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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. chandrasekhar
    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. chandrasekhar
    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. axis
    January 21, 2012 at 2:39 am

    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.

    • 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. tomas
    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.

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