Borel subspaces of polish spaces are standard borel

Apply [F1] with its AC hypothesis [F3]. It gives a Polish topology on B whose Borel sigma-algebra is the trace of that of P. The identity map from the trace measurable space to this presentation is therefore bimeasurable, proving the general assertion, including B empty.

For the instance, R is complete by [F4] and separable by [F2]–[F5], hence Polish. Each singleton rational is closed in R (a point outside it has a ball avoiding it), so Q and every subset of Q are Borel by [F2] and countable unions. Thus $\mathcal{B}(\mathbb{R}){\mathrm{\mid }}_{\mathbb{Q}}=\mathcal{P}(\mathbb{Q})$. The discrete metric $d(q,r)={\mathbf{1}}_{\{q\mathrm{\ne }r\}}$ is complete because a Cauchy sequence is eventually constant; Q itself is countable dense for this topology. Its Borel sets are again all subsets. For example the preimage of $\{1\mathrm{/}2,2\mathrm{/}3\}$ under the identity is exactly $\{1\mathrm{/}2,2\mathrm{/}3\}$ in both measurable structures. This gives the claimed explicit Polish presentation without requiring the inherited metric to be complete.