# 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.

## 2 Answers

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 $$n=1,\dots,N$$ number ny obeys the following properties:
1. 1>ny>0.
2. ny is not in H.
Then sets $$H_0, H_0+y, \dots, H_0+(N-1)y$$ are disjoint subsets of $$[0, 1+z)$$. Therefore, $$\displaystyle 1+z = m\Big( [0, 1+z) \Big) \geq m\Bigg(\bigcup_{n=0}^{N-1} (H_0 + n y)\Bigg) = N \delta.$$ Here we have a contradiction with definition of N.

• -1: Too much detail for an answer to a homeworklike question. – Charles Stewart Aug 9 '10 at 8:37
• I don't get why $M\big([0,1+z)\big) \leq m\left(\bigcup\limits_{n=0}^{N-1}(H_0 + ny)\right) = N \delta$ is true. – M.G Feb 22 '16 at 17:21
• There was a mistake in my proof: this inequality should be $\geq$, not $\leq$ (fixed above now). Clarified now. – Fiktor Jan 28 at 8:20

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.