If a set of random variables $\{X_n\}_{n \ge 1}$ is asymptotically normal is it also uniformly integrable?
Intuitively it seems like this should be true as $\{X_n\}_{n \ge 1}$ becomes 'nicer' as $n$ gets larger with the density becoming more symmetrical about the mean.
So is it true? If not, are there additional conditions we can impose on the set of asymptotically normal random variables to make them uniformly integrable?
Let $Z$ be a standard normal random variable, and $B_n$'s are i.i.d. Bernoulli$(\frac{n-1}{n})$ independent of $Z$, then define $$X_n = ZB_n +n^2 (1-B_n).$$
Note that $$ \mathbb{P}(X_n\leq x) = \left(1-\frac{1}{n}\right)\mathbb{P}(Z\leq x) + \frac{1}{n}\mathbb{P}(n^2 \leq x), $$
The last probability is either $0$ or $1$, thus $X_n$ converges to standard normal as $n \to \infty$. Moreover $\mathbb{E}(|X_n|) = n$. Now from here, conclude $X_n$'s are not UI. Also the link gives you necessery and sufficient conditions for UI.