If $\tau_x(y)=y+x$ and $\mu$ is a measure, then $x\mapsto(\mu\circ\tau_x^{-1})(B)$ is continuous

Question

Let $(E,d)$ be a metric $\mathbb R$-vector space, $\mu$ be a probability measure on $(E,\mathcal B(E))$ and $$\tau_x:E\to E\;,\;\;\;x\mapsto y+x$$ for $x\in E$.

How can we show that $$E\to[0,1]\;,\;\;\;x\mapsto(\mu\circ\tau_x^{-1})(B)=\mu(B-x)\tag1$$ is continuous for all $B\in\mathcal B(E)$?

I'm not sure how I should approach this. Let $(x_n)_{n\in\mathbb N}\subseteq E$ and $x\in E$ with $d(x_n,x)\xrightarrow{n\to\infty}$. Maybe we can show that $$1_{B-x_n}\xrightarrow{n\to\infty}1_{B-x}\tag2,$$ from which the claim clearly would follow.

$d$ is clearly (jointly) continuous; and $$\theta:E^2\to E\;,\;\;\;(x,y)\mapsto x+y$$ is (jointly) continuous as well.

Answer 1

The statement is false, by example take the Dirac measure concentrated at zero, then

$$ \lim_{n\to\infty}\delta (\{0\}-1/n)=0\neq \delta (\{0\}-0)=1 $$

Answer 2

Translation continuity, even for open, sets is a strong property. In fact, if $\mu$ is a probability measure on the Borel subsets of the real line such that $x\mapsto\mu(U+x)$ is continuous for each open set, then $\mu$ is absolutely continuous with respect to Lebesgue measure (and conversely). For this and extensions to more general translations, see the paper "A note on translation continuity of probability measures" [Ann. Probab. 20 (1992), no. 1, 410–420] by S. Zabell. If $\mu$ is absolutely continuous with a continuous density, then $x\mapsto\mu(x+B)$ is continuous for all Borel sets. I don't know what the optimal converse is.