The adaptive truncated covariance procedure is well-defined on its declared domain

The integer $k=\lfloor 3.5\beta \rfloor +1$ satisfies $1\le k\le m\mathrm{/}2$ on the declared domain. The deterministic balanced partition therefore has $k$ nonempty blocks, and each block mean is a continuous linear function of the sample. By [F2], applying the minimum-norm geometric median gives a single-valued Borel pilot.

For each finite symmetric matrix, [F3] defines its spectral truncation independently of the eigenbasis. To check continuity in the matrix, let $A_n\to A$ be symmetric and choose $R$ larger than all their operator norms. By [F4], approximate the scalar clipping function uniformly on $[-R,R]$ by a polynomial $q$ within $\eta >0$. Spectral diagonalization [F3] gives $\mathrm{\parallel }\psi (B)-q(B){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \eta$ for $B=A,A_n$; polynomial evaluation is continuous in matrix entries, so $q(A_n)\to q(A)$. Hence ${\mathrm{lim\hspace{0.17em}sup}}_n\mathrm{\parallel }\psi (A_n)-\psi (A){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 2\eta$, and letting $\eta \downarrow 0$ proves continuity. Each ${\theta }_j$ is positive and finite. The grid consists exactly of integers $0\le j<L$, where $L={\log }_2(2{\sigma }_{\max }\mathrm{/}{\sigma }_{\min })\ge 1$; it is finite by [F5] and nonempty since $j=0$ belongs to it.

The largest grid index satisfies the agreement rule vacuously, so the set of admissible indices is nonempty and has a least element $j^{\ast }$. For each fixed pair of indices, the matrix difference is Borel in the sample: the pilot is Borel by step 1.1 and the remaining finite matrix operations and truncation are continuous by step 1.2. Its operator norm is Borel as the composition with a continuous norm. Therefore the event that a given index satisfies all finitely many comparisons is Borel; taking the least among finitely many such indices is a Borel rule. The output is thus defined and Borel on the whole declared domain.