Im trying to prove a generalization of the Radon Nykodym theorem, but im having troubles even for finite measures, could someone help?

Let $\mu$ and $\nu$ two $\sigma$-finite measures in $\(X,\mathcal{F})$. If $\nu$ << $\mu$, then there exists a non-negative function $h \in L^1(X,\mu)$, such that for every function $F\in M^+(X,\mathcal{F})$, it is satisfied that $\int_X F(x)\\,d\nu =\int_X F(x)h(x)\\,d\mu$

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Dear goblin, why did you revert my corrections of spelling and formatting? –  Yemon Choi Nov 25 '12 at 2:51
Im sorry, didnt noticed –  goblin Nov 25 '12 at 3:01
It seems Theo Buehler added the tag "real" which should be eliminated. –  S. Carnahan Nov 25 '12 at 3:04
@goblin: why do you call this a generalization? This is the usual statement of the theorem, up to some incorrectness. The hypothesis "$\nu$ absolutely continuous wrto $\mu$" is denoted "$\nu < < \mu$", not " $\mu < < \nu$ ". In general, $h$ is not in $L^1(X,\mu)$ if $\mu(X)=+\infty$ (e.g. it is identically $1$ if $\mu=\nu$). –  Pietro Majer Nov 25 '12 at 7:37
Please do not crosspost at M.SE and MO (math.stackexchange.com/questions/244064/…). –  Michael Greinecker Nov 25 '12 at 9:47

If you assume you have proven the Radon-Nikodym theorem for measures $\mu,\nu$, $\nu <<\mu$, then for any $F$ in such a class, the measure $\mu_F,\nu_F$ defined by

$\mu_F(E):=\int_E F(x)\;d\mu$,

$\nu_F(E):=\int_E F(x)\;d\nu$,

then since $\nu_F << \mu_F$ (if your definition of absolute continuity is equivalent to $\nu(E)=0$ whenever $\mu(E)=0$), you can apply the Radon-Nikodym theorem for $\nu_F,\mu_F$ and achieve the desired result.

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If $h$ is a density of $\nu$ w.r.t. $\mu$ then $\int F d\nu = \int Fh d\mu$ holds for all indicator functions (just by the definitions of a density and the integral). This extends to simple measurable functions (taking only finitely many positive values) by linearity and then to all positive measurable functions by monotone convergence.

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