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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?

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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 4 '11 at 17:24
@Chris Eagle: I want to construct one. – Jonas Teuwen Feb 4 '11 at 17:32
See this link for a well-known explicit construction.… – George Lowther Feb 4 '11 at 17:36
See this previous question as well:… – Arturo Magidin Feb 4 '11 at 20:08
@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 4 '11 at 20:25
up vote 14 down vote accepted

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... ;-)

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Thanks a lot. Good idea. – Jonas Teuwen Feb 4 '11 at 21:32

There is some good stuff here:

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Thank you, nice stuff. – Jonas Teuwen Feb 4 '11 at 21:31

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