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This is an exercise taken from Mattila, Geometry of sets and measures in Euclidean space, chapter 4.

Exercise. Let $U$ be an open ball in $\mathbb{R}^n$ ($n\ge 2$) such that $d(U)=\delta$ [here $d$ stands for "diameter"]. Show that for $0\le s \le 1$, $$\tag{1} \mathcal{H}^s_\delta(U)=\mathcal{H}^s_\delta\left(\overline{U}\right)=\mathcal{H}^s_\delta(\partial U).$$

Here $\mathcal{H}^s_\delta(A)$ is the infimum of the sums $$\sum_{j=1}^\infty d^s(E_j), $$ where $\{E_j\}$ is a covering of $A$ such that $d(E_j)\le \delta$. As $\delta\downarrow 0$, $\mathcal{H}^s_\delta(A)$ tends to the Hausdorff measure $\mathcal{H}^s(A)$. The point of this exercise is to show that, even if $\mathcal{H}^s_\delta$ is a (outer) measure, it is not Borel since it fails to be additive on $\overline{U}=U\cup \partial U$.

Can you help me with this exercise? Those things are new to me and I would use a hint to start with. Thank you!

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How about first doing $U$? See that it can be covered by a single set of diameter $\delta$. That gives you a lower bound for $\mathcal H^s_\delta(U)$. –  GEdgar Jan 5 '13 at 23:01
    
@GEdgar: $U$ has diameter $\delta$ and is a cover of itself. So obviously $\mathcal{H}^s_\delta(U)\le \delta^s$. What I cannot understand is how to get from this a lower bound on $\mathcal{H}^s_\delta(U)$. –  Giuseppe Negro Jan 5 '13 at 23:19
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If $E$ is a line segment of length $\delta$, show $\mathcal H^s_\delta(E) \ge \delta^s$, Of course $s \le 1$ is important here. –  GEdgar Jan 6 '13 at 1:08
    
@GEdgar: Ok, this I can prove. If $\{E_j\}$ is a cover of $E$ then $\sum_j d(E_j) \ge \delta$, and since $s\le 1$ we have $$\sum_j (d(E_j))^s \ge \left(\sum_j d(E_j)\right)^s\ge \delta^s.$$This is why we needed $s\le 1$. The cover $\{E_j\}$ is arbitrary, so that, in particular, this inequality holds for any $\delta$-cover, from which we infer $$\mathcal{H}^s_\delta(E)\ge \delta^s.$$Now I'll try to make use of your hint to solve the exercise. Thank you!! –  Giuseppe Negro Jan 6 '13 at 13:46
    
@GEdgar: I think I finally got it. You hint was decisive in getting me started, thank you very much! –  Giuseppe Negro Jan 8 '13 at 11:52

1 Answer 1

up vote 1 down vote accepted

I'll write here a sketch of solution, based on the suggestion GEdgar kindly gave in comments above. Notation as above, let $D$ be a diameter of $U$. Then if $\{E_j\}$ is a covering of $D$ we clearly have $$\sum_j d(E_j) \ge \delta, $$ and since $0\le s\le 1$, by subadditivity we infer $\sum_j d^s(E_j)\ge \delta^s$. Since $D$ is contained in both $U$ and $\overline{U}$, this gives us a lower bound $$ \mathcal{H}^s_\delta(U)\ge \delta^s,\quad \mathcal{H}^s_\delta(\overline{U})\ge \delta^s.$$ In a somewhat similar way we can observe that if $\{F_j\}$ is a covering of $\partial U$, then $$\sum_j d(F_j)\ge \text{length of a closed polygonal path having vertices on }\partial U.$$ Since the shortest of such paths is the diameter, we infer that $\sum_j d(F_j)\ge \delta$ and so, arguing as before, $$\mathcal{H}^s_\delta(\partial U)\ge \delta^s.$$ Now $U$, $\overline{U}$ and $\partial U$ are all $\delta$-coverings of themselves, and so $$\mathcal{H}^s_\delta(U)\le d^s(U),\quad \mathcal{H}^s_\delta(\overline{U})\le d^s(\overline{U}),\quad \mathcal{H}^s_\delta(\partial U)\le d^s(\partial U).$$ We conclude that $\mathcal{H}^s_\delta(U)=\mathcal{H}^s_\delta(\overline{U})=\mathcal{H}^s_\delta(\partial U)=\delta^s$. In particular, $\mathcal{H}^s_\delta$ fails to be additive on $\overline{U}=U\cup \partial U$.


The way I see it, this unpleasant phenomenon occurs because here $U, \overline{U}$ and $\partial U$ are allowed to be covers of themselves. That's why Caratheodory construction involves taking a limit as $\delta\to 0$, that is, as the mesh of the coverings gets finer and finer. This prevents the present phenomenon from occurring.

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