# Nonmeasureable subset of ${\mathbb{R}}^2$ such that no three points are collinear?

I'm exploring the properties of sets in the plane that do not contain a set of three collinear points. In particular, I'm interested in the "largest" they can be.

Things I know so far:

Assuming the axiom of choice, one can use Zorn's lemma to show that there are maximal subsets of ${\mathbb{R}}^2$ with this property.

A circle is an easy example of such a set that is uncountable.

If a Lebesgue measurable set $S \subseteq {\mathbb{R}}^2$ has this property, then it contains at most two points on any vertical line. So the integral of $\chi_S$ in the $y$-direction is 0, and by Tonelli's theorem $m(S) = 0$. But this argument doesn't work if $S$ is nonmeasurable.

So the question arises: Is there such a set that is Lebesgue nonmeasurable? (Since there are models of ZF in which every subset of ${\mathbb{R}}$ is Lebesgue measurable, this might actually depend on the choice of set theory axioms.)

edit: One line of thought I am pursuing is thinking of Lebesgue nonmeasurable sets as sets with positive outer measure. Also, one can generalize the question to consider subsets of ${\mathbb{R}}^n$ that do not contain a set of 3 collinear points, or perhaps $n+1$ co-hyperplanar points. For these questions the case ${\mathbb{R}}^1$ is trivial, and maybe there's something different about dimensions higher than 2 à la Banach-Tarski.

• Yes, there are such nonmeasurable sets. Such a set was constructed by Sierpinski using the axiom of choice and transfinite induction (I will look te paper up later). The idea is to construct the set in such a way that it intersects every compact subset of positive measure in the plane. By regularity it must then have positive measure if it is measurable, but your argument shows that it must have measure zero. Hence it is nonmeasurable. – PhoemueX Jan 27 '15 at 8:51
• @PhoemueX Can you find this paper? – user210895 Feb 18 '15 at 21:01
• Sorry, I forgot about that. The paper is linked in my answer here: math.stackexchange.com/questions/1030183/…. But the paper is in french. A similar construction is given in this answer math.stackexchange.com/questions/1123897/sets-are-measurable. – PhoemueX Feb 18 '15 at 21:21

## 1 Answer

I didnt understand many of the things written by you, like why the integral of $\chi_S = 0$. Anyways, as far as non-measurable set of $\mathbb{R}^2$ with no three points collinear is concerned, consider this:

Let $N$ be a non-measurable set of $[0,1)$. Let $\phi:[0,1)\longrightarrow \mathbb{R}^2$ such that $\phi(t) = (\cos(2\pi t), \sin(2\pi t))$, then $\phi(N)$ is non-measurable too. Also no three points of $\phi(N)$ are collinear. See if this example helps you.

• If we are talking about Lebesgue measurability instead of Borel measurability, your example does not work, since your set clearly is a Lebesgue null set. – PhoemueX Jan 27 '15 at 13:20
• (Plane) Lebesgue measure of a circle is 0. Moreover, any subset of a measure-0 set is Lebesgue measurable, since both its outer and inner (Borel) measures are 0. Knowledge of trivial properties of the Lebesgue measure might help your PhD pursuits. – Incnis Mrsi Apr 26 '15 at 6:36