If X is $\overline{\sigma(Y)}$-measureable, then X is already $\sigma(Y)$-measureable a.s.? Let $(\Omega,\mathcal{F},\mathbb{P})$ be a complete probability space, $X$ an $E$-valued r.v. and $Y$ an $F$-valued r.v. for polish spaces $E,F$. Is the following assertion true?
If $X$ $\overline{\sigma(Y)}$-measureable, then $X$ is $\sigma(Y)$-measureable almost surely.
By $\overline{\sigma(Y)}$ we denote the completed sigma algebra. I think this is used in my script in the context of strong solutions for SDE's(see my last question). I would be very grateful for hints or a solution.
 A: 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
$\mathbb P\big\{\omega\in \Omega \mid X(\omega) \ne X_1(\omega)\big\} = 0$.
