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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 !

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  • $\begingroup$ פ.ת.ףץפתך.80פת. 0 $\endgroup$
    – Tomer Schlank
    Commented Oct 7, 2017 at 8:24

1 Answer 1

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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.

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  • $\begingroup$ Thanks a lot ! The problem is that I can't understand the calculation of such thing: By definition : $H^1 ( \partial A) = lim _{\delta \to 0} inf \{ \sum C diam(U_i) | \partial A \subseteq \cup U_i , diam(U_i ) \leq \delta \} $ Can you please detail a bit about how you deduced that $H^1 (\partial A) = 5$ in this case? Thanks a lot @! $\endgroup$
    – Jason Mraz
    Commented Jul 12, 2012 at 20:18
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    $\begingroup$ First: For a line segment of length $a$, the one-dimensional Hausdorff measure is $a$. Hausdorff measure is additive on Borel sets, so the measure for the 5-segment set is $5$. $\endgroup$
    – Gerald Edgar
    Commented Jul 12, 2012 at 21:13

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