I am corrently working on mathematical general relativity and stumbled over the following question:
The (vacuum) Einstein Hilbert action is defined to be
$$\mathcal{S}_{\mathrm{EH}}(g):=\int_{\mathcal{M}}\!R(g)\,\mathrm{d}\lambda_{(\mathcal{M},g)}$$
where $\mathcal{M}$ denotes a Lorentzian manifold of dimension $4$ and $\mathrm{d}\lambda_{(\mathcal{M},g)}$ the usual Riemann-Lebesgue measure on a given (pseudo)-Riemannian manifold $(\mathcal{M},g)$. The action is a function of the set of all Lorentzian metrics $g$ on $\mathcal{M}$ into the real numbers and $R(g)$ denotes the Ricci scalar with respect to the metric $g$.
My question (maybe its also a stubid one) is now if this action is well-defined, or in other words, is the Ricci-scalar always integrable? I think it should be, because by definition is the Ricci scalar $R(g)$ locally a combination of the metric $g$ and its derivatives. Furthermore, the space-manifold is per definition smooth and also the metric $g$ is a smooth tensor field. But is this enough to show that $\mathcal{S}_{\mathrm{EH}}$ is well-defined?