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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?

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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 $F^{-1}(B) = A$.

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