I am trying to show that the measure $\mu_1(E)=\sum_{i=1}^{n}c_i\cdot\mu_2(E\cap E_i)$ is absolutely continuous w.r.t $\mu_2$ and then compute its Radon-Nikodym derivative. Here $E_i$ are measurable sets and $c_i$ are real numbers and both $\mu_1, \mu_2$ are defined on a finite measurable space. At first, i thought that $\mu_1$ is the integral of a simple function but again i realised that $E_i$ are not necessary disjoint and so i am stack. I will appreciate your proof.
Radon-Nikodym derivative
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real-analysis
measure-theory
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0Assuming you know what to do when the $E_i$ are pairwise disjoint, why don't you replace the $E_i$ by a suitable refinement, then? – 2012-03-31
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3Look up [Johann Radon](http://en.wikipedia.org/wiki/Johann_Radon), not Random :) – 2012-03-31
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0I have edited 'Radon', thanks. Thomas, try to elaborate more or maybe you can put your proof down coz i am not so sure. – 2012-03-31
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0It doesn't really matter whether the sets $E_i$ are pairwise disjoint or not, just use that $\mu_2(E \cap E_i) = \int [E] \cdot [E_i]\,d\mu_2$ and linearity of the integral. This gives you that the desired equality holds for all measurable $E$, so $\sum c_i [E_i]$ is your Radon-Nikodym derivative $\frac{d\mu_1}{d\mu_2}$. – 2012-03-31
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0Construct sets $F_k$ such that each $E_i$ is the union of some of the $F_k$ but such that the $F_k$ are pairwise disjoint. Define new coefficients $d_k$ for $F_k$ by summing those $c_i$ for which $F_k\cap E_i$ is not empty. You can define such $F_k$ by looking at all possible intersections of the $E_i$. According to your posting there are only finitely many of them, in which case this should not be a problem at all. – 2012-03-31
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0Your explanations very brief and i am now confused. I would like to see your proof including for absolute continuity in the answer box down, t.b. and Thomas – 2012-03-31