Uniformly integrable sequence: definition and characterization 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!
 A: 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.
A: As the answer by Davide Giraudo shows, it is not enough to be bounded in $L^{1}$ in general.  However, it is almost enough (provided one is careful in one's use of almost).
Suppose $(f_{n})_{n \in \mathbb{N}}$ satisfies, for some $\epsilon \in (0,\infty]$ and $C > 0$,
\begin{equation*}
\|f_{n}\|_{L^{1 + \epsilon}(\Omega)} = \left( \int_{\Omega} |f_{n}(x)|^{1 + \epsilon} \, \mu(dx) \right)^{\frac{1}{1 + \epsilon}} \leq C.
\end{equation*}
(If $\epsilon = \infty$, use the $L^{\infty}(\Omega)$ norm...) Notice we can now apply Holder's inequality to find, for each $\mu$-measurable set $A \subseteq \Omega$,
\begin{equation} \label{E: integrability modulus}
\int_{A} |f_{n}(x)| \, \mu(dx) \leq C \mu(A)^{\frac{\epsilon}{1 + \epsilon}} \quad (*)
\end{equation}
Hence $(f_{n})_{n \in \mathbb{N}}$ is uniformly integrable.  In fact, if you want, (*) gives a uniform modulus of integrability for $(f_{n})_{n \in \mathbb{N}}$.  It's even better than that since if only depends on $C$, not the sequence.
Notice that ( * ) is somehow analogous to Holder continuity.  In the same way that a uniform Holder modulus helps get uniform convergence from pointwise convergence (using Arzela-Ascoli), (*) helps improve pointwise convergence to convergence of the integral (via Vitali's convergence theorem).
