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Let $E$ be the union of uncountably many closed disks in $\mathbb{R}^2$ with radii at least $1.$ If we let $\partial E$ denote the boundary of $E$ and $x \in \partial E,$ is $$\lim_{r \to 0} \frac{m((B(x,r)\cap \partial E) }{m(B(x,r))} <1$$ whenever the limit exists?

Here $m$ is the Lebesgue measure and $B(x,r)$ is the ball of radius $r$ centered at $x$.

Dedalus
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    Do you have any example where the limit is $>0$? – fweth Dec 26 '16 at 22:12
  • @fweth The closed upper half plane is such a union. – zhw. Dec 26 '16 at 22:19
  • Sorry, I don't get it, isn't the boundary of the upper half plane equal to $0$? Do you want $E$ to be the union of the boundaries instead of the boundary of the unions? – fweth Dec 26 '16 at 22:36
  • The boundary of the closed UHP is $\mathbb R \times {0}.$ – zhw. Dec 26 '16 at 22:39
  • Sorry, I meant the measure of the boundary of the upper half plane! – fweth Dec 26 '16 at 22:42
  • @fweth Yes, you're right, I misread the problem. – zhw. Dec 26 '16 at 23:08
  • @fwethI have no example of where this density is $>0.$ – Dedalus Dec 27 '16 at 11:27
  • Is $E$ not just $\mathbb{R}^2$? Since you can move your disk as you wish. – Student Dec 27 '16 at 11:31
  • @Cuteboy not neccesarily, take for example closed disks of radius $1$ centered at $(x,0)$ where $x \in \mathbb{R}.$ – Dedalus Dec 27 '16 at 12:06
  • @zhw: An idea I had was to use that all balls are convex and then some supporting hyperplane argument, but that gave me nothing. – Dedalus Dec 27 '16 at 12:07
  • @Dedalus ok, but I mean you used "the union", but not "a union", so it is for me not clear what you are meaning $E$. – Student Dec 27 '16 at 16:24
  • If $x$ is on the boundary, then it's on the boundary of some disk $D$ in $E$. The interior of $D$ can't be part of the boundary of $E$. So I suspect that you can show that your limit is in fact less than or equal to $\frac{1}{2}$ relatively easily. I doubt this is attainable though. No idea what a sharp upper bound for your limit would be. – Josh Keneda Dec 27 '16 at 23:21
  • The strategy I mentioned above works, but showing that the density of $D$ at a boundary point is $\frac{1}{2}$ was surprisingly tedious. I had to use this for the area formulas and then the top answer here along with the Maclauren expansion for $\sqrt{1+x}$ to explicitly compute. There's probably a better way to do it. – Josh Keneda Dec 28 '16 at 02:57
  • @JoshKeneda why must $x$ be on the boundary of some disk in E? – Dedalus Dec 28 '16 at 11:51
  • Ah! You're right. It doesn't have to be. You could just have a bunch of disks of radius one centered at $(\frac{1}{n}, -1)$ to get the origin on the boundary of $E$ without a disk next to it. Good catch; my bad. – Josh Keneda Dec 28 '16 at 15:28
  • I feel like $\partial E$ must be even $1$-rectifiable, therefore of Hausdorff dimension $1$, but I don't have a proof. This would imply that the numerator in the limit is identically zero. – Del Dec 29 '16 at 17:08

1 Answers1

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Let $D(x,r)$ denote the closed disc with center $x$ and radius $r.$ Let $E = \cup_x D(x,1),$ where $x$ runs through some set $X.$ Area measure will be denoted by $A.$

The main idea is this: No point $x_0\in \partial E$ can lie in the interior of any $D(x,1)$ with $x\in X.$ However, the boundaries of such discs must get close to $x_0.$ So fix an $r<1.$ Draw pictures. Doesn't it look like likely that the interiors of such discs will, in area, contain a fixed percentage of $D(x_0,r)?$ Answer: Yes! And this implies $\partial E$ will be missing a fixed percentage of $D(x_0,r)$ for any $r<1.$

Thus the following result seems likely: There exists $c\in (0,1)$ such that

$$\tag 1 \limsup_{r\to 0} \frac{A(\partial E \cap D(x_0,r))}{A(D(x_0,r))} \le c$$

for all $x_0 \in \partial E.$

Note that if we prove this, we've actually shown $A(\partial E) = 0.$ That's because if $A(\partial E) > 0,$ then for a.e. $x_0 \in \partial E$ we would have the limit of the fraction in $(1)$ equal to $1.$ That follows from the Lebesgue point of density theorem. But $(1)$ rules this out.

So how do we verify $(1)?$ It's better to suggest some steps rather than going through all the details. First, by scaling, we see that

$$\tag 2 \frac{A(D((0,0),r)\cap D((r,0),r))}{A(D((0,0),r))}$$

is independent of $r>0.$ Thus $(2)$ is a constant $b\in (0,1).$ Next, note that for fixed $r>0,$

$$\lim_{\epsilon\to 0^+} \frac{A(D((0,0),r)\cap D((r+\epsilon,0),r))}{A(D((0,0),r))} = b.$$

Finally, note that if $r<1,$ then $D(r+\epsilon,r)\subset D(1+\epsilon,1).$ Thus, if $r<1,$

$$\lim_{\epsilon\to 0^+} \frac{A(D((0,0),r)\cap D(1+\epsilon,1)}{A(D((0,0),r))} \ge b.$$

Use these results, the rotational symmetry available, and the preliminary remarks to see that $(1)$ holds, with $c= 1-b.$

zhw.
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