# Question about absolutely continuous measures

## Homework Statement

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

I want to construct $$\lambda$$ such that vn 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 vn(E) by the measure of the entire space. So, I need to find a more clever way of doing this.

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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!

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
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