Consider a measurable space $(\mathcal{X},\Sigma)$ with two probability measures P and Q on it such that $P<<Q$ (P is absolutely continuous with respect to Q). Under what conditions can the likelihood ratio $\frac{P(B)}{Q(B)}$ be replaced by the Radon-Nikodym derivate $\frac{dP}{dQ}$?
I'm confused because as these measures are $\sigma$-finite, they are non-atomic. But, then what does it mean to define the derivate as a density on the underlying space?
The overall context of this problem is in defining the Bayesian rule measure-theoretically, i.e. if the outcome space is discrete we know that the posterior distribution can be obtained by multiplying the likelihood with the prior over some parameter space $\Theta$: $p(\theta|x)=\frac{p(x|\theta)}{p(x)}\pi(\theta).$ Now, when the outcome space is continuous, under the measure-theoretic formalism the likelihood is not defined as $p(x)=0$. I'm wondering if it makes sense to replace $\frac{p(x|\theta)}{p(x)}$ by the corresponding Radon-Nikodym derivative.