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$\mu, \nu$ measures $(X,M)$ with $\nu \ll \mu$ and $\lambda = \mu + \nu$.

Show: $f=\frac{d\nu}{d\lambda}$ then $0 \leq f < 1$ $\mu$-a.e. and $\frac{d\nu}{d\mu} = \frac{f}{1-f}$.

In our context a measure is positive if not stated differently. So in this case $\mu,\nu$ and $\lambda$ are positiv. Let $A\in M$ with $\lambda(A)=0$ then it follows from $\lambda(A)=\mu(A)+\nu(A)=0$ that $\nu(A) =0$ and therefore it is $\nu \ll \lambda$ (and also $\mu \ll \lambda$). Then according to Radon-Nikodym it follows $\nu(A) = \int_A f d\lambda$.

That $f \geq 0$ $\mu$-a.e. follows from $\mu \ll \lambda$ and $0 < \nu(A) = \int_A f d\lambda$ for all $A$ (therefore $f \geq 0$ $\lambda$-a.e.).

For the second part I tried proof by contradiction. Assume $\mu(A) > 0$ and $f(x) > 1$ for all $x\in A$. From $\lambda(A) = \mu(A) + \nu(A)$ it follows then, that $\lambda(A) > 0$.

Then $\nu(A) = \int_A f d\lambda > \int_A 1 d\lambda = \lambda(A)$.

So that $\lambda(A) = \mu(A) + \nu(A) > \mu(A) + \lambda(A)$ which contradicts $\mu(A) > 0$.

But then how to show the last part?

Since $\nu \ll \mu$ it is (already proven) $\frac{d\nu}{d\lambda} = \frac{d\nu}{d\mu}\cdot\frac{d\mu}{d\lambda}$

So $\int_A f\cdot\frac{1}{1-f} d\mu \overset{!}{=} \nu(A) = \int_A \frac{d\nu}{d\lambda} d\lambda = \int_A \frac{d\nu}{d\mu}\cdot\frac{d\mu}{d\lambda} d\lambda$

So I need to show $1-f=\frac{d\lambda}{d\mu}$ (since $\lambda \ll \mu$ and $\mu \ll \lambda$ it is proven that $\frac{d\lambda}{d\mu} = \left( \frac{d\mu}{d\lambda} \right) ^{-1}$)

$\int_A 1-f d\mu = \mu(A) - \int_A \frac{d\nu}{d\lambda}d\mu = \cdots$

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Hm, I don't see how to get there.. I somehow got stuck in certain reasoning patterns now after trying too long.. – numaaa Feb 3 '13 at 19:49

You have to show $$\frac{f}{1-f} \stackrel{!}{=} \frac{\mathrm{d}\nu}{\mathrm{d}\mu} = \frac{\mathrm{d}\nu}{\mathrm{d}\lambda} \frac{\mathrm{d}\lambda}{\mathrm{d}\mu},$$ so the Ansatz would have to be $\frac{\mathrm{d}\mu}{\mathrm{d}\lambda} = 1-f$ (which makes sense, since $\mu = \lambda - \nu$).

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