Hi,
Let $R$ be equipped with the usual Borel structure. Let $F$ be a Borel subset and $E$ be a closed subset of $R$. Then $F+E=(f+e: f\in F, e \in E \)$ is Borel? If yes, is it true for any locally compact topological group? Thanks in advance.
Hi, Let $R$ be equipped with the usual Borel structure. Let $F$ be a Borel subset and $E$ be a closed subset of $R$. Then $F+E=(f+e: f\in F, e \in E \)$ is Borel? If yes, is it true for any locally compact topological group? Thanks in advance. 


No. Erdös and Stone showed that the sum of two subsets $E$, $F\subset\mathbb R$ may not be Borel even if one of them is compact and the other is $G_\delta$ (see "On the Sum of Two Borel Sets", Proc. Am. Math. Soc., Vol. 25, (1970), pp. 304306). Their argument works for every connected locally compact (or abelian) topological group with a complete metric. 


At least in $R^2$ it's false, and probably in R too. There exist closed subsets of $R^2$ that project to nonBorel. So if you take such a set and add it to the yaxis then you'll get a nonBorel 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. 

