Let

- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $(E_i,\mathcal E_i)$ be a measurable space
- $X_1:\Omega\to E_1$
- $X_2:\Omega\to E_2$ be $(\mathcal A,\mathcal E_2)$-measurable
- $\kappa$ be a Markov kernel with source $(E_1,\mathcal E_1)$ and target $(E_2,\mathcal E_2)$ with $$\operatorname P\left[X_2\in B_2\mid X_1\right]=\kappa(X_1,B_2)\;\;\;\text{almost surely for all }B_2\in\mathcal E_2\tag1$$

If $\mathcal E_2$ is countably generated, are we able to conclude that $$\operatorname P\left[X_2\in B_2\mid X_1\right]=\kappa(X_1,B_2)\;\;\;\text{for all }B_2\in\mathcal E_2\text{ almost surely?}\tag2$$

Since $\mathcal E_2$ is countably generated, there is a $\cap$-stable $\mathcal G_2\subseteq\mathcal E_2$ with $|\mathcal G_2|\le|\mathbb N|$ and $$\operatorname P\left[X_2\in B_2\mid X_1\right]=\kappa(X_1,B_2)\;\;\;\text{for all }B_2\in\mathcal G_2\text{ almost surely}.\tag3$$ Does this suffice to conclude?

define$Y_{B_2}$ to be equal to $Z_{B_2}$ everywhere. $\endgroup$ – Mateusz Kwaśnicki Jan 22 at 0:32