# Are measurable sets closed under projections?

For the following, let us assume that large enough sets to carry the arguments through do exist, i.e. that there are supercompact cardinals or whatever is sufficient.

I know that all projective subsets of $\mathbb{R}^n$ are Lebesgue-measurable. The projective sets are closed under complements and projections. However, because there are only $2^{\aleph_{0}}$ projective sets, most measurable sets are not projective.

So, my question is: Are projections of Lebesgue-measurable sets again Lebesgue-measurable?

• Projective sets are also closed under compliments. If you see a projective set and tell it "Oh, what a lovely definition you have!", you get nothing in return. Commented Apr 20, 2015 at 17:32

Take any non-measurable set $A\subseteq\Bbb R$, and consider $\{0\}\times A$ as a subset of $\Bbb R^2$. As a subset of $\Bbb R^2$ it is a subset of $\{0\}\times\Bbb R$ which is a Borel set which is null. So $\{0\}\times A$ is Lebesgue measurable.
But what is the projection of $\{0\}\times A$ onto $\Bbb R$?