# 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?

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.
• 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 $g \ge 0$)? – Alexander Oct 4 '14 at 21:32
• Yes.${}{}{}{}{}$ – Davide Giraudo Oct 4 '14 at 21:43