Does $\mathbb E(|X|)<\infty$ implies that $\lim_{N\to\infty}\mathbb E(|X|\,\mathbb I_{\{|X|>N\}})=0$?

Question

Let $X$ a positive random variable in $[0, \infty)$ such that $\mathbb E(X)<\infty$. It is true that $$\lim_{N\to\infty}\mathbb E(X\,\mathbb I_{\{X>N\}})=0?$$

I would say yes because I can write $$ \mathbb E(X)=\mathbb E(X\,\mathbb I_{\{X\leq N\}})+\mathbb E(X\,\mathbb I_{\{X>N\}}).$$ Therefore I can take the limit as $N\to\infty$ and get $$\mathbb E(X)=\lim_{N\to\infty}\mathbb E(X\,\mathbb I_{\{X\leq N\}})+\lim_{N\to\infty}\mathbb E(X\,\mathbb I_{\{X>N\}}).$$ I think that by the dominated convergence theroem I can say that $$\lim_{N\to\infty}\mathbb E(X\,\mathbb I_{\{X\leq N\}})=\mathbb E(\lim_{N\to\infty} X\,\mathbb I_{\{X\leq N\}})=E(X\,\mathbb I_{\{X\leq \infty\}})=\mathbb E(X).$$

Therefore I can conclude that $$lim_{N\to\infty}\mathbb E(X\,\mathbb I_{\{X>N\}})=0.$$ Is this correct?

Thank you

Answer 1

Two points :

• You have not mentioned which random variable is dominating the family $\{X\mathbb I_{\{X \leq N\}}\}$. It is true that there is one(and there is an easy candidate), so that step works out, but you should justify it.

• $\lim_{N \to \infty} X\Bbb I_{\{X \leq N\}} = X \Bbb I_{\left\{\boxed{X < \infty}\right\}}$, because if $X(\omega) = \infty$ then $X\Bbb I_{\{X \leq N\}}(\omega) = 0$ for each $N$ so the limit is zero. Everywhere else it matches. However, the integrability of $X$ ensures that $X(\omega)$ has measure zero, so the integral of $X\Bbb I_{\{X < \infty\}}$ equals that of $X$ (You need to justify this, of course).

There is an easier proof, this one is a little roundabout : what is $\lim_{N \to \infty} X\Bbb I_{\{X > N\}}$? It exists almost surely. What is the integral of that limit? Use DCT to justify interchange of limit and integral.