Does there always exist a random variable whose conditional distribution is a given distribution? Let $\{P_y|y\in \mathbb{R}^m\}$ be a family of Borel probability measures on $\mathbb{R}^n$ and $\mu$ be a Borel probability measure on $\mathbb{R}^m$
Then, does there exist random vectors $X,Y$ such that $P(X\in \cdot|Y=y)\sim P_y$ for almost every $y$ with respect to $Y_*P$, and $Y\sim \mu$?
Note that $$\int_{\mathbb{R}^m} P_y(A) d\mu(y)= \int_{\mathbb{R}^m} P_y(A) d(Y_*P)(y) =\int_{\mathbb{R}^m} P(X\in A|Y=y)d(Y_*P)(y)= (X_*P)(A)$$
So, we know what distribution $X$ follows. Now, to prove the existence of a pair of $(X,Y)$, we now need to find a suitable pair such that $P(X\in \cdot|Y=y)\sim P_y$, but how..?
 A: I do not understand why most probability text books do not mention that there is a one-to-one correspondence between conditional distributions and Markov kernels. So, studying conditional distributions is equivalent to studying Markov kernels.
Firstly, note that when a pair of random vectors are given (taking values in polish spaces), there exists a regular conditional distribution of them, which is a Markov kernel. This is a fundamental result given in most textbooks.
Now we prove the converse, which is the question I posted here.
Let $K:\mathbb{R}^m\times \mathscr{B}_{\mathbb{R}^n}\rightarrow  [0,\infty]$ be a Markov kernel and $\mu$ be a Borel probability measure on $\mathbb{R}^m$. (Euclidean spaces can be replaced by an arbitrary measurable spaces)
The question is asking whether there exists a pair $(X,Y)$ such that $P(X\in B|Y=y)=K(y,B)$ and $Y_*P=\mu$, and the answer is yes.
Define $P(A\times B):=\int_A K^*(\mathbb{1}_B) d\mu$. By Caratheodory extension theorem, we can extend $P$ to the unique (Borel) probability measure on $\mathbb{R}^m\times \mathbb{R}^n$.
Let $\pi_1,\pi_2$ be the projection maps and denote these by $Y,X$, respectively. Of course, these are random vectors.
Note that $Y_*P(A)=\int_A 1 d\mu = \mu(A)$.
Also, we have $$\int_{Y^{-1}(A)} K^B\circ Y dP= \int_A K^B d\mu = P(A\times B) = P(Y^{-1}(A)\cap X^{-1}(B)) = \int_{Y^{-1}(A)} \mathbb{1}_B \circ X dP$$.
Hence, $P(X\in B|Y=y)=K(y,B)$.
