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It is well known that the Banach-Tarski paradox does not translate to circles, i.e. a circle cannot be decomposed into finitely many pieces which can be rearranged to form two circles of equal circumference.

However what I'd like to know is can a circle be decomposed into finitely many pieces that can be rearranged to form two circles whose difference in circumference is arbitrarily small?

Thank you.

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  • $\begingroup$ Try decomposing circles into parts that are not measurable -- that's the crux of the paradox. $\endgroup$ – Trebor Apr 16 '19 at 11:39
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    $\begingroup$ I am tempted to post another answer which goes into a little more detail (in particular, if you allow a countable number of pieces, it is possible). However, this is not really my are of expertise, and Terrance Tao has already said everything that I could possibly say, and said it better. $\endgroup$ – Xander Henderson Apr 16 '19 at 12:24
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No. If you can split a figure on plane and rearrange pieces to get another figure, and both figures have well-defined area, then both have the same area.

The reason is that Lebesgue measure on bounded subset of plane can be extended to finitely additive translation invariant measure with any set being measurable. And if two sets are equicomposed, then they must have the same measure under any translation invariant finitely additive measure. In particular, if two sets are Lebesgue measurable and equicomposed, they need to have equal Lebesgue measure.

Note that it's true even if parts we decompose our sets to are not Lebesgue-measurable themselves - we just need initial and final sets to be measurable.

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    $\begingroup$ What if the area is non-measurable? $\endgroup$ – YeatsL Apr 16 '19 at 11:44

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