I am wondering about the following:

Suppose that $S$ is a non-compact hyperbolic surface of finite area. Suppose that $\lambda \subset S$ is a non-trivial, geodesic, measured lamination. Forget the transverse measure. Is there a (non-compact) geodesic lamination $\lambda'$ containing $\lambda$, and a sequence of ideal geodesic triangulations $T_i$, so that $T_i \to \lambda'$ in the Chabauty topology?

A little bit of context:

- The Chabauty topology is a generalization of the Hausdorff topology to non-compact sets. It is characterized by the condition that every point of $\lambda'$ is the limit of a sequence of points $x_i \in T_i$, and conversely every convergent sequence $x_{i_n} \in T_{i_n}$ limits to a point of $\lambda'$. See, e.g. Notes on Notes of Thurston.
- I am mainly interested in the answer in the setting where $\lambda$ is the pleating lamination on the boundary of the convex core of a quasifuchsian $3$-manifold. This places some additional hypotheses on $\lambda$: for example, it would have to be compact. But the question seems to be intrinsically $2$--dimensional, and it's not clear to me how to use compactness of $\lambda$ as a hypothesis.
- If the ideal triangulations $T_i$ are replaced by simple closed curves, the result is well-known. So one approach would be to take a sequence of closed curves $C_i$, limiting to $\lambda' \supset \lambda$, approach each $C_i$ by triangulations (twisting more and more), and then take a diagonal sequence of triangulations. But it's not clear that this diagonal sequence even converges.

Anyway, either a reference or a way to argue would be much appreciated!