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Let $\mu$ be a finite measure on $X \subseteq \mathbb{R}^n$.

Consider the Uniformly Integrable family $\{ f_n(\cdot) \}_{n \in \mathbb{N}}$ of functions $f_n : X \rightarrow \mathbb{R}_{\geq 0}$.

Consider a continuous, strictly-increasing function $\phi:\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}$, with $\phi(0)=0$ such that $\phi(f_n(\cdot))$ is integrable for all $n$.

  1. Is the family $\{ \phi( f_n(\cdot) ) \}_{n \in \mathbb{N}}$ Uniformly Integrable as well?

  2. More generally, with functions $g_n : X \rightarrow \mathbb{R}^n$ such that the family $\{ |g_n(\cdot)| \}_{n \in \mathbb{N}}$ is Uniformly Integrable, and a continuous $\psi: \mathbb{R}^n \rightarrow \mathbb{R}_{\geq 0}$, is $\{ \psi(g_n(\cdot)) \}_{n \in \mathbb{N}}$ Uniformly Integrable?

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no, they are not even necessarily integrable,e,g,$\phi(x) = x^2$ –  mike Jun 6 '12 at 23:44
Why? Counterexample? If it is not the case, under which conditions we have that UI is preserved? –  Adam Jun 6 '12 at 23:52
I see your comment. Of course, this is a trivial condition that has to be satisfied. I'll update the question accordingly. –  Adam Jun 7 '12 at 0:00

1 Answer 1

up vote 2 down vote accepted

Consider $n=1$ and $f_j=j\chi_{(0,\frac 1{j^2})}$. Fix $R>0$, when $j\geq R$ we have $$\int_{|f_j|\geq R}|f_j|dP\leq \sqrt{\mu(|f_j|\geq R)}\leq \frac 1R,$$ hence $\{f_j\}$ are uniformly integrable, and in $L^3$. We have $\lVert f_j\rVert_{L^3}=j\left(\int_{(0,\frac 1{j^2})}1\right)^{1/3}=j^{1/3}$, hence the family $\{f_n^3\}$ is not uniformly integrable.

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What additional assumptions should be made on $\phi$ to get that $\{ \phi(f_n(\cdot)) \}_n$ are UI? –  Adam Apr 2 '13 at 9:17
Moreover, in your example, are the functions $\phi(f_n(\cdot))$ integrable? –  Adam Apr 2 '13 at 9:23
Yes, $\phi(f_n)$ is bounded and the measure space is finite. For conditions about preservation of uniform integrability, I presume you already know de la Vallée Poussin's theorem, so it could be a start. –  Davide Giraudo Apr 2 '13 at 9:24
Yes, I do know that and I am not clear about your comment. In the de la Vallée Poussion Theorem, we have a scalar function $G$ such that $\lim_{x \rightarrow \infty} G(x)/x = \infty$. Here we want to use like $|f|$ is place of $G$, but it depends on the parameter $n$. –  Adam Apr 2 '13 at 9:37

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