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I saw that a Lipschitz function $f:\mathbb{R}^d \rightarrow \mathbb{R}^d$ maps measure zero set to measure zero sets.(Lusin's property) Why does a Lipschitz function $f:\mathbb{R}^d\to\mathbb{R}^d$ map measure zero sets to measure zero sets?

But, I wonder whether Lusin's property also holds for a Lipschitz function $f:\mathbb{R}^m\rightarrow \mathbb{R}^n$ ($m>n$). In other words, for any measure zero set $E$ in $\mathbb{R}^m$, is the image $f(E)$ measure zero in $\mathbb{R}^n$? Otherwise, could you give some counterexamples?

Also, please let me know if you have any related reference for $f$ mapping from $\mathbb{R}^m$ to $\mathbb{R}^n$.

bluejyellow
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    The simplest counterexample is $P(x_1, \cdots, x_m) = (x_1, \cdots, x_n)$ mapping $\mathbb{R}^m$ to $\mathbb{R}^n$ ($m > n$) .. look at the set $\mathbb{R}^n \times {0}^{m-n}$. – r9m Mar 08 '22 at 09:18

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