Converse for Fubini-Tonelli's theorem

Solution 1:

Here is a rather "simple" counter-example using Borel (or Lebesgue) $\sigma$-algebra and the Lebesgue measure $m$. Consider $[0,1]$ and let $\omega_1$ be the first uncountable ordinal. Assuming the Continuum Hypothesis, there is a bijection $o:[0,1] \to \omega_1$.

Define $E=\{(x,y)\in [0,1]^2 : o(x)<o(y)\}$. Then, for all $x\in [0,1]$, $$E_x = \{y\in [0,1] : o(x)<o(y)\} =[0,1] - \{y\in [0,1] : o(y)\leqslant o(x)\}$$ so $E_x$ is the complement (in $[0,1]$) of a countable subset. So $E_x$ is measurable and $m(E_x)=1$.

Now, for all $y\in [0,1]$, $$E_y = \{x\in [0,1] : o(x)<o(y)\}$$ so $E_y$ is a countable subset of $[0,1]$. So $E_y$ is measurable and $m(E_y)=0$.

Claim: $E$ is not measurable

Let us prove it by contradiction. Suppose that $E$ is measurable. Then $\chi_E$ is a non-negative measurable function and, by Tonelli's theorem, we have

$$ 0= \int_0^1 \left(\int_0^1 \chi_E(x,y) dx \right ) dy = \int_0^1 \left(\int_0^1 \chi_E(x,y) dy \right ) dx =1$$ Contradiction.

Remark: The set $E$ above is a Sierpinski set.

Remark 2: I assume you was looking for a counter-example using Borel (or Lebesgue) $\sigma$-algebra. Otherwise, there are simpler counter-examples.

Remark 3: You wrote: "By Fubini-Tonelli's theorem, we know that if $E\subset \mathbb{R^{n+m}}$ is measurable
then the sections $E_x=\{y\in \mathbb{R^m}: (x,y)\in E\}$ and $E_y$ are measurable".

This is not true if you are considering the Lebesgue $\sigma$-algebra. Moreover, in this case, Fubini-Tonelli's theorem implies only that $E_x$ and $E_y$ are measurable for almost every $x$ and almost every $y$ (respectively).