Is the map $(g,h)\rightarrow (g,gh)$ in a measurable group, measurable? 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?
 A: 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.
