Let $(X,\mathcal{A})$ be a measurable space and $(\mu_n)$ be a sequence of measures which satisfies $\mu_n(X)=1$. I want to show that the function defined as $$ \nu(E)=\sum_{n=1}^{\infty}2^{-n}\mu_n(E) $$ is a measure with $\nu(X)=1$.
In the proof of countable additivity part: Let $(E_j)$ be a sequence of pairwise disjoint sets in $\mathcal{A}$. $$ \nu(\bigcup_{j=1}^{\infty}E_j)=\sum_{n=1}^{\infty}2^{-n}\mu_n(\bigcup_{j=1}^{\infty}E_j)=\sum_{n=1}^{\infty}2^{-n}\sum_{j=1}^{\infty}\mu_n(E_j). $$
If I can change the order of the sum the proof will be finished. Can I do this directly or do I need something more? For example such an observation for all $i,j\in \mathbb{N}$ we have $\mu_n(E_j)\leq \mu_n(X)=1$.
