Why must metric tensor be invertible? The metric can be written as a matrix, but why must this matrix be invertible? At the points where the matrix is singular, why is the metric not defined?

Given a pseudo-Riemannian manifold $(\mathcal M,g)$ and some local tangent space $\mathrm T_p\mathcal M$ with coordinates $x^1,\dots, x^n$ , the volume form is given by
$$\mathrm dV=\sqrt{|\det g|}~\bigotimes_{i=1}^n \mathrm dx^i$$
So
$$g~\text{not invertible}\implies \det g=0\implies \mathrm dV=0$$
Which would imply our manifold is trivial, i.e, a single point.
