Globally regular sphere, proper integrals and dilute bounds

Since $r^2+b^2\ge b^2>0$, [F1] gives the unique positive root $\mathrm{\Psi }$, and [F2] makes it smooth with ${\mathrm{\partial }}_i\mathrm{\Psi }=x^i\mathrm{/}\mathrm{\Psi }$. The chain, product and quotient rules [F12] then give $H_i=-a{x}^i\mathrm{/}{\mathrm{\Psi }}^3$ and ${\mathrm{\partial }}_i{H}_i=-a({\mathrm{\Psi }}^{-3}-3(x^i{)}^2{\mathrm{\Psi }}^{-5})$, so $\mathrm{\Delta }H=-a{\sum }_{i=1}^3({\mathrm{\Psi }}^{-3}-3(x^i{)}^2{\mathrm{\Psi }}^{-5})=-a(3{\mathrm{\Psi }}^{-3}-3r^2{\mathrm{\Psi }}^{-5})=-3a{b}^2\mathrm{/}{\mathrm{\Psi }}^5<0$. The integrand $\rho H^3=3c^{2}a{b}^2\mathrm{/}[4\pi G_N(r^2+b^2{)}^{5\mathrm{/}2}]$ is nonnegative and continuous on ${\mathbb{R}}^3$, hence Borel and Lebesgue measurable by [F9]. The coupled-dust theorem [F3] applies and supplies the smooth fields and exact formal solution.

On $x^0=\text{constant}$ the metric pullback is $h_{ij}=H^2{\delta }_{ij}$, so $\det h=H^6$ and its positive volume density is $\sqrt{\det h}=H^3$. Substituting the density from step 1.1 gives $\rho H^3=3c^{2}a{b}^2\mathrm{/}[4\pi G_N(r^2+b^2{)}^{5\mathrm{/}2}]$, a nonnegative Borel integrand for [F6].

The curvature formula [F5] and strict negativity in step 1.1 give $S_g=6a{b}^2\mathrm{/}(H^3{\mathrm{\Psi }}^5)>0$. The energy-condition theorem [F4] applies because $\rho \ge 0$. Since the density formula in step 1.1 is strictly positive for every $x\in {\mathbb{R}}^3$, the set of points where it is nonzero is all of ${\mathbb{R}}^3$; no compact subset of ${\mathbb{R}}^3$ contains all those points.

If $a=\delta b$, then $H=1+z$ with $z=\delta h_0$ and $0<h_0\le 1$, since $\mathrm{\Psi }\ge b$. For $f(u)=(1+u{)}^{-2}$, $f(0)=1$, $f^{\mathrm{\prime }}(0)=-2$, and $0<f^{\mathrm{\prime }\mathrm{\prime }}(u)=6(1+u{)}^{-4}\le 6$ for $u\ge 0$ by [F12]. Fix $z>0$ and define $G_z(u):=(z-u)f^{\mathrm{\prime }}(u)+f(u)$ on $[0,z]$. The product rule [F12] gives $G_z^{\mathrm{\prime }}(u)=(z-u)f^{\mathrm{\prime }\mathrm{\prime }}(u)$ on its interior. This derivative is continuous on $[0,z]$, hence Riemann integrable by [F13]; $G_z$ is continuous there. Newton--Leibniz [F13] therefore gives ${\int }_0^z(z-u)f^{\mathrm{\prime }\mathrm{\prime }}(u)du=G_z(z)-G_z(0)=f(z)-1+2z\mathrm{.}$ The integrand lies between $0$ and $6(z-u)$. These continuous comparison functions are integrable by [F13], so [F14] and Newton--Leibniz applied to $6zu-3u^2$ give $0\le f(z)-(1-2z)\le {\int }_0^{z}6(z-u)du=3z^2\le 3{\delta }^2\mathrm{.}$ At $z=0$ the same remainder and bound are zero directly; no theorem requiring a nondegenerate interval is applied there. Since $g_{00}=-H^{-2}$, this proves the stated time-component bound.

Apply the polar formula [F7] to the integrand in step 2.1; it is constant on each sphere of radius $r$. The area factor [F8] then gives $M_{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{p}}=(3c^{2}a{b}^2\mathrm{/}{G}_N){\int }_0^{\mathrm{\infty }}{r}^2(r^2+b^2{)}^{-5\mathrm{/}2}dr$. All uses of the polar formula are under exactly [A1]'s ${\mathrm{A}\mathrm{C}}_{\omega }$ hypothesis.

The radial integrand has finite value at $r=0$, and that singleton has measure zero by [F10], so its integral over $[0,\mathrm{\infty })$ equals its integral over $(0,\mathrm{\infty })$. On $(0,\pi \mathrm{/}2)$, $\cos \theta >0$, and quotient differentiation with the sine/cosine derivatives and Pythagorean identity in [F12] gives $(b\tan \theta {)}^{\mathrm{\prime }}=b{\sec }^2\theta >0$. Also $\sin$ increases and $\cos$ decreases while remaining positive on this interval [F12], so $\tan =\sin \mathrm{/}\cos$ is strictly increasing. Their continuity and quarter-turn values give limits $0$ and $+\mathrm{\infty }$ at the interval's endpoints. For each $y>0$, choose interior points ${\theta }_{-}<{\theta }_{+}$ with $T({\theta }_{-})<y<T({\theta }_{+})$; continuity on $[{\theta }_{-},{\theta }_{+}]$ and the intermediate value theorem [F12] give a preimage, and strict monotonicity gives uniqueness. Hence $T(\theta )=b\tan \theta$ is a bijection $(0,\pi \mathrm{/}2)\to (0,\mathrm{\infty })$. Since $T^{\mathrm{\prime }}$ never vanishes, the inverse function theorem [F12] gives C1 local inverses, which agree by bijectivity to form a global C1 inverse. Thus [F10] applies, and using $\cos \theta >0$ to simplify gives ${\int }_0^{\mathrm{\infty }}{r}^2(r^2+b^2{)}^{-5\mathrm{/}2}dr=b^{-2}{\int }_0^{\pi \mathrm{/}2}{\sin }^2\theta \cos \theta d\theta$. The integrand is continuous on $[0,\pi \mathrm{/}2]$, so it is Riemann integrable and its Lebesgue integral agrees by [F13]; the endpoints have measure zero by [F10]. The function $G(\theta )={\sin }^3\theta \mathrm{/}3$ has derivative ${\sin }^2\theta \cos \theta$ on the interior by [F12], and Newton--Leibniz [F13] with $\sin 0=0$, $\sin (\pi \mathrm{/}2)=1$ yields $1\mathrm{/}3$. Thus the radial integral is $1\mathrm{/}(3b^2)$, and step 3.1 gives $M_{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{p}}=c^{2}a\mathrm{/}{G}_N<\mathrm{\infty }$.

Put $f=\rho H^3$. By step 4.1, $f\ge 0$ and $\int f=M_{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{p}}<\mathrm{\infty }$, so $f$ is real integrable under [F11]. Since $\sigma H^3=s{Q}_{s}f$ and $s{Q}_s$ is a finite real constant, [F11] gives integrability of the signed charge density and $Q_{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{p}}={\int }_{{\mathbb{R}}^3}\sigma H^{3}d{\lambda }_3=s{Q}_s{\int }_{{\mathbb{R}}^3}fd{\lambda }_3=s{Q}_s{M}_{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{p}}$.

The spatial metric is exactly $g_{ij}=H^2{\delta }_{ij}=(1+2\delta h_0+{\delta }^2{h}_0^2){\delta }_{ij}$, so the component error is ${\delta }^2{h}_0^2\mathrm{\mid }{\delta }_{ij}\mathrm{\mid }\le {\delta }^2$. Also $\rho \mathrm{/}{\rho }_{\mathrm{l}\mathrm{i}\mathrm{n}}=H^{-3}=(1+z{)}^{-3}$. By [F12], the function $P(u):=1-(1+u{)}^{-3}$ is continuously differentiable for $u\ge 0$ with $P^{\mathrm{\prime }}(u)=3(1+u{)}^{-4}$. For $z>0$, [F13] makes this derivative integrable on $[0,z]$ and gives $1-(1+z{)}^{-3}=P(z)-P(0)={\int }_0^{z}3(1+u{)}^{-4}du\mathrm{.}$ Its integrand lies between $0$ and $3$, so [F14] yields $0\le 1-\rho \mathrm{/}{\rho }_{\mathrm{l}\mathrm{i}\mathrm{n}}\le 3z\le 3\delta$. At $z=0$ both sides vanish by substitution. This proves the density bound.