$\left| X \right| = \left| X \right|{I_{\left| X \right| > a}} + \left| X \right|{I_{\left| X \right| \leqslant a}}$
so by Lebesgue monotone convergence theorem
$E\left[ X \right] = \mathop {\lim }\limits_{a \to + \infty } E\left[ {\left| X \right|{I_{\left| X \right| > a}}} \right] + \mathop {\lim }\limits_{a \to + \infty } E\left[ {\left| X \right|{I_{\left| X \right| \leqslant a}}} \right] = $
$ = \mathop {\lim }\limits_{a \to + \infty } E\left[ {\left| X \right|{I_{\left| X \right| > a}}} \right] + E\left[ {\mathop {\lim }\limits_{a \to + \infty } \left| X \right|{I_{\left| X \right| \leqslant a}}} \right] = $
$ = \mathop {\lim }\limits_{a \to + \infty } E\left[ {\left| X \right|{I_{\left| X \right| > a}}} \right] + E\left[ {\left| X \right|} \right] $ $\Rightarrow \mathop {\lim }\limits_{a \to + \infty } E\left[ {\left| X \right|{I_{\left| X \right| > a}}} \right] = 0$