In Björk (2009) a Bayes' theorem is given by
Assume that $X$ is a random variable on $(\Omega, \mathcal F, P)$ and let $Q$ be another probability measure on $(\Omega, \mathcal F)$ with Radon-Nikodym derivative $$L = \frac{dQ}{dP} \; \; \text{on } \mathcal F.$$ Let $G$ be a sigma-algebra such that $G \subset \mathcal F$. Then $$E^Q[X\mid G] = \frac{E^P[LX\mid G]}{E^P[L\mid G]}.$$
The proof is to show that the integral over any $A \in G$ of the both sides of $$E^Q[X\mid G]E^P[L\mid G] = E^P[LX\mid G]$$ is the same quantity.
Starting with the left hand side,
$$\begin{aligned} \int_AE^Q[X\mid G]E^P[L\mid G]dP &= \int_AE^P\left[L\cdot E^Q[X\mid G]\mid G\right]dP \\ &=\int_AL\cdot E^Q[X\mid G]dP \\ &= \int_AE^Q[X\mid G]dQ \\ &= \int_AXdQ. \end{aligned}$$
The first equality follows from the fact that $E[X\mid G]$ is $G$-measurable. Why the second equality follows? My guess is that he moves $L$ inside the expectation under $Q$ and then applies the law of iterated expectations. However for this $L$ has to be $G$-measurable which is not generally the case since it is defined of $\mathcal F$ which is bigger than $G$.
Try to read the second equality backwards, what it says is nothing but the definition of the conditional expectation.