10
$\begingroup$

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.

Improve this question
$\endgroup$
3
  • 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
    Commented Jul 13, 2018 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
    Commented Jul 13, 2018 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
    Commented Jul 13, 2018 at 20:28

1 Answer 1

Reset to default
4
$\begingroup$

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, $vis_C(x)> 0.99$.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ But this doesn't exclude $C$ to be non-measurable, right? $\endgroup$
    – domotorp
    Commented Jul 13, 2018 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
    Commented Jul 13, 2018 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
    Commented Jul 13, 2018 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
    Commented Jul 13, 2018 at 21:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .