# sequence of monotone measures

I have a sigma-algebra F on X and a monotone sequence of measures on F, meaning: $$\mu_n(A)\leq \mu_{n+1} (A) \forall A \in F$$ It should hold that $$\mu_1(X) <\infty$$ I want to show, that $$\mu=lim_{n \rightarrow \infty} \mu_n$$ is a measure on F. How can I do that?

Here is a first-principles approach: if $$\{A_k\}$$ is a sequence of pairwise disjoint sets then for any $$n$$ you have $$\mu_n \left( \bigcup_k A_k\right) = \sum_k \mu_n(A_k) \le \sum_k \mu(A_k)$$ so now take the limit as $$n \to \infty$$: $$\mu \left( \bigcup_k A_k\right) \le \sum_k \mu(A_k).$$ For the other direction again let $$\{A_k\}$$ be a sequence of pairwise disjoint sets and fix an index $$m$$. Then $$\sum_{k=1}^m \mu(A_k) = \lim_{n \to \infty} \sum_{k=1}^m \mu_n(A_k) = \lim_{n \to \infty} \mu_n \left( \bigcup_{k=1}^m A_k\right) \le \lim_{n \to \infty} \mu_n \left( \bigcup_{k=1}^\infty A_k\right) = \mu \left( \bigcup_{k=1}^\infty A_k\right).$$ Now let $$m \to \infty$$ to conclude $$\sum_{k=1}^\infty \mu(A_k) \le \mu \left( \bigcup_{k=1}^\infty A_k\right).$$
The first step is easy: $$\mu(\emptyset)=\lim_{n\rightarrow \infty} \mu_n(\emptyset)=0.$$ Sigma-additivity is a bit more involved. Let $$(A_k)_{k\in\mathbb{N}}\subset F$$ be pairwise disjoint and put $$A:=\cup_k A_k$$. Now let $$f_n:\mathbb{N}\rightarrow [0,\infty]$$ be $$f_n(k)=\mu_n(A_k)$$. Then $$f_n\leq f_{n+1}$$ and therefore $$\lim_{n\rightarrow}f_n=f$$ pointwise and $$f:\mathbb{N}\rightarrow[0,\infty]$$. Furthmore let $$P$$ be the counting measure on $$\mathbb{N}$$. Hence applying the monotone covergence to the integral with respect to the counting measure yields $$\mu(A)\leftarrow\mu_n(A)=\sum_{k=1}^\infty \mu_n(A_k) = \int f_n\, dP \rightarrow \int f\, dP = \sum_{k=1}^\infty \mu(A_k).$$
• @ Steven33 You need some kind of theorem, that allows you to switch the limits, i.e. $\lim_{n\rightarrow\infty}\sum_{k=1}^\infty \mu_n(A_k) = \sum_{k=1}^\infty \lim_{n\rightarrow \infty}\mu_n(A_k)$. This is typically an application of an integral theorem. So my bet is that you probably can show it without explicitly using such a theorem but you would probably prove this theorem in the process. – humanStampedist Apr 29 at 10:14