Showing $m_I = m_g$ follows from the equivalence principle
The inertial mass of an object is defined as its resistance to acceleration by $\vec{F}_{net} = m_I\vec{a}$. The gravitational mass of an object is defined as the scaling of the gravitational force an object experiences by $\vec{F}_g = m_g\vec{g}$ where $\vec{g}$ is assumed to be uniform in the following. The equivalence principle states the following: A frame at rest relative to a uniform gravitational field $\vec{g}$ is equivalent to a frame accelerating at a constant rate $\vec{g}$ relative to a completely isolated inertial frame. It is my understanding that we may demonstrate that $m_I = m_g$ by the equivalence principle. We should not have to assume that all objects fall with the same acceleration in a uniform gravitational field, rather this is what we would like to demonstrate. To this end, I produced the following.
If we let S and S' be inertial and non-inertial reference frames with $\vec{A}$ the acceleration of S' relative to S then $m_I\vec{a'} = -m_I\vec{A} + \vec{F}_1 + \vec{F}_2 \,+ ...$ where $\vec{F}_i$ are real forces. Now, according to the equivalence principle, the gravitational force $\vec{F}_g = m_g\vec{g}$ is fictitious. If we consider a frame of reference accelerating at $\vec{g}$ relative to a completely isolated inertial reference frame then an isolated object of inertial mass $m_I$ and gravitational mass $m_g$ will have acceleration $\vec{a'}$ = -$\vec{g}$ by the formula above. 766
Now, let's see what the equivalence principle says about this if we analyze the situation from a frame of reference at rest relative to the gravitational field. We must have the following for the freely falling object. $-m_I\vec{g} = m_I\vec{a'} = -m_g\vec{g}$ since the fictitious term is the gravitational force. We immediately see that $m_I = m_g$.
My question concerns my statement of the equivalence principle. Typically it is stated as follows: A freely falling frame in a uniform gravitational field $\vec{g}$ is equivalent to an isolated inertial reference frame. The problem is that if I start from there, I have to assume that the frame falls at a constant rate $\vec{g}$ in the field which is what we prove if we show that $m_I = m_g$. This exercise to show $m_I = m_g$ follows from the equivalence principle came from a book I'm reading. I'm curious if there's any way to approach the problem from the more traditional way of stating the equivalence principle or if I've essentially captured the essence of the problem. 766