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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 $k>0$.

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The set where $f(x) > 2^k$ is contained in $|x| \leq 1$ by hypothesis, i.e. the unit ball in $\mathbb{R}^d$. Next, if we ignore what happens at $0$ (a null set)

$|x|^{-a} > 2^k \Longrightarrow |x| < 2^{k+a}$

Here we had to flip the inequality since the inverse of points far from the origin are smaller than those closer. If $2^{k+a} \geq 1$, then this is just the unit ball in $\mathbb{R}^d$. If $2^{k+a} < 1$, then this is the ball of radius $2^{k+a}$ in $\mathbb{R}^d$.