For a set $C\subset \mathbb R^2$, define its visibility from a point $x$ as $vis_C(x)=\{\varphi\in \mathbb S^1\mid \exists t>0~~x+t*\varphi\in C\}$, where $\mathbb S^1$ denotes the unit circle. Say that $C$ looks $m$-big from $x$ if $\mu(vis_C(x))/2\pi=m$, where $\mu$ is the (outer) Lebesgue measure on $\mathbb S^1$.

If $C$ is at most $0.99$-big from every $x$, then is $\mu(C)=0$?

Here again $\mu$ can be the Lebesgue measure, but I'm interested in any similar results as well. My motivation is to solve this problem. The question also seems somewhat related to this one.

Note that the converse is false, as if $C$ is a line, then it looks $1/2$-big from every $x\in \mathbb R^2\setminus C$, and if we take the union of countably many horizontal lines, we can construct a set that looks $1$-big.

  • 2
    $\begingroup$ Uh, how is a line 1-big? I'm not seeing even a finite set of lines being 1-big for almost every x. Gerhard "Is This A Tau Thing?" Paseman, 2018.07.13. $\endgroup$ – Gerhard Paseman Jul 13 '18 at 20:14
  • 3
    $\begingroup$ @domotorp: Aren't lines $\tfrac{1}{2}$-big only? By the way, apparently you use $\mu$ for the Lebesgue measure both in $\mathbb{S}^1$ and $\mathbb{R}^2$. $\endgroup$ – Mateusz Kwaśnicki Jul 13 '18 at 20:22
  • $\begingroup$ Oops, thanks! I got confused because originally I wanted to post the problem with $t\in \mathbb R$ not necessarily positive. $\endgroup$ – domotorp Jul 13 '18 at 20:28

Set $C$ cannot have positive measure. Indeed, if $C$ had positive measure, then by the Lebesgue density theorem there would be a ball $B(x,r)$ such that more than 99% of $B(x,r)$ is in $C$. However, $\mu\bigl(B(x,r)\cap C\bigr)\leq vis_C(x) \cdot \mu\bigl(B(x,r)\bigr)$ by Fubini's theorem. So, $vis_C(x)> 0.99$.

  • 1
    $\begingroup$ But this doesn't exclude $C$ to be non-measurable, right? $\endgroup$ – domotorp Jul 13 '18 at 20:47
  • $\begingroup$ It does seem to say that C cannot contain a set of positive measure though. Gerhard "Looking On The Bright Side" Paseman, 2018.07.13. $\endgroup$ – Gerhard Paseman Jul 13 '18 at 21:02
  • $\begingroup$ I think the "Fubini step" should hold for outer measures, but I would need to think it through (or to check the standard references). My guess is that one should be able to prove that, not only the measure is 0, but that the Hausdorff dimension is at most 1. $\endgroup$ – Boris Bukh Jul 13 '18 at 21:05
  • 1
    $\begingroup$ There are full-outer-measure sets in the plane such that on every vertical line they have only one point: math.stackexchange.com/a/1685978/88597 $\endgroup$ – domotorp Jul 13 '18 at 21:24

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.