Pointwise product of Borel sets is Borel?

Question

If $A \subseteq \mathbb{R}$ and $B \subseteq \mathbb{R}^2$ are Borel, is $AB = \{ab:a\in A, b \in B\}$ always a Borel set? Scalar multiplication is continuous but not Lipschitz or injective, so its not obvious (to me) that $AB$ is Borel.

Answer 1

No, this is in general not true.

A result by Erdös and Stone shows that there exist Borel sets $E \subseteq \mathbb{R}$ and $F \subseteq \mathbb{R}$ such that $$E+F := \{e+f; e \in E, f \in F\}$$ is not Borel. If we define $$A := \exp(E) := \{\exp(e); e \in E\} \subseteq \mathbb{R} \quad \text{and} \quad B := \{(\exp(f),\exp(f)); f \in F\} \subseteq \mathbb{R}^2$$ then $A$ and $B$ are Borel sets. Moreover, $$A \cdot B = \{(\exp(e+f),\exp(e+f)); e \in E, f \in F\}. $$ As $E+F$ is not Borel, it follows that $A \cdot B$ is not Borel. Indeed: Suppose that $A \cdot B$ was Borel. Since the mapping

$$\mathbb{R} \ni x \mapsto T(x) := (x,x) \in \mathbb{R}^2$$

is Borel measurable, this would imply that

$$T^{-1}(A \cdot B) = \exp(E+F)$$

is Borel measurable; this, in turn, would imply that

$$\log(\exp(E+F)) = E+F$$

is Borel measurable; in contradition to our choice of $E$ and $F$.