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?