We just give a sketch of the proof. Using the Weinstein neighborhood theorem one can show that in a sufficiently small tubular neighborhood of a fibre $F_{b}$, the symplectic form is exact, i.e $\omega=-d\eta$ for some 1-form $\eta$. Notice that $\eta|_{F_{b}}$ is a closed 1-form. Define functions $a_{j}$ on $U$ by

$$a_{j}=\int_{\gamma_{j}}\eta.$$

One can show that

$$\lambda_{j}=da_{j}$$

and therefore $\lambda_{j}$ is closed. It is clear that the coordinates $a=(a_{1},...,a_{n})$ are well defined up to an integral affine transformation and therefore they define an integral affine structure on $B$ inducing the lattice $\Lambda$ in $T^{\ast}_{B}$. Finally, notice that given a section $\sigma:U\rightarrow X$ we have a covering map

$$\begin{array}[]{rcl}T^{\ast}_{U}&\rightarrow&f^{-1}(U)\\ \alpha&\mapsto&\theta_{\alpha}(\sigma(\pi(\alpha)))\end{array}$$

This map induces a diffeomorphism between $T^{\ast}_{U}/\Lambda$ and $f^{-1}(U)$. One can check that in the case $\sigma$ is Lagrangian this map is a symplectomorphism. For the proof of the last statement we refer the reader to [4]. ∎