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Let $S,T \subset \mathbb{R}^n$ be measurable sets, and suppose that there exists a measurable bijection $f\colon S\to T$ so that $$ \|f(x)-f(y)\| \;\geq\; \|x-y\| $$ for all $x,y \in S$. Does it follow that $\mu(S) \leq \mu(T)$?

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1 Answer 1

up vote 7 down vote accepted

It follows from two observations:

  • For Hausdorff meeasure your statement follows from the definition.
  • Hausdorff measure = Lebesgue measure (up to constant).
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Oh, great. Thanks! –  Jim Belk Jun 9 '11 at 22:33
    
Note that the assumption that the bijection f is measurable is not needed for this argument. –  Alex Simpson Jun 10 '11 at 7:20
    
@Alex, you need $T$ to be measurable; otherwise you can not write the inequality. –  Anton Petrunin Jun 10 '11 at 13:49
    
I wasn't questioning the measurability of S and T. But the measurability of the bijection f is an additional assumption in the question as formulated and is unnecessary. –  Alex Simpson Jun 12 '11 at 19:11

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