# Difference of elements from measurable set contains open interval

Let $A\subset\mathbb{R}$ be a measurable set s.t $,m(A)>0$. Prove that the set $$B=\{x-y\mid x,y\in A\}$$contains nonempty open interval around 0.

I thought to take an interval in $A$, $I=(x-\frac{\epsilon}{2},x+\frac{\epsilon}{2})\subset A$ and hence taking $y$ values from I we get an epsilon - neighborhood of $0$ but I'm quite not sure that I can assume the existence of such I.

How can I prove the existence of I?

• No, you cannot. There are nowhere dense measurable sets of positive measure. – Andrés E. Caicedo Dec 24 '14 at 4:09
• FIne. so can you please give a hint for alternative approach? – user65985 Dec 24 '14 at 4:10

Here is an interesting proof I came across sometime ago. It begins with a Lemma:

Let $$F$$ be a compact set and $$U$$ open such that $$F \subset U$$. Then $$\exists$$ an open neighbourhood $$V$$ of $$0$$ such that $$V+F \subset U$$

Proof: For each $$x\in F, \exists \epsilon > 0$$ such that $$(x-\epsilon,x+\epsilon) \subset U$$. Let $$V_x = (-\epsilon/2,\epsilon/2)$$, then $$F \subset \bigcup_{x\in F} x+V_x$$ Since $$F$$ is compact, $$\exists x_1,x_2,\ldots, x_n\in F$$ such that $$F \subset \bigcup_{i=1}^n x_i + V_{x_i}$$ Take $$V := \bigcap_{i=1}^n V_{x_i}$$ Then $$V+F \subset \cup_{i=1}^n V+(x_i + V_{x_i}) \subset \cup_{i=1}^n x_i+(V_{x_i}+V_{x_i}) \subset U$$

Now we may prove the result:

1. Since $$A$$ is measurable, and $$m(A) > 0$$, there is a compact set $$F \subset A$$ such that $$m(F) > 0$$. So, we may assume that $$A$$ itself is compact.

2. Since $$A$$ is measurable and compact, $$m(A) < \infty$$ and $$\exists U$$ open such that $$A\subset U$$ and $$m(U) < 2m(A) \qquad\qquad \text{(1)}$$

3. By the Lemma, $$\exists V$$ an open neighbourhood of $$0$$ such that $$V+A \subset U$$. Now, we claim that $$V \subset A-A$$. If $$v\in V$$, then it suffices to prove that $$(v+A)\cap A \neq \emptyset$$ (why?), so suppose $$v+A\cap A = \emptyset$$, then since $$v+A \subset U$$, we have $$m(U) \geq m(v+A) + m(A) \geq 2m(A)$$ This contradicts (1) and we are done.

HINT: Use the fact that for any $\epsilon>0$ there are a compact set $K\subseteq A$ and an open set $U\supseteq A$ such that $m(U)-m(K)<\epsilon$. Let

$$d=\inf\{|x-y|:x\in K\text{ and }y\in\Bbb R\setminus U\}\;.$$

• Why is $d>0$?

Fix $\delta\in(0,d)$.

• Why is $K+(-\delta,\delta)\subseteq U$?

Suppose that $|r|<\delta$ and $K\cap(K+r)=\varnothing$.

• Calculate $m\big(K\cup(K+r)\big)$ in terms of $m(K)$.
• Use the fact that $K\cup(K+r)\subseteq U$ to get an upper bound on $m\big(K\cup(K+r)\big)$ in terms of $m(K)$ and $\epsilon$.
• Derive a contradiction by choosing $\epsilon$ sufficiently small.

Another approach: WLOG, take $$A$$ to have finite measure. Let $$\mathcal C$$ be a covering of $$A$$ in disjoint open intervals such that $$\left| \bigcup_{I \in \mathcal C} I\right| < \frac 43 |A|$$ deduce that there exists an $$I \in \mathcal C$$ such that $$|A \cap I| \geq \frac 34 |I|$$. Denote $$A' = A \cap I$$. Note that $$A'$$ is bounded. Let $$I = (a,b)$$.

Now, suppose that $$d \in \Bbb R$$ is such that $$(d+A') \cap A' = \emptyset$$. Then $$2|A'| = |d+A'| + |A'| = |(d+A')\cup A'|$$ If $$d > 0$$, then $$(d+I)\cup I \subset (a,d+b)$$.

If $$d < 0$$, then $$(d+I) \cup I \subset (d+a,b)$$.

In either case, we have $$|(d + I) \cup I| \leq |d| + |I|$$. That is, for any such $$d \in \Bbb R$$, we have $$2|A'| \leq |(d+I) \cup I| \leq |d| + |I|$$ And if $$(d+A') \cap A' = \emptyset$$, then $$|(d+A') \cup A'| = 2|A'|$$.

So $$(d+A') \cap A' = \emptyset$$ implies that $$\frac 32|I| \leq 2|A'| \leq |d| + |I| \implies |d| \geq \frac 12|I|.$$ Contrapositively: if $$|d| < \frac 12 |I|$$, then $$(d + A') \cap A' \neq \emptyset$$.

That is, every $$d \in (-|I|/2,|I|/2)$$ can be written as $$x-y$$ for $$x,y \in A' \subset A$$.

• Hi why is $2|A|\leq |d|+|I|$ – Jhon Doe Sep 12 '19 at 6:10
• @JhonDoe I fixed a few typos and added some explanation; see my latest edit. – Ben Grossmann Sep 12 '19 at 8:05
• Also why is $|(d+I) \cup I| \leq 2|A'| \leq |d| + |I|$. Isn't I a super set of A'. Thanks – Jhon Doe Sep 12 '19 at 8:29
• @JhonDoe well spotted; that was another silly mistake – Ben Grossmann Sep 12 '19 at 10:10