Cfrgtkky

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 $Asubseteq X$.

I am trying to show that the set function $u$ defined on the measure as follows is also a positive measure,$$u (A) = lim_{nrightarrowinfty} u_n(A) = sup u_n(A).$$

My attempt:

Clearly $u(A) geq 0$ for any measureable $A$. Also it is clear that $u(emptyset) = 0$ since for each $nin mathbb{N}$ we have $u_n(emptyset) = 0$. Now, $u$ is also $sigma$-additive since,

$$ubig(cup_{i=1}^{infty} A_ibig)= sup_n u_nbig(cup_{i=1}^{infty} A_ibig) = sup_n Sigma_{i=1}^{infty} u_n(A_i)leq Sigma_{i=1}^{infty} sup_n u_n(A_i) = Sigma_{i=1}^{infty} u(A_i)$$

Also,

$$ubig(cup_{i=1}^{infty} A_ibig)geq u_nbig(cup_{i=1}^{infty} A_ibig) ,forall n implies ubig(cup_{i=1}^{infty} A_ibig)geq lim _{nrightarrow infty}u_nbig(cup_{i=1}^{infty} A_ibig)= lim_{nrightarrow infty}Sigma_{i=1}^{infty}u_n(A_i) $$

But

$$lim_{nrightarrow infty}Sigma_{i=1}^{infty}u_n(A_i) =Sigma_{i=1}^{infty}lim _nu_n(A_i) = Sigma_{i=1}^{infty} u(A_i)$$

These two inequalities show that $ubig(cup_{i=1}^{infty} A_ibig) = Sigma_{i=1}^{infty} u(A_i)$.

I have found this proof but struggle to understand why it follows that $$ubig(cup_{i=1}^{infty} A_ibig)geq u_nbig(cup_{i=1}^{infty} A_ibig) ,forall n implies ubig(cup_{i=1}^{infty} A_ibig)geq lim _{nrightarrow infty}u_nbig(cup_{i=1}^{infty} A_ibig)$$ aswell as why $$lim_{nrightarrow infty}Sigma_{i=1}^{infty}u_n(A_i) =Sigma_{i=1}^{infty}lim _nu_n(A_i)$$ holds. Any help greatly appreciated and thanks in advance!

The first one is easy.
If $
ubig(cup_{i=1}^{infty} A_ibig)geq u_nbig(cup_{i=1}^{infty} A_ibig)
$ for all $nge 1$, then by taking $ntoinfty$, we have
$ubig(cup_{i=1}^{infty} A_ibig)geq lim_{ntoinfty}left[u_nbig(cup_{i=1}^{infty} A_ibig)right]= lim_{ntoinfty}u_nbig(cup_{i=1}^{infty} A_ibig).$ The second one follows from monotone convergence theorem. For each fixed $N$, we have
$$
lim_{nrightarrow infty}sum_{i=1}^{infty}u_n(A_i) ge lim _{ntoinfty}sum_{i=1}^{N}u_n(A_i)=sum_{i=1}^{N}lim _{ntoinfty}u_n(A_i).
$$ Take $Nto infty$ and we get
$$
lim_{nrightarrow infty}sum_{i=1}^{infty}u_n(A_i) ge sum_{i=1}^{infty}lim _{ntoinfty}u_n(A_i).
$$ The other direction is obviously true. It holds
$$
sum_{i=1}^{infty}u_n(A_i) le sum_{i=1}^{infty}lim _{ntoinfty}u_n(A_i),quadforall nge 1.
$$ Then take limit on the left hand side.