Finite regulated Kepler shell measure is normalized

The inequalities $A\mathrm{/}R<e$ and $e\le A\mathrm{/}(16r_0)$ imply $r_0<A\mathrm{/}e<R$. At fixed $q$ with $r=\mathrm{\mid }q\mathrm{\mid }$, the momentum sublevel $h(q,p)\le -e$ is the ball $\mathrm{\mid }p\mathrm{\mid }\le \sqrt{2\mu (A\mathrm{/}r-e)}$, which is nonempty exactly for $r_0\le r\le A\mathrm{/}e$. By [F2] its momentum volume is $\frac{4\pi }{3}[2\mu (A\mathrm{/}r-e){]}^{3\mathrm{/}2}$. Integrating the resulting radial function over $q$ by [F2], then substituting $s=er\mathrm{/}A$, gives $\mathrm{\Omega }(-e)=\frac{(4\pi )(4\pi \mathrm{/}3)(2\mu {)}^{3\mathrm{/}2}{A}^3}{h_{\ast }^3}{e}^{-3\mathrm{/}2}J(\delta ),\delta =\frac{e{r}_0}{A}\mathrm{.}$ The integration interval is compact and has positive length, while the integrand is continuous and positive in its interior, so $\mathrm{\Omega }$ is finite and strictly positive.

Put $f(s)=s^{1\mathrm{/}2}(1-s{)}^{3\mathrm{/}2}$ for $0<s<1$. For $h\mathrm{\ne }0$ with $\delta +h\in (0,1)$, the oriented-integral identity gives $(J(\delta +h)-J(\delta ))\mathrm{/}h=-(1\mathrm{/}h){\int }_{\delta }^{\delta +h}f(s)ds$; its difference from $-f(\delta )$ tends to zero by continuity. At $\delta =1\mathrm{/}16$, restrict to $h<0$, which corresponds to increasing $E$ within the branch, to obtain the right derivative in $E$. Thus $J^{\mathrm{\prime }}(\delta )=-{\delta }^{1\mathrm{/}2}(1-\delta {)}^{3\mathrm{/}2}$, with that one-sided interpretation at the endpoint. Differentiating the explicit phase-volume formula with $E=-e$ now yields ${\mathrm{\Omega }}^{\mathrm{\prime }}(E)=\frac{(4\pi )(4\pi \mathrm{/}3)(2\mu {)}^{3\mathrm{/}2}{A}^3}{h_{\ast }^3}{e}^{-5\mathrm{/}2}[\frac{3}{2}J(\delta )+{\delta }^{3\mathrm{/}2}(1-\delta {)}^{3\mathrm{/}2}]>0.$ The bracket is finite and positive, so the derivative is finite and strictly positive.

Integrating the defining pushforward density over its entire source gives ${\mathrm{\Sigma }}_E(\{h=E\})=\frac{16{\pi }^2\mu }{h_{\ast }^3}{\int }_{r_0}^{A\mathrm{/}e}{r}^2\sqrt{2\mu (A\mathrm{/}r-e)}dr=\frac{16{\pi }^2\mu \sqrt{2\mu }}{h_{\ast }^3}{A}^3{e}^{-5\mathrm{/}2}{\int }_{\delta }^1{s}^{3\mathrm{/}2}(1-s{)}^{1\mathrm{/}2}ds,$ where [F2] supplies the spatial radial measure and $\sigma (S^2)=4\pi$ supplies the momentum-direction factor. Put $g(s)=s^{3\mathrm{/}2}(1-s{)}^{3\mathrm{/}2}$. Direct differentiation and integration on $[\delta ,1]$ give $\frac{3}{2}J(\delta )+{\delta }^{3\mathrm{/}2}(1-\delta {)}^{3\mathrm{/}2}=\frac{3}{2}{\int }_{\delta }^1{s}^{3\mathrm{/}2}(1-s{)}^{1\mathrm{/}2}ds\mathrm{.}$ Since $\mu \sqrt{2\mu }=\frac{1}{2}(2\mu {)}^{3\mathrm{/}2}$, this identity makes the preceding shell mass exactly the ${\mathrm{\Omega }}^{\mathrm{\prime }}(E)$ in step 2.1.

Every image point has $\mathrm{\mid }p\mathrm{\mid }=p_{\ast }(r,E)$, hence $h=\mathrm{\mid }p{\mathrm{\mid }}^2\mathrm{/}(2\mu )-A\mathrm{/}r=E$. The measure is finite by step 3.1 and has total mass ${\mathrm{\Omega }}^{\mathrm{\prime }}(E)>0$; division by this total therefore defines a probability supported on the energy surface.