## Measure zero

Prove that any subset of the real line having Lebesgue measure equal to zero can be translated into the irrationals.

**Solution: **Consider all translates of the zero measure set with a rational number. The union of all these translates has zero measure, so there exists a real number not in that set. Consider now and note that this translate of cannot contain any rational because then would be in a translate of by a rational number. Therefore is contained in the irrationals.

Alternative solution:

Define the following equivalence relation on .

. It is not hard ( but not easy either ) to prove that the set of equivalence classes of this equivalence relation has the same cardinality as . A set of representatives of the previously defined equivalence relation is called a Vitali set, and it is not Lebesgue measurable (http://math.stackexchange.com/questions/137949/the-construction-of-a-vitali-set).

If has something in common with every translate of then contains a Vitali set. But has measure zero and the Lebesgue measure is complete, so should be measurable. Contradiction.

What do you mean by this statement: can be translated into the irrationals. ?

If is the set we are talking about, being able to translate it into the irrationals means that there exists a real number such that .

thanks benni! Your collection of problems is awesome! But the only thing i don’t like in this blog is the theme!

how about this?

we see that the set of rational number Q is of measure zero. Suppose that Q can be translated into irrational, I.Then Q and I have the same cardinality. A contradiction.

The idea is that can be translated into , but the translated set need not be equal to , i.e. there exists such that .

Why can’t x+A and y+A have a rational in common?

I think there is something wrong with the proof… I’ll try and fix it. Thank you for your comment.