The set of differences for a set of positive Lebesgue measure

Solution 1:

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.

Update:

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.

Solution 2:

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

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

Let

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

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

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

Solution 3:

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.