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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?

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A variant of this can be found in this post. –  Adam Jul 1 '12 at 2:33
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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.

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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
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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
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