$Y$ is some set. $(X, \Sigma ,\mu)$ is a measure space. $\Psi : X \to Y$.
let $$\Upsilon = \{ B \in \mathscr{P}(Y)\; ; \;\; \Psi^{-1}(B) \in \Sigma \} $$
since the complement of the preimage equals the preimage of the complement and the preimage of a union equals the union of the preimages then $\Upsilon $ is closed under both complementation and countable union.
however since $\Psi$ is an arbitrary mapping and it isn't specified that $\Sigma = \mathscr{P} (X)$ I think $\Psi^{-1}(Y) \text{ isn't guaranteed to be in } \Sigma $
so is there some sort of trick to show that ? or does the problem lack of information ?
Since $X$ is the domain of the function $\Psi$ with $Y$ as its codomain, we have that $\Psi^{-1}(Y)=\{x\in X\ |\ \Psi(x)\in Y\}=X\in \Sigma$, because $\Sigma$ is a $\sigma$-algebra.