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If B is a subset of R such that
I) B' has Lebesgue measure zero
II) B is closed under addition

Show that B = R

this is my first course in measure theory. I only know that nonempty close and open subset of R is equal to R.

What do I have from i and ii? And what property of R should I use?

Martin Sleziak
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Sai
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  • Hi Sai, what is $B'$? If $B'$ is the complement of $B$, then this is false and $\mathbb{Q}$ provides a counterexample. If $B'$ is the set of limit points of $B$, then the assertion cannot be correct either as this set is all of $\mathbb{R}$ if $B=\mathbb{R}$. Do I misunderstand the problem? – Amitesh Datta Mar 03 '15 at 04:28
  • B' is the complement of B – Sai Mar 03 '15 at 04:30
  • Q cannot be used here since B has to be closed. So if B were the irrational a it would have to contain all the rationals, say -sqrt(2) + (3+sqrt(2)) – Sai Mar 03 '15 at 04:33
  • Hi @Sai, I'm not sure I understand - the condition you have is that $B$ is closed under addition, and certainly $B=\mathbb{Q}$ is closed under addition. Do you mean that the condition should be "$B'$ is closed under addition"? – Amitesh Datta Mar 03 '15 at 14:03
  • @AmiteshDatta, if $B=\Bbb Q$, (i) fails. – Martín-Blas Pérez Pinilla Mar 03 '15 at 14:12
  • Hi @Martín-Blas, sorry, I didn't notice that condition. Thanks for the clarification! – Amitesh Datta Mar 03 '15 at 15:23

2 Answers2

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Idea: using the Steinhaus's Theorem:

Let $C$, $D$ be Lebesgue measurable subsets of positive measure. Then the set $C+D$ contains an open interval.

In this case, (i) implies that $B+B$ contains an open interval. But by (ii) $B+B\subset B$, so $B$ contains an open interval $I$. By (i) again, there is $b_0\in I: -b_0\in B$. Now, $I-b_0\subset B$ is a neighborhood of 0 and...

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Martín-Blas Pérez Pinilla
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For any real $x$, $B$ and $B - x = \{b - x : b \in B\}$ have full measure so there is some $y \in B \cap B - x$. Hence $y = z - x$ for some $z \in B$. Hence $y, z \in B$ and $x = z - y$. Hence $x \in B$.

This avoids Steinhaus theorem which shows that this is also true if $B$ had just positive measure.

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  • 'Closed under addition' doesn't imply 'closed under subtraction'. Perhaps you could use $x-B$ instead of $B-x$. – TonyK Mar 04 '15 at 11:03