In different books and resources, I saw different definitions of uniformly integrable.

For example, in some books the definition is like:

Definition 1: $\{f_k\}\in L^1(E)$ is uniformly integrable if for any $\epsilon>0,$ we can find a non-negative measurable function $g$ such that $\int _{\{x\in E: |f_k(x)|\geq g(x)\}} |f(x)|dx<\epsilon, \forall k\in \mathbb{N_+}$.

In other books it's like:

Definition 2: $\{f_k\}$ is called uniformly integrable if $f_k$ is uniformly bounded on $L^1$ and there is no escape to vertical or width infinity.

i.e $\sup _k \int|f_k|<+\infty$,

$\sup_k \int_{\{x \in E: |f_k(x)|\geq M\}}|f_k(x)|dx\rightarrow 0, M\rightarrow +\infty$,

$\sup_k \int_{\{x\in E: |f_k(x)|\leq \delta\}}|f_k(x)|dx\rightarrow 0, \delta\rightarrow 0^+.$

If we already know that $E$ has finite measure, then there is another corollary:

$\{f_k\}$ is uniformly integrable iff $\sup_k ||f_k||_1<\infty,$ and for any $\epsilon >0$, there exists a $\delta>0$ such that for any $A\subseteq E, m(A)< \delta,$ we have $\int_A |f_k|<\epsilon, \forall k\in \mathbb{N_+}$.

My question is:

1. Are the definitions 1 and 2 equivalent to each other?

2. In finite measurable space, I know UI+ a.e converge (or converge in measure) $\Longleftrightarrow$ converge in $L^1$. In infinite measurable space, I don't think the ''uniformly integrable'' contains ''tight''. If we know $\{f_k\}$ is UI and converges a.e (or converges in measure), can we still get $f_k$ converges in $L^1$?


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