# Open set whose boundary is not a null set

I've just seen a theorm about a bounded set $A \subset \mathbb{R}^n$. $$\chi_A \text{ is Riemann integrable} \Longleftrightarrow \partial A \text{ is a null set}$$

Then, I wonder if there's any open set whose boundary is not a null set. Can you give me some example for that?

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You can find, for $0<\delta<1$, an open subset of $[0,1]$ of measure $\delta$ which is dense in $[0,1]$. In particular, the measure of the boundary is $1-\delta$. – Davide Giraudo Jun 10 '12 at 14:34
Isn't $(0,1)$ a simpler example than those given so far !! indeed, $(0,1)$ is an open subset of $\mathbb R$ such that $\partial (0,1)=\{0,1\}\not = \emptyset$. – palio Aug 2 '14 at 14:50

Consider $A$ a fat Cantor set, for example a set of measure $\frac13$. $A$ is closed, so its complement is open, and also $A$ is nowhere dense therefore its complement is dense. That is to say that $U=[0,1]\setminus A$ is open, so for the Lebesgue measure $m$ we have: $$m(\partial U)=m([0,1]\setminus U)=m(A)=\frac13.$$
Let $\{r_k,k\in\Bbb N\}$ an enumeration of the rational numbers of $[0,1]$. Consider the map $t\mapsto \bigcup_{n\geq 0}\left(r_n-t\cdot 2^{-n}, r_n+t\cdot 2^{-n}\right)=O_t$. Then $O_t$ is an open set, and $f\colon t\mapsto \lambda(O_t)$ is increasing, and continuous (in fact $1$-Lipschitz continuous). Hence for $\delta\in (0,1)$ you can find $t$ such that $\lambda(O_t)=\delta$, and $O_t$ is a dense subset of $[0,1]$. Its boundary is $[0,1]\setminus O_t$, which has measure $1-\delta\neq 0$.
More generally, you can find an open subset of $\Bbb R$ whose boundary has measure $m$, where $r$ is an arbitrary fixed positive number (you also may have $m=+\infty$ when you take $\{r_k\}$ an enumeration of the rational (of the whole real line) and $O:=\bigcup_{k\geq 1}(r_k-2^{-k},r_k+2^{—k})$.