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!

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    $\begingroup$ You mean both sections are measurable but $E$ is not? $\endgroup$ – Aman May 6 '16 at 12:23
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    $\begingroup$ Do you mean converse? The converse would be $E_x$ and $E_y$ measurable implies $E$ is measurable. A counterexample would then be $E_x$ and $E_y$ measurable but $E$ not measurable. Also you don't use $f$ anywhere. $\endgroup$ – parsiad May 6 '16 at 12:23
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    $\begingroup$ I think you might be interested in the last part of my answer here: math.stackexchange.com/questions/1030183/… $\endgroup$ – PhoemueX May 6 '16 at 12:32

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).


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