Show that two Markov kernels almost surely agree 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 $$\operatorname P\left[X_2\in B_2\mid X_1\right]=\kappa(X_1,B_2)\;\;\;\text{almost surely for all }B_2\in\mathcal E_2\tag1$$

Now, assume $\tilde\kappa$ is another Markov kernel with source $(E_1,\mathcal E_1)$ and target $(E_2,\mathcal E_2)$ with $(1)$. Are we able to conclude $$\tilde\kappa(x_1,B_2)=\kappa(x_1,B_2)\;\;\;\text{for }\operatorname P\circ\;X_1^{-1}\text{-almost all }x_1\in E_1\text{ and }B_2\in\mathcal E_2\tag2?$$

I guess we need to assume that $\mathcal E_2$ is countable generated, i.e. there is a $\mathcal G_2\subseteq\mathcal E_2$ with $|\mathcal G_2|\le|\mathbb N|$ and $\sigma(\mathcal G_2)=\mathcal E_2$. Then, by $(1)$, there is a $N\in\mathcal A$ with $\operatorname P[N]=0$ and $$\tilde\kappa(X_1(\omega),B_2)=\kappa(X_1(\omega),B_2)\;\;\;\text{for all }(\omega,B_2)\in(\Omega\setminus N)\times\mathcal G_2\tag3.$$ Let $$\tilde E_1:=\bigcap_{B_2\in\mathcal G_2}\left\{x_1\in E_1:\tilde\kappa(x_1,B_2)=\kappa(x_1,B_2)\right\}.$$ Since $\mathcal G_2$ is countable, $\tilde E_1\in\mathcal E_1$ and $$\operatorname P[X_1\in\tilde E_1]\ge\operatorname P\left[(\Omega\setminus N)\cap\left\{X_1\in\tilde E_1\right\}\right]=1\tag4.$$ Now, we know that a finite measure is uniquely determined by its values on a $\cap$-stable generator. So, the only remaining question is: Do we find a $\cap$-stable $\mathcal G_2$? I guess we can simply go over to $$\tilde{\mathcal G}_2:=\left\{\bigcap\mathcal H_2:\mathcal H_2\subseteq\mathcal G_2\text{ with }|\mathcal H_2|\in\mathbb N\right\}.$$
 A: If $\{C_i\}_{i=1}^{\infty}$ is a sequence of subsets then $\cup_{n=1}^{\infty} \sigma(C_1, ..., C_n)$ is a countably infinite $\pi$-system, see here: https://en.wikipedia.org/wiki/Pi-system
Thus, if a sigma-algebra is countably generated, it can be countably generated via a $\pi$-system. If two measures agree on a $\pi$-system then they agree on the entire sigma-algebra. 
So, if $\mathcal{E}_2$ is countably generated, then there is a countable $\pi$-system $\{B_2[i]\}_{i=1}^{\infty}$ that generates it. 
Fix $i \in \{1, 2, 3, ...\}$ and consider $B_2[i]$. Any two versions of the conditional expectation $E[ 1_{X_2 \in B_2[i]} | X_1]$ agree except for a set of probability measure zero. Define
$$A_i = \{ \omega \in \Omega : \kappa(B_2[i], X_1(\omega)) = \tilde{\kappa}(B_2[i], X_1(\omega))\}$$
Then $P[A_i]=1$ for all $i \in \{1, 2, 3, ...\}$ and so $P[\cap_{i=1}^{\infty}A_i]=1$.  Now $\omega \in \cap_{i=1}^{\infty} A_i$ implies that $\kappa(B_2, X_1(\omega))$ and $\tilde{\kappa}(B_2, X_1(\omega))$ agree on all sets $B_2$ of the $\pi$-system, which means they agree on all sets $B_2$ of the sigma algebra $\mathcal{E}_2$. Thus
$$ \cap_{i=1}^{\infty} A_i = \{\omega \in \Omega : \kappa(B_2, X_1(\omega))=\tilde{\kappa}(B_2, X_1(\omega)) \quad \forall B_2 \in \mathcal{E}_2\}$$
and so the probability of the latter set is 1.
