Prove: $\mu(A)=\lim_{n \to \infty} \mu_n(A)$ Is a Positive Measure

Question

Let $(X,S)$ be a measurable space. Let $\mu_1,\mu_2,...$ be a sequence of positive measures on $X$ such that for all $A\in S$ the sequence $\mu_n(A)$ is monotonic increasing.

Prove/Disprove: the function $\mu:S\to \mathbb{R}$ defined $\mu(A)=\lim_{n\to\infty}\mu_n(A)$ is a positive measure

Proof:

1. Non-negativity: Let $A\in S$ for all $n$, $\mu_n(A)\geq 0$ therefore $\mu(A)=\lim_{n\to\infty}\mu_n(A)\geq 0$

2. Null empty set: For all $n$, $\mu_n(\emptyset)=0$ therefore $\mu(\emptyset)=\lim_{n\to\infty}\mu_n(\emptyset)=0$

3. Countable additivity: let $A_i \in S$, then $$ \begin{split} \mu\left(\bigcup_{i=1}^{\infty}A_i\right) &= \lim_{n\to\infty} \mu_n \left(\bigcup_{i=1}^{\infty}A_i \right) \\ &= \lim_{n\to\infty} \sum_{i=1}^{\infty} \mu_n(A_i) \\ &=_{(1)} \sum_{i=1}^{\infty}\lim_{n\to\infty} \mu_n(A_i) \\ &=\sum_{i=1}^{\infty} \mu(A_i) \end{split} $$

I am not sure that $(1)$ is correct I know that it can be $\leq$ as it is monotonic increasing

P.S maybe because it is monotonic increasing and bounded by $\mu(A_i)$ we can write and equation?