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](https://mathoverflow.net/q/391566/485160)).


Any help would be much appreciated!



Separable C* algebras and type I states
I don't think $\pi\_\omega(A)''$ has that form. For example, take $A = M\_2$ and let $\omega$ be the normalized trace. Then $\omega = \frac{1}{2}(\psi\_1 + \psi\_2)$ where $\psi\_i(x) = \langle xe\_i, e\_i\rangle$, for $x \in M\_2$ and $\{e\_1,e\_2\}$ the standard basis of $\mathbb{C}^2$. That is, $\omega$ is the integral $\int \psi\, d\mu(\psi)$ where $\mu = \frac{1}{2}(\delta\_{\psi\_1} + \delta\_{\psi\_2})$. Then $\pi\_\omega(M\_2)'' \cong \pi\_{\psi\_1}(M\_2)'' \cong \pi\_{\psi\_2}(M\_2)'' \cong M\_2$, so $\pi\_\omega(M\_2)'' \not\cong \pi\_{\psi\_1}(M\_2)'' \oplus \pi\_{\psi\_2}(M\_2)''$.


We do have an isomorphism between $H\_\omega$ and $H\_{\psi\_1} \oplus H\_{\psi\_2}$ which takes $\pi\_\omega(x)$ to $\pi\_{\psi\_1}(x) \oplus \pi\_{\psi\_2}(x)$, but of course that doesn't make $\pi\_\omega(A) \cong \pi\_{\psi\_1}(A) \oplus \pi\_{\psi\_2}(A)$.


