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Here I will talk about measure referring to a positive measure according to Walter Rudin

Let $(X,\mu)$ a measurable space and $\{\mu_n\}$ a succession of measure in $\mu$ such that $\mu_n < \mu_{n+1}$ for all $n \in \mathbb{N}.$

If $u: \mu \to [0, \infty]$ , $u(E)=sup_{n\in \mathbb{N}} u_{n}(E)$ then $u$ is a measure in $\mu$

attempt: Given the $u_n(\emptyset)=0$ for all $n\in \mathbb{N}$, $u(\emptyset)= \sup_{n\in \mathbb{N}}=sup_{n\in \mathbb{N}} \{0\}=0$

Let's see that now that $u$ is countably additive, let $\{A_i\}$ a a countably disjoint collection of $\mu$ elements and let us see that $$u (\bigcup_{i=1}^{\infty}A_{i})= \sum_{i=1} ^{\infty} u(A_i) $$ but I have not been able to prove this equality

Thomas Andrews
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wessi
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  • Basically, you want to show that $\sup_n \sum_i u_n(A_i) = \sum_i \sup_n u_n(A_i)$. Hint: What can you say about monotone limits? – lc2r43 May 01 '21 at 16:33

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$$u(\cup_nA_n)=\sup_{k \in\ \mathbb{N}}\mu_k(\cup_nA_n)=\lim_{k \to \infty}\mu_k(\cup_nA_n)=\lim_{k \to \infty}\sum_{n \in \mathbb{N}}\mu_k(A_n)=$$ $$=\lim_{k \to \infty}\lim_{n \to \infty}\sum_{j=1}^n\mu_k(A_n)\overbrace{=}^{(*)}\lim_{n \to \infty}\sum_{j=1}^n\lim_{k \to \infty}\mu_k(A_n)=\sum_{n \in \mathbb{N}}\sup_{k \in \mathbb{N}}\mu_k(A_n)=\sum_{n \in \mathbb{N}}u(A_n)$$ The passage $(*)$ must be justified. The sequence $a_{n,k}=\sum_{j=1}^n\mu_k(A_n)$ satisfies these criteria (increasing in $n$ and $a_{n,k}-a_{n-1,k}=\mu_k(A_n)$ is increasing in $k$) so we can exchange limits.

Snoop
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