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Why does a Lipschitz function $f:\mathbb{R}^d\to\mathbb{R}^d$ map measure zero sets to measure zero sets?

It is easy to prove this statement if the domain is bounded. Is there any way to extend the argument to unbounded domains? Can anyone give me some hint, or show me a simple proof sketch?

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    $\begingroup$ If $A\subseteq \mathbb R^d$ has measure zero, so does $A_n := A \cap B_n(0)$ for $n\in \mathbb N$ and $A = \bigcup_n A_n$ ... $\endgroup$ – martini May 2 '12 at 12:33
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Let $N$ be a set of measure zero. Then for every $\varepsilon \gt 0$ we can write $N \subset \bigcup_{k=1}^\infty B_k$ where each $B_k$ is a ball of radius $r_k$ and $\sum_k \mu B_k \lt \varepsilon$. But then by Lipschitz continuity $f(B_k)$ is contained in a ball of radius $L\cdot r_k$ where $L$ is the Lipschitz constant of $f$, and thus $\mu^\ast f(B_k) \leq L^d \mu B_k$ so that $$ \mu^\ast f(N) \leq L^d \sum\nolimits_k \mu B_k \lt L^d \varepsilon. $$ As $\varepsilon \gt 0$ was arbitrary, we see that $f(N)$ must have zero outer measure and hence it is a null set.

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    $\begingroup$ I don't understand why the result for the bounded case should be any easier and the reduction is entirely unnecessary. $\endgroup$ – t.b. May 2 '12 at 12:46
  • $\begingroup$ Thanks. This is much simpler than my solution. $\endgroup$ – mysterious May 2 '12 at 13:58
  • $\begingroup$ I think $\mu^\ast f(B_k) \leq L^d \mu B_k$ is not so obvious from Lipschitz condition, may i have a proof? $\endgroup$ – mnmn1993 Nov 16 '17 at 4:52
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so let $A \subseteq \mathbb R^d$ be a set with measure zero. For $n \in \mathbb N$ let $A_n := B_n(0) \cap A$. Then $A_n$ has measure zero as $A_n \subseteq A$ for each $n$ and we have $A = \bigcup_n A_n$. From your bounded case you get that $f(A_n)$ is of measure zero for each $n$. But as \[ f(A) = f\left(\bigcup_n A_n\right) = \bigcup_n f(A_n) \] $f(A)$ is a countable union of sets of measure zero and has therefore measure zero.

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  • $\begingroup$ Thanks. But the solution given by t.b. is much simpler than mine. So I chose his answer as the correct answer. $\endgroup$ – mysterious May 2 '12 at 14:01

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