I have just read the following question about measurable and non-measurable sets. Does there exist a measurable subset of $\mathbb{R}^2$ all of whose projections are non-measurable? As with many such questions, there is an easy solution based on the general principle that measure-zero sets in the plane can be very nasty: you can just take your favourite non-measurable set in $\mathbb{R}$ and think of it as a subset of $\mathbb{R}^2.$
Just for fun, here is a meta-question: is there a way of somehow ruling out any use of this principle and thereby obtaining a more challenging question? One idea that fails miserably is to insist that the subset of $\mathbb{R}^2$ has positive measure. That fails because all you have to do is take the union of a nasty measure-zero set with a token nice set of positive measure that doesn't cause any of the projections to become measurable. And that is easy.
Here is a different idea, which comes with a warning that I've only just thought of it so the question has a very good chance of not being interesting. Let X be a measurable subset of the plane. Does there necessarily exist a measure-zero subset Y of X and a projection $\pi$ such that $\pi(X\setminus Y)$ is measurable? In case asking for Y to be a subset of X is too much of a restriction, an alternative question would be for Y to be an arbitrary set of measure zero and consider $\pi(X\Delta Y).$
Best Answer
The answer to the third paragraph is positive. Any measurable $X$ contains an $F_\sigma$ set $Z$ of the same measure. Take $Y=X\smallsetminus Z$. Since continuous images of compacts are compact, any projection of a $F_\sigma$ subset of the plane is again $F_\sigma$, hence measurable.