In Penrose's original formulation of Penrose Inequality, i.e. in Naked Singularity (1973), he speculated a thin sell matter collapsing at the speed of light, forming an ingoing null hypersurface N. Consider outgoing null hypersurface H. Along the line, he said
"Assuming cosmic censorship, we deduce the existence of an absolute-event horizon E intersecting H in a two-surface $S_1$, the surface $S_1$ lying farther out along the generators of H than $S_0$. Furthermore, the total surface area $A_1$ of $S_1$ is not less than the area $A_0$ of $S_0$."
where $S_0$ is apparent horizon.
My question is how could one deduce that area of event horizon is no less than area of apparent horizon simply from the fact that event horizon lies outside of apparent horizon? For example, the event horizon of the Schwarzschild black hole at $r=2m$ is known to be minimal, possessing a smaller area than those both outside and inside of it. To compare area, we are assuming knowledge of metric, right? But outside apparent horizon $S_0$ in Penrose's set-up, the spacetime is not Minkowski. What am I missing? Thanks in advance!