Counterexample where X is not in the Lebesgue linear space.

[mehr1methanol]

I'm trying to find a counterexample where $\lim_{n \to +\infty} P(|X|>n) = 0$ but $X \notin L$ where $L$ is the lebesgue linear space.

$∫|X|I(|X|>n)dp + ∫|X|I(|X|≤n)dp = ∫|X|dp$ therefore

$∫nI(|X|>n)dp + ∫|X|I(|X|)dp ≤ ∫|X|dp$

Suppose $∫I(|X|>n)dp = 1/(n ln n)$
Clearly the hypothesis is satisfied because $\lim_{n \to +\infty} P(|X|>n) = \lim_{n \to +\infty} ∫I(|X|>n)dp = \lim_{n \to +\infty} 1/( ln n) = 0$
But I'm not sure how to conclude $∫|X| = ∞$

 

[Stephen Tashi]

I'm trying to find a counterexample where $\lim_{n \to +\infty} P(|X|>n) = 0$ but $X \notin L$ where $L$ is the lebesgue linear space.

Make another try at stating your question. What is $X$ ? Does the use of $P(|X| > n) = 0$ imply that $X$ is a random variable? What "lebesgue linear space" are you talking about? $L_p$ space? $p = 2$?