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.


1 Answer 1


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.

  • $\begingroup$ (Minor typo in condition 2: surely you mean $A \subseteq \Omega$.) $\endgroup$
    – Nik Weaver
    Dec 1, 2014 at 4:59
  • $\begingroup$ Isn't it just the axiom of choice which excludes a "Lebesgue measure " on the power set of $[0,1]$? $\endgroup$ Dec 1, 2014 at 7:49
  • $\begingroup$ @Jochen: no, the axiom of choice excludes a translation-invariant measure on [0,1]. $\endgroup$
    – Nik Weaver
    Dec 1, 2014 at 13:06
  • $\begingroup$ I refer to a theorem of Banach and Kuratowski. $\endgroup$
    – Nik Weaver
    Dec 1, 2014 at 13:07
  • $\begingroup$ @NikWeaver, thank you for your answer - it is very helpful. While your idea seems obvious when written, I spent quite some time trying to get to an argument like this. Thank you again for your valuable help. $\endgroup$
    – Andrei K
    Dec 1, 2014 at 15:27

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