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.


1 Answer 1


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 $2^{\mathbb N}.$

  • 1
    $\begingroup$ Thank you for taking the time to write the answer. BUT in this example you gave the graph is not Borel measurable and the function is not Borel measurable. It is not what I am asking for. $\endgroup$
    – Ramiro
    Nov 15, 2020 at 14:22
  • 3
    $\begingroup$ @Ramiro: in this example the graph is Borel measurable, I added an extra "not" when transcribing... $\endgroup$
    – Harry West
    Nov 15, 2020 at 15:50
  • $\begingroup$ Ok. Thanks. I will read the example carefully. $\endgroup$
    – Ramiro
    Nov 15, 2020 at 17:48

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