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Let $\mu$ be a Radon measure on $\mathbb{R}^n$, and $\alpha \in ]0,n[$.

Show that :

$\int_{\mathbb{R}^n} \frac{1}{(\lVert x-y \rVert)^{n- \alpha}} d\mu (y) = (n-\alpha)\int _{[0,+\infty[} r^{\alpha - n -1} \mu(B(x,r)) d\lambda (r)$,

where $B(x,r)$ is the open ball of center $x$ and radius $r$ in $\mathbb{R}^n$ and $\lambda$ is the Lebesgue measure on $\mathbb{R}$.

I thought about using polar coordinates but I can't work it out.

Moreover, I can't see why the condition of Radon measure is necessary here.

Thanks in advance.

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I found a proof that makes use of this trick : $\int_X f^p d\mu = p\int_{[0,+\infty)} t^{p-1}\mu(\{x\in X: f(x)>t\}) d\mu_t$ for any natural $p\ge 1$ . Actually, the trick works for all $p > 0$.

Just set $f(y) = \frac{1}{\lVert x - y \rVert}$ and $p = n - \alpha > 0$.

Then we get : $\int_{\mathbb{R}^n} \frac{1}{(\lVert x-y \rVert)^{n- \alpha}} d\mu (y) = p \int _{[0,+\infty[} t^{p-1} \mu( \{ y : f(y) \geq t \}) dt = p \int _{[0,+\infty[} r^{1-p} \frac{1}{r^2} \mu( \{ y : f(y) \geq \frac{1}{r} \}) dr$, where we used the change of variable $t = \frac{1}{r}$ in the second equality.

After noticing that $\{ y : f(y) \geq \frac{1}{r} \} = B(x,r)$, we therefore have : $\int_{\mathbb{R}^n} \frac{1}{(\lVert x-y \rVert)^{n- \alpha}} d\mu (y) = p \int _{[0,+\infty[} r^{-p-1} \mu(B(x,r)) d \lambda (r) = (n-\alpha)\int _{[0,+\infty[} r^{\alpha - n -1} \mu(B(x,r)) d\lambda (r)$

I am not sure whether the condition of Radon measure was used or not, but I think it wasn't.

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