Hausdorff measure "invariant under rotation" proof

Question

I try solve this problem, but i can´t get a progress.

I need prove that $H_2$ is invariant under rotation (in $\mathbb{R}^2$).

Where, for $A \subset \mathbb{R}^2$,

$$H_2(A) = \lim_{\epsilon \rightarrow 0} \inf \{ \displaystyle \sum_{n \in \mathbb{N}} d(A_n) : A \subset \bigcup_{n \in \mathbb{N} } A_n , d(A_n) \leq \epsilon, \forall n \}$$

and

$d(A)$ = diameter of a set A.

Thanks for yours hint/answers.

Answer 1

A rotation in $R^2$ is an isometry: $d(x,y)=d(r(x),r(y))$. This tells you that if you rotate a set $A_n$ it will not change its diameter.

Put this next to the fact that if $\{A_n\}_n$ is a covering for $A$, then all $A_n$s rotated will be a covering for the rotated $A$, and you'll have readily proved the claim.

Edit: In trying to write out in my mind a proof in the most strict rigor, I also thought that one may want to prove first that a rotation does not increase the measure, and then point out that a rotation has an inverse which is again a rotation.