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Consider a family of functions $\{f_n\}$, where $f_n: X \rightarrow \mathbb{R}_{\geq 0 }$, and a probability measure on $X$.

Please provide an example in which all functions $f_n$ are integrable but not uniformly integrable in the "probability sense":

$$ \lim_{c \rightarrow \infty} \ \sup_n \mathbb{E} \{ f_n \mid {f_n \geq c} \} = 0$$

Here there is an example, but the family is uniformly integrale in the "probability sense".

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    I suspect you want to add a condition like $f_n$ converges in some sense. Otherwise you could take $f_n=n$ (for instance). It is a theorem that if $f_n \rightarrow f$ in probability, then $f_n \rightarrow f$ in $L^1$ if and only if the family is uniformly integrable.2012-03-11
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    I'm not completely clear about your comment: I don't want to exclude the case of having a finite number of functions in the family. Can you provide a link of the theorem are you talking about?2012-03-11
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    Any finite subset of $L^1$ is uniformly integrable. My point was that one of the reasons uniformly integrable is a useful concept is it tells you when things that converge in probability converge in $L^1$ (see http://en.wikipedia.org/wiki/Uniform_integrability#Relations_to_convergence_of_random_variables or any measure theoretic intro to probability).2012-03-11

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Take $f_n = n 1_{A_n}$ where $\mu(A_n) = 1/n$. (For instance, on $[0,1]$ with Lebesgue measure, you could take $A_n = [0,\frac{1}{n}]$.) I think this is the example that was intended in the linked answer, and it is not uniformly integrable in your sense (or any other).

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    I'm not completely clear on that. Let's apply the definition step by step: first we compute $\int_{f_n \geq c} f_n $. This is equal to $0$ if $c>n$; equal to $1$ if $c \leq n$. Taking the $\sup_n$ we have $1$ (independently from $c$) and also after taking $\lim_{c \rightarrow \infty}$. So we have $1$ instead of $0$: NON U.I. Are these steps correct?2012-03-11
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    @Adam: Yes, you've got it.2012-03-12