Uniformly integrable sequence: definition and characterization

Question

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!

Answer 1

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.

Answer 2

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).