Proving uniformly integrablility of $\{X_n\}$, if $\sup_{n\geq 1} E[f(|X_n|)]<\infty.$

Question

Suppose $\{X_n\}$ is a sequence for which there exists an increasing function $f:[0,\infty)\mapsto [0,\infty)$ such that $\lim\limits_{x\rightarrow\infty} f(x)/x= \infty$ and

$$\sup_{n\geq 1} E[f(|X_n|)]<\infty.$$

Show $\{X_n\}$ is u.i.

Attempt

By definition, $\{X_n\}$ is ui if:

$$\sup_n E[|X_n|1_{\{|X_n|>a\}}]\rightarrow0, \text{ as } a\rightarrow \infty.$$

Intuitively, I think that if $\sup_{n\geq 1} E[f(|X_n|)]<\infty$ then $\sup_{n\geq 1} E[|X_n|]<\infty \text{ almost surely}$ and then $\{X_n\}$ is ui.

I would appreciate any hint.

Answer 1

Denote $\Delta := \sup_n E[f(|X_n|)] < \infty$. Given $\varepsilon > 0$, there exists sufficiently large $M > 0$ such that $M^{-1}\Delta < \varepsilon$.

By condition, for the chosen $M$, there exists a positive number $K$ such that for all $x > K$, we have \begin{equation} \frac{f(x)}{x} > M \text{ or } x < M^{-1}f(x). \tag{*} \end{equation}

Therefore, for all $a > K$, it follows that \begin{align} & \sup_n E[|X_n|I_{\{|X_n| > a\}}] \\ \leq & \sup_n E[|X_n|I_{\{|X_n| > K \}}] \\ \leq & \sup_n E[M^{-1}f(|X_n|)I_{\{|X_n| > K\}}] \quad \text{ (by } (*))\\ \leq & M^{-1}\sup_n E[f(|X_n|)] = M^{-1}\Delta < \varepsilon. \end{align}

This completes the proof.

Answer 2

Let $M=\sup_nE[f(|X_n|]$. For any $m$, there is $a_m>0$ such that $f(|x|)>Mm|x|$ for all $|x|>a_m$. Then

$Mm|X_n|\mathbb{1}(|X|_n>a_m)\leq f(|X_n|)\mathbb{1}_{|X_n|>a_m}$ and so

$$ M\mathbb{E}[|X_n|\mathbb{1}_{|X_n|>a_m}]\leq \frac{1}{m}\mathbb{E}[f(|X_n|)\mathbb{1}_{|X_n|>a_m}] $$ for all $n$. That is, $$ \sup_n\mathbb{E}[|X_n|\mathbb{1}_{|X_n|>a_m}]\leq \frac{1}{m} $$ Uniform integrability follows. The converse is true and is a Theorem due to de la Vallée Poussin.