Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
There is a version of Rademacher's theorem initially due to Cheeger valid for quite general metric measure spaces in the sense of Gromov. Googling these key-words should lead you to a wealth of results. –  t.b. Jun 7 '11 at 10:28
    
Thanks. Since you only advised me without giving a feedback on the proof, I guess that it is correct. –  Ilya Jun 7 '11 at 10:57
    
Sorry, I haven't really looked and thought about it. I just wanted to give you this pointer to enable you to search for something that might suit your needs. –  t.b. Jun 7 '11 at 10:59
    
Ok, thanks. Then I'm still looking for the review of my proof. –  Ilya Jun 7 '11 at 11:05
    
Any point in $A'$ is either in the interior of $A'$ or on the boundary of $A'$ (true). At any point in the interior of $A'$ the gradient of $u$ exists (true). How can you deduce from that that at every point on the boundary of $A'$ the gradient of $u$ does not exist? You cannot (furthermore this happens to be false). –  Did Jun 7 '11 at 13:08

1 Answer 1

up vote 5 down vote accepted

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.

share|improve this answer
    
Thanks, just derived the same example. –  Ilya Jun 7 '11 at 16:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.