Let $X$ be a separable real normed space with a $\sigma$-algebra containing the open sets, and let $\omega \mapsto A_\omega$ be a strongly measurable family of bounded operators on $X$ (each $\omega \mapsto A_\omega x$ measurable). Then for $k \ge 1$:

• $\omega \mapsto \rho_k(A_\omega)$ is measurable, and so is $\omega \mapsto \|A_\omega\|$;
• $\omega \mapsto A_\omega v(\omega)$ is measurable for every measurable $v : \Omega \to X$ (if the $\sigma$-algebra of $X$ is Borel);
• $\omega \mapsto g(A_\omega, V)$ is measurable for every subspace $V \ne 0$, and $\omega \mapsto g(A_\omega, E(\omega))$ is measurable for every measurable $E : \Omega \to \mathcal{G}_k X$ (if the $\sigma$-algebra of $X$ is Borel);
• if moreover $X$ is complete and $E : \Omega \to \mathcal{G}_k X$ is measurable, then any $F : \Omega \to \mathcal{G}_k X$ with $F(\omega) = A_\omega E(\omega)$ for every $\omega$ is measurable.

For the norm topology on the operators, $T \mapsto \rho_k(T)$ is itself Borel measurable.