Vacuum exterior-section constraints imply the dominant energy condition

Apply the constraint identities with $K=g(\mathrm{\nabla }n,\cdot )=-k$. Ricci flatness makes both normal components of the Einstein tensor zero. The scalar identity is unchanged by $K\mapsto -k$ and yields $\mu =0$; the momentum identity changes sign and still yields $J=0$. Smoothness extends these identities to $S$.

The dominant-energy inequality is $0\ge 0$ by step 1.1. Apply the Riemannian-volume theorem to the smooth induced metric on $C$; the countable choice it assumes follows from the inherited Axiom of Choice by restricting to countable families. The nonnegative function $f=\mathrm{\mid }\mu \mathrm{\mid }+\mathrm{\mid }J{\mathrm{\mid }}_h$ is identically zero and is Borel measurable. Every nonnegative simple minorant of $f$ is zero and has simple integral zero, including when the total volume is infinite. Thus the defining supremum gives ${\int }_{C}fd{\mathrm{vol}}_h=0<\mathrm{\infty }$.

The interior $C$ is open in $\bar{C}$, so $A\cap C$ is Borel in $C$ for every Borel $A\subset \bar{C}$. Intersection with $C$ preserves disjoint countable unions, so the displayed extension is a Borel measure by countable additivity of ${\mathrm{vol}}_h$, with ${\stackrel{}{\mathrm{vol}}}_h(S)={\mathrm{vol}}_h(\mathrm{\varnothing })=0$. The same zero-function argument gives ${\int }_{\bar{C}}fd{\stackrel{}{\mathrm{vol}}}_h=0$. Thus norm integrability holds with the stated boundary convention, without imposing ${\mathrm{tr}}_{h}k=0$.