The definition I am working with:
A partial function $F:\omega^\omega\rightarrow\omega^\omega$ is said to be partial recursive iff the partial function $G:\omega^\omega\times\omega\rightarrow\omega$ s.t. $G(f,i)=F(f)(i)$ has a $\Sigma^0_1$ graph.
My definition does not mention anything about the domain of $G$, but I am assuming that $\text{dom}(G)=\text{dom}(F)\times\omega$
Under this assumption I seem to be getting that $f\in\text{dom}(F)\iff\exists i\exists j[G(f,i)=j]$, which means that $\text{dom}(F)$ is $\Sigma^0_1$.
However, later on in my notes there is a remark that says: "There is a partial recursive function $F:\omega^\omega\rightarrow\omega^\omega$ s.t. $\text{dom}(F)$ is not $\Sigma^0_2$".
So I'm making a mistake somewhere. Am I making the wrong assumption as to what the domain of $G$ should be?
Any help is appreciated, Thanks.
I guess the definition should be like this:
A partial function $F: \omega^{\omega} \to \omega^{\omega}$ is said to be partial recursive iff there is a partial recursive function $G:\omega^{\omega}\times \omega\to \omega$ s.t. for any $f$, $f\in dom(F)\Leftrightarrow\forall i(G(f,i)\downarrow)$. Then $F(f)(i)$ is defined as $G(f,i)$.
Then $f\in dom(F)$ if and only if $\forall i\exists j (G(f,i)=j)$. So $dom(F)$ is $\Pi^0_2$. But this is a weird definition.