Different definition of uniform integrability

Question

I'm reading Tao's An introduction to measure theory.And the definition of uniform integrability in this book is

$(X,{\mathfrak {M}},\mu)$ is a measure space(not necessarily finite). A sequence $f_n:X \rightarrow \mathbb{C}$ of absolutely Integrable functions is said to be uniformly Integrable if the following three statements hold

1.(Uniform bound on $L^1$ normal) One has $\sup_n\|f_n\|_{L^1(\mu)}=\sup_n\int_{X}|f_n|d\mu <+\infty$.

2.(No escape to vertical infinity) One has $\sup_n\int_{|f_n|\ge M}|f_n|d\mu\xrightarrow{M\rightarrow\infty} 0$.

3.(No escape to width infinity) One has $\sup_n\int_{|f_n|\le\delta}|f_n|d\mu\xrightarrow{\delta\rightarrow0} 0$.

However, the definition of uniform integrability in Wikipedia is

Let $(X,{\mathfrak {M}},\mu ) $ be a positive measure space. A set ${\displaystyle \Phi \subset L^{1}(\mu )}$ is called uniformly integrable if to each ${\displaystyle \varepsilon >0} $ there corresponds a ${\displaystyle \delta >0}$ such that ${\displaystyle \int _{E}|f|\,d\mu <\varepsilon } $ whenever ${\displaystyle f\in \Phi }$ and ${\displaystyle \mu (E)<\delta .}$

Tao leaves the equivalence of this two definitions when $(X,\mathfrak{M},\mu)$ is finite(namely $\mu(X)<+\infty$) as an exercise, and I've prove it. But I can't prove the equivalence of them when $\mu(X)=+\infty$. Can anyone prove or disprove it?

Any help is appreciated.

EDIT:I've also prove that the first definition implies the second one(which is also an exercise in Tao's book). So the problem becomes whether the second one implies the first one.

@Rixinner: The Wiki page on uniform integrability is now updated. It has more definitions of uniform integrability related to the classic notion introduced by Hunt. Interestingly enough, the German wiki article has been more complete than the English one by far. The definition in Tao's book is a curios definition as it is not equivalent to Hunt's: Hun't implies Tao's but not the other way around. — Mittens, May 05 '22 at 02:08

score 1 · Accepted Answer · answered Feb 28 '21 at 05:05

The condition in Wikipedia is to weak to imply 1), 2) and 3). Consider $\mathbb N$ with power set and the counting measure. Any family of functions satisfies the Wikipedia condition since $\mu (E) <1$ implies that $E$ is the empty set. You can easily write down a family of functions each of which satsifies the condition $\|f||_1=\sum |f(n) |<\infty$ but the norms are not bounded when$f$ varies over the family. [I will write down an explicit example if you ask for it].

Thanks! I've got what you mean. – Rixinner Feb 28 '21 at 06:00 — Rixinner, Feb 28 '21 at 06:00

Different definition of uniform integrability

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