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?