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

Question

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 $S=\Omega$.

Answer 1

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.