Suppose $(X,\mathcal F)$ is a measurable space and $\mu$ and $\nu_n,n\in\Bbb N$ non-negative measures on $(X,\mathcal F)$ s. t. $\mu\perp\nu_n,\forall n\in\Bbb N.$ Prove that $\displaystyle\mu\perp\sum_{n=1}^\infty\nu_n.$
My attempt:
Since $\mu$ and $\nu_n,n\in\Bbb N$ are non-negative, $$\mu\perp\nu_n,\forall n\in\Bbb N\Leftrightarrow \forall n\in\Bbb N,\exists E_n,F_n\in\mathcal F,E_n\cap F_n=\emptyset\land (\mu(F_n^c)=\nu_n(E_n^c)=0)\land (\mu(E_n)=\nu_n(F_n)=0)$$ and, for the same reason, I have to show there are disjoint $E,F\in\mathcal F$ s. t. $\displaystyle \mu(F^c)=\left(\sum_{n=1}^\infty\nu_n\right)(E^c)$ and $\mu(E)=\left(\displaystyle\sum_{n=1}^\infty\nu_n\right)(F)=0$ Let $$\begin{aligned}E:&=\bigcup_{n=1}^\infty E_n\\F:&=\bigcap_{n=1}^\infty F_n.\end{aligned}$$ Then, $$\begin{aligned}E_n\cap F_n&=\emptyset,\forall n\in\Bbb N\\\Rightarrow E_n\cap F&=\emptyset,\forall n\in\Bbb N\\\Rightarrow E\cap F&=\emptyset\end{aligned}$$ and $$\begin{aligned}\left(\sum_{n=1}^\infty\nu_n\right)\left(\left(\bigcup_{n=1}^\infty E_n\right)^c\right)&=\left(\sum_{n=1}^\infty\nu_n\right)\left(\bigcap_{n=1}^\infty E_n^c\right)\\&\le\sum_{n=1}^\infty\nu_n\left(E_n^c\right)\\&=0\\\Rightarrow\left(\sum_{n=1}^\infty\nu_n\right)(E)&=0\\\mu\left(\left(\bigcap_{n=1}^\infty F_n\right)^c\right)&=\mu\left(\bigcup_{n=1}^\infty F_n^c\right)\\&\le\sum_{n=1}^\infty\mu\left(F_n^c\right)\\&=0\\\Rightarrow\mu(F)&=0.\end{aligned}$$ Since both $\mu$ and $\displaystyle\sum_{n=1}^\infty\nu_n$ are non - negative, $\displaystyle\mu\perp\sum_{n=1}^\infty\nu_n.$
Is my asnwer valid and/or is there anything I should've done differently?
It is enough to show that there is a measurable set $E$ with $\mu(E)=0=\nu(X\setminus E)$, where $\nu=\sum_n\nu_n$.
For each $n$ there is a measurable set $E_n$ such that $\mu(E_n)=0=\nu(X\setminus E_n)$. Define $E=\bigcup_nE_n$. Then $\mu(E)\leq\sum_n\mu(E_n)=0$ and $\nu_n(X\setminus E)\leq \nu_n(X\setminus E_n)=0$. As a consequence, $\nu(X\setminus E)=\sum_n\nu_n(X\setminus E)=0$.