I'm a beginner in stochastic processes(and measure theory). I am trying to prove(or disprove) the fact that, if $\xi_n$ are uniformly integrable random variables, then they are uniformly bounded by some constant $M$, that is, $\sup_n|\xi_n| \leq M$ a.s. This is because I want to apply the Bounded Convergence Theorem. Here is my attempt at a proof:

Suppose not. Then, for any $K>0$, there exists $\delta>0$ such that \begin{equation}\sup_nP\{|\xi_n|>K\}>\delta > 0.\label{1786} \end{equation} Now, since the $\xi_n$ are uniformly integrable, there should exist $M>0$ such that \begin{equation*} \sup_n \int_{\{|\xi_n|>M\}} |\xi_n| dP < \delta. \end{equation*} Without loss of generality, we may assume $M>1$, because if the above holds for $0<M\leq 1$ then it also holds for $M>1$. But \begin{equation*} \sup_n \int_{\{|\xi_n|> M\}}|\xi_n|dP \geq M \sup_n P\{|\xi_n|>M\} > M\delta > \delta \end{equation*} which is a contradiction. This concldues the proof.

I would appreciate any help. Please understand I am new to measure theory. Thank you.

  • $\begingroup$ Uniform integrability does not imply uniform boundedness (nor does it imply domination). In fact, not being uniformly bounded (a.s.) means $P(\sup_n\lvert\xi_n\rvert>K)>0$ for any $K>0$ (the sup is inside). Nevertheless, it is true that you can replace the domination hypothesis in the dominated convergence theorem by uniform integrability. $\endgroup$
    – nejimban
    Oct 13, 2022 at 13:01
  • $\begingroup$ its not true, uniform integrability is unrelated to uniform boundedness $\endgroup$
    – Masacroso
    Oct 13, 2022 at 13:10
  • 1
    $\begingroup$ @nejimban Thank you. Ive read about the Vitali convergence theorem, and I understand the related materials now. $\endgroup$
    – Sean song
    Oct 14, 2022 at 2:23
  • 1
    $\begingroup$ @Masacroso thank you, I've found the errors in my proof now. $\endgroup$
    – Sean song
    Oct 14, 2022 at 2:23


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