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.

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