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I came along the following question in my textbook and needed a little help.

I need to find a measurable set $A \subset \mathbb{R}^2$ such that $\dfrac{m(A\cap B(0, h))}{m(B(0, h))} \rightarrow 1/3$ as $h \rightarrow 0.$

Thanks

Quinn Culver
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casey
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2 Answers2

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Take $A$ to be a sector of the unit circle with $1/3$ the area. Then you have a constant ratio of $1/3$ as $h\to 0$.

Lepidopterist
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You should check out Lebesgue's density theorem. That should put you down the right track.

Ian Coley
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  • Since the question concerns only the balls centered at a single point $0$, I don't see the relevance of the Lebesgue density theorem. That theorem tells you about limits, as in the question, for balls centered at almost all points, but the "almost" leaves plenty of room for an exception at one point --- as Lepidopterist's answer shows. – Andreas Blass Mar 18 '13 at 03:15
  • In addition to what Andreas Blass said, the Lebesgue density theorem only provides information about points with density $0$ and density $1$ -- nothing about points with density $\frac{1}{3}.$ Also, this StackExchange question is relevant. – Dave L. Renfro Mar 18 '13 at 14:08