Defining a measure $\lambda$ in terms of a sequence of measures $(\mu_n)$

Below is a problem concerning a sequence of measures. My trouble arises when trying to prove the countable additivity of our measure $\lambda$, which is defined in terms of a sequence of measures. (I have only included the portion of the proof that is giving me trouble.)

Problem: Let $(\mu_n)$ be a sequence of measures on a $\sigma$-algebra $\textbf{X}$ with $\mu_n(X)=1$ for each n. Define a function $\lambda$ by the following: $$\lambda(E)=\sum^{\infty}_{n=1}2^{-n}\mu_n(E)$$ for $E\in\textbf{X}$. Prove that $\lambda$ is a measure and that $\lambda(X)=1$

Solution [Attempt at proving that $\lambda$ is countable additive]: Let $(E_j)$ be a sequence of disjoint sets in $\textbf{X}$. Observe the following: $$\lambda(\bigcup^{\infty}_{j=1}E_j)=\sum^{\infty}_{n=1}2^{-n}\mu_n(\bigcup^{\infty}_{j=1}E_j)=\sum^{\infty}_{n=1}2^{-n}\sum^{\infty}_{j=1}\mu_n(E_j)=\sum^{\infty}_{n=1}\sum^{\infty}_{j=1}2^{-n}\mu_n(E_j)=\sum^{\infty}_{j=1}\sum^{\infty}_{n=1}2^{-n}\mu_n(E_j).$$ Then, since $$\lambda(E_j)=\sum^{\infty}_{n=1}2^{-n}\mu_n(E_j),$$ we obtain our desired result.....

My question is, of course, concerning the manipulation of those infinite sums. I knew what I needed to get at in the end and, admittedly, my approach was somewhat mechancial. Is that move "legal?" Thanks for the input. (P.S. If you can share how to make equalities scroll downward in TeX that would be great!)

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In either case, you are thinking of sequences as functions $\mathbb{N}\rightarrow\mathbb{R}$, where we consider $\mathbb{N}$ as equipped with the discrete $\sigma$-algebra and give it the counting measure $\mu$, so that the sum of a sequence is precisely its integral with respect to $\mu$.