TITLE: Every $\sigma$ finite measure is absolutely continuous with respect to a finite measure.
QUESTION [6 upvotes]: Let $\mu$ be a $\sigma$- finite measure on $(X,M)$. Prove that there exists a finite measure $\lambda$ on $M$ such that $\lambda\ll\mu$ and $\mu\ll\lambda$. 
Can anyone give me a hint on how to start on this problem?

REPLY [5 votes]: Some hints:

What would happen if you have a measure $\lambda = f\,\mathrm d\mu$? Say, if $f > 0$ everywhere? Does integral of positive function w.r.t. $\lambda$ has to be infinite if $\lambda$ is infinite?
To find such $f$, think of the fact that there exists an increasing sequence of sets on which $\lambda$ is finite, and union covers the whole space. How would you define $f$ on each of those sets?

REPLY [4 votes]: Hint: For each $n$, $\mu(E_n) <\infty$ by $\sigma$-finiteness, and define a new measure $$\mu_n = \frac{2^{-n}}{\mu(E_n)}\mu_{E_n}.$$
Rest hidden in spoilers:

Then $\|\mu_n \| = 2^{-n}$, so $\sum_{n=1}^\infty \|\mu_n\| < \infty$, so the series of the measures $\mu_n$ converges in the normed space of measures (I assume you defined this in class). Then, define $\lambda = \sum_{n=1}^\infty \mu_n$ and obtain $\mu \ll \nu \ll \mu$.