Weak steps: locally finite sums of cone bumps give a smooth step around each Cauchy level with past timelike gradient near the level

(The open past of a level is the sublevel set.) We prove $\{t<b\}=I^{-}(S_b)$. If $t(q)<b$, let $\gamma :J\to M$ be the inextendible future-directed timelike geodesic on its open maximal affine interval through $q$ with $\gamma (0)=q$; by [L6] the continuous function $s\mapsto t(\gamma (s))$ takes values tending to $-\mathrm{\infty }$ (past end) and $+\mathrm{\infty }$ (future end), so there is $s_0>0$ with $t(\gamma (s_0))=b$, i.e. $\gamma (s_0)\in S_b$, and $q=\gamma (0)\ll \gamma (s_0)$ because the geodesic is future timelike, so $q\in I^{-}(S_b)$. Conversely if $q\ll s\in S_b$ then $q\mathrm{\ne }s$ and strict monotonicity [L1] gives $t(q)<t(s)=b$. Hence $\{t<b\}=I^{-}(S_b)$, which is open, and $S_b=\{t=b\}$ is closed [L1]; moreover $J^{-}(S_b)=\{t\le b\}=\{t<b\}\cup S_b$.

(A locally finite family of cone bumps.) For each connected component $M_{\alpha }$ of $M$ fix a base point $m_{\alpha }$ and write $B_R^{\alpha }:=\{q\in M_{\alpha }:d_{h^{\mathrm{\prime }}}(q,m_{\alpha })<R\}$; the components are open and at most countable [L3], so ${\mathrm{A}\mathrm{C}}_{\omega }$ selects the base points. Since $S_b$ is closed and meets only countably many components, and each $S_b\cap M_{\alpha }$ is covered by the shells $A_m^{\alpha }:=S_b\cap (\overline{B_m^{\alpha }}\setminus B_{m-1}^{\alpha })$, $m\ge 1$ (with $B_0^{\alpha }:=\mathrm{\varnothing }$), which are closed in $M_{\alpha }$ and $d_{h^{\mathrm{\prime }}}$-bounded hence compact [L3], we obtain: $S_b={\bigcup }_{\alpha ,m}{A}_m^{\alpha }$ with each $A_m^{\alpha }$ compact. For each shell consider all admissible tuples $(x,C,h)$ supplied by [L2], with $x\in A_m^{\alpha }$ and ${\mathrm{diam}}_{h^{\mathrm{\prime }}}(C)<1$. Their sets $\{h>1\mathrm{/}2\}$ cover the shell, because a tuple exists at every center. Compactness supplies a finite list of tuples covering each shell; Countable Choice selects one such finite list for each of the countably many shells. Index every selected tuple separately, retaining repetitions of a center, by a countable index set $F$, and write its data as $(x_j,C_j,h_j)$ for $j\in F$. The family $\{C_j{\}}_{j\in F}$ is locally finite: if a tuple from shell $A_m^{\alpha }$ has $C_j\cap B_R^{\alpha }\mathrm{\ne }\mathrm{\varnothing }$, then $x_j\in C_j$ and its diameter is less than one, so $d_{h^{\mathrm{\prime }}}(x_j,m_{\alpha })<R+1$. Since the center is in that shell, $m<R+2$. Only finitely many shells satisfy this bound in a fixed component, and each supplied finitely many tuples. Components are open, so this proves local finiteness everywhere. Define $h^{+}:={\sum }_{j\in F}{h}_j$, smooth by [L4]. It is nonnegative, zero on $\{t\le a\}$, and greater than $1\mathrm{/}2$ on $S_b$. At a point $q\in \{h^{+}>0\}\cap \{t\le b\}$ only finitely many terms contribute. Each gradient is past timelike or zero by [L2], and some term is positive and therefore has a past timelike gradient. Their sum is past timelike by [L5].

(The plateau function.) Define $\beta$ and $A>0$ as in [L7] and set $\phi (u):=\{\begin{array}{ll}0,& u\le 0,\\ -1+{\displaystyle \frac{1}{A}}{\displaystyle {\int }_u^{1\mathrm{/}2}\beta (s)ds,}& 0<u<{\displaystyle \frac{1}{2}},\\ -1,& u\ge {\displaystyle \frac{1}{2}}\mathrm{.}\end{array}$ By [L7] the function $\phi$ is smooth, $\phi =0$ on $(-\mathrm{\infty },0]$, $\phi =-1$ on $[1\mathrm{/}2,\mathrm{\infty })$, $-1\le \phi \le 0$, ${\phi }^{\mathrm{\prime }}\le 0$ everywhere and ${\phi }^{\mathrm{\prime }}<0$ on $(0,1\mathrm{/}2)$; moreover $\phi$ is flat at $0$ and at $1\mathrm{/}2$.

(A neighborhood of the level.) Put $W:=\{q:h^{+}(q)>1\mathrm{/}2,\mathrm{\nabla }{h}^{+}(q)\text{ is past timelike},t(q)<c\}$. This set is open by continuity and the open timelike cone, and contains $S_b$ by step 1.2. On $W$ the inequalities and gradient property hold by definition; no point-indexed neighborhood selection is used.

(The reversed construction.) Apply the construction of steps 1.1 and 1.2 to the time-reversed manifold $(M,g,-N)$, which is again globally hyperbolic with the chronological past and future exchanged and with $t^{\mathrm{\prime }}=-t$ as volume time; its level at $-b$ is $S_{-b}^{\mathrm{\prime }}=S_b$, and the inequalities $t^{\mathrm{\prime }}\le -b$, $t^{\mathrm{\prime }}\ge -c$ translate to $t\ge b$ and $t\ge c$. By the same argument (with the dual of step 2.1 for the neighborhood of the level), with all support neighborhoods of the bumps chosen inside the open set $W\cap \{t<c\}$ (possible by the neighborhood clause of [L2]), there is a smooth $k:M\to [0,\mathrm{\infty })$ with the following properties: $k=0$ on $\{t\ge c\}$; $k>1\mathrm{/}2$ on $S_b$ and on a neighborhood of $S_b$ contained in $W$; $\mathrm{supp}k\subseteq W\cap \{t<c\}$; and $\mathrm{\nabla }k$ is future timelike at every point of $\{k>0\}\cap \{t\ge b\}$.

(The negative plateau and the weak step.) Define $h^{-}:=-1$ on $\{t\le b\}$ and $h^{-}:=\phi \circ k$ on $\{t\ge b\}$. The two formulas agree on a neighborhood of $S_b$ because $k>1\mathrm{/}2$ there by step 3.1, so $h^{-}$ is smooth; $h^{-}\le 0$, $h^{-}=-1$ on $\{t\le b\}$ and $h^{-}=0$ on $\{t\ge c\}$. Its gradient is past timelike or zero: on $\{t<b\}$ it vanishes, and on $\{t>b\}$ one has $\mathrm{\nabla }{h}^{-}={\phi }^{\mathrm{\prime }}(k)\mathrm{\nabla }k$ with ${\phi }^{\mathrm{\prime }}\le 0$ by [L7] and the chain rule, which is zero where ${\phi }^{\mathrm{\prime }}(k)=0$ or $\mathrm{\nabla }k=0$ and a nonnegative multiple of $-\mathrm{\nabla }k$, hence past timelike, where ${\phi }^{\mathrm{\prime }}<0$ and $k>0$. Let $U:=\{t<b\}\cup W$, open by step 1.1 and step 2.1, and put $u:=1+\frac{2h^{-}}{h^{+}-h^{-}}\text{ on }U,u:=1\text{ on }M\setminus U\mathrm{.}$ On $U$ the denominator never vanishes: on $\{t<b\}$ one has $h^{-}=-1$ and $h^{+}\ge 0$, so $h^{+}-h^{-}\ge 1$; on $W$ one has $h^{+}>1\mathrm{/}2$ and $h^{-}\le 0$, so $h^{+}-h^{-}>1\mathrm{/}2$. Also $\mathrm{supp}k$ is closed and contained in $W$, so every $z\in \mathrm{\partial }U$ satisfies $t(z)>b$ (a boundary point has $t\ge b$ since $\{t<b\}\subseteq U$ and $S_b\subseteq W\subseteq U$; $t(z)=b$ would put $z\in S_b\subseteq U$), hence $z\mathrm{\notin }W$, so $z\mathrm{\notin }\mathrm{supp}k$ and $k\equiv 0$ on a neighborhood of $z$; there $h^{-}=0$ and $u=1$, so $u$ is smooth at $z$ and in fact identically $1$ near it. Therefore $u$ is a smooth function on $M$ with values in $[-1,1]$ (writing $b^{-}:=-h^{-}\ge 0$ and $h^{+}\ge 0$ on $U$, one has $u=1-2b^{-}\mathrm{/}(h^{+}+b^{-})$). Its gradient on $U$ is $\mathrm{\nabla }u=2(h^{+}\mathrm{\nabla }{h}^{-}-h^{-}\mathrm{\nabla }{h}^{+})\mathrm{/}(h^{+}-h^{-}{)}^2$ by the quotient and chain rules, a sum of nonnegative multiples of vectors that are past timelike or zero ($\mathrm{\nabla }{h}^{-}$ by the previous paragraph, $\mathrm{\nabla }{h}^{+}$ on $\{t\le b\}$ by step 1.2 and on $W$ by step 2.1), hence past causal or zero by [L5]; outside $U$ it vanishes. On $S_b$ one has $k>1\mathrm{/}2$ nearby, so $\mathrm{\nabla }{h}^{-}=0$ there, while $\mathrm{\nabla }{h}^{+}$ is past timelike; hence $\mathrm{\nabla }u=2(-h^{-})\mathrm{\nabla }{h}^{+}\mathrm{/}(h^{+}-h^{-}{)}^2$ is a positive multiple of $\mathrm{\nabla }{h}^{+}$ at every point of $S_b$, so it is past timelike there, and by continuity and openness of the past timelike cone [L5] it is past timelike on a neighborhood of $S_b$. Finally $u=-1$ on $\{t\le a\}$: there $h^{+}=0$ by step 1.2 and $h^{-}=-1$, so $u=1-2\mathrm{/}(0+1)=-1$; and $u=1$ on $\{t\ge c\}$ because $U\cap \{t\ge c\}=\mathrm{\varnothing }$ by $W\subseteq \{t<c\}$. Setting $\tau :=u$ proves (a)-(d).