Radon–Nikodym derivative

Remarks

Example: let $\mu$ be Lebesgue measure on $\R$ and $\nu(A) = \int_A 2x \, \mathbf{1}_{[0,1]}(x) \, dx$. Then $\nu$ is a measure with density, and $\frac{d\nu}{d\mu}(x) = 2x$ for $x \in [0,1]$ and $0$ elsewhere, up to a null set.

Non-example: the Dirac measure $\delta_0$ has no Radon–Nikodym derivative with respect to Lebesgue measure $\lambda$. Indeed $\lambda(\{0\}) = 0$ but $\delta_0(\{0\}) = 1$, so $\delta_0$ is not absolutely continuous with respect to $\lambda$.

For every measurable $g \colon X \to [0, \infty]$, $$\int_X g \, d\nu = \int_X g \, \frac{d\nu}{d\mu} \, d\mu.$$ If $\nu \ll \mu \ll \rho$ with all three σ-finite, then the chain rule $\frac{d\nu}{d\rho} = \frac{d\nu}{d\mu} \cdot \frac{d\mu}{d\rho}$ holds $\rho$-almost everywhere.