Simple question: Why does $E(|X|) < \infty$ imply $E(|X|I_{|X|>a} )$ tends to $0$ as $a$ tends to infinity Simple question: Why does $E(|X|) < \infty$ imply $E(|X|I_{|X|>a} )$ tends to $0$ as $a$ tends to infinity?
I've seen it in a few proofs and I can't see why this is the case, I've tried a proof using Markov's inequality but am unconvinced by my reasoning.
 A: $\left| X \right| = \left| X \right|{I_{\left| X \right| > a}} + \left| X \right|{I_{\left| X \right| \leqslant a}}$
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] $
$\Rightarrow \mathop {\lim }\limits_{a \to  + \infty } E\left[ {\left| X \right|{I_{\left| X \right| > a}}} \right] = 0$
A: Write 
$$E|X| = E[|X|1_{|X| > a}] + E[|X|1_{|X| \leq a}] $$
Take limits on both sides. By Monotone convergence theorem 
$E[|X|1_{|X| \leq a}] \rightarrow E|X|$ 
You can cancel E|X| on both sides as it is $< \infty$.
Thus you arrive at your result.
A: No need to apply a monotone convergence theorem. First, fix $\delta>0$, and let $Y\leqslant X$ a simple function such that $E[|X|-Y]<\delta$ (such a function exists by definition of Lebesgue integral). We write $Y:=\sum_{j=1}^Nc_j\chi_{A_j}$, where $A_j$ are measurable and pairwise disjoin. This gives 
$$E[|X|\chi_{\{|X|> a}]<\delta+\sum_{j=1}^Nc_j\Bbb P(A_j\cap\{|X|>a\}).$$
As $a\Bbb P(A_j\cap\{|X|>a\})\leqslant \int_{A_j}|X|d\Bbb P$, we get 
$$E[|X|\chi_{\{|X|> a}]<\delta+\frac 1a\sum_{j=1}^Nc_j\int_{\Omega}\chi_{A_j}|X|d\Bbb P.$$
Taking $\limsup_{a\to +\infty}$ and using the fact that $\delta$ is arbitrary, we get the result. 
