The setup in Shao's Theorem 2.2 (Factorization theorem) is as follows:
Let $X$ be a sample from $P\in \mathcal P$, a family of probability measures on $(\mathbb R^n,\mathcal B(\mathbb R^n))$ dominated by a $\sigma-$finite measure $\mu$, i.e. $P\ll \mu$ for $P\in \mathcal P$. Suppose $T$ is a measurable function such that for every $P\in \mathcal P$, $$\frac{dP}{d\mu}(x) = g_P(T(x))h(x)$$ where $g_P$ is a function allowed to depend on $P$ and $h$ does not depend on $P$.
Then it can be shown that there exists a probability measure $$Q = \sum_{i\geq 1}c_i P_i$$ for $c_i\geq 0, \sum c_i = 1$, and $P_i \in \mathcal P$ such that $Q$ dominates $\mathcal P$, so that one has $P\ll Q\ll \mu$ for every $P\in \mathcal P$. It also follows that $$\frac{dQ}{d\mu} = \sum_{i\geq 1}c_i \frac{dP_i}{d\mu}$$
The following line is what I am confused on. Shao claims that $$\frac{dP}{dQ} = \frac{dP}{d\mu}\bigg/\sum_{i\geq 1}c_i \frac{dP_i}{d\mu} = g_P(T)\bigg/\sum_{i\geq 1}g_{P_i}(T),\quad Q-\text{a.s.}$$ The first equality is clear, it follows from the chain rule for Radon-Nikodym derivatives. I am stuck on the second equality, as substituting our hypothesis about $T$, it seems that $$\frac{dP}{d\mu}\bigg/\sum_{i\geq 1}c_i \frac{dP_i}{d\mu} =g_P(T(x))h(x)\bigg/\sum_{i\geq 1}c_ig_{P_i}(T(x))h(x)$$ But where did the $c_i$ go?