Lebesgue measurable and non-measurable sets

Question

Is it possible to construct such $A \subset [0,1]^2$ that is not Lebesgue measurable, but it's projections to coordinate axes are measurable? Similarly construct subset that is measurable, but projections aren't.

Answer 1

For the first question, take some arbitrary non-measurable subset of the square and take the union of it with the boundary of the square. Then this union is still not measurable (since it differs from a non-measurable set by a set of measure zero) and the projections are the whole intervals, so they are measurable.

For the second question, take any non-measurable subset $B$ of $[0,1]$, and take the union of $B \times \{0\}$ and $\{0\} \times B$. This union is measurable because it is contained in a set of measure zero, but the projections are not.

Answer 2

For the first question, use a two-dimensional Bernstein set. That is, construct two disjoint sets $A,B$ each of which meets every uncountable closed subset of $[0,1]^2.$ Then $A$ is nonmeasurable, and each of its projections is the whole interval $[0,1].$

For the second question, take a nonmeasurable set $S\subset[0,1]$ and let $A=\{(x,x):x\in S$; so $A$ has measure zero and its projections are nonmeasurable.