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I want to ask about Lebesgue measure for groups

I know

$m^*(\mathbb N)=0$ and $m^*(\mathbb Q)=0$

$m^*(\mathbb R)=infinite $

$m^*(${$0,1$}$)=0$

$\mathbb N$ is a normal numbers, $\mathbb Q$ is a rational numbers and $\mathbb R$ is a real numbers.

But is this information true $m^*(any\ sup\ groups\ of\ \mathbb R )=0$ ?

Such that {$0,1,2,3$} or {$1,6,8,3$} for example.

And Is this statements true or false?

1- There exists sup group from $\mathbb R$ his measures =0

2- There exists sup group from $\mathbb R$ his measures =0

I think

1- true, $m^*(\mathbb N)=0$,and $\mathbb N$ sup group from $\mathbb R$

2- false ? I did not find any sup group from $\mathbb R$ and his measures =1

Dima
  • 2,479
  • if $m^([a,b]) = b-a$ then $m^(\mathbb{N}) = 0$ since for any $\epsilon > 0$ : $\ \mathbb{N} \subset \ \bigcup_{n=1}^\infty \ [n-\frac{\epsilon}{n^2},n+\frac{\epsilon}{n^2}]$ and $\sum_{n=1}^\infty m^*([n-\frac{\epsilon}{n^2},n+\frac{\epsilon}{n^2}]) = \epsilon \frac{\pi^2}{3}$ and we can choose $\epsilon$ as small as we want. – reuns May 07 '16 at 18:11
  • Thank you, but this for intervals true ? What about groups ? – Dima May 07 '16 at 18:12
  • This may not be that helpful but $\mathbb{R}$ is also a group. – HarshCurious May 07 '16 at 18:15
  • You are right I don't notice that ! Tahnks. – Dima May 07 '16 at 18:31
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    A measurable subgroup of $\mathbb R$ with positive measure is equal to $\mathbb R$. But there are non-measurable proper subgroups of $\mathbb R$ with infinite outer measure. – GEdgar May 07 '16 at 19:33
  • Thanks, but Thats mean the second statement true ? And if true can you give me any sup group of R his measures =1 ? – Dima May 07 '16 at 20:15
  • to see why GEdgar comment is true see the following link https://en.wikipedia.org/wiki/Steinhaus_theorem – clark May 08 '16 at 04:26
  • Thanks, but I don't understand what he is mean by "is equal to $\mathbb R$" ? Is he mean every supgroup of $\mathbb R$ is equal to outer measure of $\mathbb R$ (=$infinite $) ? Or what ? – Dima May 08 '16 at 08:08
  • see http://math.stackexchange.com/questions/54163/lebesgue-outer-measure-of-0-1-cap-mathbbq for the (inner and outer) measure of $[0,1] \cap \mathbb{Q}$, then using the same argument as the one I used for $\mathbb{N}$, you'll get $m^*(\mathbb{Q})$. and no GEdgar meant that the subgroup is $\mathbb{R}$ itself – reuns May 08 '16 at 15:45

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