Results 1 to 2 of 2

Thread: Rondon Nikodym derivative

  1. #1
    Member
    Joined
    Dec 2008
    Posts
    154

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Shanks's Avatar
    Joined
    Nov 2009
    From
    BeiJing
    Posts
    374
    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].
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. contuous weak derivative $\Rightarrow$ classic derivative ?
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Apr 22nd 2011, 02:37 AM
  2. Replies: 0
    Last Post: Jan 24th 2011, 11:40 AM
  3. Radon Nikodym computation, step of a proof
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Mar 5th 2010, 10:09 AM
  4. Replies: 2
    Last Post: Nov 6th 2009, 02:51 PM
  5. Total Derivative vs. Directional Derivative
    Posted in the Advanced Math Topics Forum
    Replies: 5
    Last Post: May 30th 2008, 08:42 AM

Search Tags


/mathhelpforum @mathhelpforum