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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?

JT_NL
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  • 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). – Chris Eagle Feb 04 '11 at 17:24
  • @Chris Eagle: I want to construct one. – JT_NL Feb 04 '11 at 17:32
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    Only countable choice is need (possibly dependent choice). In fact, there are explicit descriptions of non Borel sets. Dependent choice is only needed for the proof that it is not Borel, and most treatments of measure theory implicitly assume DC. – George Lowther Feb 04 '11 at 17:34
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    See this link for a well-known explicit construction. http://planetmath.org/encyclopedia/ALebesgueMeasurableButNonBorelSet.html – George Lowther Feb 04 '11 at 17:36
  • @Chris Eagle: Could you give a reference for it being consistent that every set has measure zero? I know the first part is true, but I do not believe that the measure zero part is. Also, I do not believe the part about R being a countable union of countable sets, at least when viewed inside the model under discussion -- doesn't Cantor's argument always show that the reals are uncountable within any model you might dream up? – Noah Stein Feb 04 '11 at 17:59
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    @Noah: that was a bit misleading. What really happens is that Lebesgue measure breaks entirely, since the obvious measure on finite unions of intervals is no longer countably additive. What you define Lebesgue measure to be under these circumstances is a moot point: either it isn't a measure, or it's everywhere zero.

    Yes, the reals are always uncountable, but without countable choice a countable union of countable sets can be uncountable.

     – Chris Eagle Feb 04 '11 at 18:23
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    See this previous question as well: http://math.stackexchange.com/questions/18702/between-borel-sigma-algebra-and-lebesgue-sigma-algebra-are-there-any-other – Arturo Magidin Feb 04 '11 at 20:08
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    @Chris Eagle: Interesting. Where can I find out more about this model where Lebesgue measure breaks? – Noah Stein Feb 04 '11 at 20:14
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    @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". – Chris Eagle Feb 04 '11 at 20:25
  • This question is also related: http://math.stackexchange.com/questions/556739/what-sets-are-lebesgue-measurable – Erel Segal-Halevi Sep 26 '15 at 17:50
  • @GeorgeLowther the link is broken – Gabriel Romon Nov 16 '16 at 20:15
  • @LeGrandDODOM: Here's some alternative links. http://planetmath.org/alebesguemeasurablebutnonborelset and http://planetmath.org/sites/default/files/texpdf/41351.pdf – George Lowther Nov 17 '16 at 00:10

2 Answers2

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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... ;-)

Miguel
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There is some good stuff here:

https://mathoverflow.net/questions/32720/non-borel-sets-without-axiom-of-choice

Community
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