# Integral over $\mathbb{R}^n$ as integral of averages (Rewriting a Riesz potential operator)

I'm looking for a simple proof of the Hardy-Littlewood-Sobolev inequality from the Hardy-Littlewood Maximal operator properties.

I found an article here: http://math.mit.edu/~lguth/PolyMethod/lect30.pdf which seems to have the proof. I'm confused at a step though -- Lemma 3.1 in the notes. It says that $$\int_{\mathbb{R}^n} f(x-y) |y|^{-\alpha} d y = \int_0^\infty r^{n-\alpha-1} \frac{1}{\operatorname{Vol}(B(x,r))} \int_{B(x,r)} f(y) d y\; d r.$$

I can see that the LHS can be written in polar as $$\int_0^\infty r^{n-\alpha-1} \int_{\mathbb{S}^{n-1}} f(x - rv) dv \; d r.$$ I can't see how to go from this surface integral to an average over a ball though, without assuming more about $f$.

Any guidance would be appreciated.

If you change variables with $y=x+z$ and then use spherical coordinates you get $$\int_{B(x,r)}f(y)\,dy=\int_{B(0,r)}f(x+z)\,dz=\int_0^rs^{n-1}\int_{S^{n-1}}f(x+s\sigma)\,d\sigma\,ds.$$ Then $$\int_0^\infty r^{n-1-\alpha}\frac1{\text{Vol}\,B(x,r)}\int_{B(x,r)}f(y)\,dy\,dr\\=\frac1{\text{Vol}\,B(0,1)}\int_0^\infty r^{-\alpha-1}\int_0^rs^{n-1}\int_{S^{n-1}}f(x+s\sigma)\,d\sigma\,dsdr.$$ Now use Fubini to write the right-hand side as $$\frac1{\text{Vol}\,B(0,1)}\int_0^\infty s^{n-1} \int_s^\infty r^{-\alpha-1}dr\int_{S^{n-1}}f(x+s\sigma)\,d\sigma\,ds \\=\frac1{\text{Vol}\,B(0,1)}\frac1{\alpha}\int_0^\infty s^{n-1-\alpha} \int_{S^{n-1}}f(x+s\sigma)\,d\sigma\,ds\\=\frac1{\text{Vol}\,B(0,1)}\frac1{\alpha}\int_{ \mathbb{R}^n}|z|^{-\alpha} f(x+z)\,dz$$ Then just take $y=-z$.