Question about absolutely continuous measures

[jvalton1287]

Homework Statement

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

I want to construct \lambda such that v_n is absolutely continuous w.r.t. \lambda 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 v_n(E) by the measure of the entire space. So, I need to find a more clever way of doing this.

 

[losiu99]

There's a handy theorem:
Given \sigma-finite measure \mu, there exists a probability measure \lambda such that
\mu(A)=0 \mbox{ iff } \lambda(A)=0
Proof:
Let {A_n} be a sequence of measurable sets such that \Bigcup A_n=X and \mu(A)<\infty. Set
f=\sum_{k=1}^\infty 2^{-k}\frac{\chi_{A_k}}{1+\mu(A_k)}
From this we can obtain finite measure \mu_f. I leave to you showing it satisfies the hypothesis.
Good luck!

 

[jvalton1287]

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.

 

[berkeman]

Thread locked temporarily. This may be a question on a take-home exam.