$\tau = \left(\sum_{n = 1}^\infty f_n\right) d\nu + \sum_{n = 1}^\infty \mu_n$ the Lebesgue decomposition of $\tau$? Assume $\tau_n$ is a sequence of positive measures on a measurable space $(X, \mathcal{F})$ with $\sup_n \tau_n(X) < \infty$ and $\nu$ is another finite positive measure on $(X, \mathcal{F})$. Suppose $\tau_n = f_n\,d\nu + \mu_n$ is the Lebesgue decomposition of $\tau_n$; in particular, $\mu_n \perp \nu$. If $\tau = \sum_{n = 1}^\infty \tau_n$ is a finite measure, is$$\tau = \left(\sum_{n = 1}^\infty f_n\right) d\nu + \sum_{n = 1}^\infty \mu_n$$the Lebesgue decomposition of $\tau$?

Yes. 


*

*The composition. 
$$\tau (A) = \sum \tau_n (A) = \sum (\int_A f _n d \nu +  \mu_n(A)).$$ 
By monotone convergence $\sum \int_A f_n d \nu = \int_A \sum f_n d\nu$. Let $f=\sum f_n$. Then $f$ is measurable and nonnegative. Also since $\tau$ is finite, $f\in L^1(\nu)$.  Define the set function $\mu= \sum \mu_n$. We show this is a measure. It is clearly nonnegative. Next: 
a. $\mu(\emptyset)=0$. 
b. If $A_1,A_2,\dots$ are disjoint, then 
\begin{align*} \mu (\cup A_j)&= \lim_{N\to\infty}\sum_{n\le N} \mu_n (\cup A_j)\\
&= \lim_{N\to\infty} \sum_{n\le N} \sum_j  \mu_n (A_j)\\
&= \lim_{N\to\infty} \sum_j \sum_{n\le N} \mu_n (A_j) \\
&= \sum_j \lim_{N\to\infty} \sum_{n\le N}  \mu_n (A_j)\\
&= \sum_j \mu(A_j).
\end{align*}
We have used monotone convergence for the fourth equality, and the third equality is a statement about a finite sum of series of positive terms. 
Therefore $\mu$ is a measure. Since $\tau$ is finite, $\mu$ is also finite. 
Bottom line: $d\tau = f d\nu + d\mu$, where $f\in L^1(\nu)$ and $\mu$ is finite measure. 


*By uniqueness of Lebesgue decomposition, all that remains to show is that  $\mu\perp \nu$. Since for each $n$  $\mu_n \perp \nu$, there exists $A_n$ such that $\nu (A_n) =\mu_n (A_n^c) =0$. Let $A=\cup A_n$. Then 
$\nu(A)\le \sum_n \nu(A_n) =0$, and 
$$\mu(A^c) = \mu( \cap A_n^c) =\sum_n \mu_n (\cap A_n^c) \le \sum \mu_n (A_n^c)=0.$$ 
Therefore $\mu\perp \nu$. 

