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.
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.