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.
It is possible to construct for any $p&gt;1$ 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 $n\geqslant 1$:
$$A_n:=\left(\left(n+1\right)^{-p/(p-1)-1},n^{-p/(p-1)-1}\right),\quad 
X_n:=n^{1/(p-1)} \mathbf 1(A_n).$$
In this way, we have that 
$$\tag{*} cn^{-p/(p-1)} \leqslant \lambda(A_n)\leqslant Cn^{-p/(p-1)}$$
where $c$ and $C$ are independent of $n$. 


holds because the $A_n$ are pairwise disjoint. 
By (*), $\mathbb E\left|X_n\right|^p\leqslant Cn^{-p/(p-1)}n^{p/(p-1)}=C$. 
Since the $A_n$ are disjoint, $\sup_n\left|X_n\right|=\sum_{n=1}^{+\infty}X_n$ and by (*), $\mathbb EX_n\geqslant cn^{-1}$.
Added: This only works for functions without the restriction that they are defined on a probability space. The requirement for random variables makes my example fail, because I'm relying on the fact that the support of the functions gets smaller as you move toward $\infty$. Examples relying on the support of the function probably won't work for random variables, and it seems quite a bit harder to think of examples for this that have support on the whole real line or the positive reals and satisfy the needed integrability constraints by virtue of decaying quickly.

I'm really rusty with this stuff and I might misunderstand the question, but here is an idea.

Define 

$$ \displaystyle END(X):= \sup \textrm{ }\{x \textrm{ s.t. } X(x) &gt; 0  \}$$

Then make the sequence $X_{n}$ as:

$$ \displaystyle X_{n} = \begin{cases} 1 \textrm{ if } x \in \biggl[END(X_{n-1}), \frac{1}{n} + END(X_{n-1})\biggr] \\ 0 \textrm{ otherwise.}\end{cases} $$

for $n &gt; 0$ with $END(X_{0}) := 0$.

I think the above definition is easier to read, but another way to write it is:

$$ X_{n} = \displaystyle \begin{cases} 1 \textrm{ if } x \in \biggl[\displaystyle\sum_{j=1}^{n-1}\frac{1}{j}, \displaystyle\sum_{j=1}^{n}\frac{1}{j}\biggr] \\ 0 \textrm{ otherwise.}\end{cases}$$

With the understood definition that for $n=1$ we start off with the indicator function of $[0,1]$.

The pointwise limit for a fixed $x$, $\displaystyle \lim_{n\to\infty} X_{n}(x) = 0 \quad\forall x $, but $\sup_{n} X_{n}(x)=1$ for $x \ge 0$, so 

$$\int\sup_{n}X_{n}(x) = \int_{x=0}^{\infty}1dx = \infty.$$ 

To see that it's uniformly integrable, consider that

$$ \int X_{n}dx = \int_{END(X_{n-1})}^{END(X_{n-1})+\frac{1}{n}}dx = \frac{1}{n}.$$

For a given $\epsilon &gt; 0$, choose $n$ such that $1/n &lt; \epsilon$ and set $K = n$. The integral over $[K, \infty]$ for all $X_{m}$ with $m &lt; n$ is 0 (since those functions are zero at or before they hit $x=K$). We just showed that the integral is smaller than $\epsilon$ for $n=K$ and such an integral only becomes smaller as 
$n$ gets larger after that point. So the definition for uniform integrability should be satisfied.