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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?

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    If $G$ 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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    @Asaf: G is not a topological group.2011-08-20
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    @Davide: is it true that the projections are Borel measurable?2011-08-20
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    @Davide: I still don't see it. Are you sure about that?2011-08-20
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    @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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    Nor why the projections are measurable2011-08-20
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    The 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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    I see that now, thank you2011-08-20
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    @Davide: Why don't you collect your comments in an answer?2011-08-22
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    @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

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