Let be a set that is not measurable with respect to the n-dimensional Lebesgue measure. Show that has Hausdorff dimension n.

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- December 2nd 2011, 10:05 AMkierkegaardLebesgue measure and Hausdorff dimension
Let be a set that is not measurable with respect to the n-dimensional Lebesgue measure. Show that has Hausdorff dimension n.

- December 3rd 2011, 12:04 PMJose27Re: Lebesgue measure and Hausdorff dimension
Assuming that the Hausdorff dimension is the number , then we have for all (To see this note that the s'-Hausdorff measure of an n-cube is zero for and use subadditivity of the outer measure), and by monotonicity , so . If then , but Lebesgue measure is complete so is measurable, a contradiction.