A bounded set $F\subseteq L_{1}(\mu)$ is said to be uniformly integrable if : $\forall \epsilon$ there is a $\delta>0 $, such that $\forall$ measurable set $A$, and $\forall f\in F$ , if $\mu(A)< \delta$ , then $\int_A |f|d\mu< \epsilon$.
$F$ is called almost order bounded if for $\forall \epsilon$, there exists $u\in L_{1},u>0$ such that $F\subset [-u,u]+ \epsilon B_{1}(L_{1})$.
I want to prove that $F$ is uniformly integrable iff it is almost order bounded.
I have no idea how to start!
If the measure space is $[0,1]$, one can check that a family $\mathcal F$ which satisfies the condition $\forall\varepsilon>0,\exists \delta>0$ such that if $A$ is measurable of measure $<\delta$, then for each $f\in\mathcal F$, $\int_A|f|d\mu<\varepsilon$ satisfies the equivalent condition $$\lim_{R\to +\infty}\sup_{f\in \mathcal F}\int_{\{|f|>R\}}|f|d\mu=0.$$ More generally, such an equivalence holds if the measure space $(X,\mathcal B,\mu)$ is finite and $\mu$ doesn't contain any atom (that is, if $\mu(A)>0$, we can find $B\subset A$ such that $0<\mu(B)<\mu(A)$). This relies on the fact that for each $\delta>0$, we can write $X$ has a finite partition of sets of measure $<\delta$.
Once we have this equivalence, assume that $\mathcal F$ is uniformly integrable. Then for a fixed $\varepsilon$, take $R$ such that $\sup_{f\in\mathcal F}\int_{\{|f|>R\}}|f|d\mu<\varepsilon$. With $u:=R$, we are done.
Conversely, assume that $\mathcal F$ is almost order bounded. We need to show that if $u$ is integrable, then for all $\varepsilon>0$, there is $\delta>0$ such that if $\mu(A)<\delta$ then $\int_A |f|<\varepsilon$.