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.

  • $\begingroup$ Assuming 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? $\endgroup$ – user20266 Mar 31 '12 at 18:56
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    $\begingroup$ Look up Johann Radon, not Random :) $\endgroup$ – t.b. Mar 31 '12 at 18:59
  • $\begingroup$ I have edited 'Radon', thanks. Thomas, try to elaborate more or maybe you can put your proof down coz i am not so sure. $\endgroup$ – Bota Moses Mar 31 '12 at 19:07
  • $\begingroup$ It 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 $\frac{d\mu_1}{d\mu_2}$. $\endgroup$ – t.b. Mar 31 '12 at 19:12
  • $\begingroup$ Construct 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. $\endgroup$ – user20266 Mar 31 '12 at 19:15

First of all, absolute continuity of $\mu_1$ with respect to $\mu_2$ is clear: if $N$ is a $\mu_2$-null set then so is $N \cap E_i$, hence $\mu_1(N) = \sum_{i = 1}^n c_i \mu(N \cap E_i) = 0$ which is the very definition of $\mu_1 \ll \mu_2$. But this is irrelevant anyway, because it follows from what I say below.

To find the Radon–Nikodým derivative, just compute (writing $[E]$ for the characteristic function of $E$ and using $[E \cap F] = [E] \cdot [F]$) that for all $\mu_2$-measurable

$$\mu_1(E) = \sum_{i = 1}^n c_i \mu_2(E \cap E_i) = \sum_{i=1}^n c_i \int [E] \cdot [E_i]\,d\mu_2 = \int [E] \cdot \left(\sum_{i=1}^n c_i [E_i]\right)\,d\mu_2$$

and observe that $f = \sum_{i=1}^n c_i [E_i]$ thus satisfies the defining property of the Radon–Nikodým derivative $\frac{d\mu_1}{d\mu_2}$. If you want to write $f$ as simple function with disjoint supports, decompose the sets $E_i$ into disjoint sets $F_{ij}$, as Thomas suggested, but this is a standard step that is done at the very beginning of every course on measure theory, so I skip it.

  • $\begingroup$ Excuse me, But is $f$ non-negative in this case? $\endgroup$ – Topology Dec 22 '16 at 17:43

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