We work on a Polish group $G$ and consider finite signed measures. Let $\theta\_x\mu(B) := \mu(x^{-1}B)$. Fix a $\mu$. It is clear that $x \mapsto \theta\_x\mu$ is continuous when the codomain is given the weak\* topology when identified with a subset of $C\_b(G)^\*$. (This would be called convergence in distribution if we were working only with probability measures). Is it known whether $x \mapsto\theta\_x\mu$ is Borel measurable with the topology generated by the total variation norm on the codomain?



Is shifting $x \mapsto \theta_x\mu$ Borel measurable wrt total variation topology?
It isn't. Let $\mu =\delta\_e$ be Dirac mass at the identity, and set $F(x) = \theta\_x \delta\_e = \delta\_x$. Let $A$ be any non-Borel subset of $G$ and let $B = \{\delta\_x : x \in A\}$. Then $B$ is closed in the total variation topology (for any $x \ne y$ we have $\|\delta\_x - \delta\_y\| = 2$). But $F^{-1}(B) = A$.


