Converse for Fubini-Tonelli's theorem By Fubini-Tonelli's theorem, we know that if $E\subset \mathbb{R^{n+m}}$ and $f: \mathbb{R^{n+m}}\to \mathbb{R_{>0}}$ are measurable and $f$ integrable, then the sections $E_x=\{y\in \mathbb{R^m}: (x,y)\in E\}$ and $E_y$ are measurable.
I know that the converse is false, but I don't "see" an example where both sections are measurable but $E$ not (for lebesgue measure).
Any idea? Thanks!
 A: 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). 
