Reconciling uniform integrability definitions and their relation to tightness

Take a family $$F$$ of measurable functions on $$\mathbb{R}$$, then $$F$$ is uniformly integrable if:

Definition 1 (Royden): For each $$\epsilon >0$$, there is a $$\delta>0$$ such that for each $$f\in F$$, if a Borel set $$I$$ is such that $$\lambda(I)<\delta$$ (Lebesgue measure), then $$\int_I |f|<\epsilon$$.

Definition 2 (Billingsley): The following holds uniformly: $$\lim_{a\to\infty}\sup_n \int_{|f|\geq a} |f| =0$$.

I can easily see how definition 1 yields definition 2, but have trouble seeing the reverse direction. My guess is we look at $$(-\infty, -a]\cup [a,\infty)$$ to construct $$\delta$$ as responding to $$\epsilon$$, but unsure on the exact approach.

Next, I wanted to comment on the relation between uniform integrability and tightness (to summarize my understanding).

Here are two definitions of tightness:

1. A family of random variables $$\{Y_\alpha| \alpha\in \mathcal{S}\}$$ is tight if for $$\epsilon>0$$ there exists a compact $$K$$ such that $$\mathbb{P}(Y_\alpha \notin K)<\epsilon$$ for any $$\alpha$$
2. Tightness of a family of measures: $$\mu_n(\{x||x|>u\})\to 0$$ as $$u\to\infty$$

Using Markov's inequality, it becomes clear that uniform integrability (or even $$L_1$$-boundedness) imply tightness. See https://math.stackexchange.com/a/1153644/833246

Also, see below for relation between uniform integrability and $$L_1$$-boundedness.

We also have the following equivalence:

$$(X_i)$$ is uniformly integrable if and only if:

1. $$\sup_{i\in I} E|X_i|<\infty$$
2. For all $$\epsilon>0$$ there is $$\delta>0$$ such that for all $$B\in\mathcal{F}$$ with $$\mathbb{P}<\delta$$, $$E[X1_B]<\epsilon$$

Let us show that 1) and 2) alone do not imply each other. Indeed, uniformly integrability is needed from 1) to 2) whereas tightness is needed from 2) to 1).

Uniform integrability implies $$L_1$$-boundedness, but $$L_1$$-boundedness does not necessarily imply uniform integrability (example: $$f_n(x) = n1_{1/n}(x)$$). See https://people.math.wisc.edu/~roch/teaching_files/275b.1.12w/lect10-web.pdf

To Note:

• Tightness of a family of functions suggest, more or less, that such functions "don't escape to infinity", or rather that their "escape to infinity" is arbitrarily small and they can therefore be sufficiently approximated on a set $$E$$ of finite measure.
• Example of not tight functions: $$f_n = \mathbb{X}_{[n,n+1]}$$. To note: this seems to be uniformly integrable using Royden's definition, but on a non-finite measure space, we'd rather use Billingsley's definition under which case $$f_n$$ is also not uniformly integrable. See https://math.stackexchange.com/a/1153644/833246
• Example of not uniformly integrable (but tight) functions: $$f_n = n \mathbb{X}_{[0,1/n]}$$
• Next question: what is the relation between the aforementioned concepts and the $$L^p$$ spaces? In particular, we know that perhaps due to non-tightness, we may see functions that converge pointwise in $$L^1$$ to a $$f$$, but not weakly to it in $$L^1$$.
• Related Commented Feb 2 at 0:45

Choose $$a$$ such that $$\sup_f\int_{|f|>a} |f|<\epsilon$$. Then for any $$I$$, $$\int_I |f| \le \int_{|f|>a} |f| + \int_{(|f|\le a) \cap I} |f| < \epsilon + a \lambda(I)$$ So if we pick $$\delta = \epsilon/a$$ we get for $$I$$ with $$\lambda(I)<\delta$$ that $$\sup_f\int_{I} |f|\le2\epsilon$$, QED.
• @Calvin Khor Thanks for your explanation but I still need help understanding it if you don't mind. I understand that without the $\cap{I}$, then it would just be $\lt \epsilon$. But, when you put the $\cap$ there, does that mean that the integral is over the intersection of the two sets or that the integral is over both of the sets $|f| \le a$ and $I$. Thanks and I really appreciate your help. Commented Feb 5 at 16:52
• Also, is the $n$ in Billingsley's definition, just a typo ? That confused me also. Thanks again. Commented Feb 5 at 16:55
• @markleeds yes, the intersection symbol means the integral is over the intersection of the sets. It is simply the Lebesgue integral $\int_A |f|$ with $A= \{x \in \mathbb R : |f(x)|\le a \} \cap I$. And the $\sup_n$ I guess should be $\sup_{f \in F}$ but I didn't check Billingsley Commented Feb 5 at 17:18