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?

  • $\begingroup$ No, you cannot. There are nowhere dense measurable sets of positive measure. $\endgroup$ – Andrés E. Caicedo Dec 24 '14 at 4:09
  • $\begingroup$ FIne. so can you please give a hint for alternative approach? $\endgroup$ – 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 $x,y \in A' \subset A$.

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

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