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?
 A: U need to use the following claim, which is in essence the same as answer by saz.
If $a_{mn}$ in monotone increasing in both m and n, then 
$$\lim_{m\to\infty}\lim_{n\to\infty} a_{mn}=\lim_{n\to\infty}\lim_{m\to\infty}a_{mn}$$
Using this you will get
$$\lim_{n\to \infty}\lim_{m\to \infty} \sum_{i=1}^m \mu_n(u_i) = \lim_{m\to \infty}\lim_{n\to \infty} \sum_{i=1}^m \mu_n(u_i)=\lim_{m\to \infty} \sum_{i=1}^{m} \mu(u_i)$$
$$=\sum_{i=1}^{\infty} \mu(u_i)$$
Proof of claim:
Let $M=\sup_{m\geq 1}\sup_{n\geq 1} a_{mn}$. Then
$$a_{mn} \leq M \Rightarrow \sup_{n\geq 1}\sup_{m\geq 1} a_{mn} \leq M = \sup_{m\geq 1}\sup_{n\geq 1} a_{mn}$$
Now let $N=\sup_{n\geq 1}\sup_{m\geq 1} a_{mn}$. Similarly
$$a_{mn} \leq N \Rightarrow \sup_{m\geq 1}\sup_{n\geq 1} a_{mn} \leq N = \sup_{n\geq 1}\sup_{m\geq 1} a_{mn}$$
Thus M=N and we are done.
A: 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. $\lim = \sup$). 
