# Rondon Nikodym derivative

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