 Solution

Suppose, for contradiction, that every future null geodesic orthogonal to the compact trapped surface $T$ extends beyond $L=A/|{\theta }_0|$. Both families begin with expansion at most ${\theta }_0<0$, so part d gives a point conjugate to $T$ on every generator by affine length $L$. A generator of the achronal boundary ${\dot{J}}^{+}(T)$ cannot remain on that boundary beyond its first conjugate point, because afterward it can be deformed to a timelike curve from $T$.

The two bundles of initial null directions over compact $T$, restricted to $0\le \lambda \le L$, form a compact set. Their image under the geodesic exponential map contains all of ${\dot{J}}^{+}(T)$, so this achronal boundary is compact. Project it along a complete timelike flow onto the noncompact Cauchy hypersurface $\Sigma$. The projection is both open and closed in the connected Cauchy surface, hence would be all of $\Sigma$; compactness of the source would then make $\Sigma$ compact, a contradiction.

Therefore at least one orthogonal future null geodesic cannot extend to affine length $L$. Its maximal future development is future-inextendible with total affine lengthwhich is the incompleteness conclusion of the Penrose singularity theorem.