5
$\begingroup$

Let $E\subset \mathbb{R}^n$ have positive Lebesgue measure. What are easily interpretable sufficient conditions on $E$ to guarantee that the difference between the closure $\bar{E}$ and the interior $\operatorname{Int}(E)$ has zero Lebesgue measure?

In particular, I am interested in the following situation $-$ would the above mentioned condition be satisfied if $E$ is compact and convex?

$\endgroup$
  • 4
    $\begingroup$ It is true for compact convex sets. Look here math.stackexchange.com/questions/207609/… $\endgroup$ – MotylaNogaTomkaMazura Sep 4 '15 at 13:33
  • $\begingroup$ Thanks! This is very helpful. Would be nice to have some alternative easily understood sufficient conditions. $\endgroup$ – MerylStreep Sep 4 '15 at 13:36
  • 1
    $\begingroup$ @MerylStreep: What kind of conditions (other than compact and convex) are you looking for? I think answering the question in full generality will be difficult. What you are asking for in effect is that $\partial E$ has Lebesgue measure zero. These sets are sometimes known as "sets of continuity" of the Lebesgue measure (mostly in probability). It is known that these sets form an algebra which generates the Borel sigma algebra. $\endgroup$ – PhoemueX Sep 4 '15 at 14:12
  • $\begingroup$ @PhoemueX: can e.g. convexity be weakened to connectedness? $\endgroup$ – MerylStreep Sep 4 '15 at 14:17
  • 2
    $\begingroup$ No, not even to path-connectedness. Consider the set $(\Bbb{Q} \times \Bbb{R}) \cup (\Bbb{R} \times \Bbb{Q})$. If I am not mistaken, this should be path connected, but not fulfill your condition. The example is of course not compact, but it should be possible to get something similar. $\endgroup$ – PhoemueX Sep 4 '15 at 14:19
4
$\begingroup$

It is true for compact convex sets. Look here math.stackexchange.com/questions/207609/… -- MotylaNogaTomkaMazura

There are a few natural conditions that are weaker than convexity condition. I list them below, and comment on whether they yield the desired conclusion

Star-shaped compact sets

For these, the boundary may be of positive measure.

Let $C\subset \mathbb{R}$ be a fat Cantor set. Consider $\mathbb{R}$ as the $x$-axis in $\mathbb{R}^2$. Define a compact set $K\subset \mathbb{R}^2$ as the union of all line segments beginning at $(0,1)$ and ending at a point of $C$. This is a set of compact set of positive measure with empty interior; all these properties are inherited from $C$. It is star-shaped with respect to the point $(0,1)$.

Quasiconvex compact sets

For these, the boundary may be of positive measure.

A set $K$ is quasiconvex if there is a constant $c$ such that any two points $a,b\in K$ can be joined by a curve of length at most $c|a-b|$ lying in $K$. Unfortunately, such a set can still have the boundary of positive measure. Consider the following modification of the Sierpiński carpet: at $k$th step of construction, the square is divided into $n_k\times n_k$ equal squares and the middle one is removed. Here $n_k$ is an odd number $\ge 3$; the original construction has $n_k\equiv 3$.

The generalized Sierpiński carpet has positive area when $\sum n_k^{-2}<\infty$; specifically, the area is $\prod_k(1-n_k^{-2})$. It has empty interior. The quasiconvexity follows from the fact that when going from $a$ to $b$, having to go around a removed square multiplies the traveled path at most by the factor of $2$.

Sets with a uniform exterior or interior cone condition

For these, the boundary has measure zero.

I mean the following condition: there exists a circular cone $C$ (radius $r$, height $h$) such that for every point $p\in\partial K$, there is an isometric copy $C_p$ of $C$ that has vertex at $p$ and such that

  • $C_p\cap K=\{p\}$, for the exterior cone condition
  • $C_p\cap K^c=\{p\}$, for the interior cone condition

Note that every convex set satisfies the exterior cone condition: one can use any cone $C$ whatsoever, since there is a whole halfspace that lies outside of $K$.

Either exterior or interior condition implies that the boundary is porous and therefore has zero measure.

$\endgroup$
  • $\begingroup$ Thank you! This is very helpful. I am learning so much here. $\endgroup$ – MerylStreep Sep 6 '15 at 22:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.