Let G be a Polish space and a group such that the map $(g,h)\rightarrow gh^{-1}$ is borel measurable. It is easy to see that left and right translations and taking the inverse are all measurable. How can you show that the map $(g,h)\rightarrow (g,gh)$ is measurable?
Is the map $(g,h)\rightarrow (g,gh)$ in a measurable group, measurable?
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group-theory
measure-theory
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0If $G$ is a topological group then multiplication and inverse are continuous. In particular composition of them is continuous. In particular a function in two coordinates is continuous (in the product topology) if and only if it is continuous in both coordinates. $(g,h)\mapsto(g,gh)$ is clearly continuous, therefore Borel. – 2011-08-20
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0@Asaf: G is not a topological group. – 2011-08-20
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0@Davide: is it true that the projections are Borel measurable? – 2011-08-20
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0@Davide: I still don't see it. Are you sure about that? – 2011-08-20
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0@Davide: No, I am aware of that. I meant that i still don't see why the function is measurable... – 2011-08-20
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0Nor why the projections are measurable – 2011-08-20
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0The preimage of a basic open rectangle $U \times V\;$ is the intersection of the two sets $\{(g,h) : g \in U\}$ and $\{(g,h) : gh \in V\}$, both of which are clearly Borel by hypothesis. Since such rectangles generate the Borel $\sigma$-algebra of $G^2$, we're done. – 2011-08-20
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0I see that now, thank you – 2011-08-20
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0@Davide: Why don't you collect your comments in an answer? – 2011-08-22
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0@Rookie: it would be interesting to find a Polish space $G$ (and $G$ is also a group) such that the map $(g,h)\mapsto gh^{-1}$ is Borel-measurable but not continuous. – 2011-08-23