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is it true that if $u_n$ are equi-integrable and $f:\mathbb{R} \to \mathbb{R}$ is continuous, is $f(u_n)$ equi-integrable?

Here $u_n \in L^1(\Omega)$ where $\Omega$ is a finite measure space with Lebesgue measure.

Seems like it ought to be true?

The definition of $u_n$ being equi-integrable is: for arbitrary $\epsilon > 0$, there exists $\delta > 0$ and $S \subset \Omega$ of finite measure such that for all $k$ $$\int_{\Omega \backslash S}|u_k(x)| < \epsilon$$ and $$\int_{H} |u_k(x)| < \epsilon \text{ if } |H| \leq \delta.$$

So only the second condition needs checking since we can take $S=\Omega$.

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  • $\begingroup$ Can you recall your definition of equi-integrable? $\endgroup$ – Siminore Jun 26 '14 at 13:53
  • $\begingroup$ Yes I just edited. $\endgroup$ – delimit Jun 26 '14 at 14:02
  • $\begingroup$ Isn't the definition, $\forall \epsilon>0$ $\exists \delta>0$ such that $\forall S\subset \Omega$ with $\lambda(\Omega\setminus S)<\delta$ we have that for every $n\in\Bbb N$, $\int_{\Omega\setminus S}|u_n|\,d\lambda<\epsilon$? Or is equi-integrable not the same thing as uniformly integrable? $\endgroup$ – Alex Schiff Jun 26 '14 at 14:11
  • $\begingroup$ @AlexSchiff Hmm, I'm not sure. Maybe they are equivalent? I got my definition from andre-schlichting.de/2012/10/weak-l%C2%B9-convergence and another lecture notes. $\endgroup$ – delimit Jun 26 '14 at 14:30
  • $\begingroup$ As the author says, condition 1 is trivially true if $\lambda(\Omega)<\infty$. Condition 2 is what we need, which is what I said in my comment. $\endgroup$ – Alex Schiff Jun 26 '14 at 14:39
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This is (without further assumptions on $f$) even false if you have only one function $u$ instead of a family $(u_k)_k$.

As a counterexample, construct a family $(A_n)_n$ of pairwise disjoint subsets of $\Omega$ with $\lambda (A_n) = \lambda(\Omega)/4^n$. I ask you to believe the existence of such sets for now.

Then take $u := \sum_n n \cdot \chi_{A_n}$. It is easy to see that $u$ is integrable, and thus equiintegrable.

Now take any continuous $f$ with $f(n) = 100^n$. Then $\int_{A_n} f(u) \,dx \geq 100^n/4^n \cdot \lambda(\Omega) \rightarrow \infty$, but $\lambda (A_n) \rightarrow 0$, so that $f\circ u$ is not equi-integrable.

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  • $\begingroup$ I believe it will be true if we assume that $f$ is absolutely continuous on $\Omega$. Do you agree? $\endgroup$ – Alex Schiff Jun 26 '14 at 14:18
  • $\begingroup$ Perhaps I am mistaken. I think $100^x$ is absolutely continuous. Maybe if we impose the condition that $f$ has compact support, then the result is true. $\endgroup$ – Alex Schiff Jun 26 '14 at 14:33
  • $\begingroup$ $f$ is defined on $\Bbb{R}$, not on $\Omega$. I am not sure that any kind of continuity/smoothness is appropiate. What is needed is some kind of growth condition, like $|f(x)| \leq C \cdot (1 + |x|)$ for all $x$. $\endgroup$ – PhoemueX Jun 26 '14 at 14:33
  • $\begingroup$ Oh, my bad. I meant $f$ is absolutely continuous on $\Bbb R$. $\endgroup$ – Alex Schiff Jun 26 '14 at 14:34

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