> [!theorem] > > Let $(X, \cm)$ be a [[Measure Space|measurable space]], $\nu$ be a $\sigma$-finite [[Signed Measure|signed measure]], and $\mu$, $\lambda$ be $\sigma$-finite positive measures on $(X, \cm)$ such that $\nu$ is [[Absolute Continuity|absolutely continuous]] with respect to $\mu$, and $\mu \ll \lambda$. > > Let $g \in L^1(\nu)$ is [[Integrable Function|integrable]] with respect to $\nu$, then $g \cdot \frac{d\nu}{d\mu} \in L^1(\mu)$ is integrable with respect to $\mu$. The integral can be evaluated with respect to $\mu$ instead with the [[Lebesgue-Radon-Nikodym Theorem|Radon-Nikodym derivative]]: > $ > \int g d\nu = \int g \frac{d\nu}{d\mu}d\mu > $ > > Moreover, $\nu \ll \lambda$ and > $ > \frac{d\nu}{d\lambda} = \frac{d\nu}{d\mu} \cdot \frac{d\mu}{d\lambda} \quad \lambda\ \text{a.e.} > $ > *Proof*. First suppose that $\nu$ is a $\sigma$-finite positive measure. Let $\chi_E$ be an [[Indicator Function|indicator function]], then > $ > \int \chi_{E}d\nu = \nu(E) = \int_E\frac{d\nu}{d\mu}d\mu = \int \chi_{E}\frac{d\nu}{d\mu}d\mu > $ > which extends to any [[Simple Function|simple functions]] by linearity. Let $g \in L^+(\nu)$ and $\seq{\phi_n}$ be a [[Sequence|sequence]] of simple functions such that $\phi_n \upto g$. Then > $ > \int g d\nu = \limv{n}\int \phi_n d\nu = \limv{n}\int \phi_n \frac{d\nu}{d\mu}d\mu = \int g\frac{d\nu}{d\mu}d\mu > $ > by the [[Monotone Convergence Theorem]]. In which case, if $g \in L^1(\nu) \cap L^+(\nu)$, $g\frac{d\nu}{d\mu} \in L^1(\mu) \cap L^+(\mu)$ as well. For any $g \in L^1(\nu)$, > $ > \begin{align*} > \int gd\nu &= \int g^+ d\nu - \int g^- d\nu \\ > &= \int g^+ \frac{d\nu}{d\mu}d\mu - \int g^-\frac{d\nu}{d\mu}d\mu \\ > &= \int g\frac{d\nu}{d\mu}d\mu > \end{align*} > $ > Now suppose that $\nu$ is a signed $\sigma$-finite measure. Let $\nu = \nu^+ - \nu^-$ be its [[Jordan Decomposition Theorem|Jordan decomposition]], then > $ > \begin{align*} > \int g d\nu &= \int gd\nu^+ - \int gd\nu^- \\ > &= \int g\frac{d\nu^+}{d\mu}d\mu - \int d\frac{d\nu^-}{d\mu}d\mu \\ > &= \int g\paren{\frac{d\nu^+}{d\mu} - \frac{d\nu^-}{d\mu}}d\mu \\ > &= \int g\frac{d\nu}{d\mu}d\mu > \end{align*} > $ > Lastly, let $g = \chi_{E}\frac{d\nu}{d\mu}$, then > $ > \nu(E) = \int \paren{\chi_{E}\frac{d\nu}{d\mu}}d\mu > = \int \paren{\chi_{E}\frac{d\nu}{d\mu}}\frac{d\mu}{d\lambda}d\lambda = \int_E\frac{d\nu}{d\mu}\frac{d\mu}{d\lambda}d\lambda > $ > and since both are Radon-Nikodym derivatives, > $ > \frac{d\nu}{d\lambda} = \frac{d\nu}{d\mu}\frac{d\mu}{d\lambda} \quad \lambda\ \text{a.e.} > $ > they are equal $\lambda$-[[Almost Everywhere|a.e.]]