Four-dimensional radial measure from two polar planes

On the open set $U=(0,R)\times (0,\pi \mathrm{/}2)\times (0,2\pi )\times (0,2\pi )$ define $T(r,\theta ,\varphi ,\psi )=(r\cos \theta \cos \varphi ,r\cos \theta \sin \varphi ,r\sin \theta \cos \psi ,r\sin \theta \sin \psi )\mathrm{.}$ Let $L={\mathbb{R}}^2\setminus \{(a,0):a\ge 0\}$, the plane slit along the nonnegative horizontal ray. For $(a,b)\in L$, put $\rho =\sqrt{a^2+b^2}>0$ and define ${\mathrm{Arg}}_L(a,b)$ to be $\arccos (a\mathrm{/}\rho )$ if $b>0$, $2\pi -\arccos (a\mathrm{/}\rho )$ if $b<0$, and $\pi$ if $b=0$ (then $a<0$). The inverse and branch identities [F8]–[F10] give $\cos {\mathrm{Arg}}_L=a\mathrm{/}\rho$ and $\sin {\mathrm{Arg}}_L=b\mathrm{/}\rho$ and make this the unique angle in $(0,2\pi )$. Each non-axis branch is $C^1$ by [F9]. Near the negative horizontal axis, the same angle equals $\pi -\arcsin (b\mathrm{/}\rho )$, so it is $C^1$ there too. Positive square roots are $C^1$: continuity follows from $\mathrm{\mid }\sqrt{u}-\sqrt{v}\mathrm{\mid }\le \sqrt{\mathrm{\mid }u-v\mathrm{\mid }}$, and the difference quotient is $1\mathrm{/}(\sqrt{u+h}+\sqrt{u})$, with continuous limit $1\mathrm{/}(2\sqrt{u})$ for $u>0$. Thus all radii in these formulas are $C^1$ by [F7]. Define the open set $V=\{x:\mathrm{\mid }x\mathrm{\mid }<R,(x_1,x_2)\in L,(x_3,x_4)\in L\}$. Write $A=\sqrt{x_1^2+x_2^2}>0$, $B=\sqrt{x_3^2+x_4^2}>0$. The inverse of $T$ on $V$ is explicitly $r=\sqrt{A^2+B^2}$, $\theta =\arccos (A\mathrm{/}r)\in (0,\pi \mathrm{/}2)$, $\varphi ={\mathrm{Arg}}_L(x_1,x_2)$, $\psi ={\mathrm{Arg}}_L(x_3,x_4)$. Indeed $\cos \theta =A\mathrm{/}r$, $\sin \theta =B\mathrm{/}r$ by [F8]–[F10], and substitution into $T$ gives $x$; the same branch identities recover every parameter of $U$. Both directions are $C^1$, so $T:U\to V$ is a $C^1$ diffeomorphism. Using [F6]–[F8], The four derivative columns are pairwise orthogonal, with squared lengths $1,r^2,r^2{\cos }^2\theta ,r^2{\sin }^2\theta$; hence $\mathrm{\mid }\det DT\mathrm{\mid }=r^3\sin \theta \cos \theta$. The complement of $V$ in the closed ball is contained in the union of the boundary sphere and the four coordinate hyperplanes: the omitted angular branch cuts lie in $x_2=0$ or $x_4=0$, and the omitted endpoint planes lie in $x_1=x_2=0$ or $x_3=x_4=0$. The coordinate hyperplanes are null by [F2]. The boundary sphere is null by [F4], applied to its indicator, since its polar radial integral is supported at the singleton $\{R\}$, which has one-dimensional Lebesgue measure zero. The origin is already in the coordinate hyperplanes.

Apply [F1] to $T$ and the nonnegative Borel function $x\mapsto \phi (\mathrm{\mid }x\mathrm{\mid })$ on $V$. The null-complement calculation in step 1.1 shows that its integral over $V$ equals its integral over the closed ball, even when $\phi$ takes the value $+\mathrm{\infty }$ on a radius whose preimage in the omitted set is null. Therefore the ball integral equals ${\int }_U\phi (r)r^3\sin \theta \cos \theta drd\theta d\varphi d\psi \mathrm{.}$

The integrand in step 2.1 is nonnegative and product-measurable, so [F3] separates the four integrations. The angular factors are ${\int }_0^{\pi \mathrm{/}2}\sin \theta \cos \theta d\theta =1\mathrm{/}2$ and ${\int }_0^{2\pi }1d\varphi ={\int }_0^{2\pi }1d\psi =2\pi$; the first equality follows from [F5] with antiderivative ${\sin }^2\theta \mathrm{/}2$, whose derivative is $\sin \theta \cos \theta$ by [F6]–[F7]. The angular integrands are continuous and bounded on their compact intervals, so [F11] identifies these calculus evaluations with the Lebesgue integrals in [F3]. Their product is $2{\pi }^2$, which proves the formula in the Statement. Endpoints do not change any Lebesgue integral, and all equalities are valid as extended nonnegative integrals.