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

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.

## 2 Answers

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.

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

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

• 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 $\lambda_2( N\times \{0\})=\lambda_1(N)$ ? – Giuliano Basso Nov 19 '14 at 18:59
• Could you elaborate your hint a little bit? – Giuliano Basso Nov 22 '14 at 14:57