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Every countable union of rectangles in R2 is a Lebesgue measurable set. Is the converse true, too?

Specifically, I wonder whether the following statement is true:

Let A be a set in the unit square that is Lebesgue measurable. Then there a countable collection of rectangles and a null set such that A is equal to the union of the rectangles and the null set.

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closed as too localized by Bill Johnson, fedja, Qiaochu Yuan, Ryan Budney, Emil Jeřábek Dec 7 '11 at 11:19

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up vote 0 down vote accepted

No, see here for a counter-example (it's a variant of the Cantor set which has non null Lebesgue measure and does not contain any interval).

And if you want a counter-example for $\mathbb{R}^2$ instead of $\mathbb{R}$, just cross it with an interval.

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Yes, that should help answer it. The argument will then be that the symmetric difference of any nontrivial interval with the SVC set is not a null set. – Nahpetz Dec 7 '11 at 0:41

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