# Can a big set always look small?

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.

• 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. – Gerhard Paseman Jul 13 '18 at 20:14
• @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 $\mathbb{R}^2$. – Mateusz Kwaśnicki Jul 13 '18 at 20:22
• Oops, thanks! I got confused because originally I wanted to post the problem with $t\in \mathbb R$ not necessarily positive. – 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, $vis_C(x)> 0.99$.
• But this doesn't exclude $C$ to be non-measurable, right? – domotorp Jul 13 '18 at 20:47