Let $(\Omega_1,\Sigma_1,\mu_1)$ and $(\Omega_2,\Sigma_2,\mu_2)$ be two totally finite measure spaces (which implies that $\Sigma_1$ and $\Sigma_2$ are $\sigma$-algebras).

(As usual $\Sigma_1\times\Sigma_2$ will mean the $\sigma$-algebra generated by the set $\{ E \times F \,|\, E \in \Sigma_1 \textrm{ and } F \in \Sigma_2\}$)

Let $A\subseteq \Omega_1\times \Omega_2$ such that:

  1. for almost all $x\in \Omega_1$, $A_x=\{y\in\Omega_2 \,|\, (x,y)\in A \} \in \Sigma_2$ and the function $x \mapsto \mu_2(A_x)$ is $\Sigma_1$-measurable.

  2. for almost all $y\in \Omega_2$, $A^y=\{x\in\Omega_1 \,|\, (x,y)\in A \} \in \Sigma_1$ and the function $y \mapsto \mu_1(A^y)$ is $\Sigma_2$-measurable.

  3. $\int_{\Omega_1} \mu_2(A_x) d\mu_1 = \int_{\Omega_2} \mu_1(A^y) d\mu_2 > 0$

(Note that $A$ is not supposed to be $\Sigma_1\times\Sigma_2$- measurable)

Can we prove that there is $C\subseteq A$ such that $C$ is $\Sigma_1\times\Sigma_2$- measurable and $\mu_1\times\mu_2(C) > 0$ ?


You can not conclude that. We don't need to use the Continuum Hypothesis to show a counterexample. Here is a simple counterexample.

We will use "countable" to mean "finite or countably infinite"

Given $E \subseteq [0,1]$, we say that $E$ is full if $[0,1]-E$ is countable.

Let $\Sigma$ be the $\sigma$-algebra defined as: $\Sigma=\{ E \subseteq [0,1] \,|\, E$ is countable or full $\}$

Let $\mu$ be a the measure defined on $\Sigma$ by:

$\mu(E)=0$ if $E$ is countable;

$\mu(E)=1$ if $E$ is full.

Now, let $(\Omega_1,\Sigma_1,\mu_1)=(\Omega_2,\Sigma_2,\mu_2)=([0,1],\Sigma,\mu)$.

Let $A$ be the set $([0,1]\times [0,1]) -\{(x,x) \,|\,x\in [0,1]\}$

It is easy to see that $A$ satisfies the three conditions.

On the other hand, any set $C$ in the $\sigma$-algebra $\Sigma_1\times\Sigma_2$ is in one of the two cases below:

Case 1. $C$ is a countable set of horizontal and / or vertical "lines", each of such "lines" being either full or countable. In this case, $\mu_1\times\mu_2(C)=0$.

Case 2. $C$ is the complement (in $[0,1]\times [0,1]$) of a set in case 1, let us call such set $B$. It is easy to see that $B\cap \{(x,x) \,|\,x\in [0,1]\}$ is countable, so $\{(x,x) \,|\,x\in [0,1]\}\nsubseteq B$ and so we have that $C=[0,1]^2-B \nsubseteq A$.

So there is no $C\subseteq A$ such that $C$ is $\Sigma_1\times\Sigma_2$- measurable and $\mu_1\times\mu_2(C)>0$.

  • $\begingroup$ Thanks. Nice counterexample. It is really simpler, since it does not use the Sierpinski set (the set from Rudin 7.9c and from Halmos 31.11). $\endgroup$ – user253909 Jul 12 '15 at 17:17

No, you can't conclude that.

I'm going to show that the Continuum Hypothesis implies that there exists a counterexample. Note that I'm not claiming to have a proof that a counterexample exists! I don't see any reason to believe CH is true.

So. CH implies that a counterexample exists. If we don't believe in CH we don't have to believe in that counterexample. The funny thing is that nonetheless, whether we believe in CH or not, the fact that CH gives a counterexample shows that you cannot prove your result is true! Because that would prove that CH was false, and it's known that you cannot prove CH is false.

The example is going to be a slight modification of an example that Rudin (Real and Complex Analysis) says is due to Steinhaus. We give that other example first:

Both measure spaces are going to be $[0,1]$ with Lebesgue measure. CH shows that there exists an order on $[0,1]$ such that each element has only countably many predecessors. We're going to use "$\le$" to denote this (non-standard) order.

Let $$B=\{(x,y)\in[0,1]^2\,:\,x\le y\}.$$Now we each $B^y$ is countable, hence $m(B^y)=0$ for every $y\in[0,1]$. So Fubini shows that if $C\subset B$ and $C$ is measurable then $m^2(C)=0$.

And $[0,1]\setminus B_x$ is countable for every $x$, hence $m(B_x)=1$ for every $x\in[0,1]$.

Now for the counterexample to your question. This time both measure spaces are going to be $[0,2]$ with Lebesgue measure. Define $B\subset[0,1]^2$ as above. Let $$A=B\cup\{(y+1,x+1)\,:\,(x,y)\in B\}.$$

None of the four sets $A\cap (I\times J)$ ($I,J=[0,1],[1,2]$) contains a measurable subset of positive measure; hence $A$ does not. But both of your iterated integrals equal $1$.

  • $\begingroup$ Thanks. Yes, the set you mentioned from Rudin can also be found in Halmos (31.11). Combining one copy of such set with another copy rotated by 90 degrees, we can build a set A satisfy the hypothesis of the question, but containing no measurable subset of positive measure. $\endgroup$ – user253909 Jul 11 '15 at 18:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy