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.