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## 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 $A$ with a rational number. The union of all these translates has zero measure, so there exists a real number $x$ not in that set. Consider now $(-x)+A$ and note that this translate of $A$ cannot contain any rational because then $x$ would be in a translate of $A$ by a rational number. Therefore $(-x)+A$ is contained in the irrationals.

Alternative solution:

Define the following equivalence relation on $\Bbb{R}$.
$x \sim y \Leftrightarrow x-y \in \Bbb{Q}$. 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 $\Bbb{R}$. 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 $A$ has something in common with every translate of $\Bbb{Q}$ then $A$ contains a Vitali set. But $A$ has measure zero and the Lebesgue measure is complete, so $V$ should be measurable. Contradiction.

Categories: Measure Theory Tags: ,
1. October 7, 2010 at 7:07 am

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

• October 7, 2010 at 4:39 pm

If $A$ is the set we are talking about, being able to translate it into the irrationals means that there exists a real number $r$ such that $r+A=\{r+a : a \in A\} \subset (\Bbb{R}\setminus \Bbb{Q})$.

2. October 7, 2010 at 4:51 pm

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

3. January 21, 2012 at 2:39 am

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.

• January 21, 2012 at 11:13 am

The idea is that $\Bbb{Q}$ can be translated into $\Bbb{R}\setminus \Bbb{Q}$, but the translated set need not be equal to $\Bbb{R}\setminus\Bbb{Q}$, i.e. there exists $x \in \Bbb{R}$ such that $x+\Bbb{Q} \subset \Bbb{R}\setminus \Bbb{Q}$.

4. October 1, 2013 at 2:56 am

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

• October 1, 2013 at 11:15 am

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