Let $(X,\mathcal X,\nu)$ be a probability space, let $\{ \mu_n \}_{n\in\mathbb N}$ be a set of finite positive measures on $(X,\mathcal X)$ that are all absolutely continuous with respect to $\nu$. I am trying to think about the following $\nu$-everywhere equality (with $\mu=\sum_{n\in\mathbb N} \mu_n$) \begin{align*} \frac{d\mu}{d\nu}=\sum_{n\in\mathbb N} \frac{d\mu_n}{d\nu} \end{align*} Where it is clear that $\mu\ll \nu$, indeed for $A\in\mathcal X$ such that $\nu(A)=0$, we have $\mu_n(A)=0$ and $\mu(A)=\sum_{n} 0 = 0$.
Attempted proof : let $g=\frac{d \mu}{d\nu}$ and $f_n = \sum_{k=1}^n\frac{d\mu_k}{d\nu}$. It is clear, since $\sum_{k=1}^n \mu_k \leqq\sum_{k=1}^{n+1} \mu_k \leqq \mu$, that $f_n\leqq f_{n+1}\leqq g$ almost surely. Let $f=\lim_n f_n$ (How do I prove that this limit exists) then we know by the dominated convergence theorem and linearity of Radon Nikodym derivatives that $\mu(A)=\lim_n \sum_{k=1}^n \mu_k(A) = \lim_{n}\int_A f_n ~d\nu=\int_A f~ d\nu\leq\int_A g ~d\nu=\mu(A)$. Therefore $\int_A f ~d\nu=\int_A g ~d\nu$ together with $f\leqq g$. From there we can let $A_\varepsilon=\{ x : f(x)\leq g(x)-\varepsilon \}$, then for any $\varepsilon>0$, $\varepsilon\nu(A_\varepsilon)=\int_X \varepsilon\chi_{A_\varepsilon}~d\nu \leq \int_X (g-f)~d\nu=0$ and so $\nu(A_\varepsilon)=0$, taking the limit for $\varepsilon\to 0$ yields that $f=g$ happens $\nu$-surely.
Any idea on how to prove pointwise (or $\nu$-surely) convergence of the limit of $f_n$ ? Does the rest of the proof look fine ? Also since I am writing this proof in the proof of another thing, I would appreciate if there is a reference to it, this must exists somewhere, it feels I am reinventing the wheel.