# Proper Measurable subgroups of $\mathbb R$

If $(\mathbb{R},+)$ is a group and $H$ is a proper subgroup of $\mathbb{R}$ then prove that $H$ is of measure zero.

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I recommend you include some more details in your question. The current version is posed like a homework problem, and we are not in the business of solving others questions. If it IS he, make one of your tags homework and explain what you have tried. If it is not hw explain why you want the answer to this and what you have tried. I am thankful for your question, as this is a neat problem, but I fear for it's reception in the current form. –  BBischof Aug 4 '10 at 6:59
Further to what @BBischof said, you need to say at least a bit about why you care about the question. -1, homeworklike. –  Charles Stewart Aug 9 '10 at 8:37

As was noted by Jason DeVito if H is measurable then measure of H is 0.

From the other hand, if we suppose, that the axiom of choice holds, there is a possibility, that H is not measurable. The proof is quite simple. $\mathbb{R}$ is a vector space over $\mathbb{Q}$. Therefore there is some basis. Suppose e is one of it's elements and H is a subspace in $\mathbb{R}$, generated by others. Then H is a subgroup of ($\mathbb{R}$, +).

Lemma: H is not measurable.

Suppose H is measurable. Then as was noted above, it's measure m(H)=0. Then every set of the form $H+q e = {h+qe, h\in H}$, where $q\in Q$, has measure 0 (because it is just a shift of H). But $\mathbb{R}$ is equal to union of countable many sets with measure 0: $\mathbb{R} = \cup_{q\in\mathbb{Q}}(H+qe)$. Therefore $m(\mathbb{R})=0$. We have come to a contradiction.

Also there is simple proof of the fact, that if H is measurable, then H is of measure 0.

Lemma: If H is measurable proper subgroup of $\mathbb{R}$, then m(H)=0.

If H={0} then proposition of the lemma is obvious. Otherwise we can find positive element z in H. Suppose, $H_0 = H\cap [0,z)$. If $m(H_0)=0$ then $m(H)=0$. Otherwise $m(H_0)=\delta>0$. Let's take integer N, such that $\delta N > z+1$ (we will see later, why). Note that if x is not in H, then x/n (for every positive integer n) and -x are also not in H. Therefore, using the fact that H is proper, we can find positive x<1, such that $x\notin H$. Suppose y=x/N!. Then for every positive integer n, number ny obeys the following properties:
1. 1>ny>0.
2. ny is not in H.
Then sets H_0, H_0+y, ..., H_0+(N-1)y do not intersect. Therefore we can write $1+z = m( [0, 1+z) ) \leq m(\cup_{n=0}^{N-1} (H_0 + n y) = N \delta.$ Here we have a contradiction with definition of N.

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-1: Too much detail for an answer to a homeworklike question. –  Charles Stewart Aug 9 '10 at 8:37

Your title talks about "proper measurable subgroups of R", but the body of your post doesn't require that H be measurable. The following outline shows that if H is a proper subgroup of R and is measurable, then it must be measure 0. I'm not sure if every subgroup must be measurable or not....

Here is a rough outline of the proof:

Lemma 1: If H is a subset of R and has positive measure, then H-H = {a-b| a,b, in H} contains an interval around 0. Proof: See http://unapologetic.wordpress.com/2010/04/23/lebesgue-measurable-sets/

For Lemma 2 and 3, I'll leave the proofs to you (but I can add details if you need them).

Lemma 2: If H is a subgroup of G, then H-H=H.

Lemma 3: If H is a subgroup of the real numbers R and contains an interval around 0, then H = R.

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