# If $\text P\left[X_2\in B_2\mid X_1\right]=\kappa(X_1,B_2)$ a.s. for all $B_2$, can we select a common null-set over all $B_2$?

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?

• 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. – Mateusz Kwaśnicki Jan 22 at 0:32
• 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 $\cal{E_2}$. – Dieter Kadelka Jan 22 at 14:54
• 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 – Sinusx Jan 22 at 16:56
• @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. – 0xbadf00d Jan 23 at 18:26
• @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 $x$? – 0xbadf00d Jan 23 at 18:27