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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.

user301068
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    Yes. ${}{}{}{}$ –  Jan 03 '16 at 19:59
  • When $y=1$, $f^y\equiv 0,$ which is measurable. – Weltschmerz Jan 03 '16 at 20:04
  • @Weltschmerz: What do you want to say ? – user301068 Jan 03 '16 at 20:28
  • @user301068 You want $f^y$ to be nonmeasurable for every $y,$ right? – Weltschmerz Jan 03 '16 at 20:36
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    @Weltschmerz: no, but you are right to mention the ambiguity. I wanted $f$ such that $f$ measurable but $f^y$ is not measurable for every $y$ (i.e. that there exist $y$ such that $f^y$ is not measurable). By Tonelli theorem a function mesurable such that for all $y$ the function $f^y$ is not measurable doesn't exist since $f^y$ is measurable for almost every $y$. – user301068 Jan 03 '16 at 20:42
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    Ok. Then your example works. Be careful with your English. Change "every" to "some". – Weltschmerz Jan 03 '16 at 21:16

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