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?
No, not even with large cardinals.
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 $\Bbb R$?