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

Here is what I have done.
We need prove that lim_{a\to \infty} sup_t \int_{|X_t|\ge a}|X_t|d\mathbb P \to 0 {.
We have
\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

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