Radon-Nikodym theorem

For sigma-finite measures, $\nu \ll \mu$ implies $\nu (A)={\int }_A(d\nu /d\mu )d\mu$ for a nonnegative measurable density unique almost everywhere.

For finite positive measures $\mu ,\nu$, put $\rho =\mu +\nu$. The linear functional $g\mapsto \int gd\nu$ is bounded on real $L^2(\rho )$ by the Cauchy-Schwarz inequality. The Riesz representation theorem gives a measurable $h$ with $\nu (A)={\int }_{A}hd\rho$. Testing indicators of its negative and greater-than-one level sets shows $0\le h\le 1$ almost everywhere. Subtraction gives $\mu (A)={\int }_A(1-h)d\rho$. This constructs the densities without assuming the Radon-Nikodym theorem as a prior result.

When $\nu \ll \mu$, the Radon-Nikodym derivative is the almost-everywhere unique measurable function satisfying $\nu (A)={\int }_A(d\nu /d\mu )d\mu$.

For an absolutely continuous change of probability measure, restricting its Radon-Nikodym derivative to the current sigma-algebra gives this nonnegative uniformly integrable martingale. If $Z_T>0$ almost surely, the measures are equivalent on ${\mathcal{F}}_T$. This finite-horizon equivalence need not imply equivalence on a larger sigma-algebra.

Mutual absolute continuity is equivalent to the Radon--Nikodym derivative being finite and strictly positive almost everywhere.