Consider these definitions of tightness and integrability:
A sequence of real-valued random variables $f_n$ on a probability space $(X,\Sigma,\mu)$ is $\textit{tight}$ if for all $\epsilon$ there is a compact $K_{\epsilon}$ with $\mu(\{ x \mid f_n(x) \in K_{\epsilon} \}) > 1-\epsilon$ for all $n$.
A sequence of real-valued random variables $f_n$ on a probability space
$(X,\Sigma,\mu)$ is $\textit{uniformly integrable}$ if two conditions are satisfied:
- $\sup_{n}\int_{X}|f_n|d\mu < \infty$
- For all $\epsilon>0$ we can find a $\delta>0$ so that $\sup_n\int_{A}|f_n|d\mu < \epsilon$ whenever $\mu(A) < \delta$
I know that uniform integrability $\implies$ tightness. My first question was whether we can conclude the converse. Clearly the answer is no, since tightness does not require integrability. So my refined question is as follows:
Let $f_n$ be a collection of real-valued random variables defined on the probability space $(X,\Sigma,\mu)$, such that the $f_n$ are tight and $\sup_{n}\int_{X}|f_n|d\mu < \infty$. Can we conclude that the collection is also uniformly integrable?
Best Answer
(Edit: When first posting, I did not notice Ian's comment, so added some content).
No.
Problem is that while by tightness you require all functions to be large only on small sets, you still don't have an effective control on how large they can get on these small sets by merely requiring boundedness in $L^1$.
Example: Ian has given one.
Uniform integrability is a very strong requirement. By Vitali's theorem, it is a necessary and sufficient condition for a sequence in $L^1$ converging to $0$ a.s. (or in probability) to also converge in $L^1$ (dominated convergence is a only a sufficient condition, but much easier to verify).
You can get UI if you require a little more. For example, if you assume that $\phi$ is an nonnegative function on $[0,\infty)$ satisfying $\phi(x)/x \underset{x\to\infty}{\to} \infty$, and $\sup_n \int \phi(|f_n|) d \mu< \infty$, then it is not hard to show that the sequence is indeed UI.