# Coloring the uncountable Lebesgue-measurable sets of $\mathbb{R}$

A hypergraph $$H=(V,E)$$ consists of a set $$V$$ and $$E\subseteq {\mathcal P}(V)$$, that is, $$E$$ consists of subsets of $$V$$ of arbitrary size. Obviously, a graph is a special kind of hypergraph.

Let $$H=(V,E)$$ be a hypergraph and $$\kappa\neq \emptyset$$ be a cardinal. Then a map $$c:V\to \kappa$$ is said to be a coloring if for every $$e\in E$$ with $$|e|\geq 2$$ we have that the restriction $$c\restriction_e$$ is non-constant. The chromatic number $$\chi(H)$$ of $$H$$ is the smallest cardinal $$\kappa$$ such there is a colouring $$c:V\to \kappa$$.

Question. What is $$\chi(\mathbb{R}, E)$$ where $$E$$ consists of all uncountable Lebesgue-measurable sets?

It is continuum. The coloring with continuum many colors is clear (all points may have different color). Assume that we have $$\kappa colors. Consider the Cantor set $$K$$. All its subsets are Lebesgue measurable. If some color contains uncountably many points from $$K$$, it constitutes a monochromatic edge. So, each color contains at most $$\omega$$ points from $$K$$, and totally $$|K|\leqslant w\cdot \kappa=\kappa$$, a contradiction.
• What if one removes the measure-$0$ sets? Apr 21, 2021 at 12:25
• @LSpice then it changes dramatically: it is 2. Take a Hamel basis over $\mathbb{Q}$, denote by $f(x)$ a coordinate functional and use two colors: $\lfloor f(x)\rfloor$ is even or odd. Apr 21, 2021 at 12:28