Understanding a part of the solution given here:
Convergence of measure sequences bounded by a finite measure
Here is the part that I do not understand:
3.) $\mu\left(\bigcup_{k=1}^\infty A_k\right)=\sum_{k=1}^\infty\mu(A_k)$.
Proof: Since $\nu$ is a finite measure on $\mathcal{A}$, then for a countable collection of pairwise disjoint sets $\{A_i\}\in\mathcal{A}$ we have $$ \nu\left(\bigcup_{k=1}^\infty A_k\right)=\sum_{i=k}^\infty\nu(A_k)<\infty. $$ Now for each $n\in\mathbb{N}$ we have $\mu_n(A_k)<\nu(A_k)$, so the Weierstrass M-Test tells us that $\sum_{k=1}^{\infty}\mu_n(A_k)$ converges uniformly for each $n\in\mathbb{N}$.
My Questions are:
1- why is $\sum_{i=k}^\infty\nu(A_k)<\infty$? $\nu$ is a finite measure does not mean that the infinite series is finite, this is why I am not convinced with this statement. Could anyone help me understand this please?
2- Also, why is $\mu_n(A_k)$ considered as a sequence of functions? could anyone explain this for me please?
First question: $\sum \nu(A_k)=\nu (\cup_n A_k)<\infty$.
Second question. Define $f_n: \mathbb N \to \mathbb R$ by $f_n(k)=\mu_n(A_k)$. The series $\sum_k f_n(k)$ converges uniformly because it is dominated by the convergent series $\sum \nu(A_k)$.