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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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
1 Answers
Let $\pi_1\colon G\times G\to G$ and $\pi_2\colon G\times G\to G$ defined by $\pi_1(x,y)=x$ and $\pi_2(x,y)=y$. We will first show that a map $f\colon G\times G\to G\times G$ is measurable (for the $\sigma$-algebra $\mathcal B(G\times G)$ generated by the open subsets of $G\times G$ for the product topology) if and only if $\pi_1\circ f$ and $\pi_2\circ f$ are measurable (we put the Borel $\sigma$-algebra on $G$). We will only show the part "only if", since the first is not difficult and doesn't help to solve the problem. Let $O$ an open subset of $G^2$. Since $G$ is Polish, it's in particular separable with a metrizable topology and we can find a countable basis of the topology on $G$, namely $\left\{O_i,i\in\mathbb N\right\}$. We can write $O=\bigcup_{(i,j)\in I}O_i\times O_j$ where $I=\left\{(i,j)\in\mathbb N\times \mathbb N,O_i\times O_j\subset O \right\}$. Since $I$ is countable, we have $\begin{align}f^{-1}(O) &= \bigcup_{(i,j)\in I}f^{-1}(O_i\times O_j)\\&=\bigcup_{(i,j)\in I}\left\{(g_1,g_2)\in G^2,(\pi_1\circ f(g_1,g_2),\pi_2\circ f(g_1,g_2))\in O_i\times O_j\right\}\\ &=\bigcup_{(i,j)\in I}(\pi_1\circ f)^{-1}(O_i)\times (\pi_2\circ f)^{-1}(O_i) \end{align}$
and $f^{-1}(O)$ is measurable. Now, putting $\mathcal F:=\left\{A\in\mathcal B(G^2),f^{-1}(A)\in \mathcal B(G^2)\right\}$, we see that $\mathcal F$ is a $\sigma$-algebra which contains all the open subsets of $G^2$. We conclude that $f$ is measurable.
We put $F\colon G\times G\to G$, $F(g,h)=gh^{-1}$. To solve the problem, we have to show that the maps $f_1\colon G\times G\to G$ and $f_2\colon G\times G\to G$ where $f_1(g,h)=g$ and $f_2(g,h)=gh$ are measurable. $f_1$ is $\pi_1$ and measurable since for $B\in\mathcal B(G)$ we have $f_1^{-1}(B) =B\times G\in\mathcal B(G^2)$. @Rookie has shown that the map $f_3\colon g\mapsto g^{-1}$ is measurable and we have $f_2(g,h)=F(g,f_3(h))$. Since $(g,h)\mapsto (g,h^{-1})$ and $F$ are measurable, $f_2$ is measurable too.
Added later: we notice that the result is still true if we only take $G$ metrizable and separable (with of course the same hypothesis of measurability). We don't need the completeness of the metric.