Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mu$ be a probability measure on $X$.

Consider a family of functions $\phi_k: X \rightarrow \mathbb{R}_{\geq 0}$ such that $\sup_k \phi_k(\cdot)$ is integrable over $X$.

Let $\{X_n\}$ be an infinite sequence of compact sets such that $X_n \subset X$, $X_n \subseteq X_{n+1}$ and $X_n \rightarrow X$.

It seems to me that the following implication is true.

$$ \int_{X_n}\sup_k \phi_k(x) \mu(dx) \leq \Phi(x) \ \forall X_n \ \Rightarrow \ \int_X \sup_k \phi_k(x) \mu(dx) \leq \Phi(x)$$

Is it only a matter of applying the $\lim_{n \to \infty}$ since $\Phi(x)$ is not depending on $n$?

Is such family $\{\phi_k\}$ Uniformly Integrable?

share|cite|improve this question
A variant of this can be found in this post. – Adam Jul 1 '12 at 2:33
up vote 1 down vote accepted

We use Fatou lemma: let $f_n:=\chi_{X_n}(x)\sup_k\phi_k(x)$. It's a non-negative function, and $\lim_{n\to+\infty}f_n=\chi_x\sup_k\phi_k(x)$. We have $$\int_X\sup_k\phi_k(x)\mu(dx)=\int_X\liminf_{n\to \infty}f_n(x)\mu(dx)\leq \liminf_{n\to \infty} \int_Xf_n(x)\mu(dx)$$ and since $\int_Xf_n(x)\mu(dx)\leq \Phi(x)$ for all $n$ we get the result.

For the second question, if $\mathcal F\subset L^1(\mu)$ is such that there exist $g\in L^1(\mu)$ which satisfies $|f(x)|\leq g(x)$ for almost all $x$ and all $f\in\mathcal F$, then $\mathcal F$ is uniformly integrable. Indeed, $$0\leq \sup_{f\in\mathcal F}\int_{\{|f|\geq R\}}|f|d\mu\leq \int_{\{g\geq R\}}gd\mu$$ which converges to $0$ when $R\to +\infty$ thanks to the monotone convergence theorem.

share|cite|improve this answer
Thanks for the answer. However your second point is only sufficient. So the question is: are the given conditions already sufficient for Uniform Integrability?Integrability – Adam Apr 15 '12 at 15:42
Yes, it's a sufficient condition for uniform integrability that is true under the hypothesis you gave. – Davide Giraudo Apr 15 '12 at 15:45
A variant of this can be found in this post. – Adam Jul 1 '12 at 2:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.