Achronal boundaries are locally Lipschitz graphs, topological hypersurfaces and null for smooth volume densities

(Clause (a).) By [L1] the set $A$ is closed and achronal. If a timelike curve $\gamma$ met $A$ at two distinct parameter values $s<s^{\mathrm{\prime }}$, then $\gamma {\mathrm{\mid }}_{[s,s^{\mathrm{\prime }}]}$ would be a future timelike curve from $\gamma (s)$ to $\gamma (s^{\mathrm{\prime }})$ (or a past one after reversing), so the two points of $A$ would be chronologically related, contradicting achronality; hence at most one meeting.

(Points of $F$ and of its complement near $p$.) Since $p\in A=\mathrm{\partial }F$ we have $p\in \bar{F}$ and $p\mathrm{\notin }F$. By [L2] applied to the future set $F$, every $r\in I^{+}(p)$ lies in $F$. Moreover every $r\in I^{-}(p)$ lies outside $\bar{F}$: otherwise $r\in \bar{F}$ and $p\in I^{+}(r)$, so [L2] would give $p\in F$, a contradiction.

(Nearby points of $F$ and of $M\setminus \bar{F}$ on every vertical line.) Let $(U,x)$, $K$, $c$ be as in [L3]. Choose $a,\rho ,\eta >0$ with $\sqrt{a^2+{\rho }^2}$ smaller than the radius of $U$ and $\sqrt{c}\rho +\eta <a$; then the box $W=x^{-1}((-a,a)\times B_{\rho }(0))$ is contained in $U$. For $z\in B_{\rho }(0)$ put $q_z^{+}:=x^{-1}(\sqrt{c}\mathrm{\mid }z\mathrm{\mid }+\eta ,z)$ and $q_z^{-}:=x^{-1}(-\sqrt{c}\mathrm{\mid }z\mathrm{\mid }-\eta ,z)$, both in $W$. Computing in the flat form, ${\eta }_c(q_z^{+}-p,q_z^{+}-p)=-(\sqrt{c}\mathrm{\mid }z\mathrm{\mid }+\eta {)}^2+c\mathrm{\mid }z{\mathrm{\mid }}^2=-2\eta \sqrt{c}\mathrm{\mid }z\mathrm{\mid }-{\eta }^2<0,$ and likewise ${\eta }_c(q_z^{-}-p,q_z^{-}-p)=-2\eta \sqrt{c}\mathrm{\mid }z\mathrm{\mid }-{\eta }^2<0$. By [L3] (clause (d)) we get $p\ll q_z^{+}$ and $q_z^{-}\ll p$; with step 1.2 this gives $q_z^{+}\in F$ and $q_z^{-}\mathrm{\notin }\bar{F}$.

(Existence of a boundary point on each vertical line.) Fix $z\in B_{\rho }(0)$ and let $S_z$ be the coordinate subsegment of $L_z$ joining $q_z^{-}$ to $q_z^{+}$. The map $s\mapsto x^{-1}(s,z)$ is continuous, so $S_z$ is connected; it contains $q_z^{+}\in F$ and $q_z^{-}\in M\setminus \bar{F}\subseteq M\setminus F$. Since $F$ is open in $M$, the connected set $S_z$ cannot be covered by the two disjoint nonempty relatively open sets $S_z\cap F$ and $S_z\cap \mathrm{I}\mathrm{n}\mathrm{t}(M\setminus F)$ without meeting the complement of their union, which is $A$; hence $A\cap S_z\mathrm{\ne }\mathrm{\varnothing }$.

(Uniqueness of the boundary point on a vertical line.) Let $r,r^{\mathrm{\prime }}\in A\cap L_z$ be distinct, say $r=x^{-1}(s,z)$ and $r^{\mathrm{\prime }}=x^{-1}(s^{\mathrm{\prime }},z)$ with $s<s^{\mathrm{\prime }}$; by [L4] $r\ll r^{\mathrm{\prime }}$, contradicting the achronality of $A$ from step 1.1. Hence $A\cap L_z$ has at most one point, and by step 3.1 exactly one.

(The graphing function.) For $z\in B_{\rho }(0)$ let $\phi (z)\in (-a,a)$ be the $x^0$-coordinate of the unique point of $A\cap L_z$; it is well defined by step 4.1, and since the unique point of $A\cap L_z$ found in step 3.1 lies between $q_z^{-}$ and $q_z^{+}$ we have $\mathrm{\mid }\phi (z)\mathrm{\mid }\le \sqrt{c}\mathrm{\mid }z\mathrm{\mid }+\eta <a$. Then $A\cap W=\{x^{-1}(\phi (z),z):z\in B_{\rho }(0)\},$ because the inclusion $\supseteq$ holds by definition of $\phi$, and if $x^{-1}(s,z)\in A\cap W$ then this point lies on $L_z$ and uniqueness forces $s=\phi (z)$.

(Lipschitz bound.) Let $z,z^{\mathrm{\prime }}\in B_{\rho }(0)$ and suppose $\mathrm{\mid }\phi (z)-\phi (z^{\mathrm{\prime }})\mathrm{\mid }>\sqrt{c}\mathrm{\mid }z-z^{\mathrm{\prime }}\mathrm{\mid }$. Put $\delta :=(\phi (z)-\phi (z^{\mathrm{\prime }}),z-z^{\mathrm{\prime }})$ and assume $\phi (z)>\phi (z^{\mathrm{\prime }})$, the other case being symmetric. Then ${\eta }_c(\delta ,\delta )=-(\phi (z)-\phi (z^{\mathrm{\prime }}){)}^2+c\mathrm{\mid }z-z^{\mathrm{\prime }}{\mathrm{\mid }}^2<0$, and both $x^{-1}(\phi (z),z)$ and $x^{-1}(\phi (z^{\mathrm{\prime }}),z^{\mathrm{\prime }})$ lie in $A\cap W\subseteq U$, so [L3] (clause (d)) gives $x^{-1}(\phi (z^{\mathrm{\prime }}),z^{\mathrm{\prime }})\ll x^{-1}(\phi (z),z)$, two distinct points of $A$ chronologically related, contradicting step 1.1. Hence $\mathrm{\mid }\phi (z)-\phi (z^{\mathrm{\prime }})\mathrm{\mid }\le \sqrt{c}\mathrm{\mid }z-z^{\mathrm{\prime }}\mathrm{\mid }$.

(Clause (c).) The map $\psi :B_{\rho }(0)\to A\cap W$, $\psi (z):=x^{-1}(\phi (z),z)$, is continuous because $\phi$ is Lipschitz by step 6.1 and $x^{-1}$ is continuous; it is bijective onto $A\cap W$ by step 5.1. The restriction of the continuous projection $\pi :x^{-1}(s,z)\mapsto z$ to $A\cap W$ is its inverse: for $r\in A\cap W$ we have $r=x^{-1}(\phi (\pi (r)),\pi (r))$ by step 4.1, so $\psi (\pi (r))=r$, and $\pi$ is continuous. Hence $\psi$ is a homeomorphism onto the relatively open set $A\cap W$. For two such graph charts $\psi :B_{\rho }(0)\to A\cap W$ and ${\psi }^{\mathrm{\prime }}:B_{{\rho }^{\mathrm{\prime }}}(0)\to A\cap W^{\mathrm{\prime }}$ coming from charts $x,x^{\mathrm{\prime }}$, the transition ${\psi }^{\mathrm{\prime }-1}\circ \psi$ equals ${\pi }^{\mathrm{\prime }}\circ \psi$ on ${\psi }^{-1}(A\cap W\cap W^{\mathrm{\prime }})$, a composition of continuous maps; hence the graph parametrizations form a topological atlas on $A$, making $A$ a topological $3$-manifold and the inclusion $A\hookrightarrow M$ a topological embedding.

(Clause (d).) Assume countable choice and let $K\subseteq M$ be compact; then $A\cap K$ is compact because $A$ is closed. Cover $A\cap K$ by finitely many graph neighborhoods $W_1,\dots ,W_m$ as in (b), with charts $x_j$ and graphing functions ${\phi }_j$ defined on the balls $B_{{\rho }_j}(0)$. In the chart $x_j$ the set $A\cap W_j$ is contained in the graph of ${\phi }_j$, which is the graph of the measurable function obtained by extending ${\phi }_j$ by zero outside $B_{{\rho }_j}(0)$: that extension is continuous on the open ball and constant on its complement, hence Borel measurable by [L5]. By [L5] the graph has four-dimensional Lebesgue measure zero, so the coordinate image of $A\cap W_j$ is Lebesgue null, and the local absolute continuity of $\mu$ gives $\mu (A\cap W_j)=0$; here $A\cap W_j$ is Borel because $A$ is closed. By finite additivity and $A\cap K\subseteq {\bigcup }_j(A\cap W_j)$, $\mu (A\cap K)=0$. This proves (d).

(Conclusion.) Clauses (a), (b), (c), (d) are respectively steps 1.1, 5.1-6.1, 7.1 and 8.1.