Let $\mu, v_n$ be measures on $(X, \mathcal{B})$ with $\mu$ being $\sigma$-finite and $sup\{v_n(X)\} < \infty$. Note that $v = \sum_{n \ge 1} 2^{-n} v_n << \mu$ if and only if $v_n << \mu$. Let the Lebesgue decomposition of $v_n$ with respect to $\mu$ be $v_n = v_{n, a} + v_{n, s}$, where $\frac{dv_{n, a}}{d\mu} = f_n$.
Note that $v$ is a finite (hence $\sigma$-finite) measure, so it has a Lebesgue decomposition. I want to find the Lebesgue decomposition of $v$ with respect to $\mu$ and calculate $\frac{dv_a}{d\mu}$. I believe the Lebesgue decomposition will just be $v = v_a + v_s = \sum_{n \ge 1} 2^{-n} v_{n, a} + \sum_{n \ge 1} 2^{-n} v_{n, s}$. Then my question is this:
Are Radon Nikodym derivatives countably additive? Can I say that $\frac{dv_a}{d\mu} = \sum_{n \ge 1} 2^{-n} f_n$? Thanks.
Suppose each $v_n$ has decomposition $dv_n = f_n\,d\mu + d\lambda_n$ with $\lambda_n \perp \mu$. You are interested in the decomposition for $v = \sum_{n = 1}^{\infty}v_n$. We have $$v_n(S) = \sum_{n = 1}^{\infty}(\int_S f_n d\mu + \lambda_n(S)) = \int_S\sum_{n = 1}^{\infty}f_n\,d\mu + \sum_{n = 1}^{\infty}\lambda_n(S).$$ It is easy to show that $\lambda = \sum_{n = 1}^{\infty}\lambda_n \perp \mu$. Thus $$dv = \sum_{n = 1}^{\infty}f_n \,d\mu + d\lambda$$ is the decomposition for $v$.