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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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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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