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Let $G$ be a topological group and $\mathcal{B}$ its Baire $\sigma$-algebra (i.e. the smallest $\sigma$-algebra for which all continuous functions $G\rightarrow\mathbb{R}$ are measurable). Consider the product map $\cdot :G\times G\rightarrow G$ given by $(g,h)\mapsto g \cdot h$.

Question: Is the function $\cdot$ measurable with respect to the product $\sigma$-algebra $\mathcal{B}\otimes\mathcal{B}$ and $\mathcal{B}$?

If the answer is no, what are conditions on $G$ that guarantee that $\cdot$ is measurable?

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    $\begingroup$ Do you know: $\operatorname{Baire}(G \times G = \operatorname{Baire}(G)\otimes \operatorname{Baire}(G)$ ? $\endgroup$
    – Gerald Edgar
    Commented Apr 26 at 17:44
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    $\begingroup$ It is false, see my answer. $\endgroup$
    – Gerald Edgar
    Commented Apr 26 at 20:44

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No.

Let $G$ be a group with cardinal $ > \mathfrak c$. Let $G$ have the discrete topology. Then every subset is a zero-set, hence a Baire set (the characteristic function of the set is continuous). But in $G \times G$, the diagonal $\Delta$ does not belong to the product sigma-algebra. [See HERE.] Then $\Delta = \{(x,y) \in G \times G : xy^{-1} = e\}$ is a zero-set (hence a Baire set), but does not belong to the product sigma-algebra.

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