Let $A$ be a separable $C^*$-algebra and let $\omega$ be a state on $A$. Then there is an "orthogonal" probability measure $\mu$ on the pure state space $P(A)$ of $A$ such that $\omega(x) = \int_{P(A)} \psi(x) \, d\mu(\psi)$ [Takesaki 1, IV.6.28]).

If I understand correctly the orthogonality of $\mu$ means that the GNS representation of $\omega$ is (unitarily equivalent to) a direct integral: $$ (H_\omega,\pi_\omega) = \int_{P(A)}^\oplus (H_\psi,\pi_\psi) \, d\mu(\psi) $$ and $\Omega_\omega = \int_{P(A)}^\oplus \Omega_\psi\,d\mu(\psi)$ (see for example [Takesaki 1, IV.8.31].

But doesn't this imply that the von Neumann algebra $\pi_\omega(A)''$ takes the form $$\pi_\omega(A)'' = \int_{P(A)}^\oplus\pi_\psi(A)'' \,d\mu(\psi) = \int_{P(A)}^\oplus B(H_\psi) \,d\mu(\psi)$$ (because $\psi$ is pure one has $\pi_\psi(A)''=B(H_\psi)$). This would imply that $\pi_\omega(A)''$ is a type I von Neumann algebra since it can be written as a direct integral of type I factors, right? The argument must be wrong since not every state on a separable $C^*$-algebra is a type I state (see the answer to this question Factor states on C*-algebras).

Any help would be much appreciated!