Let $\Omega$ be an open subset in $\mathbb{R}^N$. Suppose $u \in L^1(\Omega)$ and $f: \mathbb{R} \to \mathbb{R}$ such that $f' \in L^{\infty}(\mathbb{R})$.
Is it true that $f\circ u \in L^1(\Omega)$?
I can prove it in the case $f(0) = 0$, or $\Omega$ has finite measure, but I haven't done the problem in the general case.
Can anyone help me with any idea?
It can't be true if $f(0) \not= 0$ and $\mu(\Omega) = +\infty$.
Indeed, if $\mu(\Omega) = +\infty$, one can take $f(t) = 1$ and $u = 0$. Then $u\in L^1(\Omega)$, but $f\circ u \not\in L^1(\Omega)$.
In fact, more is true. If $f(0) \not= 0$ and $\mu(\Omega) = \infty$, then there are no functions $u\in L^1(\Omega)$ such that $f\circ u \in L^1(\Omega)$. To see this, let $u\in L^1(\Omega)$ and consider the set $$ A = \left\{x\in\Omega; |u(x)| < \frac{|f(0)|}{2\|f'\|_{L^\infty}}\right\}. $$ By Markov's inequality, $$\mu(A^c)\frac{|f(0)|}{2\|f'\|_{L^\infty}} < \|u\|_{L^1},$$ so $\mu(A) = \infty$. Moreover we know that $|f\circ u(x)| > \frac{1}{2}|f(0)| / \|f'\|_{L^\infty}$ for all $x\in A$, which shows that $f\circ u\not\in L^1(\Omega)$.