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Let $\mu,\lambda$ and $\nu$ be $\sigma$-finite measures on $(X,M)$ such the $\nu\ll \mu$. Let $\mu= \nu + \lambda$. Then if $f$ is the Radon-Nikodym derivate of $\nu$ wrt $\mu$, we have $0\leq f\lt 1~\mu$-a.e. where $f = \frac{d\nu}{d\mu}$.

Approach.

Suppose to the contrary that $f\geq 1$. Then since I can infer $\nu \ll \mu$, $$\nu(E) = \int_E f~d\mu \geq \int_E 1 d\mu = \mu(E).$$ But then $\nu(E) \leq \nu(E) + \lambda(E) = \mu(E)$. So a contradiction. Thus $f\lt 1~\mu$-a.e.

Please, does this look okay?

  • 1
    The contrary is not that $f \geq 1$. The contrary is that there exists a measurable set $E$ of positive measure on which $f > 1$.2012-04-16
  • 0
    Also, be careful with your inequalities. You can certainly choose $\lambda = 0$, in which case $\frac{d \nu}{d \mu} = 1$ $\mu$ a.e.2012-04-16

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