Let $(\Omega,\mathcal F,\mu)$ be a $\sigma$-finite measure space and consider the product space $(\Omega\times \Omega,\mathcal F\otimes \mathcal F,\mu \otimes \mu)$. If $N\in \mathcal F\otimes \mathcal F$ has $\mu \otimes \mu$ measure zero, can I show that the set
$$\Delta:=\{\omega\in\Omega: (\omega,\omega)\in N\}$$
has $\mu$-measure zero?
The set $\Delta$ is the preimage of $N$ under the measurable map $\omega\mapsto (\omega,\omega)$, and so is $\mathcal F$-measurable. There is a related post here, but I wasn't able to figure out an answer for my question based on it.
Thank you for your help.
Well, no. I mean, take $([0,1], \mathcal{F},\mathcal{L})$ with $\mathcal{F}$ the Lebesgue measurables and $\mathcal{L}$ the Lebesgue measure. Then the diagonal of the square has null $\mathcal{L} \times \mathcal{L}$ measure, but $\mathcal{L}(\Delta)=1$. Or maybe I misinterpreted the question?