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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 2011 at 11:19 |
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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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