Let's recall the definition of universal measurability. Given a metric space $\left(S,\rho\right)$, we denote by $\overline{\mathscr{B}\left(S\right)}^{\mu}$ the completion of the Borel $\sigma$-field $\mathscr{B}\left(S\right)$ (generated by the open sets) with respect to the finite measure $\mu$ on $\left(S,\mathscr{B}\left(S\right)\right)$. The universal $\sigma$-field is $\mathscr{U}\left(S\right)={\displaystyle \bigcap_{\mu}\overline{\mathscr{B}\left(S\right)}^{\mu}}$, where the intersection is over all finite measures (or, equivalently, all probability measures) $\mu$.
Now my question is: if $f:S\rightarrow S$ is continuous, do we have $f$ is $\mathscr{U}\left(S\right)/\mathscr{U}\left(S\right)$-measuable? How about $S=\mathbb{R}^{d}$?
Thanks a lot in advance!
True: Let $E$ be universally measurable and let $\mu$ be any finite measure. Then $\nu =\mu \circ f^{-1}$ is a finite measure so $E=B\cup D$ where $B$ is a Borel set and $D$ is a subset of some Borel set $F$ with $\mu (f^{-1}(F))=0$. Thus $f^{-1} (E)=f^{-1} (B)\cup f^{-1} (D)$ and $f^{-1} (D) \subset f^{-1} (F)$. From this it is clear that $f^{-1}(E)$ is $\mu$ measurable. Since $\mu$ is arbitrary we are done.