Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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? –  user20266 Mar 31 '12 at 18:56
3  
Look up Johann Radon, not Random :) –  t.b. Mar 31 '12 at 18:59
    
I have edited 'Radon', thanks. Thomas, try to elaborate more or maybe you can put your proof down coz i am not so sure. –  Bota Moses Mar 31 '12 at 19:07
    
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 $\frac{d\mu_1}{d\mu_2}$. –  t.b. Mar 31 '12 at 19:12
    
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. –  user20266 Mar 31 '12 at 19:15

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.