limit of increasing sequence of measures is a measure

Solution 1:

You can use that a set function $\mu$ defined on a measurable space such that $\mu(\emptyset)=0$, $\mu$ is finitely additive and is continuous from below is a measure on that space. Also, you can define a double sequence $(s_{n,m})_{n,m\geq 1}$ given by $s_{n,m}=\sum_{k=1}^m \mu_n(E_k)=\mu_n(\bigcup_{k=1}^m E_k)$ and then show that $$\lim\limits_{n\to\infty}\lim\limits_{m\to\infty} s_{n,m}=\sup\limits_n\sup\limits_m(s_{n,m})=\sup\limits_m\sup\limits_n(s_{n,m})=\lim\limits_{m\to\infty}\lim\limits_{n\to\infty} s_{n,k}$$ Note that the equalities between the supremum and the limit is because in each index, the sequence $(s_{n,k})$ is monotonically increasing.