Limes is a measure Let $(X, A)$ be a measure space, $\mu_n: A \rightarrow [0, \infty], n \in \mathbb N$ a sequence of measures which is monotonely increasing, i.e. $\mu_n(B) \leq \mu_{n+1}(B)$ $\forall B \in A$ and all $n \in \mathbb N$.
Why is $\mu(B) := lim \mu_n(B)$ a measure on (X, A) ?  

Well, the only non-obvious thing is $\sigma$-additivity, so assume $(A_n)_{n\in\mathbb{N}}\subseteq A$ pairwise disjoint.
Then,
$$
\mu(\cup_{n=1}^{\infty} A_n)=\lim_{k\to\infty}\mu_k(\cup_{n=1}^{\infty} A_n)=\lim_{k\to\infty} \sum_{n=1}^{\infty} \mu_k(A_n)=\sum_{n=1}^{\infty} \lim_{k\to\infty} \mu_k(A_n)
$$
by the monotone convergence theorem applied  to the function sequence $f_k:\mathbb{N}\to \mathbb{R}$ given by $f_k(n)=\mu_k(A_n)$.
Now, the right-hand side is, by definition, $\sum_{n=1}^{\infty}\mu(A_n)$, proving $\sigma$-additivity. Hence, $\mu$ is a measure.
