# Is there an example of a non measurable function with a measurable graph?

Let $$(S, \Sigma)$$ be a measurable space. Let $$f: S \rightarrow \mathbb{R}$$ be function and let $$\mathcal{B}(\mathbb{R})$$ be Borel $$\sigma$$-algebra on $$\mathbb{R}$$. Let $$G(f)$$ be the graph of $$f$$, that means $$G(f)=\{(x,f(x)) \in S\times \mathbb{R} : x \in S\}$$. We write $$\Sigma \otimes \mathcal{B}(\mathbb{R})$$ to indicate the product $$\sigma$$-algebra of $$\Sigma$$ and $$\mathcal{B}(\mathbb{R})$$.

We can prove that:

1. If $$f$$ is $$\Sigma$$-$$\mathcal{B}(\mathbb{R})$$ measurable then $$G(f) \in \Sigma \otimes \mathcal{B}(\mathbb{R})$$.
2. Assuming that $$S$$ is a Polish space and $$\Sigma$$ is its Borel $$\sigma$$-algebra, if $$G(f) \in \Sigma \otimes \mathcal{B}(\mathbb{R})$$ then $$f$$ is $$\Sigma$$-$$\mathcal{B}(\mathbb{R})$$ measurable.

The proof of item 1 is rather easy. The proof of item 2 is more advanced and uses analytic sets. One such proof can found in here (section 3 proposition 6).

Question: Can we prove that if $$G(f) \in \Sigma \otimes \mathcal{B}(\mathbb{R})$$ then $$f$$ is $$\Sigma$$-$$\mathcal{B}(\mathbb{R})$$ measurable in the general case (that is, just having $$(S, \Sigma)$$ a measurable space)? If not, is there a counterexample (that is, an example where $$G(f) \in \Sigma \otimes \mathcal{B}(\mathbb{R})$$, but $$f$$ is not $$\Sigma$$-$$\mathcal{B}(\mathbb{R})$$ measurable)?

(Of course changing the $$\sigma$$-algebra in the counter-domain makes it trivial to have counter-examples).

Remark: Examples where $$G(f) \notin \Sigma \otimes \mathcal{B}(\mathbb{R})$$ and $$f$$ is not $$\Sigma$$-$$\mathcal{B}(\mathbb{R})$$ measurable are rather trivial to produce. They are not what I am asking for.

Let $$X=[0,1]$$, $$\Sigma$$ be the Borel $$\sigma$$-algebra in $$[0,1]$$, let $$A\subseteq [0,1]$$ be a set not in $$\Sigma$$. Let $$f$$ be $$\chi_A$$ (the indicator function) of $$A$$. Its is easy to see that $$G(f) \notin \Sigma \otimes \mathcal{B}(\mathbb{R})$$, and $$f$$ is not $$\Sigma$$-$$\mathcal{B}(\mathbb{R})$$ measurable.

See Srivastava, A Course on Borel Sets, in the section "Solovay's Coding of Borel Sets", beginning at "We now proceed to give an example of a function with domain coanalytic whose graph is Borel and that is not Borel measurable." Google books preview (There is a typo: it should say $$E\subseteq C\times \mathbb R\times 2^{\mathbb N}$$ instead of $$E\subseteq \mathbb R\times 2^{\mathbb N}.$$)
The example is a function $$f:C\times \mathbb R\to2^{\mathbb N}$$ where $$C$$ is a set of codes for Borel sets. Of course $$\mathbb R$$ is Borel isomorphic to $$2^{\mathbb N}.$$