Jordan non-measurable sets

A solid is the set of points $M(x,y,z)$ such that $0\leqslant z\leqslant 1$, and moreover, if $z$ is rational, $0\leqslant x\leqslant 1$, $0\leqslant y\leqslant 1$; while if $z$ is irrational, $-1\leqslant x\leqslant 0$, $-1\leqslant y\leqslant 0$. Prove that although the corresponding integral is

the volume of this solid does not exist.

Prove the Jordan non-measurability of the set:

Give a Jordan non-measurable set in $R^{3}$.

Give a Jordan non-measurable set whose closure is Jordan measurable.

Prove that the union of two disjoint sets, one of which is measurable and the other Jordan non-measurable, is a set that is not Jordan measurable.

Give two non-measurable sets whose union is measurable.

Each of the projections of a set $X \subset R^{2}$ onto the coordinate axes is a measurable set in $R^{1}$. Must the set $X$ itself be measurable in $R^{2}$?

Give a bounded non-measurable set in $R^{2}$ such that its section by any line parallel to one of the coordinate axes is a measurable set in $R^{1}$.

Give a Jordan measurable set $X$ and a function $f$ continuous on it such that the set

is Jordan non-measurable.

Give a domain in $R^{2}$ that is Jordan non-measurable.

Give an example of a bounded open set in $\mathbb {R}^{2}$ that is not a Jordan set.

Prove or disprove: Every closed and bounded set in $\mathbb {R}^{2}$ is a Jordan set.