If X is $\overline{\sigma(Y)}$-measureable, then X is already $\sigma(Y)$-measureable a.s.? [closed]

Not a solution. A careful restatement of the problem. I think.

Let $(\Omega,\mathcal F, \mathbb P)$ be a probability space, let $E$ and $F$ be Polish spaces.

Let $X : \Omega \to E$ be $\big(\mathcal F, \mathcal B(E)\big)$-measurable [where $\mathcal B(E)$ is the sigma-algebra of Borel sets in $E$].
Let $Y : \Omega \to F$ be $\big(\mathcal F, \mathcal B(F)\big)$-measurable.
Write $$ \mathcal G = \sigma(Y) = \big\{Y^{-1}(U) \mid U \in \mathcal B(F)\big\} $$ The $\mathbb P$-completion of $\mathcal G$ is $$ \overline{\mathcal G} = \big\{K \subseteq \Omega \mid \exists A,B \in \mathcal G, A \subseteq K \subseteq B, \mathbb P(B\setminus A) = 0 \big\} . $$
Assume $\mathcal F$ is $\mathbb P$-complete. That is, $\overline{\mathcal F} = \mathcal F$.
Assume $X$ is also $\big(\overline{\mathcal G},\mathcal B(E)\big)$-measurable.
To prove: There exists $X_1 : \Omega \to E$ such that $X_1$ is $\big(\mathcal G,\mathcal B(E)\big)$-measurable and $\mathbb P\big\{\omega\in \Omega \mid X(\omega) \ne X_1(\omega)\big\} = 0$.