I'm confused about measure in $R^d$. What is the measure of $E_k=\{x:f(x)>2^k\}$ when $f(x) = |x|^{-a}, |x|\leq1$, and $f(x)=0, |x|>1$, in $R^d$?
My thought is that for $k\leq0$ the set includes all $x$ such that
$\frac{1}{|x|^{a}}>2^k$
which gives us all $x$ where
$|x|\leq 2^{-k/a}$
which is some d-dimensional ball with radius $2^{-k/a}$. So then for $k\leq0$ the $m(E_k)=m(B_1)2^{-k/a+d}$ where $B_1$ is the unit ball? I was going over some old solutions and saw that I wrote the measure was $2^d$ for $k\leq0$ and $2^d 2^{-kd/a}$ for $k>0$.