I want to prove the following claim
Claim: Let $(\Omega, \Sigma, \mu)$ be a measure space. If $f : \Omega \to \mathbb{C}$ is $\mu$-measurable, then $\textrm{f}$ is $\mu$-measurable.
Here a function $f$ is called $\mu$-measurable if there exists a sequence of simple functions on $\Sigma$ $(\varphi_n)$ such that $\varphi_n \to f$ almost everywhere.
We define $\textrm{sgn}(z) := z/|z|$ for $z \neq 0$ and $\textrm{sgn}(z) := 0$ for $z = 0$.
This was just mentioned off-the-cuff in lecture as part of another proof. So I imagine it's not too difficult to see. Still, I haven't been able to prove it to myself or find anything on Stack Overflow about it.
My work so far:
I think I want to show the following:
Want to show: Let $(\varphi_n)$ be a sequence of simple functions such that $\varphi_n \to f$ almost everywhere, with $f : \Omega \to \mathbb{C}$. Then $\textrm{sgn}(\varphi_n) \to \textrm{sgn}(f)$ almost everywhere.
We let $\varphi_n \to f$ pointwise on $\Omega \setminus N$ with $\mu(N) = 0$ and take an $\omega \in \Omega \setminus N$.
It suffices to show that $|\textrm{sgn}(\varphi_n(\omega)) - \textrm{sgn}(f(\omega))| \to 0$.
Case 1: $f(\omega) \neq 0$.
$|\varphi_n(\omega) - f(\omega)| \to 0 \Rightarrow \Big||\varphi_n(\omega)| - |f(\omega)|\Big| \to 0 \Rightarrow \Big| \frac{|\varphi_n(\omega)|}{f(\omega)} - \text{sgn}(f(\omega)) \Big| \to 0$.
Not sure where to go from here. I imagine I take a subsequence $(\varphi_{k(n)})$ that with no zero elements and show that $\frac{|\varphi_{k(n)}(\omega)|}{\varphi_{k(n)}(\omega)} \to \frac{|\varphi_n(\omega)|}{f(\omega)}$ and then use this to get the convergence I want.
Case 2: $f(\omega) = 0$. Here I need to show $\text{sgn}(\varphi_n(\omega)) \to 0$. If $\exists M \in \mathbb{N}$ such that $\varphi_n(\omega) = 0$ $\forall n \geq M$, then I'm done. Otherwise, I'm not quite sure. Maybe I need to take a subsequence of nonzero $(\varphi_{k(n)}(\omega))$ and show $\text{sgn}(\varphi_{k(n)}(\omega)) \to 0$.