Existence of a function that generates a Borel set

Question

How to prove that there exists $\xi: \Omega \to \mathbb R$, which is not a random variable such that for all $x \in \mathbb R$ $\xi^{-1}(x) = \{ \omega | \xi(\omega) = x \}$ is a Borel set?

Answer 1

1. Let $\Omega=\mathbb{R}$, $\mathcal{F}=\sigma(\{\omega\})$, and $\xi(\omega):=\omega$. Then $\xi^{-1}(\{x\})=\{x\}\in\mathcal{F}$. However, for any Borel set $B$ that is not countable or co-countable $\xi^{-1}(B)\notin\mathcal{F}$ (e.g. $B=(-\infty,0]$).

2. Let $\Omega=[0,1]$, $\mathcal{F}=\mathcal{B}([0,1])$, and $\xi(\omega):=\omega+2\cdot1\{\omega\notin N\}$, where $N\subset [0,1]$ is a non-Borel set. Then $\xi^{-1}(\{x\})\in\mathcal{F}$. However, $\xi^{-1}([0,1])=V\notin \mathcal{F}$.