# Expectation and Lebesgue integration question

How might I show that if a random variable (call it Z) is such that EZ (expectation of Z) is finite (i.e. it is Lebesgue integrable), then nP(|Z|>n) tends to 0?

-

Note that $nP(|Z|\gt n)=E(Z_n)$ with $Z_n=n\mathbf 1_{|Z|\gt n}$ and that $Z_n\to0$ almost surely. Hence all that is needed is to ensure that the integral of the limit is the limit of the integrals.
Lebesgue dominated theorem tells you that, if $|Z_n|\leqslant Y$ uniformly over $n$, for some integrable $Y$, everything works fine. Surely you have some idea about a candidate for $Y$...
As in the de la Vallée-Poussin theorem, let $G$ be an increasing function on $[0,\infty)$ so that $G(x)/x\to\infty$ but so slowly that $E(G(|Z|))<\infty$. Then by Markov's inequality, $$n\,P(|Z|>n)=n\,P(G(|Z|)>G(n))\leq {n\, E(G(|Z|))\over G(n)}\to0.$$
 Thanks. Does that mean that if nP(|Z|>n) tends to 0 it is not necessarily true that EZ is finite? I.e. does that mean that the converse is false? – Fredo Baron Oct 1 '12 at 3:43 Yes, the converse is false. $E(|Z|)$ is finite if and only if $\sum P(|Z|>n)<\infty$. So if $P(|Z|>n)=1/(n\log(n))$ then $E(|Z|)=\infty$ but $nP(|Z|>n)\to 0$. – Byron Schmuland Oct 1 '12 at 3:48