Borel subsets of Polish groups

Question

Suppose that I have a polish group $G$ and two subsets $A$ and $B$ of $G$ such that: $A$ is open in $G$ and $B$ is closed in $G,$ from this, can I conclude that $AB$ is a Borel subset of $G$? if not, consider the following.

Since $B$ is closed, $B$ has a Borel transversal $T,$ that is a Borel set $T$ such that $T\cap gB$ is a unitary set, for all $g\in G,$ then can I conclude that the set $AB\cap T$ is Borel?.

Any reference will help. Thanks

Answer 1

If $A$ is open and $B$ is any set, then $$AB = \bigcup \{Ab : b \in B\}$$ is open (hence Borel). If $T$ is Borel, then $AB \cap T$ is an open subset of $T$, hence Borel.

You may already know this, but if you weaken "$A$ is open" to "$A$ is $G_\delta$", then $AB$ does not have to be Borel. This result can be found in "On the sum of two Borel sets" by Erdos and Stone (available here from JStor).