Is "the set $d_1\le|x-\mu|\le d_2$" just another way of saying $\{x \in I:d_1\le|x-\mu|\le d_2\}$?
We have $d_2<d_1$. Doesn't this mean that this set is empty?
Why is the denominator bounded from below by $d_2^\alpha+d_2^n$? Don't we have $|x-\mu|^\alpha+|x-\mu|^n\ge d_1^\alpha+d_1^n$ because of $d_2<d_1$?
