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The title might not be accurate, but here is my question:

Let $\lambda$ be the Lebesgue measure on some measurable space $(\mathbb{R}^2, \mathcal{A})$. Assume $A\in \mathcal{A}$ is such that $A = X \times Y$, where $X$ is a non-measurable subset of $\mathbb{R}$ with respect to the Lebesgue measure on some $\sigma$-algebra on $\mathbb{R}$, and $Y = \emptyset$ or $Y = \{y\}$, ${y}\in \mathbb{R}$. Is the Lebesgue measure of A identically equal to zero, or is it non-measurable? I know that generally $\lambda (A) \neq \lambda(X) \, \lambda(Y)$, still I would like to say that $\lambda(A) = 0$, since $A$ either has no extension in the space $\mathbb{R}^2$ or consists of line segments, which are null sets.

Thank you.

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I think you mean $Y = \{y\}$ instead of $Y = y$. Also, the Lebesgue measure is usually not defined/considered on the whole power set, since it is not $\sigma$-additive there.

Finally, your set $A$ is always Lebesgue measurable with measure zero, since

$$ A \subset \mathbb{R} \times \{y\}, $$

where the right hand side is a null-set. But (since the Lebesgue measure is complete) subsets of null-sets are null-sets, so that $A$ is measurable with Lebesgue measure zero.

EDIT: Now that you have edited the question, it is important to point out that the argument above only works as long as we are talking about Lebesgue measurability. If we want to have Borel measurability, the claim is no longer true.

To see this, recall from the Fubini theorem that if $A$ is Borel measurable, then so is every "section"

$$A_z = \{x \in \Bbb{R} \mid (x,z) \in A\}.$$

If you apply this to $z = y$ with $Y=\{y\}$, we get $A_y = X$, which is not measurable. This shows that the Lebesgue measure on $\Bbb{R}^2$, restricted to the Borel $\sigma$-algebra is not complete.

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    $\begingroup$ I assume the OP meant the Lebesgue outer measure on $\mathcal{P}(\mathbb{R}^2)$. $\endgroup$
    – Jason
    Feb 11, 2015 at 9:04
  • $\begingroup$ Sorry, my bad to define it on the whole power set. But the question is relevant for the Lebesgue outer measure as well, as Jason pointed out. I will edit the question with your comments in regard. Thank you for your answer, now I know how to approach it. $\endgroup$
    – Raibyo
    Feb 11, 2015 at 9:20

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