prove For non-negative r.x. $X$, $E(X)<\infty$ iff $E(XI_{X>x})\rightarrow 0$ as $x\rightarrow\infty$

Question

I found a statement that :

For non-negative r.x. $X$, $E(X)<\infty$ iff $E(XI_{X>n})\rightarrow 0$ as $n\rightarrow\infty$

Does anyone know how to prove it?

My understanding is that

for if part:

Let $XI_{X>n}=X_n$, we have $$\lim_{n\rightarrow\infty}X_n=0$$and$$0<X_n<X$$Then, by dominated convergence theorem,$$\lim_{n\rightarrow\infty}E(X_n)=E(\lim_{n\rightarrow\infty}X_n)=0$$

but how to prove the only if part?

Is it always true that $$\lim_{n\rightarrow\infty}X_n=0$$?

Answer 1

For the if part notice, $X = XI_{X\leq x} + XI_{X>x} \leq x + XI_{X>x}, \forall x>0$.

Now using the condition on the if part, choose an $x>0$ such that $E(XI_{X>x}) <\epsilon$ for some $\epsilon>0$ this will give, $E(X) < x+\epsilon<\infty.$

For the only if part using a DCT you can show, $E(XI_{X\leq x}) \to E(X).$ [Since, $XI_{X\leq x}\leq X$ and $E(X)<\infty$ and $XI_{X\leq x}\to X$ as $x\to\infty$.]

But also $E(X)= E(XI_{X\leq x})+E(XI_{X>x})$. This will give, $E(XI_{X>x}) = E(X)-E(XI_{X\leq x}) \to 0$ as $x\to\infty$.