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