Geodesic current supported on a pencil?

Consider a geodesic current $$\mu$$ on a closed surface $$\Sigma$$, as defined by Bonahon ("The Geometry of Teichmüller space via geodesic currents"). These are $$\pi_1(\Sigma)$$-invariant measures on the space of geodesics on $$\widetilde\Sigma$$. It is well-known that $$\mu$$ can only have atoms at closed geodesics. We can also look at 1-dimensional subsets, namely pencils $$P(a)$$, the set of geodesics with one endpoint at $$a \in \partial\widetilde\Sigma$$. If $$a$$ is not one of the limit points of a closed geodesic, then $$\mu(P(a)) = 0$$; see, e.g., Martelli, "An Introduction to Geometric Topology", Proposition 8.2.8.

Question. What is an example of a geodesic current without atoms so that $$\mu(P(a)) \ne 0$$ (where $$a$$ is necessarily the endpoint of a closed geodesic)?

EDIT Following Lee Mosher's request I am adding more details concerning the definition of marginals in this situation. The marginals of a probability measure on a product space are its coordinate projections. However, geodesic currents are infinite measures, and just taking their coordinate projections would lead to trivially infinite measures (or, at least, to measures which are not Radon for discrete currents). What I meant was not marginals themselves, but rather their measure classes, which make perfect sense. Namely, one should take a finite reference measure $$\lambda$$ equivalent to the geodesic current; then the measure classes of its coordinate projections do not depend on the choice of $$\lambda$$.
Here's a proof that doesn't use ergodicity. Suppose that $$\mu(\gamma)=0$$ for all elements $$\gamma \in \mathcal{G}(\tilde{\Sigma})$$. Let $$P(a)$$ be the pencil with endpoint $$a \in \partial \tilde{\Sigma}$$ and suppose, for the sake of contradiction, that $$\mu(P(a))\neq 0$$. By the stated result by Martelli, there is $$\gamma'$$, a lift of a closed geodesic in the support of $$\mu$$ ending at $$a$$. Let $$g \in \pi_1(S)$$ be an element fixing $$\gamma'$$. Let $$w$$ be a proper closed interval (not a singleton) of $$\partial \tilde{\Sigma}$$ disjoint from $$a$$ and including the starting point of $$\gamma'$$, and let $$G(a,w)$$ be the proper subset of the pencil $$P(a)$$ consisting of all geodesics starting at a point in $$w$$ and ending at $$a$$. If $$a$$ is the repelling point of $$g$$, by the dynamics of $$g$$, it follows that $$g(w) \subsetneq w$$. Otherwise, $$g^{-1}w \subsetneq w$$. Without lost of generality, suppose that $$g(w) \subsetneq w$$. Then, by the dynamics of $$g$$, we obtain $$P(a)=\bigcup_{n<0} g^n G(a,w).$$ Therefore, since $$\mu(P(a)) \neq 0$$, by countable subadditivity of $$\mu$$, it follows that $$\mu(G(a,w)) \neq 0$$. Also, by local finiteness of $$\mu$$ it follows that $$\mu(G(a,w))<\infty$$. Now, observe that by the dynamics of $$g$$, we have $$\bigcap_{n>0} g^n P(a,w)=\gamma'.$$ By continuity of $$\mu$$ from above, $$\lim_{n \to \infty} \mu(g^n G(a,w)) = \mu(\gamma')=0,$$ since $$g^n P(a,w)$$ is a sequence of nested measurable sets containing $$\gamma'$$. On the other hand, since $$\mu$$ is $$\pi_1(S)$$-invariant, $$\mu(g^nG(a,w))=\mu(G(a,w))>0,$$ yielding a contradiction for $$n$$ large enough.
• Why is $\cap_n g^n P(a)=\gamma'$? Even if this were true why would this imply $\mu(g^n P(a)) \to \mu(\gamma')=0$? (as the measure $\mu$ is infinite). – R W Sep 29 '18 at 1:19
• I think you should be looking at not the whole pencil $P(a)$ (i.e., the set of all geodesics with an endpoint at $a$), but rather some subset of it: pick a window $w$ disjoint from $a$, and consider the set $G(a,w)$ of geodesics with one endpoint at $a$ and the other in $w$. Then I believe local finiteness impless that $\mu(G(a,w)) < \infty$, while the intersection $\cap_{n > 0} g^n G(a,w)$ is a single geodesic and the union $\cup_{n < 0} g^n G(a,w) = P(a)$. – Dylan Thurston Sep 29 '18 at 18:03