A sequence of measures on a sigma algebra

Define $$ a_{j,k}=\mu_{j+1}(E_k)-\mu_j(E_k)\tag{1} $$ From the conditions above, all $a_{j,k}$ have the same sign.

Summing $(1)$ yields $$ \mu_n(E_k)=\mu_1(E_k)+\sum_{j=1}^{n-1}a_{j,k}\tag{2} $$ So we have $$ \begin{align} \mu(E_k) &=\lim_{n\to\infty}\mu_n(E_k)\\ &=\mu_1(E_k)+\sum_{j=1}^\infty a_{j,k}\tag{3} \end{align} $$ Since each $\mu_j$ is countably-additive, we can add $(2)$ yieldiing $$ \begin{align} \mu_n(E) &=\sum_{k=1}^\infty\mu_1(E_k)+\sum_{k=1}^\infty\sum_{j=1}^{n-1}a_{j,k}\\ &=\sum_{k=1}^\infty\mu_1(E_k)+\sum_{j=1}^{n-1}\sum_{k=1}^\infty a_{j,k}\tag{4} \end{align} $$ Taking the limit of $(4)$, changing the order of summation (all terms have the same sign), and then applying $(3)$, we get $$ \begin{align} \mu(E) &=\lim_{n\to\infty}\mu_n(E)\\ &=\sum_{k=1}^\infty\mu_1(E_k)+\sum_{j=1}^\infty\sum_{k=1}^\infty a_{j,k}\\ &=\sum_{k=1}^\infty\mu_1(E_k)+\sum_{k=1}^\infty\sum_{j=1}^\infty a_{j,k}\\ &=\sum_{k=1}^\infty\left(\mu_1(E_k)+\sum_{j=1}^\infty a_{j,k}\right)\\ &=\sum_{k=1}^\infty\;\mu(E_k)\tag{5} \end{align} $$ Counterexample:

I commented that the argument above assumes that $\mu_1(E)<\infty$ when the $\mu_n$ are decreasing. This condition cannot be lifted.

Let $E_k=\{k\}$ and $\mu_n(\{k\})=1$ when $k>n$ and $\mu_n(\{k\})=0$ when $k\le n$. Then $$ \mu(\{k\})=\lim_{n\to\infty}\mu_n(\{k\})=0 $$ yet $$ \mu(\mathbb{N})=\lim_{n\to\infty}\mu_n(\mathbb{N})=\infty $$

By definition, and using that each $\mu_k$ is countably additive, we have

$$\mu(E) = \lim_{k \to \infty}\mu_k(E) = \lim_{k \to \infty}\sum_{n \geq 1}\mu_k(E_n).$$

By the monotone convergence theorem, since each sequence $\{\mu_k(E_n)\}_{k}$ is increasing, this equals

$$\lim_{k \to \infty}\sum_{n \geq 1}\mu_k(E_n) = \sum_{n \geq 1}\lim_{k \to \infty}\mu_k(E_n) = \sum_{n \geq 1}\mu(E_n).$$

It would not hold true in general if the sequence of measures is decreasing.