$A+B$ Borel set if $A$ is countable, open.

Question

If $A,B \subset \Bbb{R}$ , we define $$ A+B = \{a+b \mid a \in A, b \in B\}. $$

Suppose $B$ is a Borel set, then we need to

prove that $A+B$ is a Borel set if

• $A$ is countable.
• $A$ is open.

Borel sets are the sets that can be constructed starting from the open and closed sets by repeatedly taking countable unions and intersections.

But how to apply the countable intersection/ union to this problem? Here we are taking the sum of two sets.

Since any homomorphism preserves the $\sigma$-algebra of Borel sets, I thought of considering the translation map $f:B \rightarrow A+B$ where $f(b) = a+b$. Will this do or I have to make more precise arguments?

I cannot apply the fact that difference of two Borel sets is a Borel set since I do not know whether $A$ is a Borel set or not.

Answer 1

If $A$ is countable, then $A+B=\bigcup\{a+B:a\in A\}$ is a countable union of Borel sets.

Answer 2

The case when $A$ is open is hard because people miss how easy it is.

Let $a+b \in A+B$ be arbitrary. Since $A$ is open, there exist some $\epsilon>0$ such that $(a-\epsilon,a+\epsilon) \subset A$.

Then $$(a+b-\epsilon, a+b+\epsilon) \subset A+B$$ and hence $A+B$ is open .