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 $a$ and suppose, for the sake of contradiction, that $\mu(P(a))>0$. 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 $\gamma'$. Observe that $\cap_n g^n P(a)=\gamma'$ (maybe replacing $g$ by $g^{-1}$). By continuity of measures from above, $\mu(g^n P(a)) \to \mu(\gamma')=0$ as $n$ grows, since $g^n P(a)$ is a sequence of nested measurable sets. On the other hand, since $\mu$ is $\pi_1(S)$-invariant, $\mu(g^nP(a))=\mu(P(a))>0$, yielding a contradiction for $n$ large enough.
DMG
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