Finite tie-broken projection searches are Borel measurable

Put $R_i=I-P_i$. By [F1] and [F7], $R_i$ is linear and $\mathrm{\parallel }{R}_{i}z{\mathrm{\parallel }}_2\le 2\mathrm{\parallel }z{\mathrm{\parallel }}_2$. For $y,z\in {\mathbb{R}}^n$, [F6] and [F7] give $\mathrm{\mid }\mathrm{\parallel }{R}_{i}y{\mathrm{\parallel }}_2^2-\mathrm{\parallel }{R}_{i}z{\mathrm{\parallel }}_2^2\mathrm{\mid }\le \mathrm{\parallel }{R}_i(y-z){\mathrm{\parallel }}_2\mathrm{\parallel }{R}_i(y+z){\mathrm{\parallel }}_2\le 4\mathrm{\parallel }y-z{\mathrm{\parallel }}_2\mathrm{\parallel }y+z{\mathrm{\parallel }}_2\mathrm{.}$ For fixed $z$, the last factor is bounded when $y$ is near $z$, so the squared residual and hence $q_i$ are continuous.

By [F4], a minimum score exists. Its least minimizing index is $i$ exactly on the set $A_i={\bigcap }_{j<i}\{y:q_i(y)<q_j(y)\}\cap {\bigcap }_{j>i}\{y:q_i(y)\le q_j(y)\}\mathrm{.}$ Each strict comparison is open and each non-strict comparison is closed because the score differences are continuous and [F5] applies to open and closed sets. Thus every $A_i$ is Borel by [F2]; the sets are disjoint and cover ${\mathbb{R}}^n$.

For any open $O\subseteq {\mathbb{R}}^n$, ${\widehat{m}}^{-1}(O)={\bigcup }_{i=1}^N(A_i\cap P_i^{-1}(O))$. Each $P_i^{-1}(O)$ is open by [F1, F5], so this finite union is Borel. If $B$ ranges over subsets of ${\mathbb{R}}^n$ whose inverse images under $\widehat{m}$ are Borel, those subsets form a sigma-algebra containing every open set; [F2] therefore puts every Borel $B$ in this family. By [F3], $\widehat{m}$ is Borel measurable. The index-map preimage of any set of indices is a finite union of the corresponding $A_i$, so the index map is measurable as well.