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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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    I see your comment. Of course, this is a trivial condition that has to be satisfied. I'll update the question accordingly.2012-06-07

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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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    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 $n$.2013-04-02