Putting a transverse measure on a surface foliation Let $F$ be an orientable surface with a foliation $\cal F$ with $k$-prong singularities only, for $k\geq 3$. 
Since I am looking for an invariant transverse measure on $\cal F$, assume that there is no circle leaf in $\cal F$ to which some other leaves are spiraling to and there is no infinite leaf with foliation like in the first pic. 
Even then $\cal F$ may not have an invariant transverse measure because of Denjoy blowup (pictured on the right) possibility (as pointed to me by Lee Mosher). 
Is it true however that there is a subsurface $F_0\subset F$ bounded by leaves of $\cal F$ with an invariant transverse measure on $\cal F\cap F_0$ such that $F-F_0$ is a union of topological disks? Can it be assumed that those disks are infinite bigons foliated as on the right? If not then perhaps some slightly weaker statement holds? 


 A: We cannot ensure that the region where the transverse measure vanishes is a union of foliated bigons.  In general, the union of the leaves where the transverse measure vanishes may fill up a subsurface with non-trivial topology.  Here is an example.

Take a foliation $\mathcal{F}$ of a surface $S$.  Suppose that $\mathcal{F}$ admits an invariant transverse measure of full support.  We now do a Denjoy blowup to obtain a foliation $\mathcal{F}'$ and a bigon $B$ in $S$ with $\mathcal{F}'|B$ foliated as you show.  We remove all of the leaves of $\mathcal{F}'$ meeting the interior of $B$ to obtain a transversely measured lamination $\mathcal{L}$.
We now cut a small disk out of $B$ and glue in a "handle" - a once-holed surface.  Let $S'$ be the resulting surface.  Note that $\mathcal{L}$ is again a lamination in $S'$.  To turn it into a foliation in $S'$ we first add two "linking arcs" $\alpha$ and $\beta$ running from cusp of $S' - \mathcal{L}$ to cusp of $S' - \mathcal{L}$.  Set $\mathcal{L}' = \mathcal{L} \cup \alpha \cup \beta$. So $S' - \mathcal{L}'$ consists of a single open disk, with six cusps.  So we add leaves to triangulate this disk and then add four three-pronged singularities to obtain the desired foliation $\mathcal{G}$ in $S'$.  The leaves of $\mathcal{G}$ meeting the handle have transverse measure zero, as promised.
