If I have that $\limsup_{n}E|X_n|^{r} \leq E|X|^{r}$, is that enough to show that $\{|X_n|^{r}:n\geq 1\}$ is uniformly integrable? If I have that $\limsup_{n}E|X_n|^{r} \leq E|X|^{r}$, is that enough to show that $\{|X_n|^{r}:n\geq 1\}$ is uniformly integrable? I am not sure here if the limsup condition here is as strong as if I had a uniformly bound. Does anyone have any hints as to how I can work with such a definition? Thanks.
 A: Yes, that's enough. Remember that $\limsup_n a_n = \lim_n\{\sup_{k\ge n} a_k\}$ for any sequence $\{a_n\}$ of real numbers. If the limsup is bounded above by some $A$, then there exists $N$ such that
$$
\sup_{k\ge N} a_k \le A +1.
$$
But this implies that $a_k\le A +1$ for all $k\ge N$ ( since the sup is an upper bound on every term ), so $A+1$ is a uniform bound on $a_k$ when $k\ge N$. For $k<N$, we have the bound
$$a_k \le \max\{a_1, a_2, \ldots, a_{N-1}\}.$$
The max of the above two bounds is a uniform bound for every $a_k$.
In short, we've proved that only finitely many terms of a sequence will exceed the limsup + 1 of that sequence.
A: Define $Y_n:=\left|X_n\right|^r$ and $Y:=\left|X\right|^r$. If $\limsup_n\mathbb E[Y_n]\leqslant E[Y]$, it may be not true that $\mathbb E[Y_n]$ is finite for each $n$. Even if it is the case, the best we can deduce is that $\sup_n\mathbb E\left[Y_n\right]$ is finite (see grand_chat's answer). But this is not enough for the uniform integrability, for example if $Y_n=n\mathbf 1_{(0,1/N)}$ on the unit interval endowed with Lebesgue measure.
