In my textbook Johnsonbaugh/Pfaffenberger - Foundations of Mathematical Analysis, we prove that if $A_1 \subseteq A_2 \subseteq A_3 \subseteq \dots $ with each $A_n$ in the sigma-algebra $M$, then $\mu (\cup_{n=1}^{\infty}A_n)=\lim_{n\to\infty}\mu(A_n)$. However, we also prove that $A \subseteq B \implies \mu(A)\leq\mu(B)$, where $\mu$ is the measure on $M$. I'm wondering if this alternative proof also works:
We have that $\cup_{i=1}^n A_i \subseteq A_n \subseteq \cup_{i=1}^{n+1} A_i$. Therefore $\mu(\cup_{i=1}^n A_i) \leq \mu(A_n) \leq \mu(\cup_{i=1}^{n+1} A_i)$. Taking limits, we have that $\lim_{n \to \infty}\mu(\cup_{i=1}^n A_i) \leq \lim_{n \to \infty}\mu(A_n) \leq \lim_{n \to \infty}\mu(\cup_{i=1}^{n+1} A_i)$. Therefore, $\lim_{n \to \infty}\mu(A_n) = \lim_{n \to \infty}\mu(\cup_{i=1}^{n} A_i) = \mu(\cup_{n=1}^{\infty}A_n)$.
My textbook subtracts out the inclusions and then uses pairwise disjointness to use the countable additivity of $\mu$ to establish equality, but if this way works it seems to be more natural to me. Thanks in advance!
The problem is in the very last equality in your derivation. It is not a definition of $\mu(\cup_{n = 1}^{\infty} A_n)$ - it is exactly what needs a proof.