User:Caliburn/s/mt/Sum Rule for Radon-Nikodym Derivatives

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

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


 * $\nu$ and $\mu$ are absolutely continuous with respect to $\lambda$.

Then:


 * $\ds \frac {\d \paren {\nu + \mu} } {\d \lambda} = \frac {\d \nu} {\d \lambda} + \frac {\d \mu} {\d \lambda}$

Theorem
From the Radon-Nikodym Theorem, there exists a positive $\Sigma$-measurable $g_1 : X \to \hointr 0 \infty$ such that:


 * $\ds \map \nu A = \int_A g_1 \rd \lambda$

for each $A \in \Sigma$.

Then, from the definition of the Radon-Nikodym derivative, we have:


 * $\ds \frac {\d \nu} {\d \lambda} = \eqclass {g_1} \sim$

From the Radon-Nikodym Theorem, there also exists a positive $\Sigma$-measurable $g_2 : X \to \hointr 0 \infty$ such that:


 * $\ds \map \mu A = \int_A g_2 \rd \lambda$

for each $A \in \Sigma$.

Then, from the definition of the Radon-Nikodym derivative, we have:


 * $\ds \frac {\d \mu} {\d \lambda} = \eqclass {g_2} \sim$

From Integral of Positive Measurable Function is Additive, we have:


 * $\ds \map {\paren {\nu + \mu} } A = \int_A \paren {g_1 + g_2} \rd \lambda$

for each $A \in \Sigma$.

So, from the definition of the Radon-Nikodym derivative, we have:


 * $\ds \frac {\d \paren {\nu + \mu} } {\d \lambda} = \eqclass {g_1 + g_2} \sim$

From , we have:


 * $\eqclass {g_1 + g_2} \sim = \eqclass {g_1} \sim + \eqclass {g_2} \sim$

giving:


 * $\ds \frac {\d \paren {\nu + \mu} } {\d \lambda} = \frac {\d \nu} {\d \lambda} + \frac {\d \mu} {\d \lambda}$

as required.