$X\in L_1$ then $\lim\limits_{x\to\infty} \int\limits_{{|X|>x}} XdP =0$

Question

The question is that Let $X\in L_1$ then prove $\lim\limits_{x\to\infty} \int\limits_{|X|>x} XdP =0$$

I thought if $X\in L_1$ then $E(X)=\int\limits_{\Omega} XdP =\int\limits_{|X|\leq \epsilon} XdP +\int\limits_{|X|>\epsilon} XdP$ How can i follow from there?

Answer 1

Let $x_n$ be a sequence of positive real numbers such that $x_n \to \infty$. $\int\limits_{|X| \le x_n} X dP=\int\limits_{\Omega}X 1_{|X| \le x_n} dP$. and since$|X 1_{|X| \le x_n}| \le |X|$ and converges pointwise to $X$, since $x_n\to \infty$ as $n \to \infty$ and since $X \in L^1$. Then the dominated convergence applies. Then $\lim\limits_{n \to \infty} \int\limits_{|X| \le x_n} X dP=\int\limits_{\Omega} X dP$. One more line to get the answer! (Note, $x_n$ here can be any sequence of positive numbers going to $\infty$)