Show that each sequence $(X_n)$ is UI $\forall n \in \mathbb{N}$, if $\sup_{n}\{ \mathbb{E}(|X_n|^{1+\epsilon})\}$ is finite with respect to some probability measure and for positive $\epsilon$, ideally using Holders Inequality.
For UI I need to show that $$ \sup E(|X|1_{|X|>n})\to0$$
So, my attempt is if the supremum in the question is finite, then we know
$$ \sup_{n} \mathbb{E}[|X_n|^{1+\epsilon}] = \sup_{n} \mathbb{E}(|X_nX_n^{\epsilon}|) <\infty$$
We can then use Holder's inequality:
$$ \sup_nE[|X_nX_n^{\epsilon}|] \le \sup_nE(|X_n|)\sup_nE(|X_n|^\epsilon)^{1/\epsilon} $$
So really I want to show that $\sup_nE(|X_n|)$ and $\sup_nE(|X_n|^\epsilon)^{1/\epsilon} $ are finite, but I'm not sure I'm going about this the right way to get to the result. Any advice?
For the uniform integrability of $\{X_n\}$ you need to show that $$ \sup_{n\ge 1}\mathsf{E}|X_n|1\{|X_n|>x\}\to 0 \quad\text{as}\quad x\to \infty. $$ This follows from the fact that $\sup_{n\ge 1}\mathsf{E}|X_n|^{1+\epsilon}<\infty$ for some $\epsilon>0$ because $$ \mathsf{E}|X_n|1\{|X_n|>x\}\le x^{-\epsilon}\sup_{n\ge 1}\mathsf{E}|X_n|^{1+\epsilon}. $$