# Write $|B_R|$ depending on $|B_1|$, where is my mistakes ? ($B_r$ is the ball of radius $r$).

Let $\mathcal B_R\subset \mathbb R^d$ the ball of radius $R$ and center $0$. I know that for an unspecified set $$|rA|=r^{d}|A|,$$ where $rA=\{ra\mid a\in A\}.$ Indeed, $$|rA|=\int_{rA}dx\underset{x=ry}{=}\int_A r^{d}dy=r^d|A|.$$

Since $\mathcal B_R=R\mathcal B_1$, we the should have that $$|\mathcal B_R|=R^d|\mathcal B_1|.\tag{1}$$

But :

\begin{align*} |\mathcal B_R|&=\int_0^R\int_{\partial \mathcal B_r}dxdr\\ &\underset{x=r\sigma }{=}\int_0^R r^{d-1}\int_{\partial \mathcal B_1}d\sigma dr\\ &=|\partial \mathcal B_1|\int_0^R r^{d-1}dr\\ &=\frac{R^d}{d}|\partial \mathcal B_1|, \end{align*} and thus, using (1) we would have that $$\frac{|\partial\mathcal B_1| }{d}=|\mathcal B_1|,$$ but it looks very weird. Can it be correct ? And if yes, how is it possible ?

• "Can it be correct ?" Yes it is. "And if yes, how is it possible ?" What is looking strange to you in the result? Please be specific.
– Did
Commented Nov 11, 2017 at 8:49
• @Did: It doesn't look very intuitif that $|\partial B_1|$ is bigger than $|B_1|$... that's why I thought it was wrong Commented Nov 11, 2017 at 8:57
• Well, their units are not even the same (say, square meters vs. cube meters), hence I fail to see how you could even have an intuition about their respective values...
– Did
Commented Nov 11, 2017 at 8:59

Yes of course it works ! $$|B_1|=\int_0^1\int_{\partial B_r}dxdr=\int_0^1r^{d-1}\int_{\partial B_1}dsdr=|\partial B_1|\int_0^1r^{d-1}dr=\frac{|\partial B_1|}{d}.$$

In general you have that for $B_r \subset \mathbb{R}^n$

$$|B_r| = \omega_n r^n\,, \qquad \mathcal{H}^{n-1}(\partial B_r) = n\omega_n r^{n-1}\,.$$

($\mathcal{H}^{n-1}$ essentially'' is the $(n-1)$-Lebesgue measure). Thus if you take the quotient you get

$$\frac{|B_r|}{\mathcal{H}^{n-1}(\partial B_r)} = \frac{\omega_n r^n}{n\omega_n r^{n-1}} = \frac{r}{n}\,,$$

exactly as you claim (you took $r=1$).

Your proof is correct but for writing $|\partial B_1|$ because the Lebesgue measure of a $(n-1)$-dimensional set is $0$. You only have that notational problem, the remaining steps are good.