Sufficient condition for the graph of a measurable map to be measurable

Question

Let $f:X \to Y$ be measurable map between measurable spaces w.r.t. to their corresponding $\sigma$-algebras $\Sigma_X$ and $\Sigma_Y$, resp.

If $(X,\Sigma_X)$ is a standard Borel space can we always conclude that the graph: $$ \Gamma(f) := \{ (x,y) \in X \times Y \mid y=f(x) \} $$ is measurable, i.e. an element of the product $\sigma$-algebra $\Sigma_X \otimes \Sigma_Y$?

Note that I'm interested in the case where $(Y,\Sigma_Y)$ is a general measurable space and not restricted to e.g. being countably separated, which would imply the claim (even for general $(X,\Sigma_X)$).

Edit: However, I'm willing to impose weaker conditions on $(Y,\Sigma_Y)$ than being countably separated, e.g. $(Y,\Sigma_Y)$ being allowed to be only separated or $\Sigma_Y$ containing all singletons $\{y\}$ for $y \in Y$.

The reason is the following simple counter example: $X:=Y:=\{1,2\}$ with $$\Sigma_X:=\{\emptyset,X, \{1\}, \{2\}\}, \qquad \Sigma_Y:=\{\emptyset,Y\} $$ and $f:=\mathrm{id}$. Then we have: $$ \Sigma_X\otimes\Sigma_Y = \{ \emptyset, X \times Y, \{1\} \times Y, \{2\} \times Y \}, $$ but $$\Gamma(f) =\{ (1,1), (2,2) \} \notin \Sigma_X\otimes\Sigma_Y. $$

Answer 1

The assumption that $Y$ is countably separated cannot be meaningfully weakened. The following is Proposition 2.1 of [Musial, Kazimierz. "Projective limits of perfect measure spaces." Fund. Math 110.163-188 (1980).]:

Let $(X,\mathcal{X})$ and $(Y,\mathcal{Y})$ be measurable spaces and $f:X\to Y$ a measurable function. Then the graph of $f$ is measurable in the product $\sigma$-algebra if and only if there exists a countably generated sub-$\sigma$-algebra of $\mathcal{Y}$ that contains all points of $f(X)$.