I'm trying to find a set which is Lebesgue measurable but not Borel measurable.

So I was thinking of taking a Lebesgue set of measure zero and intersecting it with something so that the result is not Borel measurable.

Is this a good approach? Can someone give a hint what set I would take (so please no full answers, I want to find it myself in the end ;-))

Also, I seem to remember that to construct a non-Lebesgue measurable set one needs to use the axiom of choice. Is this also the case for non-Borel measurable sets?

  • $\begingroup$ Are you trying just to show that such sets exist, or actually describe one in some sense? Such a sense would be quite limited, since the construction does indeed depend on choice: it's consistent with ZF that $\mathbb{R}$ is a countable union of countable sets, in which case every set of reals is Borel (and has measure 0). $\endgroup$ Feb 4, 2011 at 17:24
  • $\begingroup$ @Chris Eagle: I want to construct one. $\endgroup$
    – JT_NL
    Feb 4, 2011 at 17:32
  • 4
    $\begingroup$ See this link for a well-known explicit construction. planetmath.org/encyclopedia/… $\endgroup$ Feb 4, 2011 at 17:36
  • 1
    $\begingroup$ See this previous question as well: math.stackexchange.com/questions/18702/… $\endgroup$ Feb 4, 2011 at 20:08
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    $\begingroup$ @Noah: I don't know the original model in which $\mathbb{R}$ is a countable union of countable sets. Wikipedia cites Jech's "The axiom of choice". $\endgroup$ Feb 4, 2011 at 20:25

2 Answers 2


Bit of a spoiler: Your approach seems on the way to what I've seen done, but instead of trying to intersect your set, you might want to map a non measurable one into it using a measurable map and remember how preimages of borel sets behave.

Spoiler: your map could be one from the unit interval onto that very famous set by that very famous guy born in 1845 who suffered from depression and the dislike of many of his contemporaries... ;-)


There is some good stuff here:



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