Suppose $(u_n)_{n=1}^{\infty}$ is a sequence of increasing positive measures on a measurable space. Here increasing means that $u_n(A)\leq u_{n+1} (A)$ for any measurable set $A\subseteq X$.
I am trying to show that the set function $u$ defined on the measure as follows is also a positive measure,$$u (A) = \lim_{n\rightarrow\infty} u_n(A) = \sup u_n(A).$$
My attempt:
Clearly $u(A) \geq 0$ for any measureable $A$. Also it is clear that $u(\emptyset) = 0$ since for each $n\in \mathbb{N}$ we have $u_n(\emptyset) = 0$. Now, $u$ is also $\sigma$-additive since,
$$u\big(\cup_{i=1}^{\infty} A_i\big)= \sup_n u_n\big(\cup_{i=1}^{\infty} A_i\big) = \sup_n \Sigma_{i=1}^{\infty} u_n(A_i)\leq \Sigma_{i=1}^{\infty} \sup_n u_n(A_i) = \Sigma_{i=1}^{\infty} u(A_i)$$
Also,
$$u\big(\cup_{i=1}^{\infty} A_i\big)\geq u_n\big(\cup_{i=1}^{\infty} A_i\big) ,\forall n \implies u\big(\cup_{i=1}^{\infty} A_i\big)\geq \lim _{n\rightarrow \infty}u_n\big(\cup_{i=1}^{\infty} A_i\big)= \lim_{n\rightarrow \infty}\Sigma_{i=1}^{\infty}u_n(A_i) $$
But
$$\lim_{n\rightarrow \infty}\Sigma_{i=1}^{\infty}u_n(A_i) =\Sigma_{i=1}^{\infty}\lim _nu_n(A_i) = \Sigma_{i=1}^{\infty} u(A_i)$$
These two inequalities show that $u\big(\cup_{i=1}^{\infty} A_i\big) = \Sigma_{i=1}^{\infty} u(A_i)$.
I have found this proof but struggle to understand why it follows that $$u\big(\cup_{i=1}^{\infty} A_i\big)\geq u_n\big(\cup_{i=1}^{\infty} A_i\big) ,\forall n \implies u\big(\cup_{i=1}^{\infty} A_i\big)\geq \lim _{n\rightarrow \infty}u_n\big(\cup_{i=1}^{\infty} A_i\big)$$ aswell as why $$\lim_{n\rightarrow \infty}\Sigma_{i=1}^{\infty}u_n(A_i) =\Sigma_{i=1}^{\infty}\lim _nu_n(A_i)$$ holds. Any help greatly appreciated and thanks in advance!
Best Answer
The first one is easy. If $ u\big(\cup_{i=1}^{\infty} A_i\big)\geq u_n\big(\cup_{i=1}^{\infty} A_i\big) $ for all $n\ge 1$, then by taking $n\to\infty$, we have $u\big(\cup_{i=1}^{\infty} A_i\big)\geq \lim_{n\to\infty}\left[u_n\big(\cup_{i=1}^{\infty} A_i\big)\right]= \lim_{n\to\infty}u_n\big(\cup_{i=1}^{\infty} A_i\big).$ The second one follows from monotone convergence theorem. For each fixed $N$, we have $$ \lim_{n\rightarrow \infty}\sum_{i=1}^{\infty}u_n(A_i) \ge \lim _{n\to\infty}\sum_{i=1}^{N}u_n(A_i)=\sum_{i=1}^{N}\lim _{n\to\infty}u_n(A_i). $$ Take $N\to \infty$ and we get $$ \lim_{n\rightarrow \infty}\sum_{i=1}^{\infty}u_n(A_i) \ge \sum_{i=1}^{\infty}\lim _{n\to\infty}u_n(A_i). $$ The other direction is obviously true. It holds $$ \sum_{i=1}^{\infty}u_n(A_i) \le \sum_{i=1}^{\infty}\lim _{n\to\infty}u_n(A_i),\quad\forall n\ge 1. $$ Then take limit on the left hand side.