During a conversation I heard an assertion that I found at less dubious, but I am not able to imagine a counterexample, even if it is probably obvious to some of you.
My question: is there someone who can point out to me a counterexample?

This is the setting:
Let $(M,\omega)$ be a $2n$-dimensional symplectic manifold, and $f_1,\ldots,f_n$ independent functions on $M$ mutually Poisson-commuting, and such that the hamiltonian vector fields $X_{f_1},\ldots,X_{f_n}$ are complete.
Let $\mathcal{F}$ denote the lagrangian foliation of $M$ determined by the integrable distribution $D$ generated by $X_{f_1},\ldots,X_{f_n}$.

This is the assertion that I find dubious:
For any $x$ in $M$ there exists a local manifold $\Sigma_x$ which is lagrangian, transversal to $D$, and doesn't intersect any leaf of $M$ at two distinct points.
(I know that this condition is necessary and sufficient for the exixtence of the manifold of the leaves of $\mathcal{F}$)