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Suppose we have a collection $(A_r)_{r\in\mathbb{R}}$ of Lebesgue subsets of $\mathbb{R}$, each with Lebesgue measure $0$. Consider the set $$E=\bigcup_{r\in\mathbb{R}}\{r\}\times A_r\subset\mathbb{R}^2.$$ Is $E$ necessarily a Lebesgue subset of $\mathbb{R}^2$ and, if so, does it have measure $0$?

I was wondering this today, but didn't really get anywhere. Perhaps a counterexample could be constructed by considering an $\mathbb{R}^2$ analogue of the fat Cantor set?

Thank you.

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    $\begingroup$ If $E$ is measurable, it has measure $0$ by Fubini. But it need not be measurable, even in the special case that $A_r$ is a singleton for each $r$. See for example this question and answer: math.stackexchange.com/questions/35606 $\endgroup$
    – user83827
    Commented Mar 29, 2012 at 13:38
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    $\begingroup$ Counterexamples are also given in Gelbaum and Olmstead's Counterexamples in Analysis. See here. $\endgroup$ Commented Mar 29, 2012 at 14:03
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    $\begingroup$ A rather extreme counterexample can be found in Krzystof Ciesielski's Set Theory for the Working Mathematician (Theorem 6.1.8): CH is equivalent (assuming ZFC) to the existence of a set $A\subset \mathbb R^2$ with $\vert A \cap \mathbb R \times \{y\}\vert < \omega$ and $\vert (\mathbb R^2 \setminus A) \cap \{x\}\times \mathbb R \vert \leq \omega$ for all $x,y \in \mathbb R$. $\endgroup$ Commented Mar 29, 2012 at 14:23

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If $E$ is measurable, it has measure $0$ by Fubini. But it need not be measurable, even in the special case that $A_r$ is a singleton for each $r$. See for example this question and answer: Lebesgue Measure of the Graph of a Function – ccc Mar 29 at 13:38

Counterexamples are also given in Gelbaum and Olmstead's Counterexamples in Analysis. See here. – David Mitra Mar 29 at 14:03

A rather extreme counterexample can be found in Krzystof Ciesielski's Set Theory for the Working Mathematician (Theorem 6.1.8): CH is equivalent (assuming ZFC) to the existence of a set $A\subset \mathbb R^2$ with $\vert A \cap \mathbb R \times \{y\}\vert < \omega$ and $\vert (\mathbb R^2 \setminus A) \cap \{x\}\times \mathbb R \vert \leq \omega$ for all $x,y\in\mathbb R$. – Alexander Thumm Mar 29 at 14:23

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    $\begingroup$ I merged the answers-in-comments into a CW answer in order to remove this question from Unanswered. $\endgroup$
    – user31373
    Commented Jun 16, 2012 at 18:53

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