Let us been given a bounded domain $A\subset \mathbb{R}^n$. There is a function $u:A\to[0,1]$ such that A' = \{x\in A:u(x) = 1\} is not empty and $u\in Lip(A)$ with rate $\alpha$. Is it right that \mu(\partial A') = 0 where $\mu$ is Lebesgue measure? If not, please help me to construct a counterexample, if yes - please, help me to prove it.
Measure of the boundary
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0@Didier Piau: You right, this proof is wrong. Still, if the statement is correct? – 2011-06-07
1 Answers
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For every $x$ in $\mathbb{R}^n$ and every closed set $B\subset\mathbb{R}^n$, let $d(x,B)=\inf\{\|x-y\|;y\in B\}.$ Then $d(x,B)=0$ if and only if $x$ belongs to $B$ and, for every $x$ and $y$ in $\mathbb{R}^n$, $|d(x,B)-d(y,B)|\le\|x-y\|.$ Choose $B\subset A$ and define the function $u$ on $A$ by $u(x)=2^{-d(x,B)}.$ Then $u$ is Lipschitz continuous and A'=\{u=1\}=B. Thus, the boundary of A' may be the boundary of any closed set. In particular, it may have positive Lebesgue measure.
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0Than$k$s, just derived the same e$x$ample. – 2011-06-07