If we want to be formal, at $r=0$ there is a manifold singularity. So no tensor/scalar quantities are well defined there (i.e. our mathematical formalism does not work there). On the other hand, as physicists, we can look for different examples that behave in a similar manner. Also the (classical) electromagnetic field generated by a point-like charged particle diverges as $r \rightarrow 0$. There is however a subtle difference: a field can be ill-defined in one point without many head scratches, for gravity what is ill-defined is space-time itself, so the whole Physics becomes hard there (usually Physics is encoded mathematically with functions of position and time).

There have been attempts to formalize mathematically this issue. In Electrodynamics we study distribution (Dirac deltas) of charges why can't we use them in gravity?
Starting from the Einstein equations ($G=c=1$, $\Lambda=0$)
$$ R_{ij}-\frac{1}{2}g_{ij}R=8\pi \ T_{ij}, $$ and multiplying them by $g^{ij}$ you get the an equation for the Ricci scalar $$R=-8\pi\,T$$ where $T$ is the trace of the Energy-momentum tensor. So, if the matter is concentrated only at $r=0$ you would expect a distributional quantity that describes $T$ and therefore $R$.

In literature, it is possible to find this approach: for example in this paper Distributional Nature of the Energy Momentum Tensor of a Black Hole or What Curves the Schwarzschild Geometry ? by H. Balasin and H. Nachbagauer.

They find that for a Schwarzschild black hole of mass $M$, the Ricci tensor should have the form $$ R = 8 \pi \, M\, \delta^{(3)}(x) $$ where $x$ is the position in space-time and $\delta^{(3)}$ is the Dirac-delta function in space, namely zero everywhere but at $r=0$. If it was in a flat space-time the delta could be defined also as $\bigtriangleup\left(1/r\right)=\bigtriangledown^{2}\left(1/r\right)=\delta^{\left(3\right)}\left(x\right)$, but I am not sure in the general case.

But, **we have to be very careful** with this distributional approach since at the singularity we don't know what space-time physically is, and the mathematical theory behind it, as far as I know, for pseudo-Riemannian geometries, *is not solid*, yet.

GR describes curvature as being caused by stress-energy.

This statement is slightly wrong and is the cause of your confusion here.

Technically, in GR the stress energy tensor is the source of curvature. That is not quite the same as being the cause.

An easy analogy is with Maxwell’s equations. In Maxwell’s equations charge and current density are the sources of the electromagnetic field. However, although charges are the source of the field there exist non trivial solutions to Maxwell’s equations that involve no sources. These are called vacuum solutions, and include plane waves. In other words Maxwell’s equations permit solutions where a wave simply exists and propagates forever without ever having any charges as a source.

Similarly with the Einstein field equations (EFE). The stress energy tensor is the source of curvature, but just as in Maxwell’s equations there exist non trivial vacuum solutions, including the Schwarzschild metric. In that solution there is no cause of the curvature any more than there is a cause of the plane wave in Maxwell’s equations. The curvature in the Schwarzschild metric is simply a way that vacuum is allowed to curve even without any sources.

Now, both in Maxwell’s equations and in the EFE the vacuum solutions are not particularly realistic. Charges exist as does stress energy. So the universe is not actually described by a vacuum solution in either case. So typically only a small portion of a vacuum solution is used to describe only a small portion of the universe starting at some matching boundary. A plane wave can match the vacuum region next to a sheet of current, and the Schwarzschild solution can match the vacuum region outside a collapsing star.

So realistically, the cause of the curvature would be stress-energy that is outside of the vacuum solution, in the part of the universe not described by the Schwarzschild metric. This would be in the causal past of the vacuum region including the vacuum inside the horizon. Since it is in the causal past it can be described both as the cause and the source of the curvature, with the understanding that it is strictly outside of the Schwarzschild metric which is a pure vacuum solution in which the curvature has no source.

## Best Answer

In general relativity, in

generalmass and energy are not well-defined globally since they are not conserved. So the question is in a sense meaningless in general, but black holes may be an exception... except that thelocationof the mass still remains iffy.Locally everything is fine: the local flow of mass and energy is described by the stress-energy tensor, and there is energy conservation locally. Globally there is no good way of defining mass globally so it is conserved, in general. The fundamental problem is that energy gets its conservation from time translation invariance, and general spacetimes does not have that symmetry.

But the Schwarzschild spacetimes does! That means one can define the Komar mass. Indeed, it can be calculated by integrating a volume integral, and it does produce the "right" answer. But the volume stretches out to spatial infinity. Making "quasi-local" measures appears tricky.

Note that a collapsing material object turning into a Schwarzschild black hole seems to put all the mass into $r=0$. But this is a dynamic spacetime so the Komar mass is not defined until after the end of the collapse, not during it.

So where does this leave us? If you say "amount of mass-energy" is a measure of something conserved related to the stress-energy tensor, then it is clearly just zero in the Schwarzschild metric. The relativistic mass measures are all global, and do not correspond to any "where" of the mass. I have no idea of all the quasi-local measures.

In the end, maybe one should turn the question around: does it matter?