I have to prove
prove that $\sigma_{n-1}: 2^{S^{n-1}} \rightarrow[0,+\infty)$$$\sigma_{n-1}(E):=n m_n(\{r x: r \in[0,1], x \in E\})$$is a measure and then prove$$\int_{\mathbb{R}^n} f(x) d m_n(x)=\int_0^{\infty} r^{n-1} \int_{\mathbb{S}^{n-1}} f(r u) d \sigma_{n-1}(u) d r \quad \forall f \in \mathcal{L}^1\left(\mathbb{R}^n\right) .$$
I already proved it is a measure .I first tried for simple functions but I got stuckso if $f(x)=\chi_E(x) . \quad E \in M$$\int_{\mathbb{R^n}}f(x)dm_n=m(E)$
$\int_0^{\infty} r^{n-1} \int_{S^{n-1}} f(ru) d \sigma_{n-1}(u) d r=$$\int_0^{\infty} r^{n-1} \int_{S^{n-1}} \chi_{\frac{E}{r}}^{(u)} d \sigma_{n-1}(u) d r$$=\int_0^{\infty} r^{n-1} \sigma_{n-1}\left(\frac{E}{r}\right) dr=\int_0^1 r^{n-1} \sigma_{n-1}(\frac{E}{r} dr)+\int_1^{\infty} r^{n-1} \sigma_{n-1}\left(\frac{E}{r}\right)$
I think going on I an factorise r out , and I think that the second integral should be $0$ according to m calculation but how ?