$X_n$ are uniformly integrable if $\lim_{R \rightarrow \infty} \sup_n E[|X_n|,|X_n| \geq R] = 0$. Show that if $\sup_{n \geq 0} E[|X_n|\ln(|X_n|)] < \infty$, then $X_n$ are uniformly integrable.

My best guess is to take the converse, and show that if $\lim_{R \rightarrow \infty} \sup_n E[|X_n|,|X_n| \geq R] = ε > 0$, then $\sup_{n \geq 0} E[|X_n|\ln(|X_n|)] = \infty$. The trouble is that the best I can think to do is say that $\sup_n E[|X_n|,|X_n| \geq R] = \sup_n E[|X_n| \chi_{|X_n| \geq R}]$, and $|X_n|\ln(|X_n|) \geq |X_n|\chi_{|X_n| \geq R}$ whenever $R > e$ (since $\ln(e) = 1 \geq \chi_A$ for any set $A$), but this just barely says that $E[|X_n|\ln(|X_n|)]$ are positive, far short of saying that their supremum is infinity.

Is this even the right approach to this problem, or is there something obvious I'm missing on how to start this?


Maybe the simplest approach is to write $$|X_n|\chi\{|X_n|>R\}\leqslant \frac 1{\ln R}|X_n|\cdot\ln|X_n|,$$ then integrating $$\mathbb E[|X_n|\chi\{|X_n|>R\}]\leqslant \frac 1{\ln R}\sup_k\mathbb E[|X_k|\cdot\ln|X_k|].$$

It would work replacing $\ln$ by a non-decreasing function $\phi$ such that $\lim_{x\to \infty}\phi(x)\to \infty$.

A converse is given by de La Vallée Poussin theorem.

  • $\begingroup$ @Byron Thanks for editing. $\endgroup$ – Davide Giraudo Oct 5 '14 at 21:17
  • $\begingroup$ @user4514 Thanks for spotting this typo, it is indeed an $x$. $\endgroup$ – Davide Giraudo Mar 2 '15 at 9:33

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