Radon-Nikodym Theorem

Theorem
Let $\struct {X, \Sigma}$ be a measurable space.

Let $\mu$ and $\nu$ be $\sigma$-finite measures on $\struct {X, \Sigma}$ such that:


 * $\nu$ is absolutely continuous with respect to $\mu$.

Then there exists a $\Sigma$-measurable function $g : X \to \hointr 0 \infty$ such that:


 * $\ds \map \nu A = \int_A g \rd \mu$

for each $A \in \Sigma$.

Further, if $g_1 : X \to \hointr 0 \infty$ and $g_2 : X \to \hointr 0 \infty$ are such that:


 * $\ds \map \nu A = \int_A g_1 \rd \mu = \int_A g_2 \rd \mu$

for each $A \in \Sigma$, then:


 * $g_1 = g_2$ $\mu$-almost everywhere.

Proof
We first prove the case of $\mu$ and $\nu$ finite.

Define $\mathcal F$ to be the set of $\Sigma$-measurable functions $f : X \to \overline \R_{\ge 0}$ with:


 * $\ds \int_A f \rd \mu \le \map \nu A$

for each $A \in \Sigma$.

We show that there exists $g \in \mathcal F$ such that:


 * $\ds \int f \rd \mu = \sup \set {\int f \rd \mu : f \in \mathcal F}$

and that this $g$ has:


 * $\ds \map \nu A = \int_A g \rd \mu$

for each $A \in \Sigma$.