At least in $R^2$ it's false, and probably in R too.

There exist closed subsets of $R^2$ that project to non-Borel. So if you take such a set and add it to the y-axis then you'll get a non-Borel set too.

There's probably some general nonsense that would allow you to transfer this result to R, but I don't know it.

At least in $R^2$ it's false, and probably in R too.

There exist closed subsets of $R^2$ that project to non-Borel. So if you take such a set and add it to the y-axis then you'll get a non-Borel set too.

There's probably some general nonsense that would allow you to transfer this result to R, but I don't know it.

Edit: Actually, perhaps I do. The above argument works in $\mathbb{N}^{\mathbb{N}}$, which is isomorphic to its square and can be embedded additively into $\mathbb{R}$. I haven't checked that this works, but it feels as though it should.