# Lebesgue measurable but not Borel measurable

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?

• 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. Feb 4, 2011 at 17:34
• See this link for a well-known explicit construction. planetmath.org/encyclopedia/… Feb 4, 2011 at 17:36
• @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. Feb 4, 2011 at 18:23
• See this previous question as well: math.stackexchange.com/questions/18702/… Feb 4, 2011 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". Feb 4, 2011 at 20:25