Proof. It follows from the fact that $L$ is Lagrangian. Indeed, let us lift $L$ to $T_{Y}^{\ast}$. Then locally in a neighborhood of a connected component of $L$, one can find a smooth real function $f=f(y)$ such that $L=df$. We can write the local equation for $L$: $x_{j}^{\vee}=\partial f/\partial y_{j},1\leq j\leq n$. The connection $\nabla_{E}$ can be locally written as $\nabla_{E,0}+id_{E}\otimes(2\pi i/\varepsilon\sum_{j}\partial f/\partial y_{j}dx_{j})$, where $\nabla_{E,0}$ is the trivial flat connection on the vector bundle $E$. Since the holomorphic coordinates on $T_{Y}$ are given by $z_{j}=y_{j}+ix_{j},i=\sqrt{-1}$, one sees that the $(0,2)$-part of the curvature is equal to $curv(\nabla_{E})^{(0,2)}=const\times(\sum_{j,k}\partial^{2}f/\partial y_{j}\partial y_{k}d\bar{z_{j}}d\bar{z_{k}})=0$. The Proposition is proved. $\blacksquare$