Q:

proving an expectation limit of random variable truncation

Random variable $X,\lvert EX\rvert\lt\infty,$ we can perform a truncation$$X=X_{\le N}+X_{\gt N}$$of $X$ at any desired threshold N, where $X_{\gt N} := XI_{(\lvert X\rvert>N)}$ $\;$($I$ is the indicator funtion take {0,1})$$$$prove $E \lvert X_{\gt N}\rvert \to 0\;as\; N \to \infty$$$$$hint by Terence Tao: 'monotone convergence theorem'$$$$
the question context is proving weak LLN on Tao’s blog.  The (7) formula on this website$$$$
I can imagine the result, but could you please show me a concrete proof in real analysis? 

A:

Notice that $|X_{>N}| = |X|I_{(|X|> N)}\geq |X_{>M}|$ if $N\leq M$. This means that $\{|X_{>N}|\}$ is a monotone decreasing series of random variables indexed by $N$. We already know that $E|X| = E|X_{>0}|<\infty$. Therefore by the monotone convergence theorem, we have
$$\lim_{N\rightarrow \infty}E|X_{>N}| = E\lim_{N\rightarrow \infty}|X_{>N}| =E 0 = 0 .$$
Here $\lim_{N\rightarrow \infty}|X_{>N}| = 0$ almost surely is implied by the fact that $P(|X|<\infty)=1$ (since $E|X|<\infty$).