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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?

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The contrary is not that $f \geq 1$. The contrary is that there exists a measurable set $E$ of positive measure on which $f > 1$. –  copper.hat Apr 16 '12 at 5:53
    
Also, be careful with your inequalities. You can certainly choose $\lambda = 0$, in which case $\frac{d \nu}{d \mu} = 1$ $\mu$ a.e. –  copper.hat Apr 16 '12 at 5:55
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Why don't you just use the fact that since $\nu(E) + \lambda(E) = \mu(E)$ for all measurable $E$, then $\nu(E) \leq \mu(E)$. Writing this in integral form, we have $\int_E \frac{d \nu}{d \mu} d \mu \leq \int_E 1 d \mu$. From this you should be able to conclude that $\frac{d \nu}{d \mu} \leq 1$ $\mu$-a.e.

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But then I wand $\frac{d\nu}{d\mu}\lt 1$ –  Jay Apr 16 '12 at 12:33
    
As I pointed out above, your hypotheses include the case where $\lambda=0$, so, sorry, you can want it all you like, but unless you add additional hypotheses, the $\leq$ is all you can expect. –  copper.hat Apr 16 '12 at 15:00
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