# Rondon Nikodym derivative

• January 8th 2010, 03:02 PM
Kat-M
Rondon Nikodym derivative
Let $\mu,\nu$ be measures on $(X,F)$ such that $\nu << \mu$ and $\lambda = \mu+\nu$. Show that $f= \frac{d\nu}{d\lambda}$ exists and satisfies $0 \leq f \leq 1$ a.e $\lambda$.
(Hint: $\mu_1 << \mu_2$ and $\mu_2 << \mu_1$ implies $\frac{d\mu_1}{d\mu_2}=(\frac{d\mu_2}{d\mu_1})^{-1}$.
i do not have any idea how to do this. any help would be appreciated.
• January 9th 2010, 01:47 AM
Shanks
since $v<,therefore $v<<\lambda$.
the Radon Nikodym Theorem asserts that $f=\frac{dv}{d\lambda}$exist.
similarly,since $u<<\lambda, g=\frac{du}{d\lambda}$exist.
Therefore, $d\lambda=du+dv=gd\lambda+fd\lambda$
therefore $1=g+f$ a.e.[\lambda]
since g is nonnegative, which implies $0\geq f\geq 1$, a.e.[\lambda].