Question about absolutely continuous measures

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  • #1

Homework Statement

Suppose we're given some sigma-finite measures v1, v2, v3,...

I want to construct [tex]\lambda[/tex] such that vn is absolutely continuous w.r.t. [tex]\lambda[/tex] for all n.

2. The attempt at a solution

So far, I've tried thinking of making an infinite weighted (weighted by 2^(-n)). The problem is that sigma-finiteness requires that I can't divide vn(E) by the measure of the entire space. So, I need to find a more clever way of doing this.

Answers and Replies

  • #2
There's a handy theorem:
Given [tex]\sigma[/tex]-finite measure [tex]\mu[/tex], there exists a probability measure [tex]\lambda[/tex] such that
[tex]\mu(A)=0 \mbox{ iff } \lambda(A)=0[/tex]
Let [tex]{A_n}[/tex] be a sequence of measurable sets such that [tex]\Bigcup A_n=X[/tex] and [tex]\mu(A)<\infty[/tex]. Set
[tex]f=\sum_{k=1}^\infty 2^{-k}\frac{\chi_{A_k}}{1+\mu(A_k)}[/tex]
From this we can obtain finite measure [tex]\mu_f[/tex]. I leave to you showing it satisfies the hypothesis.
Good luck!
  • #3
I don't quite understand this solution. How are all of the original measures v1, v2, ... related to this mu?

I need to show that all of those initial measures are absolutely continuous w.r.t. to mu.
  • #4
Thread locked temporarily. This may be a question on a take-home exam.

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