Suppose that $X \ge 0$ and $\mathbb{E}[X] < \infty$. Show that $$\lim_{n \to \infty} \mathbb{E}[X\mathbb{I}_{x>n}] = 0.$$
My Current Approach
Note that $$ X\mathbb{I}_{x>n}(\omega) = \begin{cases} X(\omega), & \text{if } X(\omega) > n\\ 0, & \text{if } X(\omega) \le n \end{cases} $$ We want to show that $$ \lim_{n \to \infty} \mathbb{E}[X\mathbb{I}_{x>n}] = \mathbb{E} \left[\lim_{n \to \infty} X\mathbb{I}_{x>n}\right]. $$ So it suffices to check conditions of the dominated convergence theorem.
First we want to show that $$ \lim_{n \to \infty} X\mathbb{I}_{x>n} = 0 \quad \text{a.s.} $$ Let $\omega \in \Omega$. Then, $X(\omega) \in [0, \infty)$.
Could you please help me finish this? Thank you.
Intuitively speaking, as $n \rightarrow \infty$, the indicator function of the interval $(n,\infty]$ will tend to the indicator function of the singleton $\{\infty \}$. As X is integrable, so do its products with indicator functions, this implies that those functions are finite almost everywhere. That proves the limit you want.
To verify the conditions of the dominated convergence theorem, we use the fact that X is integrable and greater or equal that its product with an indicator function; this allow us to use X as the dominating function needed for the theorem to work.