# 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 $$\tilde{\mathcal G}_2:=\left\{\bigcap\mathcal H_2:\mathcal H_2\subseteq\mathcal G_2\text{ with }|\mathcal H_2|\in\mathbb N\right\}.$$

• If you have a countable numer of sets, the union of sigma algebras over all finite collections of the sets is a pi-system that is also countable. – Michael Jan 25 at 3:13
• @Michael Do you think that anything is wrong with my $\tilde{\mathcal G_2}$? This should be a countable $\cap$-stable system wtih $\sigma(\tilde{\mathcal G_2})=\sigma(\mathcal G_2)$ too. – 0xbadf00d Jan 29 at 12:59
• No but I only see your last comment in retrospect, I assume what you mean by $\cap$-stable is what I mean by $\pi$-system. I had written a comment that we seem to be doing similar things but I deleted it since I take a closer look and your $\pi$ system seems a bit different from mine. I still wonder then why you are aksing this question if you feel you have already answered it. I also wonder what you mean by equation (1) because it looks like $P[X_2 \in B_2 | X_1]$ is just a third kernel, in which case (1) already says that any two kernels differ by a set of measure 0. – Michael Jan 30 at 4:39
• Anyway since I did not quite know what your question was, or what (1) meant, I did not initially follow your discussion after equation (2) and I interpeted your question to mean you have two kernels with properties as in my answer. I suspect I interpreted correctly. – Michael Jan 30 at 4:43

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.