Uniform integrability of rescaled sample mean 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 $B$.
 A: 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)
