Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We say a set $A \subseteq \mathbb{R}^n$ contains the pattern of a finite set $B \subseteq \mathbb{R}^n$ if there exists a shift $t \in \mathbb{R}^n$ and scale $s > 0$ such that $t+sB \subseteq A$. I've read that if $A$ has positive measure, then for any choice of finite set $B$, $A$ contains the pattern of $B$, but proofs are never provided (they say it is "clear"), and I don't know how to show this rigorously.

I was able to show this in the easier case when $B \subseteq \mathbb{R}$ is of the form $\{x,x+y,x+2y\}$ by using the Lebesgue density theorem and considering a sufficiently small ball around a density point intersected with $A$, say with measure at least 3/4 of that of the ball (anything strictly larger than 1/2 will do) and making a measure argument, but I needed the symmetry of the set $B$ around the center $x+y$ in order for my argument to work. How would one prove this fact in full generality?

share|cite|improve this question
Good description. But I don't see where you'd need the symmetry. If $A$ contains greater than a fraction $1-1/\#B$ of a ball $D$, then consider whether the intersection $\bigcup_{b\in B} ((A\cap D)-sb)$ can be empty. Here we take $s$ extremely small, so that $D-sb$ is almost the same as $D$. – Greg Martin Mar 11 '13 at 20:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.