It is well known, that if we have an interval $[a,b] \subset \mathbb{R}$ and a continuous curve $c \colon [a,b] \to \mathbb{R}^2$ with $c(t) \neq 0$ for all $t$, then there exist continuous functions $r \colon [a,b] \to (0,\infty)$ and $\varphi \colon [a,b] \to \mathbb{R}$ so that $$c(t) = r(t)(\cos(\varphi(t)), \sin(\varphi(t)))$$ holds for all $t \in [a,b]$. This motivates the following question:
If $(X, \mathcal{A}, \mu)$ is a measure space and $f \colon X \to \mathbb{C}$ is measurable, do there exist measurable functions $r \colon X \to [0, \infty)$ (note that $r$ is allowed to be 0) and $\varphi \colon X \to \mathbb{R}$ so that $$f(x) = r(x)(\cos(\varphi(x)) + i \sin(\varphi(x)))$$ I know that $\varphi$ (if existent) certainly will not be unique and that $r$ can indeed be found quite easily by $$r(x) = |f(x)|$$ So the main struggle is with finding $\varphi$.
Can we prove that there always exists such a measurable $\varphi$?
Edit: As pointed out in the comments, the statement holds for simple functions $f = \sum \alpha_k \chi_{A_k}$, by simply setting $r = |f|$ and $\varphi = \sum \varphi_k \chi_{A_k}$, where the $\varphi_k \in \mathbb{R}$ are chosen such that $f(x) = r(x) \exp(i\varphi_k)$, since then by construction $$f = r \cdot \exp(i\varphi)$$ Now it is clear, that if we have an arbitrary measurable function $f$, then we can approximate $f$ by simple function, thus by what we have seen before, we may find a sequence of simple functions $\{\varphi_n\}$ such that $$r(x)\exp(i\varphi_n(x)) \to f(x)$$ for all $x \in X$. However, this does not guarantee, that there exists some $\varphi \colon X \to \mathbb{R}$ so that $\varphi_n(x) \to \varphi(x)$ and $r(x)\exp(i\varphi(x)) = f(x)$ for all $x$.
Let $$\varphi(x) = \begin{cases} \arccos\left(\mathrm{Re}\left(\frac{f(x)}{\lvert f(x) \rvert}\right)\right) &\text{if $f(x)\neq 0$ and $\mathrm{Im}(f(x))\geq 0$,} \\ -\arccos\left(\mathrm{Re}\left(\frac{f(x)}{\lvert f(x) \rvert}\right)\right) &\text{if $f(x)\neq 0$ and $\mathrm{Im}(f(x))< 0$,} \\ 0 &\text{otherwise.}\end{cases}$$
Then $\varphi$ is measurable as a composition of measurable functions, and one has $f(x)=r(x) e^{i\varphi(x)}$.