Measure of the intersection of a ball and a compact subset

Do you know a large class of compact subset $K$ of $\mathbb{R}^d$ such that for each such compact $K$, there exists a $r>0$ with $\inf_{x \in K} \lambda^d(B_r(x) \cap K) > 0$, where $\lambda^d$ is the $d$-dimensional Lebesgue measure, and $B_r(x)$ is the ball of centre $x$ and radius $r$ for the euclidean metric.

I believe that compact domains with a smooth or piecewise smooth boundary satisfy this property, but I don't have a proof.

Any reference is welcome, thanks !

Every compact set of nonzero measure has this property.

By Heine-Borel theorem, every compact set is bounded, so it has diameter bounded by some $r$, and then trivially for any $x\in K$ we have $\lambda(B_r(x)\cap K)=\lambda(K)$ which is certainly bounded away from zero if $K$ is not null.

A possibly more interesting question would be if you had considered $\inf_{r>0}\inf_{x\in K}$ or the reverse $\inf_{x\in K}\inf_{r>0}$ divided by the measure of the ball. Those kinds of questions are related to geometric measure theory, for example see the related Lebesgue's density theorem.

• Note that the $r$ seems to be only dependent on the family of sets, i.e. you need a uniformly bounded family of compact sets, such as the one I wrote down in my answer. Good point on the other remark though. – AlexR Dec 5 '14 at 14:44
• @AlexR: I think the statement is quite clearly $\forall K \exists r$, not the opposite. – tomasz Dec 5 '14 at 15:56
• Oh, yes I didn't notice that and assumed $\exists r\forall K$ because the former is (as we both agree) pathological. – AlexR Dec 5 '14 at 16:05

Compact sets are bounded in $\mathbb R^d$, thus $r = R$ where $K \subset B_R(0)$ is sufficiently large to get the infimum equal to $\lambda^d(K)$.
Also using smooth ($C^\infty$) peaks, the infimum can get arbitrarily small for fixed $r$ and variable $K$.

This means for $R>0$ given, the family $\{K \subset B_R(0) | K\text{ is compact and } \lambda^d (K) > 0\}$ should be a good start.

• There are compact sets with zero measure, though. I would add the condition that $\lambda^d(K) \gt 0$. – Arthur Dec 5 '14 at 13:53
• @Arthur Good point. Added. – AlexR Dec 5 '14 at 14:42