Lebesgue Measure of open ball in $\mathbb{R}^n$

Question

I want to know what is the Lebesgue Measure of open ball in $\mathbb{R}^n$. Because I have to prove that $m(B_r(x))= r^nm(B_1(0))$ , where $B_r(x)=\{ y \in \mathbb{R}^n: \vert \vert {x-y} \vert \vert <r \}$. And I need the Measure of open ball

Answer 1

The formula you want to prove is an essential ingredient of the calculation of the measure of the unit ball. Things go roughly like this:

• Show "integration in polar coordinates" in $\mathbb R^n$: with $X_{n-1}\subset\mathbb R^n$ the unit ball, $$\tag1\int_{\mathbb R^n}f\,dm=\int_0^\infty \int_{S_{n-1}}r^{n-1}f(ru)\,d\sigma(u)\,dr,$$ where $\sigma$ is the Borel measure on $S_{n-1}$ given by $\sigma(A)=n\,m(\tilde A)$, where $\tilde A=\{ru:\ 0<r<1, u\in A\}$.

• Show that $m(B_1(0))=\frac{\sigma(S_{n-1})}n$

• Show $$\tag2\int_{\mathbb R^n}e^{-|x|^2}\,dm=\pi^{n/2}.$$

• Show $$\tag3\int_{\mathbb R^n}e^{-|x|^2}\,dm=\sigma(S_{n-1})\,\Gamma(n/2).$$

• Show $m(B_r(0))=r^n\,B_1(0)$.

• Conclude: $$m(B_r(0))=r^n\,m(B_1(0))=\frac{r^n}n\,\sigma(S_{n-1})=\frac{\pi^{n/2}\,r^n}{n\,\Gamma(n/2)}.$$