Compact-tube timelike length bound

Fix one of the stated Lipschitz curves and a parameter $q$ for which it is future timelike. Subdivide its compact parameter interval into finitely many closed subintervals on each of which the curve has Lipschitz coordinates and its image lies in a coordinate neighborhood where the smooth function ${\tau }_q(t,x)=t-u_q(x)$ is Lipschitz. Such neighborhoods exist by [F4]; continuity and compactness give the finite subdivision. Each coordinate function is absolutely continuous by [F4], so its derivative exists almost everywhere, is measurable, and is bounded almost everywhere by its Lipschitz constant, as follows by taking limits of its difference quotients. The composite ${\tau }_q\circ \gamma$ is Lipschitz on each subinterval, hence absolutely continuous. At every point where all coordinates are differentiable, the chain rule gives $({\tau }_q\circ \gamma {)}^{\mathrm{\prime }}=d{\tau }_q(\dot{\gamma })$. Applying the absolutely continuous FTC on these subintervals and telescoping gives ${\int }_a^{b}d{\tau }_q(\dot{\gamma })ds={\tau }_q(\gamma (b))-{\tau }_q(\gamma (a))$.

In the product coordinates $T={\mathrm{\partial }}_t$, the Killing equation for $g$ gives ${\mathrm{\partial }}_t{g}_{\mu \nu }=0$, so the components of $T^{\mathrm{♭}}=g(T,\cdot )$ are independent of $t$. By [F8], the components of $W_0$ and the scalar $\sigma$ are also independent of $t$. The displayed formulas therefore make $D_q$ and the components of $W_q$ independent of $t$; $d_q=D_q(p)$ is a fixed scalar, not a function of the flowing point. The coefficients of $A_q$ and consequently those of $g_q=(D_q\mathrm{/}4)A_q^{\ast }g$ are independent of $t$. Thus each translation ${\varphi }_s^T(t,x)=(t+s,x)$ is a $g_q$-isometry. It also preserves the declared future cones: for a future timelike vector $v$ at $z$, the vectors $v_s=(d{\varphi }_s^T{)}_{z}v$ remain timelike, and $g_q(v_s,Y_q({\varphi }_s^{T}z))$ is continuous in $s$, is negative at zero, and cannot vanish because both vectors are timelike. It stays negative for every $s$. This uses the time orientation in [F7], without requiring $N$ itself to be stationary.

Extend $n_q$ by the differential of the stationary flow and $N_q$ by constant transport along each orbit. Step 1.2 shows that the transported normal remains the future unit normal to each translated spacelike graph. Since the flow preserves $T$ and $g_q$, its lapse is still $-g_q(T,n_q)=N_q$, so the graph lapse bound in [F2] holds on every translate. The level sets of ${\tau }_q$ are the stationary translates of the graph. Their normal covector is a multiple of $g_q(n_q,\cdot )$; evaluating on $T$ gives $d{\tau }_q=-g_q(n_q,\cdot )\mathrm{/}{N}_q$. For any future timelike tangent $v$, decompose $v=a{n}_q+w$ with $w\perp n_q$. Then $a=-g_q(n_q,v)>0$ and $g_q(w,w)\ge 0$, so $\sqrt{-g_q(v,v)}\le a=N_{q}d{\tau }_q(v)$ and $d{\tau }_q(v)>0$. By step 1.1 these inequalities hold almost everywhere along the curve.

Fix a future $g_{q_{\ast }}$-timelike Lipschitz curve in the tube with $g_{q_{\ast }}(\dot{\gamma },\dot{\gamma })\le -ϵ$ almost everywhere for some $ϵ>0$. Its coordinate velocity is essentially bounded on a finite chart subdivision, as in step 1.1. Compact uniform metric convergence therefore makes $g_{q_j}(\dot{\gamma },\dot{\gamma })<0$ almost everywhere for all large $j$. Use the supplied orientation $Y_q=A_q^{-1}N$ from [F7], which varies jointly smoothly for this single-primitive family. On the compact curve image set $c=\min \sqrt{-g_{q_{\ast }}(Y_{q_{\ast }},Y_{q_{\ast }})}>0$. Normalize $e=Y_{q_{\ast }}\mathrm{/}\sqrt{-g_{q_{\ast }}(Y_{q_{\ast }},Y_{q_{\ast }})}$ and decompose a future tangent $v=ae+w$ with $w\perp e$. As in step 2.1, $a>0$ and $-g_{q_{\ast }}(v,v)=a^2-g_{q_{\ast }}(w,w)\le a^2$, so $-g_{q_{\ast }}(v,Y_{q_{\ast }})\ge c\sqrt{ϵ}$. Uniform convergence of $g_{q_j}(\cdot ,Y_{q_j})$ to $g_{q_{\ast }}(\cdot ,Y_{q_{\ast }})$, together with the coordinate velocity bounds, then gives $g_{q_j}(\dot{\gamma },Y_{q_j})<0$ almost everywhere for all large $j$. Thus the tangents remain future directed in the declared orientations. For a fixed piecewise $C^1$ timelike curve in $\tilde{Q}$ with nonzero tangent on each closed piece, compactness supplies the same margins and boundedness, giving persistence on every piece, including one-sided endpoint tangents.

Integrating step 2.1 and using $0<N_q\le B_{\delta }$ and the endpoint identity from step 1.1 yields $L_{g_q}(\gamma )\le B_{\delta }{\int }_a^{b}d{\tau }_q(\dot{\gamma })ds=B_{\delta }({\tau }_q(\gamma (b))-{\tau }_q(\gamma (a)))$. The two endpoint time coordinates lie in $[-\tilde{A},\tilde{A}]$, and both endpoint spatial coordinates lie in $\tilde{K}$. Thus the difference is at most $2\tilde{A}+2H_{\delta ,\tilde{K}}$. The bounds in [F2] hold for every $q\in [\delta ,1]$, so this proves the bound for each $q_j$ and directly for $q=q_{\ast }$, for every future $g_{q_{\ast }}$-timelike Lipschitz curve in $\tilde{Q}$. No uniform negative tangent-norm margin is used in this direct estimate.

Apply step 3.2 to the fixed curve for all sufficiently large $j$, as permitted by step 3.1. On the finite chart subdivision, define the length integrands to be zero on the common null set where coordinate derivatives do not exist or the required tangent conditions fail. They are measurable: the coordinate velocities are measurable by step 1.1, the curve and metric coefficients are continuous, and the square root is continuous on the nonnegative reals. Uniform metric convergence gives almost-everywhere convergence of these integrands to the $g_{q_{\ast }}$ integrand. On each chart piece choose finite constants $M$ bounding the coordinate velocity almost everywhere and $C$ bounding the absolute quadratic forms of all sufficiently late metrics and the limiting metric on the compact curve image. Then $0\le \sqrt{-g_{q_j}(\dot{\gamma },\dot{\gamma })}\le \sqrt{C}M$ almost everywhere. The finitely many constant bounds give one integrable majorant on $[a,b]$. Dominated convergence [F5] gives $L_{g_{q_j}}(\gamma )\to L_{g_{q_{\ast }}}(\gamma )$, and the same uniform bound passes to the limit. This proves the conclusions, including the piecewise $C^1$ case.