The causal domain of dependence of a partial Cauchy hypersurface is an open causally convex manifold on which the hypersurface is Cauchy

Fix $s\in S$. There is no timelike loop through $s$: if $s\ll s$, openness gives $s\ll s^{\mathrm{\prime }}$ for all $s^{\mathrm{\prime }}\in S$ sufficiently near $s$, and the topological hypersurface has points $s^{\mathrm{\prime }}\mathrm{\ne }s$ arbitrarily near $s$, contrary to acausality. Thus no two points of $S$, equal or distinct, are chronologically related. In particular $I^{+}(S)$ and $I^{-}(S)$ are disjoint by push-up.

We explain the arclength use in [L2]. For an avoiding one-sided causal curve, restrict to the half ending at the prescribed point. Its $h$-speed is locally integrable and positive almost everywhere, so its accumulated $h$-length is continuous strictly increasing. The inverse parameter gives a curve which is $1$-Lipschitz for $d_h$, because distance is at most original subcurve length; hence it is locally Lipschitz in charts. To verify the tangent and the arclength identity without assuming absolute continuity of the inverse parameter, let $\lambda (t)$ denote accumulated length. At almost every original parameter, both the chart curve and $\lambda$ are differentiable and ${\lambda }^{\mathrm{\prime }}(t)=\sqrt{h({\gamma }^{\mathrm{\prime }}(t),{\gamma }^{\mathrm{\prime }}(t))}>0$. Their exceptional set has measure zero. An absolutely continuous real function sends a measure-zero set to a measure-zero set: cover that set by intervals of arbitrarily small total length and apply the defining AC modulus to bound the total lengths of their image intervals. Thus its image under $\lambda$ is null, locally and hence on a countable exhaustion. At each remaining new parameter $u=\lambda (t)$, the inverse difference quotient is $1\mathrm{/}{\lambda }^{\mathrm{\prime }}(t)$, and the chain difference quotient of the reparametrized chart curve is ${\gamma }^{\mathrm{\prime }}(t)\mathrm{/}{\lambda }^{\mathrm{\prime }}(t)$. It is nonzero future causal and has $h$-norm one. This proves both admissibility and $h$-arclength parametrization. If the available one-sided length were finite, the $1$-Lipschitz curve would be Cauchy at that end and completeness would give a point limit, contrary to inextendibility. Thus these reparametrizations have full half-line domains, as required by [L2]. This argument changes parameters and never asserts an extension preserving an arbitrary original parameter.

Edgelessness supplies an open neighborhood $O$ of $s$ in which every timelike curve from $I_O^{-}(s)$ to $I_O^{+}(s)$ meets $S$. Take a temporal coordinate box $B=(-a,a)\times B_{\rho }$ with closure in $O$, centered at $s$. First choose $a>0$ small, and then $\rho >0$ small enough that all points $(-a,z)$ and $(a,z)$ belong respectively to $I_O^{-}(s)$ and $I_O^{+}(s)$; the narrower constant cone of [L1] ensures this. Every vertical segment between these endpoints therefore meets $S$, and has only one meeting by step 1.1. Write it $(f(z),z)$. The same narrower cone implies $\mathrm{\mid }f(z)-f(z^{\mathrm{\prime }})\mathrm{\mid }\le C\mathrm{\mid }z-z^{\mathrm{\prime }}\mathrm{\mid }$: otherwise the two graph points are joined by a timelike straight segment in $O$, contrary to step 1.1. Consequently $S\cap B$ is exactly a continuous Lipschitz graph; the sign of $t-f(z)$ separates the box into two open sides. A point above the graph belongs to $I^{+}(S)$ by its vertical segment; a point below belongs to $I^{-}(S)$.

A future causal segment starting at $s$ and remaining in this box cannot have an endpoint $q$ below the graph: the vertical segment from $q$ to the graph point $r$ gives $s\le q\ll r$, so $s\ll r$, excluded by step 1.1. If $q\in S$ and $q\mathrm{\ne }s$, acausality excludes it. If $q=s$ but the segment is nondegenerate, choose an intermediate point $q^{\mathrm{\prime }}$ different from $s$ sufficiently near its start; it cannot lie below the graph or on $S$, so lies above it. Its vertical predecessor $r^{\mathrm{\prime }}\in S$ satisfies $r^{\mathrm{\prime }}\ll q^{\mathrm{\prime }}\le s$, again excluded by step 1.1. The same argument applies even if the rest of the loop leaves the box. Thus no nondegenerate causal loop passes through $S$, no causal curve meets $S$ twice, and every sufficiently short future segment from $S$ lies strictly above its local graph. Reversing time puts short past segments strictly below. Every causal curve through $s$ therefore crosses its local graph, and uniform approximations of a sufficiently short compact crossing segment also meet that graph by continuity of $t-f(z)$ and the intermediate value theorem.

If $q\in J^{+}(S)\setminus S$, take a causal curve $s\to q$ and a point $r$ on a sufficiently short initial segment. Step 3.1 puts $r$ above the graph. Its vertical predecessor $u\in S$ gives $u\ll r\le q$, hence $q\in I^{+}(S)$. Dually $J^{-}(S)\setminus S\subset I^{-}(S)$. For $p\in D^{+}(S)\setminus S$, the past half of the maximal timelike geodesic through $p$ is past-inextendible by [L3], and hence meets $S$; its timelike segment from that meeting to $p$ puts $p\in I^{+}(S)$. The dual argument proves the past-side inclusion. Also $S\subset D^{+}(S)\cap D^{-}(S)$ immediately.

Let $p\in D^{+}(S)\setminus S$. If $p$ were not interior to $D^{+}(S)$, choose $p_k\mathrm{\notin }{D}^{+}(S)$ converging to $p$ and past-inextendible curves ending at $p_k$ and avoiding $S$. Step 1.2 and [L2] give a past-inextendible causal limit $\alpha :(-\mathrm{\infty },0]\to W$ with $\alpha (0)=p$. It meets $S$ at a negative parameter, by $p\in D^{+}(S)$ and $p\mathrm{\notin }S$. By step 3.1 a short compact segment around that meeting crosses the local graph strictly. Uniform convergence on that segment forces the avoiding approximants to meet $S$, contradiction. Therefore $D^{+}(S)\setminus S$ is open; time reversal gives the dual result.

Let $s\in S$. If $s$ were not interior to $D(S)$, take $p_k\mathrm{\notin }D(S)$ converging to $s$. At each $p_k$ choose a past-inextendible avoiding half and a future-inextendible avoiding half, concatenate them, and use step 1.2 to parametrize the two halves by signed arclength with $p_k$ at zero. Apply [L2] first to their past halves and then to the future halves of that subsequence. The two limits join at $s$ to an inextendible causal curve, and the common subsequence converges uniformly on each compact interval of the joined domain. Its short crossing segment through $s$ from step 3.1 forces the avoiding curves to meet $S$, contradiction. Thus $S\subset \mathrm{Int}D(S)$; together with step 4.2 and step 4.1 this proves (a).

A causal segment whose endpoints belong to $D^{+}(S)$ stays in $D^{+}(S)$. Indeed, if an intermediate $r$ were outside $D^{+}(S)$, a past-inextendible avoiding half ending at $r$ concatenated with the remaining segment $r\to q\in D^{+}(S)$ forces that remaining segment to meet $S$ at $u$. The initial endpoint $p\in D^{+}(S)$ belongs to $S\cup I^{+}(S)$ by step 4.1, whereas $p\le u$ puts $p\in S\cup I^{-}(S)$ by the dual of that step. Disjointness forces $p\in S$, and then the nondegenerate segment $p\to r\to u$ meets $S$ twice, excluded by step 3.1. This is a contradiction. The time reverse proves causal convexity of $D^{-}(S)$.

For a segment $p\to q$ in the remaining mixed case $p\in D^{-}(S)$, $q\in D^{+}(S)$, an intermediate $r\mathrm{\notin }D(S)$ has both past- and future-inextendible avoiding halves. Concatenating the past half with $r\to q$ forces a meeting of $S$ strictly after $r$, and concatenating $p\to r$ with the future half forces a meeting strictly before $r$. This contradicts step 3.1. The opposite mixed case $p\in D^{+}(S)$, $q\in D^{-}(S)$ is impossible for a nondegenerate segment: step 4.1 yields $p\in S\cup I^{+}(S)$, $q\in S\cup I^{-}(S)$; if either endpoint is outside $S$, push-up gives a chronological relation between points of $S$, excluded by step 1.1, and if both are in $S$, step 3.1 excludes the segment. Along with step 5.2 this proves (b).

Let $\gamma :I\to D(S)$ be inextendible there and avoid $S$. Step 4.1 and connectedness of its image put it entirely in one of the disjoint open sets $I^{+}(S)$ and $I^{-}(S)$. Take the latter case; the other reverses time. Then every value belongs to $D^{-}(S)$. Fix $t_0\in I$. If the future half from $t_0$ has no point limit in $W$, that half is future-inextendible in $W$ and avoids $S$, contradicting $\gamma (t_0)\in D^{-}(S)$. Therefore the future end converges to some $y\in W$. If $y\in S$, then $y\in D(S)$ contradicts inextendibility in $D(S)$.

Suppose $y\mathrm{\notin }S$ and $y\mathrm{\notin }{D}^{-}(S)$. There is a future-inextendible avoiding half starting at $y$. A causal closed segment from $\gamma (t_0)$ to $y$ can be obtained as follows: use a smooth increasing diffeomorphism to reparametrize the future half of $\gamma$ to a finite interval with excluded upper end; this preserves local absolute continuity on its interior, and the endpoint-extension assertion of [L3] adjoins its limit $y$ as a causal endpoint. The resulting closed segment avoids $S$, including its endpoint. Concatenate it with the avoiding half from $y$. It is future-inextendible in $W$, avoids $S$, and starts at $\gamma (t_0)\in D^{-}(S)$, contradiction. Hence $y\in D^{-}(S)\subset D(S)$, which again contradicts inextendibility of $\gamma$ in the open submanifold. Every such curve therefore meets $S$, and step 3.1 supplies uniqueness of the parameter visit. This proves (c).