As far as I know, given a positive measure space, $f\in L^1(\Omega)$ is an uniformly integrable function if for all $\varepsilon>0$, $\delta>0$ exists such that
$$\int_A |f| \, d\mu <\varepsilon,$$
with $\mu(A)<\delta$.
It means, for me, that $f\in L^1(\Omega)$ is an uniformly integrable if its integral over a “small” set it is itself “small”.
I guess that, if $(f_n)_n\subset L^1(\Omega)$, thus $(f_n)_n$ is uniformly integrable sequence if
$$\int_A |f_n| \, d\mu <\varepsilon \quad\forall n\in\mathbb{N},$$
$\mu_(A)<\delta$.
${\bf My\; question\; is}$: there exists a characterization which allow to say that
$$\sup_n \int_{\Omega} |f_n| d\mu <+\infty,$$
thus $(f_n)_n$ is a uniformy integrable sequence? If yes, could anyone please give me a reference or a hint for the proof?
Thank you in advance!
Best Answer
The only probability spaces for which uniform integrability is equivalent to boundedness in $L^1$ are the probability spaces $(\Omega,\mathcal F,\mu)$ for which there exists a positive $\delta$ such that each $A\in\mathcal F$ has either probability $0$ or $\mu(A)>\delta$.
Indeed, if such a $\delta$ does not exists, it is possible to find a sequence of sets $(A_n)$ whose probability is in $(0,1/n)$. Let $f_n=\mathbf{1}_{A_n}/\mu(A_n)$. Then $(f_n)$ is bounded in $L^1$ but not uniformly integrable.
If there exists positive $\delta$ such that each $A\in\mathcal F$ has either probability $0$ or $\mu(A)>\delta$, then each sequence which is bounded in $L^1$ is actually bounded in $L^\infty$ (write down the essential supremum to see this) hence uniformy integrable.