A question about measure theory.

Question

If $A \subseteq \mathbb{R}^2$ is a set such that, if we fix $\bar{y} \in \mathbb{R}$ we have $|\{ (x,\bar{y}) \in A\}|=\omega$, and if we fix $\bar{x} \in \mathbb{R}$ we have $|\mathbb{R} \backslash \{ (\bar{x},y) \in A \}|= \omega.$ If $\chi_A$ is the characteristic funcion of $A$, I think that $\int_0^1 \chi_A \, dx= 0$ because numerable set have measure zero, but $\int_0^1 \chi_A \, dy = 1$, because it's complementar set have measure zero. It is right?

Answer 1

There is a small error in your question: $\chi_A$ is a function with two variables, and you integrate your function just once. I believe the following statement is what you have intended: $\int_0^1 \chi_A(x,y)dx=0$ and $\int_0^1 \chi_A(x,y)dy=1$.

Except for this point, your reasoning is correct. $\chi_A$ is the standard counterexample of a generalization of Fubini's theorem (Fubini's theorem without the measurability of the function over the product space.) It would be interesting that, however, the existence of your $A$ is equivalent to the continuum hypothesis:

Theorem. $A$ exists iff $2^{\aleph_0}=\aleph_1$.

Proof. Assume that $2^{\aleph_0}=\aleph_1$. Let $\prec$ is a well-order of $\mathbb{R}$, whose order-type is $\aleph_1$. Let consider $$A_0 = \{(x,y)\in \mathbb{R}^2 \mid x\prec y\},$$ then $|\{x\mid (x,y)\in A_0\}|\le \aleph_0$ and $|\mathbb{R}\setminus\{y\mid (x,y)\in A_0\}|\le \aleph_0$. Let $A$ be the set which is obtained by the following modification:

• If $\{x: (x,y)\in A_0\}$ is finite, add countably many $(z,y)$, and
• If $\{y: (x,y)\in A_0\}$ is cofinite, remove countably many $(x,z)$.

(We must ensure that adding points does not make co-countable vertical section to co-finite ones. Since there are countably many finite horizontal sections of $A_0$, we may evade this situation by choosing $z$ with no duplication. The same caution is also applied to the removing process.)

On the other hand, assume that your $A$ exists. Assume the contrary that $2^{\aleph_0}\ge \aleph_2$. Let $B\subseteq\mathbb{R}$ be a subset of cardinality $\aleph_1$.

Since every vertical section $A^x=\{y\mid (x,y)\in A\}$ of $A$ is co-countable, the complement of $\bigcap_{x\in B} A^x$ has cardinality $\aleph_1$. Especiallty, if $y\in \bigcap_{x\in B} A^x$, then $B\subseteq A_y=\{x\mid (x,y)\in A\}$. A contradiction.