# Measure of the boundary

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.

-
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

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.