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mehr1methanol

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

[itex]∫|X|I(|X|>n)dp + ∫|X|I(|X|≤n)dp = ∫|X|dp [/itex] therefore

[itex]∫nI(|X|>n)dp + ∫|X|I(|X|)dp ≤ ∫|X|dp[/itex]

Suppose [itex]∫I(|X|>n)dp = 1/(n ln n) [/itex]

Clearly the hypothesis is satisfied because [itex] \lim_{n \to +\infty} P(|X|>n) = \lim_{n \to +\infty} ∫I(|X|>n)dp = \lim_{n \to +\infty} 1/( ln n) = 0[/itex]

But I'm not sure how to conclude [itex]∫|X| = ∞[/itex]

[itex]∫|X|I(|X|>n)dp + ∫|X|I(|X|≤n)dp = ∫|X|dp [/itex] therefore

[itex]∫nI(|X|>n)dp + ∫|X|I(|X|)dp ≤ ∫|X|dp[/itex]

Suppose [itex]∫I(|X|>n)dp = 1/(n ln n) [/itex]

Clearly the hypothesis is satisfied because [itex] \lim_{n \to +\infty} P(|X|>n) = \lim_{n \to +\infty} ∫I(|X|>n)dp = \lim_{n \to +\infty} 1/( ln n) = 0[/itex]

But I'm not sure how to conclude [itex]∫|X| = ∞[/itex]

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