Pair of laminations that fill on a closed surface Let $S$ be a hyperbolic surface of genus $g \geq 2$.
A discrete geodesic lamination on $S$ is a set of disjoint, simple, closed geodesics.
Let $L_{1}$ and $L_{2}$ be two discrete geodesic laminations on $S$. We say that $L_1$ and $L_2$ fill $S$, if $S \setminus (L_{1} \cup L_{2})$ is a disjoint union of open topological disks.
Or equivalently, let $i(\alpha,L_{i})$ be the intersection number of the closed curve $\alpha$ with the leaves of the lamination $L_{i}$, then for any closed curve $\alpha$, $i(\alpha,L_{1}) + i(\alpha,L_{2}) > 0$ (including the closed curves that belong to $L_1$ and $L_2$, and recall that if $\alpha \in L_{i}$, then $i(\alpha,L_{i}) = 0$).
My question is: If we take two discrete geodesic laminations $L_1$ and $L_2$ on $S$, such that $L_{1}$ and $L_{2}$ have no common leaf. Can we add leaves to $L_1$ and leaves to $L_2$, to obtain two laminations $L^{'}_{1}$ and $L^{'}_{2}$ (then $L_{i} \subset L^{'}_{i}$) such that $L^{'}_{1}$ and $L^{'}_{2}$ fill $S$ ?
Since every geodesic lamination can be completed to $3g-3$ closed curve (then if $\alpha \notin L$, $i(\alpha,L) > 0$), the question remains to the following: Let $\alpha$,$\beta_{1}$,...,$\beta_{k}$ be disjoint, simple, closed geodesics in $S$. Can we find a simple closed geodesic $\gamma$ such that $i(\gamma,\alpha) >0$ and $\forall j$,$i(\gamma,\beta_{j}) = 0$.
 A: The answer is "no".  For consider the case where $L_1 = L_2$ is a single simple closed geodesic.

Now you've added the hypothesis that $L_1$ and $L_2$ have no common leaf, the answer becomes "yes".  Here is a sketch.
Let $X_i = S - L_i$.  If $X_1$ (say) is a union of pants then $L_1$ is a pants decomposition and so is maximal. Suppose not. Then we pick a lamination $\lambda_1 \subset X_1$ which restricts in each non-pants component $X' \subset X_1$ to a filling lamination.  That is, a closed subset of $X'$ which is a union of simple geodesics, none of which are loops, so that $X' - \lambda_1$ is a union of disks and peripheral annuli.  Furthermore $\lambda_1$ admits a transverse measure (so has no leaves that spiral about a component of $L_1$). Filling laminations exist by Thurston's theory of pseudo-Anosov maps.
We do the same for $L_2$: that is, we pick a lamination $\lambda_2$ that lives in and fills all non-pants components of $X_2$.  Now define $\Lambda_i = L_i \cup \lambda_i$
Exercise: Since $L_1$ and $L_2$ share no leaves the same holds of $\Lambda_1$ and $\Lambda_2$.
We deduce that there is a definite angle $\epsilon$ so that, at every point of $\Lambda_1 \cap \Lambda_2$, the two laminations intersect with an angle of at least $\epsilon$.  In particular $\Lambda_1 \cup \Lambda_2$ fills $S$.
We now choose sufficiently close Hausdorff approximations of $\Lambda_i$ by pants decompositions $L_i'$.  We deduce that $L_i'$ contains $L_i$.  Also $L_1'$ and $L_2'$ fill, as desired.
