Dear all,


Is there any possible way to construct a set $A \subseteq \mathbb{R}^n $ for which $ H^{n-1} (\partial A) > Leb^ + (A ) $?


Where $ H^{n-1} (\partial A) $ is the Hausdorff measure of the boundary of $A$ and:
$ Leb^{+} (A) = \lim\_{\epsilon \to 0 } \frac{ Leb(A\_ \epsilon) - Leb(A) }{\epsilon} $ , 
$A\_\epsilon := \{ x \in \mathbb{R} ^n | d(x,A) \leq \epsilon \} $ =Minkowski's content with respect to Lebesgue measure. 


Thanks in advance ! 



Different Measures On R2 
So, $A$ is just a "set"? Say $A$ is an open square minus a mid-line:
$$
A = \{ (x,y): 0 \lt x \lt 1 \text{ and (} 0 \lt y \lt 1/2 \text{ or } 1/2 \lt y \lt 1\text{)}\}
$$
Then $\partial A$ consists of 5 line-segments of length $1$ (the 4 sides and the mid-line), so $H^1(\partial A) = 5$. But $\mathrm{Leb}^+(A) = 4$, missing the mid-line.


