# Tightness + Integrability does not imply uniform integrability?

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?

• I don't think it is good enough because the second integrability condition depends on how $K_\epsilon$ grows as $\epsilon$ shrinks. For example, let $X$ be a uniform random variable on $[0,1]$ and $f_n(x)=n \chi_{[0,1/n]}(x)$. Then $f_n(X)$ is tight with $K_\epsilon=[0,1/\epsilon]$ and the integrals are all $1$, but this is pretty much the standard example of a non-uniformly integrable sequence. This is very much like the difference between a.s. convergence and convergence in probability. – Ian Apr 23 '17 at 0:44
• @Ian: Or rather, it's exactly like the difference between convergence in distribution and convergence in $L^1$. You've given the classic example, which everyone learns when first studying modes of convergence, of a sequence that converges a.s. (hence also in probability and in distribution) but not in $L^1$. – Nate Eldredge Apr 24 '17 at 15:14
• @NateEldredge Good point (though I think you mean it is the difference between convergence in probability and convergence in $L^1$; convergence in distribution is another matter entirely). I guess I came up with that comment because I had recently answered math.stackexchange.com/questions/2244981/… and I had the whole group of relationships on my mind. – Ian Apr 24 '17 at 15:30
• No, I meant what I said. Tightness is intimately related to convergence in distribution; every sequence converging in distribution is tight, and we have Prohorov's theorem that a tight sequence has a subsequence converging in distribution. And uniform integrability is likewise related to $L^1$ convergence - every sequence converging in $L^1$ is ui, and we have Vitali's theorem as mentioned by Fnacool below. (Strictly speaking, to make the analogy exact, we should speak of weak $L^1$ convergence; Dunford-Pettis plays the role of Prohorov.) – Nate Eldredge Apr 24 '17 at 17:29
• @NateEldredge I was trying to say that there is something inherently "weak" built into convergence in distribution, in the sense that convergence in distribution does not even depend on the probability spaces being the same. A.s. convergence, convergence in probability, and convergence in $L^p$ are all by comparison "strong" in the sense that they require reference comparison between pointwise values of $X_n$ and $X$. Switching over to weak $L^1$ completes the analogy. – Ian Apr 24 '17 at 19:30

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 $L^1$.
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.