Suppose $f\colon\mathbb{R}^{\omega}\longrightarrow\mathbb{R}$ is a function such that $$G(f):=\{(x,y)\in\mathbb{R}^{\omega}\times\mathbb{R}\mid f(x)=y\}$$ is a Borel set. Does it necessarily follow that $f$ is a Borel function?

## 1 Answer

Yes. This immediately follows from Lusin's separation Theorem. Note that $\mathbb{R}^\omega$ is Polish.

Let $B \subseteq \mathbb{R}$ be a Borel set. The set $f^{-1}(B)=\{x \in \mathbb{R}^\omega \colon \exists y \,\,\, y \in B \land (x,y) \in G(f) \}$ is $\Sigma_1^1$ (analytic). However, $f^{-1}(B)=\{x \in \mathbb{R}^\omega \colon \forall y \,\,\, (x,y) \in G(f) \rightarrow y \in B \}$ is also $\Pi_1^1$ (co-analytic). Therefore, $f^{-1}(B)$ is Borel.