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 2


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.


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
    $\begingroup$ -1: Too much detail for an answer to a homeworklike question. $\endgroup$ Aug 9, 2010 at 8:37
  • $\begingroup$ 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. $\endgroup$
    – M.G
    Feb 22, 2016 at 17:21
  • $\begingroup$ There was a mistake in my proof: this inequality should be $\geq$, not $\leq$ (fixed above now). Clarified now. $\endgroup$
    – Fiktor
    Jan 28, 2019 at 8:20

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