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Dear all,

I have recently found the following discussion: http://mathoverflow.net/questions/102061/hausdorff-measure-and-minkowskis-content-boundary-measures regarding different boundary measures on $\mathbb{R}^n $ .

The discussion made me wondering:

Is there any example in the opposite direction?

i.e. an example of a set $A \subseteq \mathbb{R} ^n $ such that $ Leb^+ (A) > H^{n-1} (\partial A) $ ?

Thanks in advance !

flag	asked Jul 22 at 18:11 Jeremy Young 91●3

Tapio Rajala · Accepted Answer · 2012-07-22 19:22:04Z UTC

3

Take $[0,1]\times \{0\} \subset \mathbb{R}^2$.

answered Jul 22 at 19:22

Tapio Rajala
1,885●1●6●12

	Dear @Tapio: Thanks for your answer, but I don't think I understand it: We have $H^1(\partial [0,1]\times0 )=H^1([0,1]\times0 ) = 1 $ . But I can't understand how to calculate the minkowski's content of this thing- $Leb ( [0,1] \times 0 ) = 0 $ since we are in $\mathbb{R}^2 $ . But what is the Lebesgue measure of an $\epsilon$ extension of this thing? Thanks ! – Jeremy Young Jul 23 at 6:57
	@Jeremy, $\mathop{Leb}^+=2$. – Anton Petrunin Jul 23 at 7:53
	@Anton: Can you please explain this calculation? Thanks! – Jeremy Young Jul 23 at 11:24
	@Jeremy: An $\epsilon$ extension of that set is very simple, and the computation of its area is completely elementary. Just draw a picture of it. – Mark Meckes Jul 23 at 12:14
	@Mark: I agree that an $\epsilon$ extension of this set is pretty simple to calculate in the $\mathbb{R} $ case (and it's equal to $2\epsilon$ there indeed, which gives $Leb^+ =2 $ as you said. But since we're looking at the $\mathbb{R}^2 $ case, the $\epsilon$ extension we are looking at is actually some kind of an ellipse, which I have no idea how to calculate its volume. I hope I made my misunderstandings clear now. Thanks a lot ! – Jeremy Young Jul 23 at 15:57

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Anton Petrunin · Answer 2 · 2012-07-23 07:51:27Z UTC

4

$A=\mathbb{Z}\subset \mathbb R^2$. In this case $\partial\mathbb{Z}= \mathbb{Z}$ and $\mathop{Leb}^+(\mathbb Z)=\infty$ and $H^1(\mathbb Z)=0$.

You can get a bounded example of the same type. Take a countable nowhere dense set $A$ in the unit disc such that the $\varepsilon$-neighborhood of $A$ contains a disc of radius $\sqrt[3]{\varepsilon}$.

edited Jul 23 at 7:51

answered Jul 22 at 19:23

Anton Petrunin
16.4k●41●110

1

But $\partial \mathbb{Q} = \mathbb{R}$. Perhaps you mean $\mathop{Leb}^+(\mathbb Z)=\infty$ and $H^1(\mathbb Z)=0$? – Tapio Rajala Jul 22 at 19:44

Ups, you are right, now it is corrected. – Anton Petrunin Jul 22 at 20:13

Thanks @Anton, but I was thinking about this question in the context of boundary measures...In this case, since $\partial \mathbb{Z} $ is empty , it doesn't give me an answer. Thanks anyway! – Jeremy Young Jul 23 at 6:52

@Jeremy, $\partial \mathbb Z=\mathbb Z$. – Anton Petrunin Jul 23 at 7:46

yes. you are right... sorry and thank you very much! – Jeremy Young Jul 23 at 11:24

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