Let $P=\sum_{n=1}^{\infty} a_n P_n$ be a probability measure where $a_n >0$, $\{P_n\}$ be a sequence of probability measures and $\sum_{n=1}^{\infty} a_n=1$. If $P \ll \nu$ where $\nu$ is a sigma-finite measure, is it possible to show that $\frac{dP}{d\nu} = \sum_{n=1}^{\infty} a_n \frac{d P_n}{d \nu}$ ?
I know that by the Radon–Nikodym theorem, the last statement is true for a finite sum but I am not really sure how to prove this infinite sum.
This should just be the monotone convergence theorem. If $\{\mu_n\}$ is a sequence of $\sigma$-finite measures, $\mu = \sum_n \mu_n$, and $\mu \ll \nu$ for some $\sigma$-finite measure $\nu$, then $\mu_n \ll \nu$ and all $n$ and for any measurable set $E$ you have $$\int_E \frac{d\mu}{d\nu}\, d\nu = \mu(E) = \sum_n \mu_n(E) = \sum_n \int_E \frac{d\mu_n}{d\nu} d\nu = \int_E \sum_n \frac{d\mu_n}{d\nu} \, d\nu$$ because $d\mu_n/d\mu \ge 0$ for all $n$.
Since $E$ is arbitrary it follows that $$\frac{d\mu}{d\nu} = \sum_n \frac{d\mu_n}{d\nu}$$$\nu$-almost everywhere.