## Expanding Measurable Sets

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

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 2011 at 22:33 Note that the assumption that the bijection f is measurable is not needed for this argument. – Alex Simpson Jun 10 2011 at 7:20 @Alex, you need $T$ to be measurable; otherwise you can not write the inequality. – Anton Petrunin Jun 10 2011 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 2011 at 19:11