I wanted to construct such a counterexample in the following way:

Take as measurable space $\mathbb{R}^2$ with the product of Lebesgue $\sigma$-algebras in $\mathbb{R}$ and fit a Vitali set in the diagonal (let's call it $V$), then all $x$ and $y$-sections are either singletons or empty sets, hence measurable.
Now, if $V$ itself were measurable then its 45° clockwise rotation would also be measurable, but its 45° clockwise rotation is the Vitali set in the $x$-axis wich can't be meadurable because one of its $y$-sections (namely the one corresponding to $y=0$) is the Vitali set on $\mathbb{R}$, which is not measurable.
In conclusion, $V$ is a non-measurable set in the product $\sigma$-algebra with measurable sections.

Is my argument correct? I'm not sure if the claim "if it's measurable, then its rotation must also be measurable" holds in this case

  • $\begingroup$ $V$ is a subset of the diagonal which is a null set, isn't it? $\endgroup$ – d.k.o. Jun 27 '17 at 15:40
  • $\begingroup$ @d.k.o. yes, with Lebesgue measure on $\mathbb{R}^2$. But the product $\sigma$-algebra here is not the Lebesgue $\sigma$-algebra but the Borel $\sigma$-algebra on $\mathbb{R}^2$, which is not complete with Lebesgue measure, right? $\endgroup$ – la flaca Jun 27 '17 at 15:50
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    $\begingroup$ Your claim is true since every rotation $\Phi$ is a homeomorphism of $\mathbb{R}^2$. You can then show that $\Phi(B)$ is Borel iff $B$ is. $\endgroup$ – Nate Eldredge Jun 27 '17 at 17:31
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    $\begingroup$ You must carefully distinguish between $1$-dimensional and $ 2$-dimensional Lebesgue sets. If $S\subset \mathbb R$ then $S\times \{0\}$ is a subset of the $2$-dimensional Borel set $\mathbb R\times \{0\} ,$ and the $2$-dimensional measure of $\mathbb R\times \{0\}$ is $0.$ So $S\times \{0\}$ is a $2$-dimensional Lebesgue set of measure $0 .$ $\endgroup$ – DanielWainfleet Jun 28 '17 at 0:13
  • $\begingroup$ @DanielWainfleet I understand that but the product of Lebesgue $\sigma$-algebras in $\mathbb{R}$ is not the Lebesgue $\sigma$-algebra in $\mathbb{R}^2$, it is the Borel $\sigma$-algebra in $\mathbb{R}^2$, which is not complete with respect to Lebesgue measure. $\endgroup$ – la flaca Jun 29 '17 at 1:53

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