Geodesic current supported on a pencil?

Question

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)?

Answer 1

No - this is impossible because any ergodic geodesic current either has purely non-atomic marginals or corresponds to a closed geodesic. Indeed, the quoted result from Martelli implies that if one of the marginals contains atoms then it has to be concentrated on the orbit of an endpoint of a periodic geodesic (actually, this is true for a compact negatively curved manifold in any dimension without requiring that the curvature be constant). The Birkhoff ergodic theorem in combination with the asymptotic convergence of the geodesics with the same endpoint then implies that the invariant measure coresponding to this current is the one concentrated on the periodic geodesic (this is the so-called "Hopf argument").

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 $\lambda$.

Answer 2

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.