convergence of increasing sequence of measures Let $(X,A)$ be a measurable space. Show that if $(\mu_n)$ is an increasing sequence of measures, then $\mu(A) = \lim_{n\rightarrow \infty} \mu_n(A)$ defines a measure on  $(X,A)$.
i) $\mu(\emptyset) = 0$ since its true for all $\mu_n$.
ii) $\mu (\cup u_i) = \lim_{n\rightarrow \infty} \sum_i \mu_n(u_i)$ by definition. But how can I move the limes inside?

Hint
$$\lim_{n \to \infty} \sum_{i=1}^{\infty} \mu_n(u_i) = \sup_{n \in \mathbb{N}} \sup_{k \in \mathbb{N}} \sum_{i=1}^k \mu_n(u_i) = \sup_{k \in \mathbb{N}} \sup_{n \in \mathbb{N}} \sum_{i=1}^k \mu_n(u_i) = \ldots$$
The first equality follows from the fact that the sequence of measures is increasing (i.e. $\lim = \sup$). 
