If $u \in L^2(\Omega)$, then $\text{sign}u \in L^2(\Omega)$?

Question

If $\Omega$ is a bounded domain and $u$ is in $L^2$, why is $\text{sign}(u) \in L^2?$

I am only stuck with the measurabilituy part. the integral is obviously finite on a bounded domain.

Answer 1

$sign^{-1}$ maps 1 and -1 (measurable point sets) to the half intervals [0, $\infty$) and (-$\infty$, 0) respectively, which are measurable sets. These may be open or closed at 0, either way sign is measurable. Since f is measurable Ran(f) is measurable and $f^{-1}$ of either of these half lines (which are measurable sets) intersected with Ran(f) must be measurable. The composition of sign and f is then measurable.