Not Borel measurable set

Question

Do there exist non-Borel measurable sets?

I think not on $\mathbb{R}$ because open sets, closed sets, compact sets, and so on are Borel since they generate Borel.

Answer 1

If you use Axiom of choice, for all$y\in (a,b)$, there exists wellorder$<$ and $$A:=\{(x,y) \in (a,b)^2| x<y\}$$ So assume continnum hypothesis, $A_y:=\{x\in (a,b)|x<y \}$ is countable set. Then, by $$\int \int 1_A dx dy =0$$, $$\int \int 1_A dy dx=1$$ So, by Fubini's theorem, $A$ is not measurable.