If $ X $ and $ Y $ are metric spaces, $ f: X \rightarrow Y $ is lipschitzian and $ H^k $ is the Haussdorf measure, it is easy to check that $ f(A) $ is $H^k $-measurable whenever $ A $ is $H^k $-measurable. This implies that if
$ p: R^{m+k} \rightarrow R^k $
is the projection onto the first k coordinates then $ p(A) $ is $ L^k $-measurable whenever $ A $ is $ H^k $-measurable, where $ L^k $ is the Lebesgue measure of $ R^k $.
I suspect that it is possible to construct a $ L^2 $-measurable set $ A \subset R^2 $ such that the projection onto the first coordinate is not $ L^1 $-measurable.
Any suggestion for this? Thank you
Posting the comments as an answer:
Yes, it is possible. An easy way is to start with a non-measurable subset $B$ of $\mathbb R$, and let $A=B\times\{0\}$. This example has measure $0$, but this does not matter: For an example with positive measure, take the union of $A$ and a solid rectangle on the side. The projection is the disjoint union of an interval and the non-measurable set $B$.
On the other hand, if $A$ is "topologically simple", its projection will be measurable: If $A$ is Borel, or analytic (the continuous image of a Borel set), then its projection will be analytic, and all analytic sets are measurable.