I'm looking for a function $f:\mathbb R^2\longrightarrow \mathbb R$ which is measurable but such $f^{y}$ defined by $f^{y}(x)=f(x,y)$ is not measurable for every $y$. Does $$f(x,y)=\chi_{\mathcal N\times \{0\}}(x,y)$$ where $\mathcal N$ is a non measurable set in $\mathbb R$ work ? I think it does since $$f(]-\infty ,\alpha [)=\begin{cases}\emptyset & \alpha \leq 0\\ (\mathbb R\times \mathbb R^*)\cup(\mathcal N^c\times \{0\})&0< \alpha\leq 1\\ \mathbb R^2&\alpha >1 \end{cases}$$ where are all measurable set in $\mathbb R^2$. (The fact that $\mathcal N^c\times \{0\}$ is measurable comes forme that $m_{\mathbb R^2}(\mathcal N^c\times \{0\})=0$ where $m_{\mathbb R^2}$ is the Lebesgue measure un $\mathbb R^2$).
Thanks to confirm or not my example.