Let $\mathbb{P}$ be a complete probability measure on the measurable space $(\Omega,\mathcal{F})$. Define the measure $\mathbb{Q}$ on measurable sets $A \in \mathcal{F}$ by $$ \mathbb{Q}(A)\triangleq (1-\mathbb{P})I_{\{\mathbb{P}(A)\neq 0\}}. $$
Clearly $\mathbb{Q}$ is absolutely continuous with respect to $\mathbb{P}$. Therefore, the Radon-Nikodym theorem implies that it has a density $\frac{d\mathbb{Q}}{d\mathbb{P}}$.
Unless im missing something obvious, is there a concrete expression for$\frac{d\mathbb{Q}}{d\mathbb{P}}$?
$Q$ is not a measure. If $A$ and are disjoint sets with $P(A)>0$ and $P(B)>0$ then $Q(A\cup B)=1-P(A\cup B)=1-P(A)-P(B)$ whereas $Q(A)+Q(B)=2-P(A)-P(B)$.