Measured geodesic laminations have either discrete or Cantor set local cross-sections

Question

I'm reading through Kerckhoff's paper "The Nielsen Realization Problem": https://www.jstor.org/stable/2007076.

In section 1, after he defines measured geodesic laminations, he makes the following comment (I included the full paragraph for context):

A geodesic lamination on a hyperbolic surface $M$ is a closed subset $\mathcal L\subset M$ which is the union of geodesics and which is foliated in the following sense: there are open sets $U_i$ covering $\mathcal L$ with continuous maps $\psi_i:\mathcal L\cap U_i\subset U_i\to (0,1)\times B_i\subset\mathbb R^2$ taking $\mathcal L\cap U_i$ to horizontal arcs $(0,1)\times y$, $y\in B_i$, in the place such that the overlap maps preserve the horizontal property, i.e. $\psi_i\circ\psi_j^{-1}(x,y)$ is of the form $(f(x,y),g(y))$.

The geodesic laminations of interest here will also be required to possess a positive Borel measure $\mu$ on its local leaf space $B_i$ which is invariant under co-ordinate change. Equivalently, $\mu$ can be defined as a measure on arcs transverse to $\mathcal L$ as support and to be finite on compact arcs. Now $\mu$ is required to have all of $\mathcal L$ as support and to be finite on compact arcs. The existence of the transverse measure on all of $\mathcal L$ restricts the local behavior in that a local cross-section is either discrete or a Cantor set.

I'm having trouble seeing why this (the local cross-section is either discrete or a Cantor set) is true and cannot find any other reference to this fact.

Can someone give a brief explanation as to why this is true or point me to where I can find a proof/discussion of this fact?

Thank you in advance.

Answer 1

The statement you want is given in Proposition 7 of Bonahon's expository article "Geodesic laminations on surfaces". He reduces the proof to a result of Birman-Series from "Geodesics with bounded intersection numbers on surfaces are sparcely distributed". They prove that the union of all simple geodesics in a hyperbolic surface (of finite type) has Hausdorff dimension one. Of course, this result is much stronger than what you need... I believe that Birman-Series prove this by analysing various collections of (almost geodesic) train tracks.

Bonahon also refers to a paper of Thurston's: "Minimal stretch maps between hyperbolic surfaces" as giving a different proof of the Birman-Series result.

Answer 2

Let $(\mathcal{L}, \mu)$ be a connected measured lamination which is not a simple closed geodesic. Then there's no atomic measure on $\mathcal{L}$, which means that for each point $p \in \mathcal{L}$, $\mu(p)=0$. This can be proved by recurence of laminations.

Let $\gamma$ be a simple arc that intersects a connected measured lamination $(\mathcal{L}, \mu)$ transversely, and assume that $\mathcal{L}$ is not a simple closed geodesic. Then $\gamma \cap \mathcal{L}$ has the following properties: it is closed(by definition), has no interior points (by properties of hyperbolic geometry) or isolated points(since there's no atomic measure). Those properties will imply $\gamma \cap \mathcal{L}$ is homeomorphic to a Cantor set, see https://en.wikipedia.org/wiki/Cantor_space.