# Confusion on Uniform Integrability of Random Variables

We have the definition that a random process, $$X_n$$, is (1st power) uniformly integrable if

$$\lim_{M\to\infty}\sup_n\mathbb{E}(|X_n| ; |X_n|\geq M)=0$$.

My question is whether the following four statements are true and (intuitively) why this is or isn't the case:

1. "$$|X_n|$$ is bounded for all $$n$$ $$\implies$$ $$X_n$$ is 1st power uniformly integrable."

My guess: it's true since if it's bounded there must exist some $$M\in \mathbb{R}$$ such that $$M>|X_n|$$, and so $$\lim_{M\to \infty}\mathbb{I}\{|X_n|\geq M\}=0$$

1. "$$|X_n| < \infty$$ $$\implies$$ $$X_n$$ is 1st power uniformly integrable."

My guess: intuitively to me it feels like this should be true by similar reasoning to (1); if $$|X_n|$$ is finite then as $$M\to \infty$$ we get $$\mathbb{I}\{|X_n|\geq M\}=0$$? And hence $$\lim_{M\to\infty}\sup_n\mathbb{E}(|X_n| ; |X_n|\geq M)=0$$? I think I'm wrong, however.

1. "$$\mathbb{E}|X_n| < \infty$$ $$\implies$$ $$X_n$$ is 1st power uniformly integrable."

My guess: I think this is wrong because $$\mathbb{E}|X_n| < \infty$$ doesn't imply that $$|X_n|$$ is finite and so $$\lim_{M\to \infty}\mathbb{I}\{|X_n|\geq M\}$$ is not necessarily $$0$$?

1. "If $$\forall \epsilon>0, \exists\delta>0$$ such that $$\mathbb{P}(|X_n|\geq M)<\delta$$ implies $$\mathbb{E}(|X_n| ; |X_n|\geq M)<\epsilon$$ $$\implies$$ $$X_n$$ is 1st power uniformly integrable."

My guess: I think this is not true since I think that this doesn't necessarily mean that we can find $$\delta$$ such that $$\sup_n\mathbb{E}(|X_n| ; |X_n|\geq M)<\epsilon$$? My question would also be - if we add in either the condition that $$\mathbb{E}|X_n|<\infty$$ or that $$|X_n|<\infty$$ then does this imply uniform integrability?

1. It depends actually whether the bound is allowed to depend on $$n$$ or not. If $$\sup_{n\geqslant 1}\lvert X_n\rvert$$ is finite, then your arguement works. But if we only assume that $$\lvert X_n(\omega)\rvert\leqslant c_n$$ for almost every $$\omega$$, uniform integrability may fail: consider $$X_n(\omega)=n$$ if $$0<\omega<1/n$$ and $$0$$ otherwise on $$(0,1)$$ endowed with Lebesgue measure.

2. and 3. are covered by the previous counterexample.

For 4., the thing is that there are probability spaces such that $$\mathbb P(A)$$ is either $$0$$ or bigger than some $$\delta_0$$, like $$\{a,b\}$$ with $$\mathbb P(\{a\})=\mathbb P(\{b\})=1/2$$. In this case, the implication will be always satisfied with $$\delta<1/2$$, because the only way to have $$\mathbb{P}(|X_n|\geq M)<\delta$$ is that $$\{|X_n|\geq M\}$$ is empty. Hence a non-uniformly integrable sequence can also satisfy the implication.