What is meant with a $\sigma $-finite measure $\mu $ can be replaced by finite measure in the sense that $d \bar {\mu}=w d \mu $? 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!

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.