# $\delta$- hausdorff measure of open ball

Let U open ball in $$\mathbb{R}^n$$, $$n \ge 2$$, such that diameter $$d(U)=\delta$$. Let $$0 \le s \le 1$$, we need to prove that $$H_{\delta}^s(U)= H_{\delta}^s(\partial U)= H_{\delta}^s(\bar{U})$$.

Here $$H_{\delta}^s(A)= \inf \{ \sum_{i} d(E_i)^s : A \subset \cup E_i, d(E_i) \le \delta \},$$ and $$d(B)$$ is just the diameter of $$B$$.

I am thinking since $$\partial{U} \subset \bar{U}$$ and $$U \subset \bar{U}$$, we have $$H_{\delta}^s(\partial U) \le H_{\delta}^s(\bar{U})$$ and $$H_{\delta}^s(U) \le H_{\delta}^s(\bar{U})$$. I also know that $$H_{\delta}^s(U) \le \delta^s$$. Any hint how to continue?

• Can you give the definition of $H^s_\delta$? If it is what I think it is, the result is false for $n=2$ and $s=1$ (the measure of the boundary should be finite, but not the measure of $U$ or its closure). Apr 30, 2021 at 5:55
• I added the definition.
– user752801
May 1, 2021 at 1:19
• Are the $E_i$ allowed to be closed? (I doubt it, but it would make things a lot easier!) May 9, 2021 at 12:23
• Hint. For $H_\delta^s$ you must cover your set $A$ by countably many sets $E_i$ of diameter ${} \le \delta$. In fact, all three of your sets have diameter exactly $\delta$. So you can cover by a single set. Nov 12, 2021 at 2:04

## Claim 1: $$H_δ^s(U) ≥ δ^s$$

For this claim, suppose that $$n \in \mathbb N$$ (i.e., we also allow $$n = 1$$).

Let $$(E_i)_{i}$$ be a cover of $$U$$ by open balls.

(Assume there are countably many, otherwise the sum of diameters is necessarily infinite),

Then, Letting $$C_n$$ denote the ratio between an $$n$$ dimensional ball's radius and its diameter (e.g., $$C_2 = π/4$$), we compute:

% \begin{align} % \sum_i \text{diam}(E_i)^s &= (C_n)^{-s} \sum_i \text{area}(E_i)^{s/n} \\ % & ≥ (C_n)^{-s} \left(\sum_i \text{area}(E_i)\right) ^{s/n} \qquad(\text{since s/n \in (0,1]})\\ % & ≥ (C_n)^{-s} \left(\text{area}\left(\bigcup_i E_i\right)\right) ^{s/n} \qquad(\text{σ-subadditivity of area measure})\\ % &≥ (C_n)^{-s} \left(\text{area}(U)\right) ^{s/n} \qquad(\text{monotonicity of area measure, since the E_i cover U})\\\\ % &=\text{diam}(U)^s = δ^{s} % \end{align} %

The first inequality can be viewed as a monotonicity of norms statement. I.e., for any sequence $$(x_n)_{n\in\mathbb N}$$,

$$\|(x_n)\|_{l^{n/s}} \leq \|(x_n)\|_{l^1},$$

which is discussed in more generality here: How do you show monotonicity of the $\ell^p$ norms? . The second and third are monotonicity of the area measure.

Taking infs gives the result.

## Claim 2: $$H_δ^s(\overline U) ≤ δ^s$$

Consider the following cover of the closed ball by open balls:

Consider placing an open ball of radius $$δ/2$$ with centre a distance $$ε< δ/2$$ from the centre of $$U$$, as pictured below. You can then try and cover the remaining points (a deeper shade of blue) by any open ball of radius greater than $$A_ε$$, where $$A_ε>0$$ is the distance shown in the diagram.

Some cheeky Pythagoras on these triangles (dropping the perpendicular of the isosoles triangle) tells you that

$$A_ε^2 = \left(\frac δ2 + \frac {3ε}2\right)^2 + \left(\frac δ2\right)^2 - \left(\frac ε2\right)^2 = \frac{3 δ}{2} ε + 2 ε ^2.$$

Therefore, for $$ε$$ sufficiently small, the doubled-in size ball has diameter $$4A_\epsilon ≤ δ$$, and hence the cover by this and the original ball gives

$$H^s_δ(\overline U) \leq δ^s + 4^{s}A_ε^s,$$

and taking $$ε → 0$$ gives the result.

That is, so far we've shown (with the OP observations)

$$δ^s ≤ H^s_δ(U) ≤ H^s_δ(\overline U) ≤ δ^s,$$

and also that $$H^s_δ(∂U) ≤ H^s_δ(\overline U) ≤ δ^s.$$

## Claim 3: $$H^s_δ(∂ U) ≥ δ^s$$

Consider the projection $$π$$ onto the first $$n-1$$ coordinates, $$π: \mathbb R^{n} → \mathbb R^{n-1}$$.

$$π$$ decreases distances (and therefore decreases the Hausdorff outer measure), and takes spheres onto closed balls of the same radius (both of these can be proven directly from the definition). Hence,

$$H^s_δ(∂ U) ≥ H^s_δ(π (∂ U)) = H^s_δ(U'),$$

where $$U' \in \mathbb R^{n-1}$$ is a closed ball of radius $$δ$$.

Since Claim 1 applies for all $$n \in \mathbb N$$, we can apply it to the interior of $$U'$$ to get that $$H^s_δ(U') ≥ δ^s$$. I.e.,

$$H^s_δ(∂ U) ≥ H^s_δ(U') ≥ δ^s,$$

as required.