Consider a sequence $T_i$ of tetrahedra in $\mathbb H^3$ whose vertices tend to the vertices of a regular ideal tetrahedron $T$ in $\partial \mathbb H^3$. Then $$Vol(T_i)\to v_3.$$
This should follow from Lebesgue dominated convergence if $T_i\subseteq T_{i+1}$ for (almost) all $i$, since, calling $\nu$ the volume form on $\mathbb H^3$, $$|\nu\chi_{T_i}|\leq|\nu\chi_T|$$ so the integrals converge.
I think one can always suppose to be in this case by moving the $T_i$ by isometries: is this true? Is there a formally satisfying way to see it?
Thank you in advance.
There are some explicit formulae for volumes of hyperbolic tetrahedra in terms of dihedral angles which are not just continuous but real-analytic functions, say, one by Ushijima (Theorem 1.1):
A volume formula for generalized hyperbolic tetrahedra.
(See also references to earlier works that he gives in the paper.) Dihedral angles, in turn, depend continuously (actually, real-analytically) on the vertices. Hence, volume is a real-analytic function on the vertices.