Let $s>0$ and $0 < \delta \leq \infty$. For a set $E \subset R^n$ define

$$ H_{s}^{\delta} (E)=\inf \left\{ \sum_i r_{i}^{s}\right\},$$

where the infumum is taken over all coverings of $E$ by balls $B_i$ with diameter $r_i$ no exceeding $\delta$.

Denote by $ H_{s}(E) = \displaystyle\sup_{\delta > 0} H_{s}^{\delta}(E) = \displaystyle\lim_{\delta \rightarrow 0}H_{s}^{\delta}$ the $s$ Hausdorff measure of the set $E\subset R^n$ .

I know this result : if $H_s(E) < \infty$ then $H_r (E) =0$ for $r>s$.

I am trying to prove : if $0<s<n$ and the interior of $E$ is nonempty then $H_s (E) = \infty $ .

I believe that this is the solution for my question :

if $H_s (E) < \infty$ then $H_n(E) = 0$. but this is a contradtion because :

interior of $E$ is nonempty implies $H_n (E) >0$ but I don't know how to prove this last implication..

Someone can help me ?

thanks in advance =)

  • $\begingroup$ Do you already know that $H_n$ is a multiple of the (Borel-)Lebesgue measure? $\endgroup$ – Daniel Fischer Jul 20 '13 at 22:08
  • $\begingroup$ no ... . I only know the result that I write in my question and I know the s hausdorff measure is a borel regular measure .. $\endgroup$ – math student Jul 20 '13 at 22:12
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    $\begingroup$ Pity. If you had that, it'd be trivial. I don't remember how difficult it is to prove that, or at least, that an open $n$-box has finite nonzero $H_n$-measure, unfortunately. $\endgroup$ – Daniel Fischer Jul 20 '13 at 22:16
  • $\begingroup$ @DanielFischer . thanks four your comentaries =) $\endgroup$ – math student Jul 20 '13 at 23:22
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    $\begingroup$ Well, you need to show $H^n(U) > 0$ for SOME nonempty open set. This is an interesting fact, and not just something from fiddling with the definitions. A good way to do it (say with $U$ the unit cube in $\mathbb R^n$) is to use knowledge about Lebesgue measure. Bound $H_n^\delta$ from below using Lebesgue measure. $\endgroup$ – GEdgar Jul 21 '13 at 12:55

It's inevitable that in order to prove something about measures on $\mathbb R^n$, you have to use some measure-theoretic fact about $\mathbb R^n$. What's so special about $\mathbb R^n$ that makes $H^n(U)>0$ for nonempty open $U$?

Fact. There exists a Borel measure $\lambda$ on $\mathbb R^n$ that satisfies $0<\lambda(B(x,r))\le Cr^n$ for every ball $B(x,r)$, where $C$ is a constant.

Sketch: construct $\lambda$ on $\mathbb R$ (Lebesgue measure), then take the product.

Once you have the fact, the rest is easy. For any cover of the unit ball $B(0,1)$ with $B(x_i,r_i)$ we have $$\sum r_i^n\ge C^{-1}\sum \lambda(B(x_i,r_i)) \ge C^{-1}\lambda(B(0,1))$$

  • $\begingroup$ where I can find a proof of the fact ? thanks for your answer =) $\endgroup$ – math student Jul 22 '13 at 1:32
  • $\begingroup$ @LeandroTavares Read anything with the words construction of lebesgue measure. $\endgroup$ – 40 votes Jul 22 '13 at 1:50

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