Wave boundary data determine integrals along maximal geodesics shorter than T

First use [F1] with both potentials equal to a single real $q$. Its left side is zero, so for every $h,k$ in the boundary input space ${\int }_{\mathrm{\Sigma }}k{\mathrm{\Lambda }}_q^{+}hd{S}_{g}dt={\int }_{\mathrm{\Sigma }}h{\mathrm{\Lambda }}_q^{-}kd{S}_{g}dt\mathrm{.}$ This is a bilinear transposition identity and uses exactly the classical forward and terminal solutions, whose endpoint terms vanish. Therefore the right side of [F1] for $q_1,q_2$ equals ${\int }_{\mathrm{\Sigma }}k({\mathrm{\Lambda }}_{q_1}^{+}-{\mathrm{\Lambda }}_{q_2}^{+})hd{S}_{g}dt$, which is zero under the assumed equality of forward maps. Hence ${\int }_Q(q_1-q_2)uvd{V}_{g}dt=0$ for every admitted forward $q_1$ wave and terminal $q_2$ wave.

Choose $t_0\in (0,T-L)$ and insert the exact waves of [F2] into step 1.1. Their boundary traces belong to the full input space for each $\lambda$, so the zero identity applies without a limit or density assumption on boundary data. The product convergence of [F2], with $A(t,x)=q_1(x)-q_2(x)$, gives $0={\lim }_{\lambda \to \mathrm{\infty }}{\int }_Q(q_1-q_2)u_{1,\lambda }{u}_{2,\lambda }d{V}_{g}dt={\int }_0^L(q_1-q_2)(\gamma (s))ds\mathrm{.}$ The test is legitimate either by [F2]'s closed-cylinder conclusion or by a temporal cutoff equal to one on all approximate packets; its omitted exact-product terms contain vanishing correction factors. Since the ray was arbitrary subject to $L<T$, every claimed ray integral is determined.