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Quite a while ago, I heard about a statement in measure theory, that goes as follows:

Let $A \subset \mathbb R^n$ be a Lebesgue-measurable set of positive measure. Then we follow that $A-A = \{ x-y \mid x,y\in A\}$ is a neighborhood of zero, i.e. contains an open ball around zero.

I now got reminded of that statement as I have the homework problem (Kolmogorov, Introductory Real Analysis, p. 268, Problem 5):

Prove that every set of positive measure in the interval $[0,1]$ contains a pair of points whose distance apart is a rational number.

The above statement would obviously prove the homework problem and I would like to prove the more general statement. I think that assuming the opposite and taking a sequence $\{x_n\}$ converging to zero such that none of the elements are contained in $A$, we might be able to define an ascending/descending chain $A_n$ such that the union/intersection is $A$ but the limit of its measures zero. I am in lack of ideas for the definition on those $A_n$.

I am asking specifically not for an answer but a hint on the problem. Especially if my idea turns out to be fruitful for somebody, a notice would be great. Or if another well-known theorem is needed, I surely would want to know. Thank you for your help.

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up vote 16 down vote accepted

Here's an attempt at a hint for the first result you ask about: Assume without loss of generality that $A$ has finite measure. Let $f$ be the characteristic function of $A$ and let $\tilde{f}$ be the one of $-A$. The convolution $g = f \ast \tilde{f}$ is continuous and $0$ is in the support of $g$.

Added later: One nice standard application is that every measurable homomorphism $\phi: \mathbb{R} \to \mathbb{R}$ is continuous. For more on that and related matters have a look at these two MO-threads:

  1. On measurable homomorphisms $\mathbb{C} \to \mathbb{C}$.
  2. On measurable automorphisms of locally compact groups

They might elucidate what is mentioned in another answer.


What I wrote above is the way I prefer to prove this.

Another approach is to appeal to regularity of Lebesgue measure $\lambda$ (used in $1$ and $2$ below).

  1. Since $A$ contains a compact set of positive measure, we can assume $A$ to be compact right away (as $B-B \subset A - A$ if $B \subset A$).
  2. There is an open set $U \supset A$ such that $\lambda(U) \lt 2 \lambda(A)$.
  3. Since $A$ is compact there is $I = (-\varepsilon, \varepsilon)^{n}$ such that $A + x \subset U$ for all $x \in I$.
  4. Since $\lambda (U) \lt 2\lambda(A)$ we must have $\lambda((A + x) \cap A) \gt 0$.

This is of course very closely related to the argument given by Chandru1 below.

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I wonder if I am thinking right. $$ g(x) = (f*f)(x) = \int_{\mathbb R^n} 1_A(y)\cdot 1_A(x-y) dy$$ But wouldn't the function $$ g(x) = \int_{\mathbb R^n} 1_A(y)\cdot 1_A(x+y) dy$$ help more, since its support conincides with $A-A$? – Peter Patzt May 13 '11 at 16:46
@Peter: yes, you're right of course. I'll edit my answer – t.b. May 13 '11 at 17:04
@Theo,@Peter: You can ever refer, Michael Artin's Algebra book for a prof of the result which Theo mentions. – user9413 May 13 '11 at 17:30
Thinking about the continuity of $g$, it seems to me not to far off the original problem. It seems to be well-known as wikipedia indicates, I do not see why $g$ must be continuous right away. – Peter Patzt May 13 '11 at 22:47
@Peter: It follows from Hölder and the fact that translation is continuous on $L^1$. Here's one link that you may be able to adapt. Feel free to ask if you need further information, I can add it to my answer. – t.b. May 13 '11 at 22:49

Solution found in this link:

Assume $A$ is the set contained in [0,1] with positive measure, say $m(A) > 0$ with $m$ the Lebesgue measure on $[0,1]$. Let Q be the set of all rational numbers in $[0,1]$ Since $\mathbb{Q}$ is countable, it can be presented as

$$\mathbb{Q}=\{p_1, p_2, ..., p_n, ...\}$$


$$A_n = A+ p_n = \{x+p_n\mid x\in A\}.$$

If there exists a pair of integers $n$ and $m$ such that $A_n$ and $A_m$ intersect, then the claim of this proposition is proved. If no such pair exists, then the set of $\{A_n\}$ are all disjoint. Since the union of this family of sets is contained in $[0,2]$ and since $m(A) > 0$, we have

$$2 = m([0,2]) \geq m( \bigcup A_n ) = \sum\limits_{n \in \mathbb{N}} m(A_n) =\infty\cdot m(A) = \infty .$$

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Can you turn your answer into a hint? The OP specifically asks for hints and not a solution. – t.b. May 13 '11 at 15:52
So ... is the exercies in the Kolmogorov book just after the section with the construction of a non-measurable set using this "rational difference" method? – GEdgar May 13 '11 at 15:57
My opinion is that hints and purposefully incomplete answers harm the site and that we mathematicians should always strive to provide solutions with whatever informative clarity and elegance we can muster. – JDH May 13 '11 at 17:00
@JDH: I agree with you in principle. However, I think solving a problem, when the OP specifically asks for hints in his question, doesn't answer the question and is a disservice to the OP's intentions. One can always add a complete solution a bit later on if one is unable or unwilling to give hints. – t.b. May 13 '11 at 17:17
@Theo, my opinion continues to be that hints and incomplete partial solutions are harmful, even when they are requested by the OP. This site exists mainly for readers other than the OP; if the OP desires only a hint, then let him or her just skim the answer if necessary, and meanwhile, the rest of us may enjoy beautifully written illumintating answers that strike at the mathematical heart of the matter. How could one prefer it any other way? – JDH May 14 '11 at 1:06

If you need to find information on the subject, the first proof of the fact that the set of differences contains a neighbourhood of the origin is (for Lebesgue measure on the line) due to Steinhaus. There is a substantial collection of generalizations of the result.

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