Example: $\{X_n \}$ is uniformly integrable but $E[ f(X_n) ]=\infty$ with $f(x)=\omega(x)$

Question

I am looking for a counterexample where the sequence of random variables is $\{X_n \}$ is uniformly integrable, that is \begin{align} \lim_{ k \to \infty} \sup_{n \ge 0} E[ |X_{n}| 1_{ |X_n| \ge k} ]=0 \end{align} and yet for every non-negative and non-decreasing function $f(x)$ and $f(x)=\omega(x)$ we have that \begin{align} \sup_{n \ge 0} E[f(|X_n|)]=\infty. \end{align}

Where $f(x)=\omega(x)$ means that $ \lim_{x \to \infty} |f(x)|/|x|=0$.

Answer 1

I believe that you actually meant $\lim_{x\to+\infty} |f(x)|/x = +\infty$.

Still, there is no such counterexample as, in fact, de la Vallée Poussin condition for uniform integrability is a criterion. Here is the proof of the inverse conclusion:

Let $\{k_m, m\ge 1\}$ be such that $\sup_{n\ge 1}\mathbb E\left[|X_n|\mathbf{1}_{|X_n|\ge k_m}\right]\le 2^{-m}$; without loss of generality we can assume that $k_{m+1}> k_m$ for all $m\ge 1$. Setting $f(x) = x\sum_{m=1}^\infty m\mathbf{1}_{[k_m,k_{m+1})}(x)$, we have for any $n\ge 1$, \begin{gather*} \mathbb E[f(|X_n|)] = \sum_{m=1}^\infty m\,\mathbb E\left[|X_n|\mathbf{1}_{[k_m,k_{m+1})}(|X_n|)\right]\le \sum_{m=1}^\infty m 2^{-m}, \end{gather*} so $\sup_{n\ge 1} \mathbb E[f(|X_n|)]\le \sum_{m=1}^\infty m2^{-m}<\infty$, as required.

Answer 2

Actually, the conditions on $\{X_n\}$ and $f$ imply that $\sup Ef(|X_n|) <\infty $: Since $f(x)<x$ for $x$ exceeding some number $x_0$ and $f(x) \leq f(x_0)$ for $x \leq x_0$ we see that $f(x) \leq x+c$ for all $x$ where $c=f(x_0)$. Thus $Ef(|X_n|) \leq c+E|X_n|$. Uniform integrability implies that $\{E|X_n|\}$ is bounded.