One uses the same arguments as in the proof of Proposition 6.9. Suppose there is an extension of $f^{+}:X^{+}\rightarrow B^{+}$ to a smooth Lagrangian fibration $\tilde{f}^{+}$ defined on a neighborhood $W\subseteq X$ of $Z$ such that $\tilde{f}^{+}|_{\mathfrak{U}}=f|_{\mathfrak{U}}$. Then one may compute the period lattice of $\tilde{f}^{+}$; this gives a smooth function $H$ extending the function $H^{+}$ in Proposition 6.23. Assuming that also $f^{-}$ has been extended to $\tilde{f}^{-}$ so that $\tilde{f}^{-}|_{\mathfrak{U}}=f|_{\mathfrak{U}}$, one may verify that the period map $\Theta^{+}:T^{\ast}U/\penalty\Lambda_{H}\rightarrow W^{\#}$ gives the required equivalence between $\mathcal{F}$ and $\mathcal{F}_{u,H}$ where $u=\tilde{f}^{-}\circ\Theta^{+}$.

To extend $f^{+}$, notice that $f_{\mathfrak{U}}=f|_{\mathfrak{U}}$ is smooth so, tautologically, $f_{\mathfrak{U}}$ is an extension of $f^{+}$ to $\mathfrak{U}$. It remains to extend $f^{+}$ away from $\mathfrak{U}$. Let $\mathfrak{U}^{\prime}\subset\mathfrak{U}$ and define $f^{\circ}:X^{\circ}\rightarrow B$ as in (62). Denote $Z^{\circ}=Z\cap X^{\circ}$ and by $\bar{Z}^{\circ}$ its $S^{1}$ quotient with $\bar{f}^{\circ}:\bar{Z}^{\circ}\rightarrow\Gamma$ the reduced fibration. Then $\bar{f}^{\circ}$ is a smooth Lagrangian cylinder fibration.

The coisotropic neighborhood theorem allows us to identify a neighborhood of $Z^{\circ}$ inside $X^{\circ}$ with a neighborhood $V$ of $\{0\}\times S^{1}\times\bar{Z}^{\circ}$ inside $\mathbb{R}\times S^{1}\times\bar{Z}^{\circ}$ ($t$ will denote the $\mathbb{R}$ coordinate). Moreover, since $\bar{Z}^{\#}$ can be identified with $T^{\ast}\Gamma/\bar{\Lambda}_{H}$ (see Remark 6.24), $\bar{Z}^{\circ}$ can be identified with a subset of $T^{\ast}\Gamma/\bar{\Lambda}_{H}$ of the type $\bar{Z}^{\circ}_{L}$ for some positive $L$ (see Example 6.25). The pullback of $f^{\circ}$ under these identifications gives a piecewise smooth Lagrangian fibration on $V\subset\mathbb{R}\times S^{1}\times\bar{Z}^{\circ}_{L}$

where $V^{+}=V\cap\{t\geq 0\}$, $V^{-}=V\cap\{t\leq 0\}$ and $u^{\pm}$ is the restriction to $V^{\pm}$ of a $C^{\infty}$ map. The set $Z^{\circ}\cap\mathfrak{U}$ where $f^{\circ}$ is smooth, corresponds (under the above identifications) to the interior of $Z^{\circ}_{L}-Z^{\circ}_{L^{\prime}}$ which we denote $C_{L,L^{\prime}}$, where $L^{\prime}<L$. Notice that the map $g$ above is then smooth along $C_{L,L^{\prime}}$, in particular the Taylor expansions in $t$ of $u^{+}$ and $u^{-}$ coincide along $C_{L,L^{\prime}}$. With the same arguments used in the proper case one can show that $u^{\pm}$ can be smoothly extended to a Lagrangian fibration $\tilde{u}^{\pm}$ beyond $V^{\pm}$ (cf. Proposition 6.9 above, or [2] Proposition 6.3 for more details). In fact with a little more care one can do this so that along $\mathbb{R}\times C_{L,L^{\prime}}$, where an extension already exists, namely $g$ itself, we have $\tilde{u}^{\pm}|_{\mathbb{R}\times C_{L,L^{\prime}}}=g|_{\mathbb{R}\times C_{L,L^{\prime}}}$. The map $\tilde{u}^{+}$ gives the required extension $\tilde{f}^{+}$ of $f^{+}$, where the last observation guarantees that $\tilde{f}^{+}|_{\mathfrak{U}}=f|_{\mathfrak{U}}$. ∎