqrt{2\pi}},\\ \lim_{n\to\infty}\mathsf{E}[T_n^21_{(|T_n|\le k)}] =\lim_{n\to\infty}\int_{(|t|\le k)}t^2\,\mathrm{d}F_n(t) =\int_{(|t|\le k)}t^2\phi(t)\,\mathrm{d}t ,\\ \lim_{n\to\infty}\mathsf{E}[T_n^21_{(|T_n|> k)}] = \lim_{n\to\infty} (1-\mathsf{E}[T_n^21_{(|T_n|\le k)}]) =\int_{(|t|> k)}t^2 \phi(t)\,\mathrm{d}t. \tag{2} \end{gather*} Next, we will prove that $\{ T_n^2,\; n\ge 1\}$ is uniformly integrable, i.e. \begin{equation*} \lim_{k\to\infty}\sup_{n}\mathsf{E}[T_n^21_{(|T_n|> k)}]=0, \tag{3} \end{equation*} For given $\delta>0$, there exists an $k_1>0$ such that \begin{equation*} \int_{\{|t|>k_1\}}t^2\phi(t)\,\mathrm{d}t<\frac{\delta}{2}. \tag{4} \end{equation*} Due to (2) and (4), there exists $n_1$ such that \begin{equation*} \sup_{n\ge n_1}\mathsf{E}[T^2_n1_{(|T_n|> k_1)}]<\delta. \tag{5} \end{equation*} For $k>k_1$, \begin{align*} &\sup_{n}\mathsf{E}[T^2_n1_{(|T_n|\ge k)}] \le \sup_{n\le n_1}\mathsf{E}[T^2_n1_{(|T_n|\ge k)}] + \sup_{n> n_1}\mathsf{E}[T^2_n1_{(|T_n|\ge k)}]\\ & \qquad \le \sup_{n\le n_1}\mathsf{E}[T^2_n1_{(|T_n|\ge k)}] + \sup_{n> n_1}\mathsf{E}[T^2_n1_{(|T_n|\ge k_1)}] \\ & \qquad \le \sup_{n\le n_1}\mathsf{E}[T^2_n1_{(|T_n|\ge k)}] + \delta \qquad (\text{using (5)}) \end{align*} Now let $k\to \infty$ and $\delta\to0 $ successively, (3) holds.
(please refer to Y. S. Chow & H. Teicher, Probability Theory, 3Ed, Springer Verlag, 1997, p.278, Cor.8.1.8)