The approximating Hausdorff measure is not Borel 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!
 A: 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.
