# Volume of the $n$-ball

The following is a simple and beautiful proof, shown to me to my great delight while I was in high school, that the ${n}$-ball of radius ${r}$ has ${n}$-volume $\displaystyle \frac{\pi^{n/2} r^n}{\Gamma(\frac{n}{2}+1)}$

Although I expect most readers will know it, I believe that everybody should see it. I don’t know the history of it, and would be interested in learning.

Suppose that the ${n}$-ball ${B^n}$ of radius ${1}$ has ${n}$-volume ${V_n(1)}$. Then, by considering linear transformations, the ${n}$-ball ${rB^n}$ of radius ${r}$ has ${n}$-volume ${V_n(r) = V_n(1) r^n}$. Moreover, differentiating this with respect to ${r}$ should produce the surface ${(n-1)}$-volume of the ${(n-1)}$-sphere ${S^{n-1} = \partial B^n}$: thus we expect ${\text{vol}^{(n-1)}(S^{n-1}) = V_n(1) n r^{n-1}}$.

Now consider the integral $\displaystyle I = \int_{\mathbf{R}^n} e^{-\pi|x|^2}\,dx.$

We will compute ${I}$ in two different ways. On the one hand, $\displaystyle I = \int_{-\infty}^{+\infty}\cdots\int_{-\infty}^{+\infty} e^{-\pi x_1^2 -\cdots - \pi x_n^2}\,dx_1\cdots dx_n = \left(\int_{-\infty}^{+\infty} e^{-\pi x^2}\,dx\right)^n = 1.$

On the other hand, the form of ${I}$ suggests introducing a radial coordinate ${r=|x|}$. Computing this way, $\displaystyle I = \int_0^\infty e^{-\pi r^2} (V_n(1) n r^{n-1})\,dr = \frac{n V_n(1)}{\pi^{n/2}} \int_0^\infty e^{-t^2} t^{n-1}\, dt\qquad\quad$ $\displaystyle \qquad\quad= \frac{n V_n(1)}{2 \pi^{n/2}} \int_0^\infty e^{-s} s^{n/2-1}\,ds = \frac{n V_n(1)}{2 \pi^{n/2}} \Gamma(n/2) = \frac{V_n(1)\Gamma(n/2+1)}{\pi^{n/2}}.$

# Inducing a Haar measure from a quotient

Suppose that ${G}$ is a locally compact group and that ${H}$ is a closed subgroup with an ${H}$-left-invariant regular Borel measure ${\mu_H}$ such that ${G/H}$ possesses a ${G}$-left-invariant regular Borel measure ${\mu_{G/H}}$. For instance, ${G = \mathbf{R}}$, ${H=\mathbf{Z}}$, and ${G/H= S^1}$. The following is how you then induce a Haar measure on ${G}$. (Technically, it’s easier to construct Haar measure on compact groups, so this extends that construction slightly.)

For ${f\in C_c(G)}$, define ${T_H(f) : G\rightarrow \mathbf{C}}$ by $\displaystyle T_H(f)(x) = \int_H f(xh) d\mu_H.$

Let ${K=\text{supp}f}$. Then ${f(xh)}$, as a function of ${h}$ is supported on ${x^{-1} K}$, so the above integral is finite. Moreover, if ${x,y\in U}$ and ${U}$ is compact, then $\displaystyle |T_H(f)(x) - T_H(f)(y)| = \left| \int_H (f(xh)-f(yh))d\mu_H\right| \leq \mu_H(H\cap U^{-1} K) \sup_h |f(xh)-f(yh)|.$

Because a continuous function on a compact set is uniformly continuous (in the sense that there exists a neighbourhood ${V}$ of ${e}$ such that ${gh^{-1} \in V}$ implies ${|f(g)-f(h)| <\epsilon}$), ${T_H(f)}$ is continous. Since ${T_H(f)}$ is ${H}$-right-invariant, ${T_H(f)}$ descends to a continuous function ${\hat{T}_H(f)}$ defined on ${G/H}$. Moreover, if ${q}$ is the quotient map ${G\rightarrow G/H}$, then ${\hat{T}_H(f)}$ is supported on ${q(K)}$, so ${\hat{T}_H:C_c(G)\rightarrow C_c(G/H)}$. Finally, define ${\lambda: C_c(G)\rightarrow \mathbf{C}}$ by $\displaystyle \lambda(f) = \int_{G/H} \hat{T}_H(f) d\mu_{G/H}.$

This linear functional ${\lambda}$ is positive (in the sense that ${f\geq0}$ implies ${\lambda(f)\geq0}$), so the Riesz representation theorem guarantees the existence of a regular Borel measure ${\mu_G}$ on ${G}$ such that $\displaystyle \lambda(f) = \int_G f d\mu_G$

for all ${f\in C_c(G)}$. It is now a simple matter to check that ${\mu_G}$ is ${G}$-left-invariant.