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How to construct a set $E \subseteq \mathbb{R}^n$ satisfying the following two conditions:

(i) $E$ is Lebesgue measurable;

(ii) $E$ is not a Borel set.

(Here a Borel set is a member of Borel $\sigma$-algebra, which is defined by the $\sigma$-algebra generated by the collection of all open sets in $\mathbb{R}^n$.)

ScienceAge
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  • It will be enough to construct a set $E\subseteq\mathbb R$ which satisfies (ii), that $E$ is not a Borel set. For $n\gt1$, the set $E$ is already a Lebesgue null set which is not a Borel set. For $n=1$, first, we can assume $E$ contains no rational numbers. (If $E$ is not a Borel set, neither is $E\setminus\mathbb Q$.) There is a continuous injective function $f:\mathbb R\setminus\mathbb Q\to\mathcal C$ where $\mathcal C$ is the Cantor set. Now $f(E)$ is a Lebesgue null set and is not a Borel set. – bof Jul 24 '19 at 04:33
  • @Bach Sorry. It seems that the "Borel measurable" does not meet the requirement "Borel sets". – ScienceAge Jul 24 '19 at 09:08
  • @bof Thanks for your answer and I will check it later. – ScienceAge Jul 24 '19 at 09:09
  • @ScienceAge They are the same thing, perhaps it's not good to use the term "Borel measurable" there, but you can take a look at it. – Bach Jul 24 '19 at 09:13
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    And see the many reference on the question mentioned by @Bach . A memorable example is of a Borel set in the plane, whose projection onto the $x$-axis is not a Borel set (but that projection is a Lebesgue measurable set). – GEdgar Jul 24 '19 at 10:57

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