# Uniformly Integrable Random Variables are Uniformly Bounded Almost Everywhere?

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 $$$$\sup_nP\{|\xi_n|>K\}>\delta > 0.\label{1786}$$$$ Now, since the $$\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 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.

• 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. Oct 13, 2022 at 13:01
• its not true, uniform integrability is unrelated to uniform boundedness Oct 13, 2022 at 13:10
• @nejimban Thank you. Ive read about the Vitali convergence theorem, and I understand the related materials now. Oct 14, 2022 at 2:23
• @Masacroso thank you, I've found the errors in my proof now. Oct 14, 2022 at 2:23