# Does a product measure on a product space constructed from two sub-fields of the same space determine a measure on the underlying space?

Let $\mathcal{A}_1,\mathcal{A}_2$ be $\sigma$-algebras on $\Omega$. Let $P$ be a probability on $\mathcal{A}_1$ and let $Q$ be a Markov kernel from $\mathcal{A}_1$ to $\mathcal{A}_2$. Set $K:=P\otimes Q$, so that $K$ is a probability measure on $\mathcal{A}_1\otimes\mathcal{A}_2$. Does there exist a probability measure $K'$ on $\sigma\left(\mathcal{A}_1,\mathcal{A}_2\right)$, that satisfies $$K'\left(D\cap E\right)=K\left(D\times E\right)$$ for all $D\in\mathcal{A}_1$, $E\in\mathcal{A}_2$?

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I've modified a text a bit (hopefully clarifying) if you don't mind. – Ilya May 23 '13 at 7:32
What are your thoughts? First, forget about the structure of $K$ and think of the following. You define $K':\mathscr A\to [0,1]$ for any $D\cap E$ where $D\in \mathscr A_1$, $E\in \mathscr A_2$ and $\mathscr A := \sigma(\mathscr A_1, \mathscr A_2)$. What are the properties of the collection $\{D\cap E\}$, e.g. is it a $\pi$-system (closed under intersections?). Furthermore, $$K'(D\cap E) = \int_D Q(x,E)P(\mathrm dx).$$ Did you check whether it satisfies necessary conditions of being a measure? – Ilya May 23 '13 at 7:35
@Ilya: Firstly i'd have to check consistency, namely that if $D_1,D_2\in\mathcal{A}_1$ and $E_1,E_2\in\mathcal{A}_2$ with $D_1\cap E_2=D_2\cap E_2$, $K\left(D_1\times E_1\right)=K\left(D_2\times E_2\right)$. I don't know how to do that. – Evan Aad May 23 '13 at 7:39
try to do this for a couple of cases when $\Omega$ is finite, say it has $3$ elements. – Ilya May 23 '13 at 7:43

The answer is: no. $K'$ is not even necessarily consistent.

Set $\Omega:=\left\{0,1\right\}$, and $\mathcal{A}_1:=\mathcal{A}_2:=\mathcal{P}\Omega$.

Define

$$P(0):=P(1):=\frac{1}{2}$$

$$Q\left(\omega,B\right):=P(B)$$

Then $K=P^2$.

Set $D_1:=D_2:=E_1:=\left\{1\right\}$, $E_2:=\Omega$.

Then $D_1\cap E_1=D_2\cap E_2$, but

$$K\left(D_1\times E_1\right)=\frac{1}{4}\neq\frac{1}{2}=K\left(D_2\times E_2\right)$$

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Well-done :) ${}$ – Ilya May 23 '13 at 8:19
@Ilya: Thank you very much! – Evan Aad May 23 '13 at 8:22