TITLE: Laminations as a limit of ideal triangulations
QUESTION [7 upvotes]: 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!

REPLY [5 votes]: It turns out that the answer to the question is "yes". Saul Schleimer and I needed this result for a paper that we just finished writing, so we ended up sorting it out. The full argument is written down in Lemma A.6 in the Appendix of this paper, so what follows below is an outline. 
Take a sequence of closed curves $C_i$, which limits to $\lambda$ in the measure topology. This sequence has a subsequence (which I will still call $C_i$) limiting to $\lambda' \supset \lambda$ in the Chabauty topology. Now, each $C_i$ is contained in the Chabauty limit of a sequence of triangulations $T_{i,j}$. This means that one can take a representative triangulation $T_{i,j(i)}$ that is very close to $C_i$, where "very close" can be quantified (say, closer than distance $1/i$) because the Chabauty topology is metrizable. Now, the sequence $T_{i,j(i)}$ will converge to a lamination $\lambda'' \supset \lambda' \supset \lambda$.