Uniform integrability and stopping times

Question

I want to know whether there is any example where $X_n$ is uniformly integrable, $N$ is a stopping time and $E[X_N] =\infty$? Or uniform integrability of $X_n$ implies that $E[X_N]< \infty$?

Answer 1

This is not true. Moreover, it seems pretty obvious to me that the fact like this cannot be true without any further assumption like the martingale property; however, the requirement that $N$ should be a stopping time (with respect to the filtration of $X$, I presume) makes it harder to find a counterexample.

Let $\Omega$ be $[0,1]$ with the Lebesgue measure and $X_n(\omega) = \frac{2^n}{n} \mathbf{1}_{(2^{-n},2^{-n+1}]}(\omega)$. Then $$\sup_{n} E[|X_n|\mathbf{1}_{|X_n|>C}] = \sup_{n:\frac{2^n}n>C} \frac1{n} = \frac{1}{\inf\{n: 2^n/n>C\}}\to 0,C\to\infty,$$ so the sequence is uniformly integrable (alternatively, one can use the de la Vallée-Poussin theorem with $V(x) = x\log x$). However, setting $N = \inf\{n\ge 1: X_n>0\}$, $$ E[X_N] = E\left[\sum_{n=1}^\infty \frac{2^n}{n} \mathbf{1}_{(2^{-n},2^{-n+1}]}(\omega)\right] = \sum_{n=1}^\infty \frac1n = +\infty. $$