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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 $$P(a)=\cup_{n<0} g^n G(a,w).$$ 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 $$\cap_{n>0} g^n P(a,w)=\gamma'.$$ 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.