Let $(\Omega,\mathcal{\Sigma,\mathbb{P}})$ be a complete probability space.
I have to show that $$\lim_{\alpha \to \infty} \sup_{n \geq 1}\int_{\{|X_n| \geq \alpha\}}|X_n|\, d\mathbb{P} = 0$$
assuming title's assumptions and let $\alpha > 0$
my attempt :
$| X_n| \leq \sup_{n\geq 1}|X_n| \leq [\sup_{n\geq 1}|X_n|]^p$
therefore
$$\int_{\{|X_n| \geq \alpha\}}|X_n|\, d\mathbb{P} \leq \int_{\{|X_n| \geq \alpha\}} [\sup_{n\geq 1}|X_n|]^p\, d\mathbb{P} \leq \mathbb{E}[ [\sup_{n\geq 1}|X_n|]^p] < \infty $$
so for every $n \geq 1$
$$\int_{\{|X_n| \geq \alpha\}}|X_n|\, d\mathbb{P} < \infty $$
however I'm not sure this completes the proof because although it's finite it could possibly depend on $n$ and taking the supremum with $n$'s in an expression can cause a blow up.
any hints or help will be greatly appreciated, thanks !
We can use the fact that $$ \int_{|X_n|\ge \alpha}|X_n|d\Bbb P\le\frac{1}{\alpha^{p-1}}\int_{|X_n|\ge \alpha}|X_n|^pd\Bbb P. $$ If $\sup\limits_{n\in\Bbb N}\|X_n\|_p=M<\infty$ (our case is a special case of this), then we have $$ \int_{|X_n|\ge \alpha}|X_n|d\Bbb P\le\frac{1}{\alpha^{p-1}}M^p,\quad\forall n\in\Bbb N. $$ This gives $$ \sup_{n\in\Bbb N}\int_{|X_n|\ge \alpha}|X_n|d\Bbb P\le\frac{1}{\alpha^{p-1}}M^p $$ and we get the desired conclusion by taking $\alpha\to\infty.$