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

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

• 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.