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 $S=\Omega$.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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