Anand Deopurkar
Posted on Friday, 30 April, 2004 - 07:43 am:
I have just started reading Lebesgue Theory. Can anyone give me an example of a set which is not measurable with respect to the usual Lebesgue Measure?
David Loeffler
Posted on Friday, 30 April, 2004 - 09:17 am:
The usual construction is something like this. Divide the reals in [0,1] into equivalence classes, where two reals are equivalent if they differ by a rational. Then define a set S by picking one element from each clas.

For any rational q, define Sq={a+q:a Î S }, where addition is done modulo 1.

Now suppose S is measurable. Then each of the Sqs is also measurable with measure equal to that of S; so if this measure is zero, then the measure of

È
q Î QÇ[0,1] 
 Sq
would be zero. Since this union is just [0,1] itself, with measure 1, this is a contradiction.

However, this is a disjoint union, so if S has measure greater than zero, the measure of the union would have to be infinity; so we have a contradiction here as well.

So the set S is not measurable.

David

Anand Deopurkar
Posted on Friday, 30 April, 2004 - 10:11 am:
Wow !
David Loeffler
Posted on Friday, 30 April, 2004 - 10:40 am:
It's a very nice trick, as it doesn't actually use any technicalities about the properties of Lebesgue measure: for any measure on that is countably additive and translation-invariant, this proof will give a nonmeasurable set.

The slightly subtle point about the proof is whether or not the set S exists at all. To justify the construction you need to invoke the Axiom of Choice. It is difficult to do most parts of analysis without using choice somewhere, but it has all sorts of troublesome consequences (look up the Banach-Tarski paradox, if you're interested in such things; there's a very good book on it by Stan Wagon).

David
Anand Deopurkar
Posted on Friday, 30 April, 2004 - 08:06 pm:
Yes. I realized it needs AoC when you said that choose one member from each equivalence class. But I dont think that there should be any hesitation in taking AoC because I read that if Set theory is consistent without AoC then it is consistent with AoC too.
By the way what is Banach-Tarski Paradox? I read somewhere that using AoC we can prove that a sphere can be decomposed and the parts rearranged to get two spheres of the same radius. Is it the same paradox? (It is only an apparent paradox I guess...)
James Cranch
Posted on Friday, 30 April, 2004 - 08:14 pm:
Yes, that's the one.

Your statement about consistency strength is right too. But that doesn't mean we should necessarily accept Choice: it may be that there's a different axiom, which contradicts it, but which gives nicer results in some circumstance.