Talk:Measurable group

The assertion that "Every topological group $${\displaystyle (G,{\mathcal {O}})}$$ can be taken as a measurable group" is false. The problem is that if $$\mathcal{B}(G)$$ are the Borel sets for $$G$$ and $$\mathcal{B}(G\times G)$$ are the Borel sets for $$G\times G$$, it does not necessarily occur that $$\mathcal{B}(G\times G)=\mathcal{B}(G)\otimes \mathcal{B}(G)$$. Thus, the continuity of the group operations is not enough to ensure the measurability of certain sets. If, however, the group has a countable basis, it is true that $$\mathcal{B}(G\times G)=\mathcal{B}(G)\otimes \mathcal{B}(G)$$ and then the statement becomes true.