1

If a set has a "small enough" Lebesgue measure, can we say that the set of points "close enough" to it will have a "small enough" Lebesgue measure too? Details below.

Let $\lambda$ denote the Lebesgue measure.

Suppose $E \subseteq \mathbb{R}^2$. $E$ is compact and convex, $\lambda(E)<\infty$. Further, $A \subseteq E$, $\lambda(A)<\epsilon\lambda(E)$ for some $\epsilon>0$.

What is the tightest upper bound on the Lebesgue measure of the following set? $$B:=\{x \in E:d(x,A)\leq \delta\}$$

where $d(x,A):=\inf\limits_{y \in A} d(x,y)$ where $d(\cdot,\cdot)$ is the Euclidean metric on $\mathbb{R}^2$. Clearly, $B \supseteq A$.

I would like to buttress that we know nothing more about $A$ than the fact that it has a small enough Lebesgue measure. In other words, it is allowed to be all sorts of "not nice".

For the sake of completeness, I would also like to add that question closest to this question I found is this.

Canine360
  • 1,574
  • 4
    You aren't really asking any concrete question so it's hard for me to say, but wouldn't a dense countable subset of $E$ contradict any nice upper bound? If $A$ however has a nice boundary, you could maybe try to relate it's length to $B\backslash A$? – Boxonix Oct 31 '24 at 15:42
  • 1
    The closure of $A$ can be as big as $E$ itself, even when $A$ is countable. Where did this question come from? – Brian Moehring Oct 31 '24 at 15:43
  • That is exactly my confusion @BrianMoehring – Canine360 Oct 31 '24 at 15:45
  • Related though incomplete for these purposes – Greg Martin Oct 31 '24 at 15:48
  • 1
    Again, where did the question come from? If it's your creation, perhaps there's a natural assumption you're missing. If it's a textbook, perhaps we can find errata. Absent that info, however, I don't have the insight to help. – Brian Moehring Oct 31 '24 at 15:50
  • @BrianMoehring I'm stuck in an economics problem and it's a part of that so yes it's almost my creation, and I think we need more assumptions on the nature of $A$, as you correctly observed. – Canine360 Oct 31 '24 at 16:02
  • 2
    If $A$ is closed in $E$ then you get at least $\lambda(B_\delta)\to\lambda(A)$ for $\delta\to 0$ (where $B_\delta={x\in E: d(x,A)\le\delta$). But I doubt that this qualitative result can be turned into a quantitative form for general $A$. Note that $B_\delta$ is the same for $A$ and its closure. – Jochen Oct 31 '24 at 16:06
  • Seems to be related: Measure of a "tightened" set? – Dave L. Renfro Oct 31 '24 at 20:16
  • Take $E=[0,1]^2$. So, $E \subseteq \mathbb{R}^2$, $E$ is compact and convex, and $\lambda(E)<\infty$. Take $A = [0,1]^2 \cap (\Bbb Q \times \Bbb Q)$. So, $A \subseteq E$, $\lambda(E)=1$, and $\lambda(A)=0<\epsilon = \epsilon\lambda(E)$ for some $\epsilon>0$. For any $\delta >0$, $$B:={x \in E:d(x,A)\leq \delta} = E$$ So, without any additional conditions on $E$ or $A$, the tightest upper bound on the Lebesgue measure of $B$ is $\lambda(E)$. – Ramiro Oct 31 '24 at 21:17
  • @Jochen This is super helpful. Thank you so much! – Canine360 Nov 02 '24 at 16:21

0 Answers0