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.


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.

  • $\begingroup$ Thank you. I think this works. $\endgroup$ – Giuliano Basso Nov 20 '14 at 12:40

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

  • 1
    $\begingroup$ The intersection is equal to $ N\times \{0\} $ and is also Lebesgue measurable with respect to the 2 dimensional Lebesgue measure. But I dont't see the contradiction. Is it true that $\lambda_2( N\times \{0\})=\lambda_1(N)$ ? $\endgroup$ – Giuliano Basso Nov 19 '14 at 18:59
  • $\begingroup$ Could you elaborate your hint a little bit? $\endgroup$ – Giuliano Basso Nov 22 '14 at 14:57

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