I have one question about uniform integrable r.v.
"Let a filtration $\displaystyle \mathb F=(\mathcal F_t)_{t\ge 0}$; X is random variable $\displaystyle \mathcal F_{\infty}$-measurable integrable random variable. Then $\displaystyle X_t=E(X|\mathcal F_t)$ is uniform integrable r.vs."

Here is what I have done.
We need prove that $\displaystyle lim_{a\to \infty} sup_t \int_{|X_t|\ge a}|X_t|d\mathbb P \to 0 {$.
We have
$\displaystyle \int_{|X_t|\ge a}|X_t|d\mathbb P=\int_{|X_t|\ge a}|\mathbb E(X|\mathcal F_t)|d\mathbb P \le \int_{|X_t|\ge a}\mathbb E(|X| |\mathcal F_t) d\mathbb P $

$\displaystyle \int_{|X_t|\ge a}\mathbb |X| d\mathbb P$
I have no idea from this.
Help me. Thanks.