Lemma A.1 — local uniqueness within a single outcome subspace's refinements

\emphFix an outcome subspace $\mathcal H_i$ with $\dim\mathcal H_i\ge2$. Among all assignments of the form $\mu([\psi],P_i)=f(\|\psi_i\|)$ with $f$ continuous (or merely monotonic), the only choice consistent with additivity over orthogonal refinements of $\mathcal H_i$ (condition A.2.3) is $f(x)=kx^2$ for a constant $k$.

Proof. By the Pythagorean theorem, an orthogonal refinement $\mathcal H_i=\bigoplus_\alpha \mathcal H_{i,\alpha}$ satisfies $\|\psi_i\|^2=\sum_\alpha\|\psi_{i,\alpha}\|^2$. Writing $g(t):=f(\sqrt t)$ for $t\ge0$, additivity becomes $g\big(\sum_\alpha t_\alpha\big)= \sum_\alpha g(t_\alpha)$ for arbitrary non-negative $t_\alpha$ summing to $t$ (realizable because $\dim\mathcal H_i\ge2$ allows continuously varying splits). This is Cauchy's functional equation on $\mathbb R_{\ge0}$; under continuity (or monotonicity), $g(t)=kt$, hence $f(x)=kx^2$. $\blacksquare$

Remark (what this does and does not establish). Lemma A.1 fixes the exponent to 2 only for outcome subspaces of dimension $\ge2$, and only for consistency of $f$ within one subspace's own internal refinements. It does not by itself show that a single function $f$ can be chosen consistently across every possible resolution of the identity on $\mathcal H$ (i.e. across every choice of measurement context) — the cross-context/global-consistency step that is the actual technical heart of Gleason's theorem, and the reason that theorem requires $\dim\mathcal H\ge3$ and is known to fail at $\dim\mathcal H=2$. Nor does it treat mixed states. \textbfThe claim that tensor-product composition singles out $p=2$ (as opposed to normalization, addressed separately in Lemma A.2 below) is false: for product states, $\|(\psi\otimes\phi)_{ij}\|=\|\psi_i\|\, \|\phi_j\|$, so $\mu_p(\psi\otimes\phi)=\mu_p(\psi)\mu_p(\phi)$ for every $p>0$, not only $p=2$.