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I don't understand the following. Before proving the Radon-Nikodym theorm, Rudin proves a lemma that say that $\mu $ is a $\sigma $-finite measure on a $\sigma $-algebra, then there is a function $w \in L ^1(\mu) $ such that $0<w<1 $.

Then he says that the point with this is that a $\sigma $-finite measure $\mu $ can be replaced by finite measure in the sense that $d \bar {\mu}=w d \mu $?

I believe what is meant with $d \bar {\mu}=w d \mu $ is that for any integrable function $f $, we have that $\int fd \bar {\mu}=\int (fw) d \mu $?

What is the significans of integrating with respect to a finite measure, and how does the fact that $0<g<1 $ relate to the fact that $\bar {\mu } $ gives a finite measure?

Thanks in advance!

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The fact that $\bar{\mu}$ is a finite measure comes from integrability of $w$ with respect to $\mu$.

The point of the statement is to reduce the proof of Radon-Nikodym theorem to the case of finite measures. The fact that $w$ is positive allows us to do the reduction to the finite case of both involved measures.

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  • $\begingroup$ Am I correct that you mean that if we let $f $ be the constant function equal to 1, then $\bar {\mu } (X)=\int f d\bar {\mu } )=\int (fg)d \mu< \infty$, since $fg)=g $ and $g $ is integrable (assuming $g \ge 0 $)? $\endgroup$ – Alexander Oct 4 '14 at 21:32
  • $\begingroup$ Yes.${}{}{}{}{}$ $\endgroup$ – Davide Giraudo Oct 4 '14 at 21:43
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dμ1 =wdμ2 means: μ1(E)=∫wdμ2 on E for everey measurable set E.In this case μ1 is finite because: |μ1(E)|=|∫wdμ2 |<= ∫|w|dμ2 <∫dμ2 =μ2(E) (1) Pay attention to this important point that w is positive and w<1 in (1).

best wishes.

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