Edit: I have just figured a much easier way. So, I edited the answer.
Let $\mathcal{V} = \{V_n : n = 1, 2, \dotsc\}$ be a countable base for the topology of $Y$.
For each $V_n$, choose a negligible $E_n \subset X$ such that $f^{-1}(V_n) \setminus E_n \in \mathcal{F}$. It may happen that $\bigcup E_n \not \in \mathcal{F}$. But since it is a negligible set, there is a negligible $Z \in \mathcal{F}$ such that $\bigcup E_n \subset Z$.
Fix some $y \in Y$, and then define $ g(x) = \left\{ \begin{array}{} f(x), & x \not \in Z \\ y, & x \in Z \end{array} \right. $
Notice that for any $V_n \in \mathcal{V}$, if $y \not \in V_n$, $ \begin{align*} g^{-1}(V_n) &= f^{-1}(V_n) \setminus Z \\&= (f^{-1}(V_n) \setminus E_n) \setminus Z \in \mathcal{F}. \end{align*} $ And if $y \in V_n$, $ \begin{align*} g^{-1}(V_n) &= f^{-1}(V_n) \cup Z \\&= (f^{-1}(V_n) \setminus E_n) \cup Z \in \mathcal{F}. \end{align*} $ That is, $g^{-1}(\mathcal{V}) \subset \mathcal{F}$. All open sets of $Y$ are (countable) union of elements in $\mathcal{V}$. Therefore, $\mathcal{V}$ generates the $\sigma$-algebra of Borel sets $\mathcal{B}$. And so, $g$ is $\mathcal{F}$-measurable. In fact, $ g^{-1}(\mathcal{B}) = g^{-1}(\sigma(\mathcal{V})) = \sigma \left(g^{-1}(\mathcal{V})\right) \subset \mathcal{F}. $
Since it is evident that $g$ and $f$ are equal almost everywhere, the proof is complete.