We have two probability spaces $(\Omega\_1,\mathcal{F\_1},P\_1)$ and $(\Omega\_2,\mathcal{F\_2},P\_2)$. Is it possible to construct probability space $(\Omega=\Omega\_1\times\Omega\_2,\mathcal{F},P)$ such that:


1. $\forall A\_1\in\mathcal{F\_1}, A\_2\in\mathcal{F\_2}$ set $A\_1\times A\_2\in\mathcal{F}$ and its measure equals $P\_1(A\_1)P\_2(A\_2)$
2. $\mathcal{F}$ contains all sets $A\subseteq\Omega$ such that $\forall \omega\_1\ \{\omega\_2:(\omega\_1,\omega\_2)\in A\}\in\mathcal{F\_2}$ and $P\_2(\{\omega\_2:(\omega\_1,\omega\_2)\in A\})$ is constant for all $\omega\_1$
3. $P$-measure of such sets from (2) equals $P\_2(\{\omega\_2:(\omega\_1,\omega\_2)\in A\})$ (does not depend on $\omega\_1)$


In other words, how to add to ($\Omega\_1\times\Omega\_2,\mathcal{F\_1}\times\mathcal{F\_2},P\_1P\_2)$ all sets from $\Omega$ with constant cross-$\Omega\_2$ measure?


One can assume $\Omega\_1=[0,1]$ with Lebesgue measure and $\Omega\_2=[0,1]^\Gamma$ (infinite-dimensional cube). 


This is needed for my research in Economics. Any advice on this problem will be greatly appreciated. Links to papers would be the best help. I am struggling with this question for several months already. Please, share any thoughts.



Extend product sigma-algebra to cross-constant sets
Consider the case $\Omega\_1 = \Omega\_2 = [0,1]$, equipping both factors with Lebesgue measure. Let $S$ be *any* subset of $[0,1]$. Then the set $$(S \times [0,1/2]) \cup (S^c \times (1/2,1])$$ is "cross-constant" in your sense --- each vertical slice is measurable and has measure $1/2$. So all such sets would belong to $\mathcal{F}$. But also $[0,1]\times [0,1/2]$ belongs to $\mathcal{F}$, and taking intersections with sets of the previous form yields $S \times [0,1/2] \in \mathcal{F}$. For every $S \subseteq [0,1]$.


Defining $\mu(S) = 2\cdot P(S \times [0,1/2])$ for all $S \subseteq [0,1]$ would give us a countably additive extension of Lebesgue measure to all subsets of $[0,1]$. Whether such a thing exists is set-theoretic. I believe the continuum hypothesis implies that such $\mu$ cannot exist. In any case, there is certainly no "natural" solution to the problem you pose.


