# 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?

• 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. planetmath.org/encyclopedia/… – George Lowther Feb 4 '11 at 17:36
• See this previous question as well: math.stackexchange.com/questions/18702/… – 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