Let $\mu$ and $\nu$ be probability measures on a measure space $(\mathcal{X},\mathcal{M})$. We observe that both $\mu$ and $\nu$ are absolutely continuous with respect to the measure $\frac{\mu+\nu}{2}$, and hence $\mu$ admits a density (Radon-Nikodym derivative) $p = \frac{d\mu}{d\left(\frac{\mu+\nu}{2}\right)}$ and $\nu$ admits a density (Radon-Nikodym derivative) $q = \frac{d\nu}{d\left(\frac{\mu+\nu}{2}\right)}$ with respect to $\frac{\mu+\nu}{2}$.
Before I continue, I use three properties of the Radon-Nikodym derivative that I believe are true (but I'm not sure): $\frac{d(\mu+\nu)}{d\kappa} = \frac{d\mu}{d\kappa} + \frac{d\nu}{d\kappa}$, if $C \geq 0$, then $\frac{d(C\mu)}{d\kappa} = C\frac{d\mu}{d\kappa}$, and $\frac{d\mu}{d\mu} = 1$. Now I do the following calculation:
$$1 = \frac{d\left(\frac{\mu+\nu}{2}\right)}{d\left(\frac{\mu+\nu}{2}\right)} = \frac{1}{2}\frac{d\mu}{d\left(\frac{\mu+\nu}{2}\right)} + \frac{1}{2}\frac{d\nu}{d\left(\frac{\mu+\nu}{2}\right)} = \frac{p+q}{2}$$
Is this above calculation correct? It seems bizarre to me that $\frac{p(x)+q(x)}{2} = 1$ for both $\mu$- and $\nu$-almost every $x \in \mathcal{X}$. If it is correct, does anyone have an explanation for what is going on?