MEMS gyro measured angular velocity vector
Notations I use:
Consider a world frame ${O_0, \vec{i}_0, \vec{j}_0, \vec{k}_0}$ coordinate system and a rigid body $B$ with a coordinate system ${O_1, \vec{i}_1, \vec{j}_1, \vec{k}_1}$ attached to its center of mass $O_1$. Suppose $$\begin{bmatrix} \vec{i}_1\ \vec{j}_1\ \vec{k}_1\end{bmatrix} = R(\theta, \chi,\phi) \begin{bmatrix} \vec{i}_0\ \vec{j}_0\ \vec{k}_0\end{bmatrix} $$ Then for a point $P_1 \in B$ one has: $$ \overrightarrow{O_1 P_1} = \begin{bmatrix} x_1 &y_1 &z_1 \end{bmatrix} \cdot \begin{bmatrix} \vec{i}_1\ \vec{j}_1 \ \vec{k}_1\end{bmatrix} = \begin{bmatrix} x_1 &y_1 &z_1 \end{bmatrix} R(\theta,\chi, \psi) \begin{bmatrix} \vec{i}_0\ \vec{j}_0 \ \vec{k}_0\end{bmatrix} \ \begin{bmatrix} x_0 &y_0 &z_0 \end{bmatrix} \begin{bmatrix} \vec{i}_0\ \vec{j}_0 \ \vec{k}_0\end{bmatrix} $$ therefore
$$\begin{bmatrix} x_0\ y_0 \ z_0\end{bmatrix} = R^T(\theta, \chi, \phi) \begin{bmatrix} x_1\ y_1 \ z_1\end{bmatrix} $$
Problem statement:
It is known that $$ \dot{R}^T = [\omega] \cdot R^T$$ where $[\omega]$ is the skew-symmetric form of the angular velocity vector $\vec{\omega}$, in world coordinate system.
I am using a MEMS gyroscope to measure orientation, and according to this I think the output of the gyro (assuming a $3$ axis sensor) is the vector $\vec{\omega}$, BUT in body coordinate system ... Is this right? Then what I am actually measuring is $\vec{\omega}_B = R^T \vec{\omega}$ ...
Further more if I want to use DCM algorithm to get the rotation matrix then using the fact that $[R^T \omega] = R^T [\omega] R$ we get $$ \dot{R}^T = - R^T [\omega] = - R^T[\omega] R R^T = -[R^T \omega] R^T = -[\omega_B] R^T$$ where $\omega_B$ is the output of the MEMS gyro ... Is this correct?
Is seems, as suggested in the answer below, that actually $ \omega = R^T \omega_B$ hence the differential equation for $R$ is
$$ \dot{R}^T = [R^T\omega_B] R^T = R^T [\omega_B]$$ 101