Let $\mu$ and $\nu$ be two $\sigma$ -finite measures on the same measurable space , such that $\nu\ll \mu$ (i.e. $\nu$ is absolutely continuous with respect to $\mu$ .) Then there exists a measurable function $f$ , which is nonnegative and finite, such that for each , $$ \nu(A)=\int_A fd\mu $$ This function is unique (any other function satisfying these conditions is equal to $f$ $\mu$ -almost everywhere,) and it is called the Radon-Nikodym derivative of $\nu$ with respect to $\mu$ , denoted by $f = \frac{d\nu}{d\mu}$ .

Remark. The theorem also holds if $\nu$ is a signed measure. Even if $\nu$ is not $\sigma$ -finite the theorem holds, with the exception that $f$ is not necessarely finite.

Some properties of the Radon-Nikodym derivative

Let $\nu$ , $\mu$ , and $\lambda$ be $\sigma$ -finite measures in .

1. If $\nu \ll \lambda$ and $\mu\ll\lambda$ , then $$ \frac{d(\nu+\mu)}{d\lambda} = \frac{d\nu}{d\lambda}+\frac{d\mu}{d\lambda}\;\; \mu\mbox{-almost everywhere} $$
2. If $\nu\ll\mu\ll\lambda$ , then $$ \frac{d\nu}{d\lambda}=\frac{d\nu}{d\mu}\frac{d\mu}{d\lambda} \;\; \mu\mbox{-almost everywhere} $$
3. If $\mu\ll\lambda$ and $g$ is a $\mu$ -integrable function, then $$ \int_\Omega gd\mu = \int_\Omega g\frac{d\mu}{d\lambda}d\lambda $$
4. If $\mu\ll\nu$ and $\nu \ll\mu$ , then $$ \frac{d\mu}{d\nu}=\left(\frac{d\nu}{d\mu}\right)^{-1} $$