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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?

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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 $$\sum_{k=1}^\infty \mu(A_k) \le \mu \left( \bigcup_{k=1}^\infty A_k\right).$$

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  • $\begingroup$ Thanks Umberto. You didn't need any theorem:) $\endgroup$ – Steven33 Apr 29 at 14:17
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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 $$\mu(A)\leftarrow\mu_n(A)=\sum_{k=1}^\infty \mu_n(A_k) = \int f_n\, dP \rightarrow \int f\, dP = \sum_{k=1}^\infty \mu(A_k). $$

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  • $\begingroup$ Can I do that without using integral theorems? $\endgroup$ – Steven33 Apr 29 at 10:03
  • $\begingroup$ @ 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. $\endgroup$ – humanStampedist Apr 29 at 10:14
  • $\begingroup$ We aren't so far in this subject. I am not allowed to use such theorems. But thank you very much for your elegant proof:) $\endgroup$ – Steven33 Apr 29 at 14:19

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