Assume that $N\subset \mathbb{R}$ is not Lebesgue measurable. Is then the set $ N\times \mathbb{R} $ also not Lebesgue measurable?

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Assume that $N\subset \mathbb{R}$ is not Lebesgue measurable. Is then the subset $ N\times \mathbb{R} \subset \mathbb{R}^2 $ also not Lebesgue measurable?

Im not sure whether this is true or not. The problem is that $\mathcal{L}(\mathbb{R}^2) \neq \mathcal{L}(\mathbb{R})\otimes \mathcal{L}(\mathbb{R})$ but $\mathcal{L}(\mathbb{R}^2)$ is equal to the completion of $\mathcal{L}(\mathbb{R})\otimes \mathcal{L}(\mathbb{R})$, so we may find a Lebesgue null set $M\in \mathcal{L}(\mathbb{R})\otimes \mathcal{L}(\mathbb{R})$ such that $N\times \mathbb{R} \subset M$ and therefore we may obtain that $ N\times \mathbb{R}$ is Lebesgue measurable.

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You can use Fubini's theorem to show that if E is a measurable subset of plane then almost every (but not necessarily every) vertical and horizontal section of E is a measurable subset of line.

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HINT: Suppose $N \times \mathbb{R}$ is Lebesgue measurable. What can you say about $(N \times \mathbb{R}) \cap ( \mathbb{R} \times \{ 0 \} )$?