Regular conditional distribution of $Y$ given $X=x$ in Klenke's book

Question

In his book "Probability theory", Klenke uses the following definition of transition kernel:

and if in $ii)$ the measure is a probability measure for all $\omega_1$ then $K$ is called a stochastic kernel.

Later on he defines regular conditional distribution of $Y$ given $\mathcal F$ as follows:

After the last line in the image above, he added "(the function from the factorization lemma with an arbitrary value for $x \notin X(\Omega)$) is called a regular conditional distribution of $Y$ given X.", which perplexes me a lot.

My attempt to understand this line:

First, I think he forgot to assume that the map $ \omega \mapsto K_{Y|\sigma(X)}(\omega, B)$ should be $\mathcal F$-measurable for any fixed $B \in \mathcal E$.

Next, I assume the previous statement holds and fix one $B \in \mathcal E$. Since $K_{Y|\sigma(X)}(\cdot, B)$ is a version of $ P(Y \in B | \sigma(X)) $ and is $\sigma(X)$-measurable too, by factorization theorem, there is a measurable function $\kappa(x, B)$ from $(E',\mathcal E')$ to $\mathbb R$ such that $\kappa(X(\omega), B) = K_{Y|\sigma(X)}(\omega, B) $ for all $\omega$. This shows, for every fixed $x \in X(\Omega))$, $B \mapsto \kappa(x,B)$ is a measure since it equals to $ K_{Y|\sigma(X)}(\omega_x, B) $ for any $B \in \mathcal E$ for some $\omega_x \in X^{-1}(x)$.

We still have to check if it is a measure for a fixed $x \notin X(\Omega)$. If I understand correctly his last line, we need to modify the map $x \mapsto \kappa(x,B)$ in the following way. Fix one probability measure $\mu$ on $\mathcal E$, for any fixed $B$, define

$$ \forall x \in E', \qquad \kappa'(x,B) := \kappa(x,B) 1_{x \in X(\Omega)} + \mu(B)1_{x \notin X(\Omega)}. $$

Now we can easily check that for any fixed $x \in E'$, $B \mapsto \kappa'(x,B)$ is a probability measure. HOWEVER, measurability is now an issue! Unless $X(\Omega) \in \mathcal E'$, which rarely happens, then the map $x \mapsto \kappa'(x,B)$ is likely not $\mathcal E'$ measurable.

I think he missed this in this book. Is there anyway to resolve this issue?

My attempt to fix this issue: Suppose there is a set $N \in \mathcal E'$ such that $P^X(N)=1$ ($P^X$ is the distribution of $X$ on $(E',\mathcal E')$) and that $ N \subset X(\Omega)$ then we can define

$$ \forall x \in E', \qquad \kappa'(x,B) := \kappa(x,B) 1_{x \in N} + \mu(B)1_{x \notin N}. $$

The measurability is now resolved and it still holds that

\begin{align} \int_{E} \kappa'(x,B) P^X(dx) &= \int_{N} \kappa(x,B) P^X(dx) \\\\ &= \int_{E} \kappa(x,B) P^X(dx) \\\\ &= \int_{\Omega} \kappa(X(\omega),B) P(d\omega) \\\\ &= \int_{\Omega} K_{Y|\sigma(X)}(\omega,B) P(d\omega) \\\\ &= \int_{\Omega} P(Y \in B | \sigma(X)) P(d\omega) \\\\ &= P(Y \in B) \end{align} and so this new $\kappa'(\cdot,\cdot)$ is indeed a stochastic kernel from $(E',\mathcal E')$ to $(E, \mathcal E)$. However, I believe such a set $N$ does not always exists....

My question is: Is there a way to fix this issue? or is there other approach to this regular conditional distribution?

I borrow the images from this post. And consulted several related questions such as this, or this (Fabio forgot the check measurability here!), and this one too. However, they give no answers to my question.

Thank you for your suggestions and comments.

Answer 1

I also think his construction of $\kappa_{Y,X}(x,A)$ is a bit strange.

Instead of making sense of the strange composition $\kappa_{Y,\sigma(X)}(X^{-1}(\omega),A)$, I would declare $\kappa_{Y,X}$ as the stochastic kernel from $(E', \mathcal{E}')$ to $(E, \mathcal{E})$ such that

$$ \int_{\Omega} \mathbf{1}_B(Y(\omega)) \mathbf{1}_A(X(\omega)) \, \mathbf{P}(\mathrm{d}\omega) = \int_{\mathcal{E}'} \kappa'_{Y,X}(x, B) \mathbf{1}_A(x) \, \mathbf{P}(X \in \mathrm{d}x) \tag{1}\label{e:def} $$

for all $A \in \mathcal{E}'$ and $B \in \mathcal{E}$, or equivalently,

$$ \mathbf{P}(Y \in B, X \in A) = \int_{\mathcal{E}'} \mathbf{P}(Y \in B \mid X = x) \mathbf{1}_A(x) \, \mathbf{P}(X \in \mathrm{d}x) $$

where the notation $\kappa'_{Y,X}(x, B) = \mathbf{P}(Y \in B \mid X = x)$ is adopted for better readability and $\mathbf{P}(X \in \cdot) = (\mathbf{P}\circ X^{-1})(\cdot)$ is the distribution of $X$.

If the kernel $\kappa'_{Y,X}$ as defined above exists, then $\kappa_{Y,\sigma(X)}$ also exists and satisfies

$$ \kappa'_{Y,X}(X(\omega), B) = \kappa_{Y,\sigma(X)}(\omega, B) $$

holds for $\mathbf{P}$-a.e. $\omega$ for all $B \in \mathcal{E}$. Indeed, this follows from the change of variables formula applied to the right-hand side of $\eqref{e:def}$:

$$ \int_{\mathcal{E}'} \kappa'_{Y,X}(x, B) \mathbf{1}_A(x) \, \mathbf{P}(X \in \mathrm{d}x) = \int_{\Omega} \kappa'_{Y,X}(X(\omega), B) \mathbf{1}_A(X(\omega)) \, \mathbf{P}(\mathrm{d}\omega). $$

Answer 2

Amazingly, in an attempt to find a rectification, I tumbled on the book "Conditional measures and applications" by M. M. Rao. It turns out that my approach to fix the problem with an $N \in E'$ such that $P(X\in N) = 1$ and $N \subset X(\Omega)$ is called perfectness (of the measure space $(\Omega, \mathcal A)$) in the literature (which I know nothing about before). See page 133 of the book mentioned above.

Also the assumption $X(\Omega)$ is measurable turns out to be a crucial assumption in Doob's theorem too, see page 129 of the same book.

Conclusion: there is no way to fix the issue unless we make these assumptions!