A sequence of measures on a sigma algebra 
Let $X$ be a set and $\mathcal{A}$, a sigma algebra of subsets of $X$. Let $\{\mu_n\}$ be a sequence of measures of $\mathcal{A}$ such that $\mu_{n+1}(E)\geqslant \mu_n(E)$ for every $E\in \mathcal{A}$. Let $\lim_{n\rightarrow \infty} \mu_n(E) = \mu(E)$. I want to show that $\mu$ is a measure on $\mathcal{A}$.
  Would the conclusion still be true if $\mu_{n+1}(E)\leqslant \mu_n(E).$  

My Attempt: 
Let $E_n$ be a sequence of pairwise disjoint sets in $\mathcal{A}$. Then, I have to show $$\mu\left(\cup_n E_n\right)=\sum_n \mu(E_n),$$  where the $E_n$ are pairwise disjoint.
Let $E=\cup_n E_n$. Then $\mu(E)=\sum_{n=1}^{\infty}\mu(E_n)$. For one inclusion, we have that $$\begin{align*}
\mu(E) & \geqslant \lim_{k\rightarrow \infty}\sum_{n=1}^N \mu_k(E_n)\\ 
& =\sum_{n=1}^N\mu(E_n). \\
\end{align*} $$ 
But this holds true for any $N$. Thus $\mu(E)\geqslant \sum_{n=1}^\infty \mu(E_n)$.  
Is the above inclusion okay? I can't can up with a way to tackle the other inclusion. Perhaps someone will be kind enough to give a hint?   
 A: 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
$$
\mu(\mathbb{N})=\lim_{n\to\infty}\mu_n(\mathbb{N})=\infty
$$
A: 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.
