A question about measure theory.

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 $$A_0 = {(x,y)in mathbb{R}^2 mid xprec y},$$ then $|{xmid (x,y)in A_0}|le aleph_0$ and $|mathbb{R}setminus{ymid (x,y)in A_0}|le aleph_0$. 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 $Bsubseteqmathbb{R}$ be a subset of cardinality $aleph_1$.

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