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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?

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  • $\begingroup$ I think the question is ill-posed: there is an uncountable family of functions $Y_{B_2} := P[X_2 \in B_2 | X_1]$, each member of which is defined up to a null set and equal a.s. to $Z_{B_2} = \kappa(X_1, B_2)$. There's no reason to think that $Y_{B_2} = Z_{B_2}$ outside a common null set, but at the same time it is perfectly OK to define $Y_{B_2}$ to be equal to $Z_{B_2}$ everywhere. $\endgroup$ – Mateusz Kwaśnicki Jan 22 at 0:32
  • $\begingroup$ Sorry, I cannot see the difference between (1) and (2). I think that (3) implies (2). Fix $\cal{G}$. Then $\kappa(X_1,B_2)$ are representants of the equivalence class $P[X_2 \in B_2 | X_1]$ for $B_2 \in \cal{G}$, hence for all $B_2$ in the Dynkin system generated by $\cal{G}$. This is just $\cal{E_2}$. $\endgroup$ – Dieter Kadelka Jan 22 at 14:54
  • $\begingroup$ Do you mean in (2) 'almost surely for all $B_2$'? As it is written, it could be understood that (2) is the same as (1), as was also noted by @DieterKadelka $\endgroup$ – Sinusx Jan 22 at 16:56
  • $\begingroup$ @DieterKadelka $(1)$ means that for any fixed $B$ there is a null set such that the identity holds. $(2)$ means that there is a null set such that the identity holds for all $B$. It's clear that $(3)$ implies $(1)$ and that's what your argument yields. $\endgroup$ – 0xbadf00d Jan 23 at 18:26
  • $\begingroup$ @Sinusx Don't we usually write (a) property $p(x)$ holds almost surely for all $x$, when we mean that for any fixed $x$ there is a null set such that $p(x)$ holds and (b) property $p(x)$ holds for all $x$ almost surely, when we mean that there is a null set such that $p(x)$ holds for all $x$? $\endgroup$ – 0xbadf00d Jan 23 at 18:27

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