Let $f: D\to \mathbb{C}$ be (Lebesgue) measurable. Prove there exists a measurable function $u : D\to \mathbb{C}$ so that $|u(x)| = 1$ for all $x\in D$ and $f = u|f|$.
I think the following definition should work: $u(x) = 1$ when $f(x) = 0$ and otherwise $u(x) = \frac{f}{|f|}$. I know that since $f$ is measurable, $|f|$ is also measurable, and if $f$ were always nonzero, then $\frac{f}{|f|}$ would be measurable. But the issue is that f might be zero. I tried considering open subsets of $\mathbb{C}$ and I observed that $u^{-1}(\{1\}) = f^{-1}([0,\infty))$, but that didn't seem to help much.
The set $E=f^{-1}(\{0\})$ is measurable. So you can define $$ u=1_E+\frac f{|f|}\,1_{D\setminus E}. $$