Let $(\Omega,\mathcal{F})$ be a measurable space. Let $\left\{ \mu_{n}\right\} _{n\geq1}$ be a sequence of finite measures on $(\Omega,\mathcal{F})$. Then there exists a probability measure $\lambda$ such that $\mu_{n}\ll\lambda$. For example, we can consider $$\lambda(\cdot)={\displaystyle \sum_{n=1}^{\infty}\frac{1}{2^{n}}\frac{\mu_{n}(\cdot)}{\mu_{n}(\Omega)}}.$$
What is we extend it to the case where $\left\{ \mu_{n}\right\} _{n\geq1}$ are $\sigma$-finite?
Best Answer
Suppose you show that $\mu_n <<\nu_n$ for some finite measure $\nu_n$ for each $n$. Then you can find a p.m. $\lambda$ with $\nu_n <<\lambda$ for all $n$ which implies $\mu_n <<\lambda$ for all $n$.
So consider any sigma finite measure $\mu$. There exist disjoint sets $A_n$ such that $\mu(A_n) <\infty$ for all $n$ and $\bigcup_n A_n=\Omega$. Let $\nu(E)=\sum \frac 1 {2^{n}} \frac {\mu_n(E\cap A_n)} {1+\mu_n( A_n)}$. This gives a finite measure $\nu$ with $\mu <<\nu$.