Uniformly integrable sequence tending to 0 a.s. but with $\mathbb{E}(\sup_n|X_n|) = \infty$ I am trying to find a uniformly integrable sequence of random variables $(X_n: n \in \mathbb{N})$ such that both $X_n \to 0$ almost surely and $\mathbb{E}\left(\sup_n|X_n|\right) = \infty$. I think this is probably a well known counterintuitive result in the theory, but I haven't come across it before, so any help would be great.
 A: It is possible to construct for any $p>1$ a sequence $(X_n)_{n\geqslant 1}$ of random variables such that

*

*the sequence $(X_n)_{n\geqslant 0}$ converges to $0$ almost surely;

*the sequence $\left(\mathbb E\left|X_n\right|^p\right)_{n\geqslant 1}$ is bounded and

*the random variable $\sup_n\left|X_n\right|$ is not integrable.

Indeed, consider the unit interval endowed with the Borel $\sigma$-algebra and Lebesgue measure. Define for a sequence $(\delta_n)_{n\geqslant 1}$ which decreases to $0$ and will be specified later
$$A_n:=  (\delta_{n+1},\delta_n) $$
and $X_n=\left(\delta_n-\delta_{n+1}\right)^{-1/p}\mathbf{1}(A_n)$.

*

*holds because the $A_n$ are pairwise disjoint.


*We have  $\mathbb E\left|X_n\right|^p=1$ by construction.


*Since the $A_n$ are disjoint, $\sup_n\left|X_n\right|=\sum_{n=1}^{+\infty}X_n$ and $\mathbb EX_n\geqslant \left(\delta_n-\delta_{n+1}\right)^{1-1/p}  $.
Therefore, it suffices to choose the sequence $(\delta_n)_{n\geqslant 1}$ such that $\sum_{n} \left(\delta_n-\delta_{n+1}\right)^{1-1/p}=\infty$. Take $\delta_n=\sum_{k\geqslant n}k^{-q}$ for some $q\in (1,p/(p-1))$. Then $\left(\delta_n-\delta_{n+1}\right)^{1-1/p}=n^{-q\left(p-1\right)/p}$ and $q\left(p-1\right)/p<1$.
