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If $E$ and $F$ are sets of positive Lebesgue measure on $\mathbb{R}$, prove that some translate of $F$ intersects $E$ in a set of positive measure.

Since each are approximated from below by compact sets, I can just assume that $E$ and $F$ are compact. I've proven the case where $F$ is an interval or a union of intervals. I assume it should be done with interval covers of $E$ and $F$, but I can't make it work.

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2 Answers 2

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Consider the integral $$\int m(E\cap (F+y))dy=\iint\mathbb{1}_E(x)\cdot \mathbb{1}_F(x+y)dxdy$$ By the substitution $x\mapsto x$, $y\mapsto z-x$ it becomes $$\iint\mathbb{1}_E(x)\cdot \mathbb{1}_F(z)dxdz=m(E)\cdot m(F)>0$$ This implies that there exists at least one $y$ such that $m(E\cap(F+y))>0$. In fact, we know that there is a set of positive measure of such points.

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  • $\begingroup$ This is a slick proof. I was wondering if there is a nice solution without integrals. $\endgroup$
    – user164976
    Jul 18, 2014 at 12:26
  • $\begingroup$ This is a slick proof. As a matter of fact, I was wondering if there is a nice solution without integrals. But until now, I can't find it...Oh my god... $\endgroup$
    – David Chan
    May 15, 2015 at 8:17
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Perhaps there is a stronger version states that one can shift $F$ by a rational number $q$ such that the intersection is of positive measure.

Apply Lebesgue's density theorem on both $E$ and $F$. We find that there exists $x$ and $\epsilon$ such that $m(E\cap B(x,\epsilon/2))>3\epsilon/4$. For the same reason there exists $y$ such that $m(F\cap B(y,\epsilon/2))>3\epsilon/4$; take a smaller $\epsilon$ if necessary. Now we choose a rational $q$ from $B(x-y,\epsilon/4)$.

Consider intersections $$E':=E\cap B(x,\epsilon/2)\subset B(x,\epsilon/2),\qquad F':=(F+q)\cap B(y+q,\epsilon/2)\subset B(y+q,\epsilon/2).$$ By how we take $x,\epsilon$ it is clearly that $E'$, comparing to $B(x,\epsilon/2)$, loses $1/4$ of its mass. Also, $F'$, comparing to $B(y+q,\epsilon/2)$, loses $1/4$ of its mass. Forming the intersection $E'\cap F'$ will loses another 1/4 by how we take $q$. Combining these three inequalities and Boole's inequality we see that $E'\cap F'$ is of positive measure and so is $E\cap F$.

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